text:
User talk:Alexf

 Hello Alexf, 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.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 17:35, 8 September 2008 (UTC)

Template Babel pt not working properly?
Sorry, I'm a template dunce! Wikiversity is tiny, so most questions are asked at Colloquium. Also, if you do IRC chat, there are often technically-minded people at freenode #wikiversity-en. Good luck! --JWSchmidt 23:49, 8 September 2008 (UTC)
 * Will do. Thanks. -- Alexf 00:23, 9 September 2008 (UTC)
 * Have a look now, it is not perfect yet. If you wanna improve go to: Template:User pt-3 and also add more here: Category:User pt, Erkan Yilmaz uses the Chat (try) 00:42, 9 September 2008 (UTC)
 * Excellent! Thanks. -- Alexf 09:50, 9 September 2008 (UTC)

title: User talk:Alexf


text:
User:Eml4500c.f08.gravy.mmm/Matlab

 Summery of Matlab tutorial by Kermit Sigmon

Statements expressions and variables
Matlab uses operators, functions and variable names (which are case-sensitive) to complete tasks. When an operation has been completed the results are displayed in matrix form, unless the line ends in a semicolon. The semicolon will suppress the output but the operation will still be carried out. To stop matlab mid operation hit CTRL and C, or CTRL and break.

The workspace saves all variables and their values. The command “who” or “whos” will displace the variables in the workspace. To clear a variable the command is “clear” followed by the variable name, if the variable name is omitted then all variables will be cleared.

Saving a session
To save the variables from the workspace the command is “save”. Save will write the variable in a file called matlab.mat. To load the saved session the command is “load”.

Hardcopy
Saving input from the command window to an alterable and printable file can be done using the command diary. To start recording the command is “diary” then the file name. If you don’t give a file name the default file name is diary. To end recording the command is “diary off”.

Planer plots
The syndics for plot is “plot (x,y)” with x and y being the variables to be plotted. An interval and mesh size can be set by “variable name= lower limit: mesh size: upper limit”.

“fplot (‘function file’,[lower limit, upper limit])” is the command to graph functions. Function file is a m-file that contains the function to be graphed.

Only one figure is active at a time but you can switch between the figures using “figure(n)” where n is the figure number wanted. “gcf” command gives the active figure’s number.

Multiple plots on one graph can be made using “plot(x, y1,x,y2,…)”Other ways are, declaring a matrix of functions like U=[function1, function2 ] and using plot(x,U). Or using the “hold” command to freeze the current plot and all plots after will be placed on the same plot. To unfreeze the plots, the command is  “hold off”.

3-D line plots
To create 3-d plots the command is “plot3(x,y,z)”.

3-D mesh and surface plots
To create 3-D wire mesh surface plots the command is “mesh(z)” and three dimensional facetted surface plots “surf(z)”where z is a matrix of points in the z direction.

To plot a function over a rectangle use first the command “[x,y]=meshgrid(xx,yy)” where xx and yy are matrixes of the sides of the rectangle. x and y result in matrixes whose columns equal xx and yy respectably. Next the command mesh is used with z= exp(-x.^2-y.^2)

The color of the shadings can be changed using the command “shading” followed by the setting. The settings are faceted, interp ( for interpolated), and flat. The color profile is changed using the command “colormap” followed by the color. The color codes are hsv, hot, cool, jet, pink, copper, flag, gray, and   bone.

Labeling the figure
The commands Title, xlabel, ylabel and zlabel, can be used to label the figure. An example is Title(‘Title name’).

gtext(‘The Spot’), command places text where on the figure the mouse is clicked, while the text command places the text at coordinates.

Axis scaling
The following commands are used to change the scaling on the axis.

axis([x lower limit, x upper limit, y lower limit, y upper limit ]), to set scaling.

axis(axis), freezes scaling for subsequent graphs.

axis auto, resets the scale to the default.

Variable= axis, the variable will be set to the current scaling.

axis square, makes the same scale for both axes.

axis equal, make the same scale and tic marks for both axes.

axis off, turns off scaling and tic marks axis on turns them on.

Line types
To change the line type, mark type and color use the tables below. The syntax is Plot(x,y,’color line type mark type’) so Plot(x,y,’b:*’) is a blue dotted line with star markers.

'''

Graphics hardcopy
The command “print” will send the current graphic to the default printer. The command “print ” then the file name will save and overwrite the graphic to a file, but ”print -append” then the file name will add the graphic to the end  of a file.

To save a graphic that is not the current one, the command is “print –deps –f#” then the filename and # the figure number.

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

title: User:Eml4500c.f08.gravy.mmm/Matlab


text:
User:Kunt

IM GOING TO .... YOUR .....!

title: User:Kunt


text:
User talk:Kunt

Thank you for experimenting with Wikiversity. Your test was recorded, and it has been reverted or removed. Please use the sandbox for further testing. You may wish to learn more about what Wikiversity is to help you contribute. Consider registering if you are keen on our project. Erkan Yilmaz uses the Chat (try) 18:14, 8 September 2008 (UTC)

title: User talk:Kunt


text:
User:Eml4500.fall08.gravy.sms

test

title: User:Eml4500.fall08.gravy.sms


text:
User talk:Eml4500.fall08.gravy.sms

 Hello Eml4500.fall08.gravy.sms, 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 18:39, 8 September 2008 (UTC)

title: User talk:Eml4500.fall08.gravy.sms


text:
User:Eml4500.f08.Ateam.Slevinski

HOMEWORK 1

HOMEWORK 2

HOMEWORK 3

HOMEWORK 4

HOMEWORK 5

HOMEWORK 6

HOMEWORK 7

title: User:Eml4500.f08.Ateam.Slevinski


text:
User:EML4500.f08.FEABBQ.Kurth

FEABBQ.Kurth
Welcome to Eml4500.f08.FEABBQ!!

Link to group page: Eml4500.f08.FEABBQ

FEABBQ Team Members:

 * Esmail Hamdan
 * Jose Jayma
 * Grant Stark
 * Nik Vitt
 * William Kurth
 * Andy Kobyljanec

EML 4500 Class Homepage:
User:Eml4500.f08

title: User:EML4500.f08.FEABBQ.Kurth


text:
User:Eas4200c.f08.spars.rondon

=Homework 1= See: Homework 1 =Homework 2= See: Homework 2 =Homework 3= See: Homework 3

title: User:Eas4200c.f08.spars.rondon


text: User:Wm.jones

title: User:Wm.jones


text:
User talk:Remi/Learning links

http://www.eurekalert.org/pub_releases/2008-09/uond-nps090808.php

http://www.knowledgefoundation.com/indexkf.php

http://redgreenandblue.org/2008/09/10/100-billion-us-investment-to-create-2-million-green-jobs/

title: User talk:Remi/Learning links


text:
User talk:EAS4200c.f08.blue.c

 Hello EAS4200c.f08.blue.c, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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 * 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 20:30, 8 September 2008 (UTC)

title: User talk:EAS4200c.f08.blue.c


text:
User talk:Eas4200c.f08.spars.rondon

 Hello Eas4200c.f08.spars.rondon, 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.
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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 20:33, 8 September 2008 (UTC)

Assignment
I'll do the Ch 2 review since I've already read a lot of it. --Eas4200C.f08.spars.mcgilvray 15:27, 25 September 2008 (UTC)

title: User talk:Eas4200c.f08.spars.rondon


text:
User talk:EML4500.f08.FEABBQ.Kurth

 Hello EML4500.f08.FEABBQ.Kurth, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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 * 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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 * 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 20:34, 8 September 2008 (UTC)

title: User talk:EML4500.f08.FEABBQ.Kurth


text:
User talk:Eml4500.f08.Ateam.Slevinski

 Hello Eml4500.f08.Ateam.Slevinski, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 20:38, 8 September 2008 (UTC)

title: User talk:Eml4500.f08.Ateam.Slevinski


text:
User talk:Wm.jones

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 20:38, 8 September 2008 (UTC)

title: User talk:Wm.jones


text:
User talk:Stark

 Hello Stark, 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.
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 * 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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 * 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 20:39, 8 September 2008 (UTC)

title: User talk:Stark


text:
User:Eml4500.f08.bike.sweeting

Normal  0         false   false   false                             MicrosoftInternetExplorer4

Homework 1-Individual Part
EML4500 Summary: Aug. 25, 2008 – Aug. 29, 2008

Finite Element Analysis with Dr. Loc Vu-Quoc

Lectures are held at the University of Florida on Monday, Wednesday, and Friday from 4:05-4:55 PM in Florida Gym 270.

EML4500 teaches the fundamentals of the finite elemental method used in the analysis of truss, beam, and solid structures. The finite element method is taught through matrix algebra with the derivation of the stiffness matrix. Local and global coordinate systems are emphasized, as well as 3-D truss and beam structures and their linear stability. Coding and plotting is another area of interest in this course, as Matlab programming will be used in order to solve many of the problems. Applications of the finite element method are discussed, as well as its uses in mechanical, aerospace, and biomedical engineering fields.

The text book that will be used throughout the semester is: M. Asghar Bhatti, Finite Element Analysis With Mathematica and Matlab Computations and Practical Applications: Fundamental Concepts, John Wiley & Sons, Inc. Copyright 2005

Students will be required to work in groups of 5 to 6 under a team name that is 4 to 6 characters long. Cooperative learning techniques will be used, and group member evaluations will be held at the end of the semester. These evaluations will directly impact the homework grade of that individual, approximately 31% of his/her overall grade. The rest of the student’s individual grade is comprised of 3 exams, each accounting for 23%.

Each group will be required to use Wikiversity in order to submit homework and projects. Each member must create his/her own personal class account where his/her portion of the work will be posted. Each members name and UFID number, as well as his/her Wikiversity class username, is to be submitted to the instructor on hard copy.

Homework and project load will be divided evenly among the team members at their discretion. Once each group member has completed his/her assigned task the assignment will be posted as a whole on the group leader’s user namespace. This should be done close to the homework deadline in order to keep other groups from stealing ideas or plagiarizing. Another safety concern is that users may change information on a students Wikiversity page. In order to prevent this type of fraud from occurring the student must save an archived version of the assignment and email that address, not the address of the updated page, to the TA.

Homework 2-Individual Part(9/17)
Step 4 in the finite element analysis process begins with the reduction of the global force-displacement relations. This reduction is done by first eliminating the known degrees of freedom(DOF’s). $$ d_{1}=d_{2}=d_{5}=d_{6}=0 $$

Deleting the corresponding columns in the global stiffness matrix and the force matrix results in the following relationship.

$$ \left[ \begin{array}{cc} k_{33} & k_{34}\\ k_{43}& k_{44} \end{array} \right]_{2x2} $$ $$\begin{Bmatrix} d_{3}\\d_{4} \end{Bmatrix}_{2x1}=\begin{Bmatrix}F_{3}\\F_{4} \end{Bmatrix}_{2x1}$$

In order to find the inverse of the stiffness matrix one must first calculate the determinant as follows. det $$k = k_{33}k_{44}-k_{34}k_{43} $$

The inverse of the stiffness matrix can then be calculated as shown below.

$$k^{-1} = \frac {1}{det(k)}$$$$ \left[ \begin{array}{cc} k_{44} & -k_{34}\\ -k_{43}& k_{33} \end{array} \right] $$

If done correctly the stiffness matrix multiplied by its inverse will yield the identity matrix, I. $$I= \left[ \begin{array}{cc} 1 & 0\\ 0& 1 \end{array} \right] $$

For the two beam truss system shown below, the displacement matrix is calculated by multiplying the inverse matrix by $$\begin{Bmatrix}{0} \\ {P} \end{Bmatrix}$$, where P=7.

$$\begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix}=k^{-1}\begin{Bmatrix}{0}\\{7}\end{Bmatrix}$$

det $$k=(3.0625)(2.6875)-(-2.175)(-2.175)=3.50$$

Therefore, $$ k^{-1}=\frac {1}{3.50}$$$$ \left[ \begin{array}{cc}2.6875&2.175\\ 2.175&3.0625\end{array} \right]=\left[ \begin{array}{cc}.7678&.6214\\ .6214&.8750\end{array} \right]$$

so $$\begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix}=\left[ \begin{array}{cc}.7678&.6214\\ .6214&.8750\end{array} \right]\begin{Bmatrix}{0}\\{7}\end{Bmatrix}=\begin{Bmatrix}{4.352}\\{6.1271}\end{Bmatrix}$$

Step 5 consists of computing the reactions, which can be done by two different methods. Method 1 is to use the element force-displacement relations and method 2 is the Euler cut principle. The force-displacement relation corresponding to the two beam truss system is shown below.

$$k_{4x4}^{(e)}d_{4x1}^{(e)}=f_{4x1}^{(e)}$$ where $$e=1,2$$

The free body diagram of element 1(e=1) is shown below, along with the corresponding force-displacement relation. $$k^{(1)}d^{(1)}=f^{(1)}$$ where $$k^{(1)}$$ and $$d^{(1)}$$ are known.

$$d^{(1)}=\begin{Bmatrix}{0}\\{0}\\{4.352}\\{6.1271}\end{Bmatrix}$$ and $$f^{(1)}=\begin{Bmatrix}{f_{1}^{(1)}}\\{f_{2}^{(1)}}\\{f_{3}^{(1)}}\\{f_{4}^{(1)}}\end{Bmatrix}$$

The free body diagram of element 2(e=2) is shown below, along with the corresponding force-displacement relation. $$k^{(2)}d^{(2)}=f^{(2)}$$ where $$k^{(2)}$$ and $$d^{(2)}$$ are known.

$$d^{(2)}=\begin{Bmatrix}{4.352}\\{6.1271}\\{0}\\{0}\end{Bmatrix}$$ and $$f^{(2)}=\begin{Bmatrix}{f_{1}^{(2)}}\\{f_{2}^{(2)}}\\{f_{3}^{(2)}}\\{f_{4}^{(2)}}\end{Bmatrix}$$

Homework 4-Individual Part(10/6)
The elements of a two bar truss system are shown below by Figures 1 and 2.

The connectivity array, denoted "conn," is shown below for the two bar truss system.

conn = $$ \left[ \begin{array}{cc} {1} & {2} \\ {2}& {3} \end{array} \right]$$

Each row of the connectivity array represents an element(i.e. row 1 = element 1, row 2 = element 2), and each column represents a local node number(i.e. column 1 = local node 1, column 2 = local node 2). The numbers within the connectivity array correspond to the global node numbers, such that conn(e,j) = global node number of local node j of element e.

The location matrix master array, denoted "lmm," is shown below for the two bar truss system.

lmm = $$ \left[ \begin{array}{cccc} {1} & {2} & {3} & {4} \\ {3} & {4} & {5} & {6} \end{array} \right]$$

Somewhat similar to the connectivity array, the location matrix master array has rows that correspond to the element number and columns that represent the local degree of freedom(dof) number. The numbers within the location matrix master array correspond to the global dof number, or equation number, in the matrix K. Generally, lmm(i,j) = global dof number(or equation number) for the element stiffness coefficient corresponding to the jth local dof number.

Method 2 to derive the stiffness matrix $$K_{4x4}^{e}$$ is to transform a system with 2 dof's into a system with 4 dof's so that the transformed matrix is a 4x4, and hopefully invertible.

Homework 4-Individual Part(10/15)
For the justification of assembly of the element stiffness matrix into the global stiffness matrix, consider the two-bar truss system.

Let the global stiffness matrix be represented by K and the element stiffness matrix be represented by $$k^{(e)}$$, where e = 1,...,n and n is any number of element.

Recall the element force-displacement relationship $$k^{(e)}_{4x4}d^{(e)}_{4x1}=f^{(e)}_{4x1}$$ as well as the second method of the Euler cut principle shown below. This method involves an equilibrium analysis of node 2.

Also recall the free body diagram of the two bar truss system in question. The free body diagrams of elements 1 and 2 are shown below, taking note of the naming process for the element degrees of freedom.

It is possible to relate the global degrees of freedom to element degrees of freedom for both element 1 and element 2. This is done below for node 2.

$$

\textrm{Node\ 2:\ }d_{3}=d_{3}^{(1)}=d_{1}^{(2)}\textrm{,\ }d_{4}=d_{4}^{(1)}=d_{2}^{(2)}

$$

The equilibrium of node 2 can then be analyzed by cutting each element and relating the forces as shown below.

$$\sum F_x = 0 = -f_3^{(1)} - f_1^{(2)} = 0 $$

$$\sum F_y = 0 = P - f_4^{(1)} - f_2^{(2)} = 0 $$

Homework 5-Individual Part(10/31)
The free body diagram of section dx can be seen below in Figure 1.

Summing the forces in the x direction yields

$$\sum F_x = 0 = -N(x,t) + N(x+dx,t) + f(x,t)dx - m(x){\ddot{u}}dx $$

$$ = \frac{\partial N(x,t)}{\partial x}dx + h.o.t + f(x,t)dx - m(x){\ddot{u}}dx $$ ⇨ $$ Equation \left(1\right)$$

The dx terms are canceled out of this equation, as well as the high order terms(h.o.t). Let this equation be call equation (1).

Why are the high order terms neglected?

Recall Taylor series expansion: $$ f(x+dx) = f(x) + \frac{df(x)}{dx}dx + \frac{1}{2}\frac{d^{2}f(x)}{dx^{2}}dx^{2} + ... $$ where $$\frac{1}{2}\frac{d^{2}f(x)}{dx^{2}}dx^{2}$$ is considered a h.o.t and is therefore neglected.

From equation (1) the equation of motion(EOM) of the system can be derived. Let the EOM be called equation (2).

$$ \frac{\partial N}{\partial x} + f = m{\ddot{u}} $$ ⇨ $$ Equation \left(2\right)$$

Equation (3), the constitutive relation, can then be determined.

$$ N\left(x,t\right) = A(x)\sigma(x,t) $$ ⇨ $$ Equation \left(3\right)$$ where $$ \sigma\left(x,t\right) = E(x)\epsilon(x,t) $$ and $$ \epsilon(x,t) = \frac{\partial u(x,t)}{\partial x} $$

Substituting equation (3) into equation (2) yields the partial differential equation of motion, equation (4) shown below.

$$ \frac{\partial}{\partial x}[A(x)E(x)\frac{\partial u}{\partial x}] + f(x,t) = m(x){\ddot{u}} $$  ⇨   $$ Equation \left(4\right)$$

There are 2 initial conditions(displacement and velocity), and therefore 2 boundary conditions(2nd order derivative with respect to x).

For Figure 2 the boundary conditions are $$ u\left(0,t\right) = 0 = u(L,t) $$

For Figure 3 the boundary condition at 0 is also $$ u\left(0,t\right) = 0 $$, but the boundary condition at L can be derived as follows

$$ N\left(L,t\right) = F(t) $$ where $$ N = A\left(L\right)\sigma(L,t)$$ and $$\sigma\left(L,t\right) = E(L)\epsilon(L,t)$$ and $$ \epsilon(L,t) = \frac{\partial u(L,t)}{\partial x} $$

The boundary condition at L can therefore be simplified to

$$ \frac{\partial u(L,t)}{\partial x} = \frac{F(t)}{A(L)E(L)} $$

Homework 6-Individual Part(11/3)
Figure 1 shows the loaded elastic bar system in question.

The initial conditions for the system in Figure 1 at t = 0 are

$$ u\left(x,t=0\right)=\overline{u}(x) $$ for the known displacement function

and

$$ \frac{\partial u}{\partial t}(x,t=0)={\ddot{u}}(x,t=0)=\overline{V}(x) $$ for the known velocity function.

The partial differential equation for the continuous principle of virtual work(PVW) of dynamics for the elastic bar in Figure 1 can be seen below in Equation 1.

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

The discrete equation of motion can then be shown by Equation 2, where $$ \underline{M} $$ is the mass matrix. Equation 2 represents a multiple defree of freedom system(MDOF). Note that a single degree of freedom system(SDOF) has a very similar equation of motion, shown be Equation 3.

MDOF System ⇨ $$ \underline{M}\underline{\ddot{d}}+\underline{k}\underline{d}=\underline{F} $$ ⇨ $$ Equation \left(2\right)$$

SDOF System ⇨ $$ m\ddot{d}+kd=F $$ ⇨ $$ Equation \left(3\right)$$

The integral shown in Equation 4 was used to derive Equation 2 from Equation 1. The term $$ w(x) $$ represents the weighting function.

$$ \int_{x=0}^{x=L} w(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)dx=0 $$ for all possible $$ w(x) $$ ⇨ $$ Equation \left(4\right)$$

To go from Equation 1 to Equation 4 is a trivial task, but going from Equation 4 to Equation 1 is not trivial. This process is shown below.

For simplicity, let $$ \left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)=g(x) $$

Equation 4 then becomes $$ \int_{x=0}^{x=L} w(x)g(x)dx=0 $$ for all $$ w\left(x\right) $$

Since Equation 4 holds for all $$ w\left(x\right) $$, select $$ w\left(x\right)=g\left(x\right) $$,

then Equation 4 becomes $$ \int_{x=0}^{x=L}g^2dx=0 $$ ⇨ $$ g\left(x\right)=0 $$

Homework 6-Individual Part(11/5)
Recall the steps for and integration by parts:

$$ \left(rs\right)^{'}=r^{'}s+rs^{'} $$

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

$$ \int \left(rs\right)^{'}=\int r^{'}s + \int rs^{'} $$

Which yields a final result of $$ \int r^{'}s=rs-\int rs^{'} $$

Now Recall the Cont. Principle of Virtual Work(PVW) shown below.

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

In the first term, let $$ s\left(x\right)=w(x) $$ and $$ r\left(x\right)=(EA)\frac{\partial u}{\partial x} $$

Solving using integration by parts yields

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

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

where $$ \left(EA\right)(L)\frac{\partial u}{\partial x}(L,t) = N(L,t)=F $$

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

Now lets consider the model problem with 2 boundary conditions shown below.

At $$ x=0 $$ select $$ w(x) $$ such that $$ w(0)=0 $$. This means that the system is kinematically admissible.

Motivation: discrete PVW applied to the equation shown below.

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

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

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

Since $$ \underline{w} $$ can be selected arbitrarily, select $$ \underline{w} $$ such that $$ w_{1}=w_{2}=w_{5}=w_{6}=0 $$ so to eliminate equations involving unknown reactions. Simply put, rows 1, 2, 5, and 6 are eliminated. The result is shown below in Equation 1.

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

$$ \underline{K}_{2x2}\underline{d}_{2x1}=\underline{F}_{2x1} $$ ⇨ $$ Equation \left(1\right) $$

Back to the principle of virtual work, the unknown reaction is

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

The equation then becomes

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

The final equation can be written as

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

Homework 7-Individual Part(11/21)
$$ \left[\epsilon\right] = \frac{[du]}{[dx]} = \frac{L}{L} = 1 $$

$$ \left[\sigma\right] = \left[E\right] = \frac{F}{L^2} $$

$$ \left[A\right] = L^2, [I] = L^4 $$

$$ \left[\frac{EA}{L}\right] = [\tilde{k}_{11}] = \frac{\frac{F}{L^2}{L^2}}{L} = \frac{F}{L} $$

$$ \left[\tilde{k}_{11}\tilde{d_1}\right] = [\tilde{k}_{11}\tilde{d_1}] = F $$

$$ \left[\tilde{k}_{23}\tilde{d_3}\right] = [\tilde{k}_{23}][\tilde{d_3}] $$ where $$ \tilde{d_3} = 1 $$

$$ = \frac{\left[6\right][E][I]}{L^2} $$

$$ = \frac{\left(1\right)\left(\frac{F}{L^2}\right)(L^4)}{L^2} = F $$

The element force displacement relation in global coordinates from the element force displacement relation in local coordinates yields:

$$ \underline{k}^{(e)}_{6x6}\underline{d}^{(e)}_{6x1} = \underline{f}^{(e)}_{6x1} $$

with $$ \underline{k}^{(e)}_{6x6} = {\underline{\tilde{T}}^{(e)}_{6x6}}^T\underline{\tilde{k}}^{(e)}_{6x6}\underline{\tilde{T}}^{(e)}_{6x6} $$

from $$ \underline{\tilde{k}}^{(e)}_{6x6}\underline{\tilde{d}}^{(e)}_{6x1} = \underline{\tilde{f}}^{(e)}_{6x1} $$

The derivation of $$ \tilde{K}^{(e)} $$ from PVW results in the equation shown below, focusing on the bending effects.

$$ \frac{\partial^{2}}{\partial x^{2}}\left((EI)\frac{\partial^{2}v}{\partial x^{2}}\right) - f_{t}(x) = m(x)\ddot{v} $$

Homework 7-Individual Part(12/1)
The interpolations of the applied internal forces $$ N_{5}\left(\tilde{x}\right) $$ and $$ N_{6}\left(\tilde{x}\right) $$ are shown below.

Recall the equation of the displacement matrix in the tilde coordinate system shown below, where $$ \underline{d}_{6x1}^{(e)} $$ is known after solving the finite element system.

$$ \underline{\tilde{d}}_{6x1}^{(e)} = \underline{\tilde{T}}_{6x6}^{(e)}\underline{d}_{6x1}^{(e)}$$

Now we will look at the computation of $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$.

$$ \underline{u}\left(\tilde{x}\right) = u(\tilde{x})\tilde{i} + v(\tilde{x})\tilde{j} = u_{x}(\tilde{x})\tilde{i} + u_{y}(\tilde{x})\tilde{j} $$

Computing $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$ required using the equations shown below.

$$u(\tilde{x})=N_1(\tilde{x})\tilde{d}_1+N_4(\tilde{x})\tilde{d}_4$$

$$v(\tilde{x})=N_2(\tilde{x})\tilde{d}_2+N_3(\tilde{x})\tilde{d}_3+N_5(\tilde{x})\tilde{d}_5+N_6(\tilde{x})\tilde{d}_6$$

The computation of $$ u_{x}\left(\tilde{x}\right) $$ and $$ u_{y}\left(\tilde{x}\right) $$ from $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$ is shown below.

$$\begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \underline{R}^{T}\begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix}$$

$$ \begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix} = \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix}\begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} $$ where $$ \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix} $$ is represented by $$ \underline{N}\left(\tilde{x}\right) $$ and $$\begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix}$$ is denoted $$ \underline{\tilde{d}}^{(e)} $$

$$\begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \underline{R}^{T}\underline{N}\left(\tilde{x}\right)\underline{\tilde{T}}^{(e)}\underline{d}^{(e)} $$

title: User:Eml4500.f08.bike.sweeting


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File talk:I want a pony!.jpg

I think this Image:I want a pony!.jpg is on the source page declared as CC by-nd 2.0. Unfortunately at Wikiversity it is now under licence: cc-by-2.0 Erkan Yilmaz uses the Chat (try) 21:38, 8 September 2008 (UTC)

title: File talk:I want a pony!.jpg


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User:Eml4500.f08.bottle.butler

Homework
User:Eml4500.f08.bottle.butler/HW 1 User:Eml4500.f08.bottle.butler/HW 2 User:Eml4500.f08.bottle.butler/HW 3 User:Eml4500.f08.bottle.butler/HW 4 User:Eml4500.f08.bottle.butler/HW 5 User:Eml4500.f08.bottle.butler/HW 6 User:Eml4500.f08.bottle.butler/HW 7

title: User:Eml4500.f08.bottle.butler


text:
University of Florida/Eml4500/f08.FEABBQ.Jayma/HW1

Jose's Contribution to HW#1
This is me finishing what Grant started...

Section 19 - Sparse Matrix Computations
Matlab typically handles matrices of which the contents are mostly nonzero. These matrices are considered "dense". A matrix with mostly zero entries is considered "sparse" and is worth defining for the possibility of ignoring the zero values and operating on the nonzero values only. The benefit of this is to reduce the memory footprint of the matrix and the computation time. Suppose a matrix is created by the following command, and that the matrix is filled with mostly zeros by the following command, the output of which looks like The nonzero values of the matrix can be easily found with the command, followed by the output, The following commands will return the matrix to it's previous form and check it's storage mode, respectively Sparse matrices can also be generated directly as opposed to calling the sparse function on a full ("dense") matrix. A sparse diagonal matrix can be generated with the  command. The output is similar to calling the. are the sparse analogs of, respectively. A more specific approach would be to define the nonzero values of the sparse matrix individually. For example (followed by output) The nonzero entities were stored in the row vector. Another example of directly defining the nonzero values of a sparse matrix is The output of Matlab arithmetic and functions will be in one of the two storage modes (either Full(F) or Sparse(S)). The following defines the output results of the manipulation done to either full or sparse matrices


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User:Eml4500.f08.ateam.mcnally/Week 2

Friday September 5, 2008
Assigned reading from Fundamental Finite Element Analysis And Appllications: With Mathematica and MATLAB Computations by M. Asghar Bhatti.
 * Chapter 1: Big Picture
 * Section 1.1 Discretization and Element Equations
 * 1.1.1 Plane truss element
 * 1.2 Assembly of Element Equations. See Example 1.4: Five bar truss
 * 1.4 Element Solutions and Model Validity
 * 1.4.1 Plane truss element

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: 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.

 * 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: The Element Picture
 * 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: 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: Elimination of the known dofs to reduce the global Force Displacement Relation Step 5: Compute the element Forces from the now known ‘’’d’’’ element stresses
 * 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 6: Compute the unknown reaction Forces from the now known element forces and element stresses

For further explanation please see the example on A team group member Pedro Rivero under the heading weekly notes summary.

title: User:Eml4500.f08.ateam.mcnally/Week 2


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User:EAS4200c.f08.blue.b

Lecture 5, September 5th, 2008 Marc Pelletier

Lecture 5

Lecture 12, September 22nd, 2008 Marc Pelletier

Lecture 12

Lecture 18, October 6th, 2008 Marc Pelletier

Lecture 18

Lecture 30, November 5th, 2008 Marc Pelletier

Lecture 30

Lecture 36, November 19th, 2008 Marc Pelletier

Lecture 36

title: User:EAS4200c.f08.blue.b


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User:EML4500.f08.RAMROD.E

/HW1  /HW2  /HW3  /HW4  /HW5  /HW6

/HW7

title: User:EML4500.f08.RAMROD.E


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User:Alexf

I am an Administrator (Custodian in Wikiversity parlance) in the English Wikipedia. I visit here rarely so if you need to Talk to me please visit me at User:Alexf

title: User:Alexf


text:
User:YS~enwikiversity

I am YS. A forth year international medical student and author of the prep4md blog. I am interested in science, psychology, web 2.0, networking and "eduinnovation".

title: User:YS~enwikiversity


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University of Florida/Eml4500/f08.FEABBQ/HW1

Homework 1
BACK TO FEABBQ

Due 09/12/08 at 17:00 EDT (i.e., 21:00 UTC)

 * Matlab Assignment (current version). Archived version for submission.

 * Lecture Notes (current version). Archived version for submission.

Homework 1 Contributing Members

 * J.Jayma EML4500.f08.FEABBQ.Jayma 22:02, 16 September 2008 (UTC)

 * Esmail Hamdan EML4500.f08.FEABBQ.Hamdan16:35,17 September 2008 (UTC)

 * Andy Koby 16:31, 17 September 2008 (UTC)
 * William Kurth EML4500.f08.FEABBQ.Kurth 20:04, 19 September 2008 (UTC)

 * Grant Stark EML4500.f08.FEABBQ.Stark 20:24, 19 September 2008
 * Nik Vitt Eml4500.f08.FEABBQ.vitt 20:27, 19 September 2008  (UTC)


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User:Eas4200c.f08.vqcrew.c

Homework 1

Homework 2

Homework 3

Homework 4

Homework 5

Homework 6

Homework 7

title: User:Eas4200c.f08.vqcrew.c


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User:Eas4200c.f08.vqcrew.c/Homework 1

Administrative Considerations for EAS4200C
The general structure of the course is formulated to cultivate a collaborative setting in academia for students to prepare for working in similar situations in industry. This is achieved through assigning primarily group-based tasks, chiefly in the form of regular homework reports created and submitted using a MediaWiki-based platform (in this case, Wikiversity). The intent is that groups of 5 or 6 students will collaborate on a regular basis to create the reports, which contain summaries of important lecture topics as well as traditional homework problems and solutions. These reports are advantageous not only through their fostering of teamwork and collaborative abilities but also by encouraging proper citation of references and sources, which is a key skill in industry. As an incentive to fully exploit the value of this resource, the reports constitute 31% of the final grade in the course, the remainder of which is split evenly amongst 3 exams.

Aside from collaboration, the emphasis in the course is to develop both an understanding of mechanics and good problem-solving skills. A key to this development is an ability to formulate problems and subsequently judge the correctness of a solution to these problems. Approaching the course in this fashion avoids the old “ad-hoc” methods of structural analysis. Additional information about the aims of the course and its structure can be found at the course website.

Overview of Using Wikipedia Responsibly
During lecture, Dr. Vu-Quoc pointed out several of the advantages of using Wikipedia, how to use it responsibly and how to avoid vandalism. Various administrative issues were discussed such as how to gain access to the Wikiversity page, the format for usernames and where and how to post information on the group's user namespace. Examples were given of proper and improper ways to post and edit material and suggestions were offered such as using the hide feature for derivations of equations, when applicable. Dr. Vu-Quoc went over pages that he had created to provide an idea of what was expected. He also showcased examples of vandalism to stress the importance of not editing unrelated articles so as to avoid drawing attention to ourselves. He also illustrated the process of how to retrieve a vandalized page in the case, which would be helpful if one's page should be vandalized to restore it to the original form.

He additionally explained the importance of not creating actual Wikipedia articles for the class assignments, but rather to edit only the user namespace. The process of homework would be done this way to avoid Wikipedia's review process when a Wikipedia page is created. He went over the usefulness of drop-boxes and displayed the simplicity of copying and pasting from one's user namespace to the group's namespace. Additionally, he explained the importance of turning in link that came from the history rather than the actual page to avoid any confusion. This would remedy the problem of someone changing something in the article after the homework has been submitted.

Preface of the Text
In the preface of the first edition of Mechanics of Aircraft Structures written by C. T. Sun, the author first explains that the text is written for third and fourth year aeronautical engineering students. The author assumes that the reader has already taken the mechanics of materials course in their curriculum. The text is written for a one semester course.

The author then proceeds to outline that the book will deviate from the traditional structures text's "ad hoc techniques" of problem solving. The industry has shifted to using finite elemental methods for structural analysis. The text explains that it will emphasize the understanding of structures so that numerical data given by modern finite elemental methods will be interpreted correctly. Another difference is that the author derives the governing equation via displacement rather than stress and strain.

Secondly, fracture mechanics are discussed in the text, giving the student exposure to this topic prior to graduate school. The author states that fracture mechanics have been a useful tool in the past thirty years of aircraft structures. The topic of fracture mechanics is explained by "Griffith's concept of strain energy release rate." The author feels that calculating the strain energy release rate won't be a difficult problem for the third and fourth year students to calculate. Calculating this also reveals the change of the stored strain energy before and after the fracture.

Lastly, with the recent development of composite structures in industry, it gives an introduction to the properties of composite materials. Until now, traditional metals have been studied in the curriculum, and relative to the metals previously studied, composites are significantly different. The text covers the material properties of composites such as the stress-strain curves of anisotropic and further covers the analysis of symmetric laminates of composite materials. The author emphasizes that this is merely an introduction to the subject on composite materials, and isn't a substitute for a composites course.

The preface also mentions that Chapter 7 will also include buckling concepts at an elementary level. The author states that postbuckling strengths of bars and panels are commonly used in aircraft structures and this chapter exposes the reader to the concept without the necessity of more advanced courses.

In the second edition the author corrected typographical errors and add to the materials in chapters 3, 4 and 6. Chapters 1 and 2 were not modified and in the latter chapters he covers new topics including an introduction to the effect of plasticity on fractures, Timoshenko beam theory and other principles as well.

Introduction to Aerospace Structures
The governing equations in aerospace structures and structural composition in general are governed by partial differential equations. These equations can be solved by the finite elemental method (FEM), which is used throughout the engineering discipline as well as structural composition.

The emphasis in this course is to develop problem solving skills while understanding mechanics. The finite elemental method will numerically solve the partial differential equations so it will be the job of the future engineer to interpret the numerical results to judge the correctness of the results. In addition, the ability to formulate problems and avoid the more traditional method of structural analysis will be emphasized throughout the course.

The main consideration during the aircraft design process is to reduce weight while maintaining strength in the structure. Strength, stiffness, and toughness all play important roles in the overall performance of an aircraft. The role of a design engineer is to create a balance of these properties that best satisfies all of the requirements of the aircraft.

 * Strength - the resistance of a material to deformation and rupture.
 * The yield stress, $$\displaystyle \sigma_Y$$, of a material is the point at which the material will begin to plastically and permanently deform.
 * The ultimate stress, $$\displaystyle \sigma_u$$, of a material is the critical point at which the material fails catastrophically.
 * Stiffness - the ability of a material to resist deformation caused by forces acting on the object.
 * The strain, $$\displaystyle \epsilon$$, of a material is the measure of the deformation caused by an applied force.
 * Young's modulus, $$\displaystyle E$$, is the measure of a material's stiffness. Young's modulus is the ratio of the stress applied over the strain caused in response, $$\displaystyle E = \sigma/\epsilon$$.  Figure 1 displays the stress-strain curve for steel.  Young's modulus represents the slope of the curve in the linear, or "elastic", region.  In this region, no permanent or "plastic" deformation has taken place and the material will return to its original dimensions.
 * Toughness - the overall resilience of a material and resistance to fracture. Toughness is measure in units of energy per unit volume a material can withstand before failure.

Examples of Materials and Their Properties
 Glass  - high stiffness, low toughness. Unlikely to bend or deform, yet vulnerable to fracture.

 Plastics  - low stiffness, high toughness. Easily deformed, but resistant to rupture.

 Metals  - Aluminum has high toughness, but is weak in terms of stiffness. Aluminum will not resistant deformation under high loads. Steel alloys are better suited for a situation requiring high stress handling capabilities.

All of these materials have their strengths and weaknesses. All of these materials can be used safely and effectively as long as these constraints are recognized. In the search for highly stiff, highly tough materials, the strengths of different materials were combined and their weaknesses diminished. Materials known as composites are formed from a network, or "matrix", of strong, flexible fibers held in place by stiff epoxy. The union of these materials creates a new, strong and lightweight material ideal for the aerospace industry.

In order to optimize performance, aerodynamic considerations must be made during the design process. Even though a material is structurally robust, there many be other factors that make that material detrimental to the overall function of the aircraft. For example, a material with vulnerability to cold or corrosion could not be used on the outer skin of an aircraft. The shape of major features of the aircraft must function both structurally and aerodynamically. The fuselage must be strong enough to withstand years of stress associated with pressurizing and depressurizing the aircraft while also maintaining a small cross-sectional area to keep aerodynamic drag to a minimum. The wings must be shaped to create lift to allow the aircraft to fly, but they must also act to support the weight of the fuselage.

Further considerations can be found at the online notes.

Problem 1.1
Given Figure 2, find the optimum ratio $$\frac{b}{a}$$  to maximize the load bearing capability of the thin-walled beam.

The assumptions for this problem are as follows:
 * 1) The cross-sectional perimeter is constant, i.e. $$L=2(a+b)=Constant$$
 * 2) The torque $$T$$ is equal to the bending moment $$M$$, i.e. $$T = M$$
 * 3) The maximum allowable bending normal stress is twice the maximum allowable shear stress, i.e. $$\sigma_{allow} = 2\tau_{allow}$$

The assumption that $$t$$ is very small compared to $$a$$ and $$b$$ is also valid.

Since the cross-section walls are very thin, the shear stress can be assumed to be uniform along the walls, as shown in the shear flow diagram in Figure 3. The shear distribution and resultant shear vector $$V$$ for a small piece of the wall is also shown. This distribution is also mechanically equivalent to the linear distribution shown in Figure 4, where $$\tau =\frac{V}{t}$$.

The total torque ($$T$$) can be described as the sum of the torques on each wall of the box beam as follows: $$T = T_{AB}+T_{BC}+T_{CD}+T_{DA}$$

$$T_{AB}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{BC}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

$$T_{CD}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{DA}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

Therefore, $$T=2t\tau ab$$, and $$\tau =\frac{T}{2abt}$$.

There are two cases that must be analyzed in order to solve the problem:
 * 1) Assume the bending normal stress (σ) reaches $$\sigma_{allow}$$ first, then verify that $$\tau \le \tau_{allow}$$.
 * 2) Assume the shear stress (τ) reaches $$\tau_{allow}$$ first, then verify that $$\sigma \le \sigma_{allow}$$.

Case 1

Case 1 begins by recalling the formula for bending normal stress: $$\sigma =\frac{Mz}{I}$$, where $$M$$ is the bending moment, $$z$$ is the ordinate of a point on the axis perpendicular to the bending neutral axis, and I is the second area moment of inertia defined by $$I=\iint_A z^2 \ dy\, dz$$. Figure 5 shows each item pictorially.

HW1 Contributing Team Members
The following students contributed to this report: John Saxon Eas4200c.f08.vqcrew.c 03:26, 19 September 2008 (UTC) Adam Edstrand Eas4200c.f08.vqcrew.a 03:43, 19 September 2008 (UTC) Javier Stober Eas4200c.f08.vqcrew.b 13:22, 19 September 2008 (UTC) Darin Toscano Eas4200c.f08.vqcrew.d 04:02, 19 September 2008 (UTC) Kevin Klauk Eas4200c.f08.vqcrew.E 12:21, 19 September 2008 (UTC)

title: User:Eas4200c.f08.vqcrew.c/Homework 1


text:
User talk:EML4500.f08.RAMROD.E

 Hello EML4500.f08.RAMROD.E, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 00:10, 9 September 2008 (UTC)

title: User talk:EML4500.f08.RAMROD.E


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User talk:Eml4500.f08.bottle.butler

 Hello Eml4500.f08.bottle.butler, 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 00:11, 9 September 2008 (UTC)

title: User talk:Eml4500.f08.bottle.butler


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User talk:EAS4200c.f08.blue.b

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Talk:Novel economic mechanisms

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User:Eml4500.f08.ateam.didomenico/hw 1

A-Team Teampage

EML 4500 Notes: September 5, 2008
Read: Chapter 4 – Trusses, beams, frames

Chapter 1: Big Picture

Section 1.1: Discretization

Section 1.1.1: Plane truss elements

Section 1.2 Assembly of element equations
 * Example: Five-bar truss

Section 1.4 Element solution and model validity

Section 1.4.1 Plane truss elements

Steps to solve simple truss system described previously
1. Global picture

The global picture consists of a diagram of the entire structure detailing all of the nodes and nodal forces relevant to the problem. A global coordinate system is selected. At the structure level there are two components to be concerned about. These are the global degrees of freedom, or displacement degrees of freedom, and the global forces. Both are divided into known parts and unknown parts. In the previous truss example, the known parts consist of the applied force and the fixed degrees of freedom for nodes 1 and 3. These two nodes are fixed, therefore the degrees of freedom for each is constrained to zero. The unknown part is solved using the Finite Element Method (FEM). FEM is a numerical method used to obtain solutions to ordinary and partial differential equations. In the previous truss system, the reactions forces are unknown.

2. Element picture

An element picture is made of each individual part of the structure. These elements are considered "free" and a free-body diagram with a local coordinate system is made for each. Included in the element picture are both the element degrees of freedom and the element forces. These can be described either in the global coordinate system or the local coordinate system.
 * Element degrees of freedom: $$d_1,\ d_2,\ d_3,\ d_4,$$
 * Element forces: $$f_1,\ f_2,\ f_3,\ f_4$$

3. Global force-displacement relationship

The global force-displacement relationship is shown through matrices. These matrices consist of:
 * Element stiffness matrix in global coordinates ($$k$$) . The dimensions of this matrix are $$n$$ x $$n$$ where $$n$$ is the number of both known and unknown displacement degrees of freedom.
 * Element displacement matrix in global coordinates ( $$d$$) . The dimensions of this matrix are $$n$$ x $$1$$.
 * Element force matrix in global coordinates ($$f$$) . The dimensions of this matrix are $$n$$ x $$1$$.

To assemble the element stiffness matrices, the element displacement matrices, and the element force matrices into global force-displacement relationship matrix multiplication is applied in the form of:

 * $$kd= f$$

This is considered a free-free system where $$k$$ is singular, meaning $$k$$ is not invertible, or $$k^{-1}$$ does not exist.

4. Elimination of known degrees of freedom

In order to make the method more manageable, the global force-displacement relationship is reduced by eliminating the known degrees of freedom. This makes the stiffness matrix non-singular, or invertible. The form is:

 * $$KD= F$$

$$K$$ is the new stiffness matrix with dimensions $$m$$ x $$m$$, where $$m$$ is the number of unknown displacement degrees of freedom. Therefore, $$m$$ < $$n$$. Similarly, $$D$$ is the new displacement matrix with dimensions $$m$$ x $$1$$ and $$F$$ is the new force matrix with dimensions $$m$$ x $$1$$.

To compute the displacement, the equation is rearranged:

 * $$D = K^{-1}F$$

5. Compute element forces

Solving for $$D$$ above, the element stresses can now be determined.

6. Compute reactions

From the element stresses, the reaction forces can be determined.

Example: Truss problem

 Data

Element length:

L(1) = 4

L(2) = 2

Young's Modulus:

E(1) = 3

E(2) = 5

Cross section area:

A(1) = 1

A(2) = 2

Inclination angle:

θ(1) = 30°

θ(2) = -45°

1. Global picture

Global degrees of freedom: $$d_1,\ d_2,\ d_3,\ d_4,\ d_5,\ d_6$$

n = global node number

Note: When numbering the displacement degrees of freedom, it is useful to follow the order of the global node number. For each individual node, number the displacement degrees of freedom in the order of the global coordinate axes.

Global forces: $$f_1,\ f_2,\ f_3,\ f_4,\ f_5,\ f_6$$

Global force matrix

 * $$\begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix}$$

Global stiffness matrix

 * $$\begin{bmatrix}

K_{11}&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&K_{66} \\ \end{bmatrix} $$

Global displacement matrix

 * $$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}$$

To solve for the unknowns, the matrices are arranged according to the equation above.

 * $$\begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} = \begin{bmatrix}

K_{11}&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&K_{66} \\ \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}$$

title: User:Eml4500.f08.ateam.didomenico/hw 1


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User:Eas4200c.f08.blue.a/Homework1

EAS 4200c Homework Report 1 Team Blue
Lecture Note Re-writes

Notes

Homework Problems - NONE FOR REPORT 1

Homework Problems

title: User:Eas4200c.f08.blue.a/Homework1


text:
User:Eas4200c.f08.aero6.ortega/hw1

EAS4200c Learning Method
Homework reports will be collaborative efforts of teams of 6. Each team will use wiki software to present class activities including lectures, homework problems and additional contributions. This collaborative style allows students to assist and learn from each other while discussing the course material. The method also encourages a larger class collaboration as the best of each set of homework reports will be selected for the entire class to review. Each student will be evaluated by all of his or her teammates. This will be used to ensure that each student receives a deserved portion of their team's grade.

Additionally, students will be graded on traditional tests. These exams will be taken individually without outside resources.

The following is an outline of material presented in the text, Mechanics of Aircraft Structures by C.T. Sun

Preface to First Edition
Three main tasks of the text include:
 * 1. FEA modeling of problems and judging correctness of numerical results. Realizing new FEA codes have recently revolutionized the field of structural analysis in all mechanical applications.
 * 2. Fracture mechanics
 * 3. Composite properties in aircraft structures, brief study of laminates and the calculation of Young’s modulus derived from directional displacements as opposed to isotropic alloy mechanics for the purpose of describing torsion and bending problems.

In addition, Griffith’s Criterion is introduced for describing the relation between the strain energy release rate and crack extension. Also, buckling and post-buckling of bars and panels utilized in aircraft structures is introduced.

Preface Notes Second Edition modifications include material on the following topics: a Warping, Boundary Constraint Effects, Saint-Venant’s Principle, Shear Lag, Timoshenko Beam Theory, Plasticity and Fracture.

Introduction
Weight is the main consideration in aircraft structure design, unlike the structures of most civil engineering applications. Also, an aircraft’s structural design is modeled after the primary concern of the components, such as lift/drag considerations. Thus, limiting structural design options and resorting to material of the highest strength to weight ratios. Aircraft components typically follow the shell-like monocoque or stiffened shell semimonocoque structure. In the past, aluminum and titanium alloys have been the material of choice in aircraft design due to their high stiffness/weight and strength/weight ratios. Today, fiber-reinforced composites are replacing much of the weight in designs. Basic structural loads of an aircraft include axial, bending or torsional.

Axial Member
Axial members are employed to support loads applied, as the name would imply, in the axial direction.

Recall $$\sigma = E\epsilon\!$$ (Eqn 1.1 pg.2)

where

 * $$\sigma\!$$ is the "resulting uniaxial stress" from the load specified above (Mechanics of Aircraft Structures, C.T. Sun)
 * $$E\!$$ is the Young’s modulus and
 * $$\epsilon\!$$ is the normal strain in the loading direction and

Therefore $$F = A\sigma = EA\epsilon\!$$ (Eqn 1.2 pg.2)

where

 * $$F\!$$ is the total axial force and
 * $$A\!$$ is the cross sectional area

$$EA\!$$ is termed axial stiffness and is independent of the shape of the cross-section, rather determined by Young’s modulus and the cross-sectional area as illustrated in Figure 1. A large value of $$E\!$$ corresponds to high stiffness. For brittle materials the stress vs. strain curve is similar to that of Figure 2. For metals and other ductile materials, the stress vs. strain curve is similar to that of Figure 3. Channels are used in place of rods to increase bending stiffness.

Buckling strength is enhanced by adding lateral supports such as ribs and frames.

Shear Panel
These are thin sheets of material that carries in-plane shear loads. The following formula can be used to describe the shear force in the x-direction, $$V_x\!$$, for a shear panel under constant shear stress $$\tau\!$$ as illustrated in Figure 4. $$V_x = \tau ta = G\gamma ta\!$$ (Eqn 1.3 pg.4)

where

 * $$G\!$$ is shear modulus,
 * $$\gamma\!$$ is shear strain and
 * $$t\!$$ is thickness.

Similarly, curved plates may have their shear stress loads decomposed into $$V_x\!$$ and $$V_y\!$$ as follows and detailed in Figure 5.

$$V_x = \tau ta\!$$ (Eqn 1.4 pg.5)

$$V_y = \tau tb\!$$ (Eqn 1.5 pg.5)

and therefore the resultant force components have the following relationship

$$\frac{V_x}{V_y} = \frac{a}{b}\!$$ (pg.5)

Due to the ratio of geometric value in the $$x/y$$ directions and the corresponding shear stresses, a flat plate is the most efficient in providing shear force per unit of material.

Bending Member
These are members that carry a moment (Beams). Beams can also carry axial loads in tension or compression. Bending moments depend of beam deflection as shown below,

$$M = -EI\frac{d^2\omega}{dx^2}\!$$ (Eqn 1.6 pg.5)

where

 * $$M\!$$ is the bending moment,
 * $$I\!$$ is the moment of inertia and
 * $$\omega\!$$ is the previously mentioned beam deflection.

Where $$EA\!$$ was earlier termed as the axial stiffness, again the new term $$EI\!$$ is called the bending stiffness. Bending stresses lead to transverse stresses, however bending still plays the dominant role. Again, optimizing the cross-sections of beams is the key to increase bending stiffness. To utilize a materials full capacity, the cross-section should move material as far from the neutral axis as possible to take advantage of the linearity of the bending stress distribution over depth in the elastic range. An example of this is shown in Figure 6.

Torsion

Torque induced shear stress $$\tau\!$$ is also linearly distributed along the cross-section, only now it is distributed in the radial direction as shown in Figure 7.

$$T=GJ\theta\!$$ (Eqn 1.10 pg.7)

where

 * $$T\!$$ is the applied torque,
 * $$J\!$$ is the torsional constant and
 * $$\theta\!$$ is the angle of twist per unit length.

$$GJ\!$$ is known as the torsional stiffness and is dependent on the inner and outer radii for hollow cylinders, but generally depends upon the wall thickness. Thin walled structures are very efficient torsional members. To illustrate this, Figure 8 depicts three circular rods, each with equal cross-sectional areas, but with varied wall thicknesses.

It can be easily seen from the equation above that for a given material, the greater the torsional constant, $$J\!$$, the more torque, $$T\!$$, a beam can withstand. For a cylindrical beam, $$J\!$$ is defined by the following equation:

$$J = \frac{1}{2}\pi(b^4-a^4)\!$$ (Eqn 1.11 pg.8)

where

 * $$b\!$$ is the outer radius of the beam, and
 * $$a\!$$ is the inner radius of the beam.

Therefore,

 * $$J_{leftmost} = \frac{1}{2}\pi(1 cm)^4 = \frac{\pi}{2} cm^4\!$$
 * $$J_{middle} = \frac{1}{2}\pi((5.1 cm)^4-(5 cm)^4) = \frac{51.5\pi}{2} cm^4\!$$
 * $$J_{rightmost} = \frac{1}{2}\pi((10.05 cm)^4-(10 cm)^4) = \frac{201.5\pi}{2} cm^4\!$$

This helps show the advantages of thin walled beams in high torque operations, as the rightmost tube has a torsional constant which is two orders of magnitude greater than that of the solid member.

Load Transfer
Box beams due well to illustrate the beam and torsional members of aircraft substructure design. Loads generally are caused by air-pressure, landing gears, power-plants and seats, etc. Box beams are this sheets (shear panel) and longitudinal stringers (axial members) that distribute loads to major load-carrying members to avoid excessive deflection. Ribs collect all transverse loads from the stringers and transfer them to two wide-flange beams (spars).

Wing and Fuselage Structures

Main wing function is to carry air and power-plant loads to the fuselage. The wing itself acts as a beam and torsional member with an outer form designed by aerodynamic considerations. Spars are heavy beams that run span-wise to take transverse shear and bending loads. Wing ribs are planar structures that hold stringers to the desired contour and improve compressive ability of the wing. They are supported by span-wise spars. Subsonic aircraft have relatively thin skins and utilize spars and stringers as the main bending resistance. These wings may be composed of simple spars or a combination of spars and stringers. Supersonic wings on the other hand have thinner airfoils, at the same time requiring thicker skins to withstand high surface air-loads and improve bending resistance of the wing. To improve efficiency, stiffeners may be manufactured as integral parts of the wing. The fuselage must be designed to handle concentrated loads from wings and landing gears primarily. Also, payload and internal pressures must be supported. Again, stringers run along the length of the fuselage and rings maintain the shape of the fuselage and shorten the stringer length, increasing its bending resistance.

Aircraft Materials
These materials include metal alloys, such as steel, aluminum and titanium, and fiber-reinforced composites of either polymer, metal or ceramic matrices. Costs and properties of materials dictate their usage in design. Costs include manufacturing, maintaining and of course initial bulk. Properties vital to performance are density, strength, stiffness, durability, damage tolerance and corrosion resistance. Steel alloys are denser and corrosion prone so they are used for highly loaded critical points and must be coated. Aluminum alloys range from higher strength to higher toughness; yet both are light-weight and serve vital tasks at different sections of the aircraft. Titanium is lighter and stronger than steel but much more expensive and therefore used mostly in military aircraft. They also withstand higher temperatures than aluminum (350o F versus 1000o F).

 * Materials with high stiffness and high strength - steel alloys, titanium alloys, and aluminum alloys

 * Materials with high stiffness and low toughness - glass (toughness is ability to resist fracture)

 * Materials with low stiffness and high toughness - plastic, nylon, aluminum

 * Materials with high stiffness and high toughness - composites

Fiber-reinforced composites usually employ unidirectional fibers of high tensile strength in a matrix of polymer, metal or ceramic material. These sections form thin laminae with a stacking pattern of different fiber orientations to produce laminates with excellent material properties and multidirectional load capabilities. Ceramic composites in particular have excellent heat resistance, fatigue life, damage tolerance and corrosion resistance. One difference between fiber reinforced composites and metal matrix composites is that the metal matrix composite has a metal matrix, with some type of reinforcement. The fiber reinforced composite can have several different types of matrices.

Contributing Team Members
The following students contributed to this report:

Scott Chastain Eas4200c.f08.aero6.chastain 19:52, 12 September 2008 (UTC)

Marlana Behnke Eas4200c.f08.aero6.behnke 19:57, 12 September 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 20:18, 12 September 2008 (UTC)

Eduardo J. Villalba Eas4200c.f08.aero6.villalba 20:40, 12 September 2008 (UTC)

Felipe Ortega Eas4200c.f08.aero6.ortega 18:51, 19 September 2008 (UTC)

References

Figure 2- http://en.wikipedia.org/wiki/Image:Stress_v_strain_A36_2.png

Figure 3- http://en.wikipedia.org/wiki/Image:Stress_v_strain_brittle_2.png nq

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User:Eml4500.f08.Ateam.paquette

 Homework Contributions

/HOMEWORK 1/

/Homework1/

/Homework2/

/Homework3/

/Homework4/

/Homework5/

/Homework6/

/Homework7/

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User:Eml4500.f08.Ateam.paquette/HOMEWORK 1

MATLAB Tutorial: Part 2

Managing M-files

The !-feature allows for editing or printing to be done without having to exit MATLAB. Using the command !ed will let the named file be edited and then returned to MATLAB where you left off. Many commands control the working directory. Pwd returns the name of the present working directory. Cd changes the working directory. Dir or ls lists the contents of the working directory, but what lists only the M-files. Files named matlab are accessible from any directory. M-files have to be in a directory MATLAB can excess.

Comparing efficiency of an algorithms: flops, tic and toc

The efficiency of an algorithm can be measured by the number of floating point operations (flops) performed and the elapsed time. The total number of flops performed is kept track of by the function flops. This number can be reset by the command flops(0). Tic and toc are stopwatch times that can return the elapsed time. Tic starts the timer and toc returns the elapsed time.

Output format

The output on display is controlled by a variety of format commands, but the computations are done in double precision. Format compact suppresses most of the blank lines so more information can be added on the screen, while format loose returns the display back.

Hardcopy

The diary command causes the next code written to be written on the name diskfile until the command diary off is given. After this is done the file can be edited and printed on the local system.

Graphics

MATLAB can create planar plots of curves, 3-D plots of curves, 3-D mesh surface plots, and 3-D faceted surface plots.

Planar plots

The command plot produces a liner x-y plot. Plot(x,y) will draw an x-y plot if the vectors x and y are the same length. There can be many different graphic figures, but only one at a time will be call “current.” The command figure(1) will make figure 1 the current figure if it was not before, and gcf will return the number of which figure is current. An example of a linear x-y plot is shown below.

 * x = -1.5:.01:1.5; y = exp(-x.^2); plot(x,y)

Fplot is used to plot the graph of a function. Plots can be made of parametrically defined curves as well.

 * t=0:.001:2*pi; x=cos(3*t); y=sin(2*t); plotx,y)

All graphs can be given titles, labeled axes, and text within the graph.

 * Title: graph title

 * xlabel: x-axis label

 * ylabel: y-axis label

 * gtext: place text on graph using mouse

 * text: place text at specified coordinates

Axes are auto-scaled, but this can be overwritten with the command axis. This command should always be given after the plot command. Multiple plots can also be displayed on a single graph.

 * x = 0: .01:2*pi; y1 = sin(x); y2 = sin(2*x); y3 = sin(4*x); plot(x,y1,x,y2,x,y3)

This can also be done with the hold command. It freezes the current on screen graphics so other plots can be put on the same graph, but the axes may become rescaled. Colors and line types can be added to a plot as well. The code plot(x,y,’r--‘) shows a red dashed line.

Graphics hardcopy

The command print is the easiest way to send a high-resolution copy of the current graphics figure to the default printer. The printopt M-file identifies the default setting used by the print command. Print filename saves the figure to the filename given in the default format. If there is no extension in the filename then the default file format will be assigned. A file will be overwritten if the filename already exists. The option –append will append the current graphics figure into the existing file so more than one figure can be saved in each file. The command code is print –append filename. This command can be overwritten with the option –f3.

3-D line plots

Plot3 makes a curve in three dimensions. If x, y, and z are defined vectors of the same size, then plot3(x,y,z) will create a 3-D curve passing through the coordinates of the respective elements. A zlabel can still be added to this plot. The helix shown below was created using the code:

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

3-D mesh and surface plots

A three dimensional wire mesh surface plot with the elements of matrix z is created with the command mesh(z). The function meshgrid allows for a graph to be drawn over a rectangle by allowing matrices to be created. Surface plots are created using surf(z).

z=e-x^2-y^2 can be drawn over the square [-2,2]x[-2,2] with the following codes:
 * xx = -2: .2:2;

 * yy = xx;

 * [x,y] = meshgrid(xx,yy);

 * z = exp(-x.^2 – y.^2);

 * mesh(z)

Surf(z) of the same code is shown as well.

There are three different settings for the color shading of surfaces: faceted, interpolated, and flat. The command shading should be done after the surf command. Different color profiles can be produced by the command colormap.

Handle Graphics.

MATLAB contains many different graphic functions. The user is able to control almost all of the graphics allowing for detailed plots to be made.

Sparse Matrix Computations

MATLAB provides the ability to save storage space, so larger problems can be solved, by only storing the nonzero elements of the matrix. Matlab has the storage mode full, the default, and sparse. The sparse matrix is stored as a linear array of its nonzero elements. The output of a matrix in sparse mode is a list of nonzero entries and their row and column indices. A matrix is sparse if any constituent block is sparse.


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User:Egm6321.f09.Team06.sada/HW1

Problem 1
Derive the Equation 1 (Lecture 1) 1st order total time derivative of f where f is a function of (Y1(t),t).

Y1(t) is the Nominal motion

f(s,t)

s=X1 = axial co-ordinate

t=time

Evaluating at s=Y1(t), we have f(Y1(t),t)

To find the 1st derivative of function f(Y1(t),t)

$${d\over dt}f(Y^1(t),t)=$$$$\frac{\partial f}{\partial s}+\frac{\partial f}{\partial t}={\partial f\over\partial s}(Y^1(t),t)\dot{Y^1}(t)+\frac{\partial f}{\partial t}(Y^1(t),t)(1)$$

1st Derivative $$=\Bigg[{d\over dt}f(Y^1(t),t)={\partial f\over\partial s}(Y^1(t),t)\dot{Y^1}(t)+\frac{\partial f}{\partial t}(Y^1(t),t)\Bigg]$$

Problem 2
Derive the Equation 2 (Lecture 1) 2nd order total time derivative of function f.

1st Derivative $$=\Bigg[{d\over dt}f(Y^1(t),t)={\partial f\over\partial s}(Y^1(t),t)\dot{Y^1}(t)+\frac{\partial f}{\partial t}(Y^1(t),t)\Bigg]$$

$$={\partial f\over\partial s}(Y^1(t),t)\ddot{Y^1}(t)+\dot{Y^1}(t)\Bigg({\partial \over\partial s}{\partial f\over\partial s}(Y^1(t),t)\dot{Y^1}(t))+{\partial \over\partial t}{\partial f\over\partial s}(Y^1(t),t)\Bigg)+{\partial \over\partial s}{\partial f\over\partial t}(Y^1(t),t)\dot{Y^1}(t))+{\partial \over\partial t}{\partial f\over\partial t}(Y^1(t),t)$$

2nd Derivative $$=\Bigg[{d^2\over dt^2}f(Y^1(t),t)={\partial f\over\partial s}(Y^1(t),t)\ddot{Y^1}(t)+{\partial f\over\partial ss}(Y^1(t),t)\dot{Y^1}^2(t)+2{\partial f\over\partial st}(Y^1(t),t)\dot{Y^1}(t)+{\partial f\over\partial tt}(Y^1(t),t)\Bigg]$$

title: User:Egm6321.f09.Team06.sada/HW1


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lot of
Use clear instead of a lot of  .Crochet.david 07:39, 21 November 2008 (UTC)

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University of Florida/Eml4500/f08.FEABBQ/notes3

Trusses, Matrix Method
The Truss is a commonly studied structure and serves as a good example for illustrating many concepts, such as the Finite Element Method (FEM). While a truss is usually defined by three or more beams comprising one or more triangular elements, the analysis of the structure in Figure 1 below follows that of a truss due to its constraints at points 1 and 3. Figure 1

The structure above is statically determinate, and therefore does not require the FEM. However it was the example used in class to demonstrate a few key concepts. The first, although a review from statics, was the free body diagram (FBD). First the global FBD, which includes the entire structure, was drawn. This was followed by a local FBD of each beam. In addition, a numbering scheme was introduced to avoid confusion in more difficult problems to be encountered later. This includes using numbers enclosed by: circles for nodes (joints), and triangles for beams.

In addition, the force displacement relation based on the model of a 1-D spring was introduced. Also, it was shown how matrices can be used to solve multiple degree of freedom (dof) systems.


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title: User:Eml4500.f08.echo.sell.br


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User talk:Highhi~enwikiversity

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 21:33, 9 September 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Highhi. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Highhi~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:38, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:11, 19 April 2015 (UTC)

title: User talk:Highhi~enwikiversity


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University of Florida/Eml4500/f08.FEABBQ/9 3 Notes

Trusses, Matrix Method
The Truss is a commonly studied structure and serves as a good example for illustrating many concepts, such as the Finite Element Method (FEM). While a truss is usually defined by three or more beams comprising one or more triangular elements, the analysis of the structure in Figure 1 below follows that of a truss due to its constraints at points 1 and 3. Figure 1

The structure above is statically determinate, and therefore does not require the FEM. However it was the example used in class to demonstrate a few key concepts. The first, although a review from statics, was the free body diagram (FBD). First the global FBD, which includes the entire structure, was drawn. This was followed by a local FBD of each beam. In addition, a numbering scheme was introduced to avoid confusion in more difficult problems to be encountered later. This includes using numbers enclosed by: circles for nodes (joints), and triangles for beams.

In addition, the force displacement relation based on the model of a 1-D spring was introduced. Also, it was shown how matrices can be used to solve multiple degree of freedom (dof) systems.


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University of Florida/Eml4500/f08.FEABBQ/9-3 Notes

Trusses, Matrix Method
The Truss is a commonly studied structure and serves as a good example for illustrating many concepts, such as the Finite Element Method (FEM). While a truss is usually defined by three or more beams comprising one or more triangular elements, the analysis of the structure in Figure 1 below follows that of a truss due to its constraints at points 1 and 3. Figure 1

The structure above is statically determinate, and therefore does not require the FEM. However it was the example used in class to demonstrate a few key concepts. The first, although a review from statics, was the free body diagram (FBD). First the global FBD, which includes the entire structure, was drawn. This was followed by a local FBD of each beam. In addition, a numbering scheme was introduced to avoid confusion in more difficult problems to be encountered later. This includes using numbers enclosed by: circles for nodes (joints), and triangles for beams.

In addition, the force displacement relation based on the model of a 1-D spring was introduced. Also, it was shown how matrices can be used to solve multiple degree of freedom (dof) systems.


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User:Eml4500.f08.ateam/hw 1/overview

EML4500 - Week 1 Summary

Overview: EML4500 is an upper level engineering course in the field of Finite Element Analysis and Design. The course, taught by Dr. L. Vu-Quoc at the University of Florida is an overview of the Finite Element Method and its applications. Throughout the course of this semester we will be introduced to the different aspects of the FEM and be applying these to solve different engineering problems. The curriculum will be presented to us through lectures from the professor and homework assignments that will assist in our understanding of the material. To ease in the understanding of the material we will work in small groups that will collaborate on weekly homework reports to be submitted using Wikipedia.

Wiki Homework Reports: As stated previously, we will be using Wikipedia (MediaWiki, WikiVersity,...) to submit our weekly HW reports. Working in small teams, we will be collaborating on the current HW questions and lecture notes to form a weekly HW report. These reports will consist of two parts: the HW problem statement/solution and a summary of that week's lecture notes. For the problem statement and solution we will work with our team to come up with a solution to the problem to the best of our ability and form this into the report. After proof-reading and, if needed, editing each of our team members contribution to the report, our team leader will submit a final copy using our Wiki page. The second half of our weekly reports will consist of a summary of the prevous week's lecture notes. For this section, we will take the lecture notes, rewrite them in our own words and also add any comments or suggestions to clarify and topics that we might not fully understand. In addition to the notes we will also redraw any figures or examples that would assist in our learning of the material.

Pros/Cons of Using Wikipedia: Using Wikipedia to work together and submit our HW reports has its pros and cons. Working in our teams will help our understanding of the material, but also presents some difficulties. This is where using Wiki will help. With this, we will be able to collaborate no matter where we are. As Dr. Vu-Quoc stated, "Even if a group member is in Australia, they will still be able to collaborate with the team." This will be beneficial further in the semester when our schedules keep the entire team from being able to meet in person. As with most good things there is the bad part. For our situation, that problem is the "vandalizing" of Wikipedia pages. Wikipedia is a free online encyclopedia that can be edited by anyone on the internet. This is an issue for us when it comes to submitting our weekly reports. Anyone with malicious intent could go to our page and edit our work to provide incorrect results or even just vandalize it with inappropriate content as seen by an example in class. To counteract this issue, we will be submitting the archived version of our report. Once a page on Wikipedia has been archived, that version can not be changed. The following versions might have been edited, but the archived copy will remain the same.

title: User:Eml4500.f08.ateam/hw 1/overview