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User:Eml4500.f08.wiki1.aguilar/TeamHW1

EML4500 Fall 2008 Notes
These class notes cover lectures from the first day of class up to the material covered in the lecture of Friday, Sept. 5th 2008. Brooks Handy is this section's primary author.

Syllabus
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 EML 4500 Finite Element Analysis and Design (FEAD) [http://clesm.mae.ufl.edu/wiki.pub/index.php/Main_Page |Dr. L. Vu-Quoc|] Fall 2008

Course description
Fundamentals of the finite element (FE) method and applications to structural, solid, and bio-mechanics. Principle of virtual work. Derivation of the stiffness matrices for truss, beam, and solid elements. Coordinate transformation for structural elements in 3-D. Direct stiffness method. Static analysis of FE systems with a matrix linear algebra approach, internal structure of a FE code, coding and plotting. Linear stability analysis of truss and beam structures. Applications to aerospace structures and biomechanics. Other topics such as axisymmetric solid elements, dynamic analysis, substructuring analysis, etc. could be taught, depending on the class and the instructor.

Course objective
This course is intended to be the first Finite Element Method (FEM) course for all engineering students. Its objective is to develop a basic understanding of the fundamentals of the FEM through careful derivation of the stiffness matrix from first principles. Fundamental concepts of linear algebra are taught, e.g., inversion and transpose of matrix product, matrix partitioning and solution by static condensation, determinant, eigenvalue problems, etc. For trusses and beams, coordinate transformation of the degrees of freedom from local to global coordinates and the direct stiffness method to assemble element matrices into global matrix is emphasized. The 3-D aspect is emphasized for all truss and beam elements. Euler-Bernoulli beam theory with shear correction, in 3-D, is taught in detail. The connection of linear algebra to the FEM is emphasized, e.g., rigid-body modes and the singularity of the element stiffness matrix. For solid elements, an introduction to the principle of minimum potential energy and the principle of virtual work taught. Commercial codes can be used to train students on practical aspects of FE analysis. Some aspects of buckling analysis and dynamic analysis, and their connection to linear algebra, are addressed.

Topics
* 3-D truss, beam, and solid structures. Beams with shear correction. * Matrix algebra and its application to finite element solution. * Construction of a FE code, coding and plotting. * Linear stability (buckling) analysis of truss and beam structures. * Applications to mechanical and aerospace engineering, biomechanics, and other engineering    areas (e.g., circuit analysis, heat conduction, etc.) * Class projects: Students will "design" and solve open-ended problems that involve FE model    building and analysis of complex problems (e.g., prostheses, organs, airframes, automobiles,     off-shore platforms, etc.). * Other advanced topics: axisymmetric solid elements, dynamic analysis, nonlinear analysis,    substructuring analyis, etc. depending on mutual interest between the students and the     instructor.

Text and Other Resources
M. Asghar Bhatti, Finite Element Analysis With Mathematica and Matlab Computations and Practical Applications: Fundamental Concepts John Wiley & Sons (February 4, 2005) ISBN: 0471648086 Fri, 07 Jul 2006.

Recommended
* J.S. Przemieniecki, Theory of Matrix Structural Analysis, Dover, 1985, (McGraw-Hill, 1968). ISBN 0-486-64948-2 (Paperback). Library call no. 624.171 P973t. * Y.W. Kwon and H. Bang, Finite Element Method using MATLAB, 2nd edition, CRC Press, 2000. ISBN 0-8493-9653-0. Library call no. TA347.F5 K86 1997 (first edition). * Matlab Guide to Finite Elements: An Interactive Approach, Peter Issa Kattan, Springer Verlag   (February 15, 2003), ISBN: 3540438742.

Web pages of related courses
Fall 2006: EML 4500 Finite Element Analysis and Design. Fall 2003: EML 4500 Finite Element Analysis and Design. Spring 2000: EAS 4210C Finite Element for Engineering Mechanics Spring 1999: EAS 4210C Aerospace Structures 2

Grade determination:
Tentatively, homework/project and class participation including bonuses (31%), exam1 (23%), exam2 (23%), exam3 (23%). Adjustments to the weights could be made at the end of the course.

Homework:
There will be HW assignments, which are to be solved following Cooperative Learning Techniques. HW should be thought of as mini projects, which include "hand solution" with the help of Matlab and the use of the Matlab codes that come with the textbook. Students will also be asked to develop their own Matlab codes. For a tutorial on how to use Matlab, see [|Matlab] matters. See the [|course policy] for more details.

Course outcome:
Upon a successful completion of this course, a student is expected to: The above outcomes are consistent with the Departmental and College of Engineering Educational Objectives and meet the following ABET criteria:
 * 1) Have an understanding of matrix linear algebra and its role in the solution of FE systems.
 * 2) Master the derivation of various structural elements in 3-D (truss, beam), and the associated transformation of the stiffness matrix from local coordinates to global coordinates by the Principle of Virtual Work.
 * 3) Know how to correct for shear deformation in Euler-Bernoulli beam.
 * 4) Have a notion of the derivation of the stiffness matrix of solid elements, and their use in a commercial FE code.
 * 5) Know that the concepts of the FEM are also applied to fields other than structural and solid mechanics, e.g., circuit analysis, heat conduction, fluid mechanics, electromagnetics, etc.
 * 6) Develop skills to tackle open-ended problems in FE analysis and design.
 * 7) Develop communication skills, presentation skills, report-writing skills, team work (cooperative learning).
 * 1) Ability to apply knowledge of mathematics, science, and engineering.
 * 2) Ability to identify, formulate, and solve engineering problems.
 * 3) Ability communicate effectively.
 * 4) Ability to use the techniques, skills, and modern engineering tools necessary for engineering practice.

This page was last updated on: Tue, 12 Aug 2008, 17:56:54 EDT

 * }

Class Structure
The class is structured to focus on teamwork. Everyone in the class is broken up into groups of 5-6 members. In order to facilitate members working together, Wikiversity plays a vital role in assignments and projects.

Through Wikiversity, the members are able to access other group members information and everyone is able to work on an assignment or project. Wikipedia and class guidelines provide for a safe and effective method for project submittal and significantly reduce the chances of data loss.

Labeling Conventions
Bar Element 1: Global Free-Body Diagram

The structure should be labeled using numbers in triangles to identify elements, numbers in circles to identify global nodes, and numbers in squares to identify local nodes (in local free body diagrams). Draw free body diagrams to identify all forces and reactions. Draw a free body diagram for each element. Identify forces using Fie to describe the ith internal force on element e. Identify displacements with die using the same labeling convention.

Describe elements as springs
Next we must understand that each element can be described as a spring, under force F, displacement d, and with a spring constant K. The governing equation is then F=Kd. When considering a spring with 2 free ends, it can be analyzed using a matrix equation F=Kd.

6 Steps of the FEM
The Finite Element Method (FEM) is a six step process. The first step is to draw a global picture at a structural level including the global degrees of freedom (DOF) and global forces. The displacement DOFs can be partitioned into a known part (constraints) and an unknown part (must be solved using FEM). The global forces can also be partitioned into a known part (applied forces) and an unknown part (reactions). The second step is to draw individual element pictures including DOFs and forces in either the global coordinate system, or a local coordinate system. The third step is determine the global force-displacement relationships. Assemble the element stiffness and force matrices using global coordinates in the form Kd=F where K is a nxn matrix, d is a nx1 matrix, and F is a nx1 matrix. The fourth step is to eliminate the known DOFs. The stiffness matrix is non-singular and therefore invertable. We reduce the equation to  d = K -1 F . The fifth step is the compute the element forces from the known d and determine the element stresses. The sixth step is to compute the reactions.

MATLAB Tutorial
This section is based on the MATLAB primer tutorial that was written by Mermit Sigmon of the University of Florida. This section had multiple primary authors and was separated as follows: Sections 1-5: Bryan Ambrosio Sections 6-9: Michael Aguilar Sections 10-12: Lyndsey Schaet Sections 13-17: Clayton Brannon Sections 18-19: Benjamin Oatley

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 Sections 1-5

Section 1 How to Access MATLAB
After logging onto a computer with MATLAB installed, run the system command matlab or click on the desktop icon. To exit MATLAB, enter quit or exit in the MATLAB command window, or click on the File menu and select Exit MATLAB.

When the Editor in MATLAB is opened, it is convenient to have the Editor and Command windows opened separately. The specifics of the Editor are covered later on in this tutorial.

Section 2 Entering Matrices
MATLAB is based around operating with matrices. In fact, MATLAB is an abbreviation for Matrix Laboratory. The program can interpret these matrices as either scalars or vectors.

Matrices can be created by entering an explicit list of elements, generating built in statements and functions, creating a disk file with a local editor, or loading external data files or applications.

The following are all examples for creating matrix A, which is a 3x3 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]

Notice that the elements in the rows of the matrix can be separated by commas or blank spaces.

The following is an example of how to enter a number in exponential form.

x = 2.34e-9

Notice when the number in entered, there are no spaces in between the e.

MATLAB is capable of doing computations with complex numbers. The following are examples of how to create complex matrices.

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

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

Notice when the complex numbers were listed inside the matrix, no blank spaces were used. It should also be noted that either i or j can be used as the imaginary unit. Also, if i or j are assigned as variables to represent a new value, they will not retain their value as an imaginary number. For example if the user enters the command

i = 5

then i will no longer be equal to 'sqrt(-1)', it will be equal to 5.

If a large matrix must be entered, creating an ASCII file is a good option because it can make it easier to correct errors. If an ASCII file is created and named data.ext, simply type load data.ext into the MATLAB command line and press enter. This command will read the file and save it as the variable data. MATLAB also has built in functions such as rand, magic, and hilb. If rand(n) is typed into the command line, MATLAB will create a n x n matrix with randomly generated numbers distributed uniformly between 0 and 1. If rand(m,n) is typed into the command line, MATLAB will create a m x n matrix with the same criteria as before. If magic(n) is typed into the command line, MATLAB will create an integral n x n matrix which is a magic square. This means all the rows, columns, and diagonals have a common sum. If hilb(n) is typed into the command line, MATLAB will create a n x n Hilbert matrix.

Individual entries in a matrix can be easily referenced. For example,

A(2,3)

selects the value in the second row, third column of matrix A. Also

X(3)

Would select the third coordinate of vector X.

Section 3 Matrix and Array Operations
The following matrix operations are available in MATLAB:

+	addition -	Subtraction ^	power ‘	conjugate transpose \	left division /	right division
 * multiplication

Since a scalar value entered into MATLAB is stored as a 1 x 1 matrix, these operations also apply to those values. If MATLAB is told to perform an operation that is not possible for a specific matrix, an error message will appear on the screen.

Matrix division can sometimes be confusing in MATLAB. The difference in right and left division is shown below.

x = A\b is the solution of A*x = b

x = b/A is the solution of x*A = b (another way of writing this could be x = (A’\b’)’ )

It is also important to note that addition and subtraction operate entry wise, but multiplication, power, and left and right division operate as matrix operations. These matrix operations can operate as entry wise operations by placing a period in front of them. An example is

[1,2,3,4].*[1,2,3,4]	or 	[1,2,3,4].^2

Both of these operations will give [1,4,9,16] as the solution.

Section 4 Statements, Expressions, and Variables and Saving Sessions
MATLAB uses expression language. This means when values are typed in, they are entered in the form of variable = expression or just as expression. For example,

x = 5*4

The evaluation of this expression is created as a matrix and displayed on the screen, such as

x = 20

If a variable and equal sign are not used, the result is automatically saved as the variable ans.

The user can also decide to continue the statement to the next line by using three or more periods (…) followed by enter, or several statements can be placed on a single line by separating them with commas (,) or semicolons (;).

If the variable is rather large, it may be useful to suppress the statement. By placing a semicolon (;) at the end of a statement, the assignment is carried out, but the result will not be printed on the screen.

It is also important to note that MATLAB is case sensitive. For example, alpha is not the same as Alpha.

In order to see a list of variables that are currently being used in the workspace, type who (or whos) in the command line. If a variable needs to be cleared from the workspace, enter clear variablename'. If clear is entered into the command line, all nonpermanent variables will be cleared.

If the unit round off needs to be changed, type in eps (epsilon) to have MATLAB use a round off of 10^-16. This can be useful in specifying tolerances for convergence of iterative processes.

Sometimes a computer will enter a loop that does not stop or takes too long to calculate. These computations can be stopped in MATLAB by pressing CTRL-c.

When MATLAB is closed, all of the variables in the workspace are deleted. If these variables need to be saved, enter save in the command line before closing the program. These variables are written to a disk file named matlab.mat. When MATLAB is started again, enter the command load and the variables will be restored to the workspace.

Section 5 Matrix Building Functions
The following are convenient matrix building functions:

eye		identity matrix zeros 		matrix of zeros ones		matrix of ones diag		create or extract diagnols triu		upper triangular part of a matrix tril		lower triangular part of a matrix rand		randomly generated matrix hilb		Hilbert matrix magic		magic square toeplitz	see help toeplitz

For example, zeros(m,n) creates a m x n matix of zeros, and zeros(n) produces a n x n matrix of zeros. If A has already been defined as a matrix, then zeros(size(A)) will create a matrix of zeros that is the same size as matrix A.

If x has already been defined as a vector, diag(x) creates a diagonal matrix with x down the diagonal. If A has already been defined as a square matrix, then diag(A) creates a vector consisting of the diagonal of A.

Matrices can also be built from blocks consisting of other matrices. For example, if A is defined as a 3 x 3 matrix, then B = [A, zeros(3,2); zeros(2,3), eye(2)] will be 5 x 5 matrix.

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 Sections 6-9

Section 6 For, While, If, and Relations
As with other computer programming languages, MATLAB has the flow control statements For, While, If, and Relations.

For
For loops are started with the command. For example, one of the most simple examples of a for-loop is:

for j=1:4, j end Here, a loop is cycled through, setting the variable j to 1, then 2, then 3, and then 4, and showing each of these outputs in the command window.

Another example, for a given "n", the statement: x=[]; for i=1:n, x=[x,i^2], end which is the same as x=[]; for i=1:n x=[x,i^2] end will produce a certain "n" vector. The statement, x=[]; for i=n:-1:1, x=[x,i^2], end produces the same vector but in the reverse order.

The statements for i=1:m for j= 1:n H(i,j) = 1/(i+j-1); end end H generate an m-by-n Hilbert matrix.

Note: Using the semicolon in the code stops MATLAB from printing unwanted intermediate results.

MATLAB allows for any matrix to be used in place of 1:n. The variable just consecutive assumes the value of each column of the matrix.

While
While loops are done in the exact same way using the command. The general form of the  loop is:

relation

 * statements

As expected, the  loop will continue to be executed as long as the relation remains true.

A basic  loop example is given below. n=0; while n<5 n=n+1; n end This loop cycles through, increasing the variable n by 1 until the relation is no longer true.

If
The  command allows for various if-then statements to be executed. The general form of the  loop is:

 * relation

 * statements

As expected, the  command executes only when the relation is true.

A basic  command example is given below. n=0; if n==0 n=2; n end Here, we set n=0, so the relation is true and theprint statement exectues accordingly. If the code was repeated, n would now be equal to 2, so nothing would be printed.

Multiple branching is also possible. This is done with  and   commands, as shown below. n=x; if n>0 m='Positive'; elseif n<0 m='Negative'; else m='0'; end

We can see how the output changes depending on the value of x.

Relations
MATLAB uses typical programming relationship operators: <    less than >    greater than <=   less than or equal to     >=    greater than or equal to     ==    equal ~=   not equal Note: The assignment statement "=" is different from the "==" relationship operator.

You can also use these logical operators in conjunction with the relationship operators: &    and |    or     ~     not

When applied to scalars, the relation is the scalar 1 if the relation is true, and 0 if the relation is false. For example, executing 1<2 prints "ans = 1", however, if you enter 2<1, the result is "ans = 0".

These relationship operators can also be applied to matrices. If the are the same size, the output is a matrix of 0's and 1's giving the value of the relation between corresponding entries.

Note: if you wish for a statement to be executed if matrices A and B are equal, you could type:

if A == B  statement end

but the seemly obvious opposite below would not execute as expected:

if A ~= B  statement end This statement would only execute if each of the corresponding entries was different. So, alternatively, to execute a statement from two matrices that were even one entry different, you would use:

if A == B else statement end

Section 7 Scalar Functions
Many MATLAB functions are used mainly on scalars. A list of the most common include: sin            asin             exp                    abs               round cos            acos             log (natural log)      sqrt              floor tan            atan             rem (remainder)        sign              ceil

Note: The operate element-wise when applied to a matrix.

Section 8 Vector Functions
Other MATLAB functions operate essentially on a vector (row or column), but act on an m-by-n matrix (m$$>=$$2) in a column-by-column fashion to produce a row vector containing the results of their application to each column.

For example: m=[1,3;2,4]; mean(m) gives the result: ans =

1.5000   3.5000 As you can see, the mean of 1 and 2 is 1.5 and the mean of 3 and 4 is 3.5.

This means that in order to find the maximum entry in a matrix A, we use max(max(A)) rather than max(A).

Note: Row-by-row action can be obtained by using the transpose; for example, mean(B')'.

Some of these vector functions are listed below: max         sum          median          any min         prod         mean            all sort                     std

Section 9 Matrix Functions
MATLAB is especially useful when using any of its various matrix functions. Some of the most useful include: eig      eigenvalues and eigenvectors chol     cholesky factorization svd      single 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, infinite-norm cond     condition number in the 2-norm rank     rank

Note: Some of these functions may have single or multiple output arguments. While  produces a column vector containing the eigenvalues of A,   produces a matrix U whose columns are the eigenvectors of A and a diagonal matrix D with the eigenvalues of A on its diagonal.

For example, if m=[1,2,3;4,5,6;7,8,9], then [u,d] eig(m) outputs:

u =

-0.2320  -0.7858    0.4082        -0.5253   -0.0868   -0.8165        -0.8187    0.6123    0.4082

d =

16.1168        0         0              0   -1.1168         0              0         0   -0.0000
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 Sections 10-12

Section 10 - Command Line Editing and Recall
This section discusses the ease of editing commands using the arrow keys on the keyboard and recalling command lines in Matlab without having to write M-files (section 12).

Editing commands is easy in Matlab. The cursor can be placed anywhere in the command line using the mouse, left/right arrow keys, and backspace. This simple method of editing does not exist in other programming languages such as Fortran. On the PC's that most of us use, the Home, End, and Delete keys can also be used to place the cursor to make revisions. Using the up/down arrow keys allows the user to scroll through past command lines. This allows the user to simply copy and paste a command they want to use again. Once pasted, the user can also alter the command line. This is very useful if you were using a long linear regression equation and only one constant or variable needed to be replaced.

Section 11 - Submatricies and Colon Notation
This section covers the use of colon notation for matricies. It also discusses subscripting by integral vectors. Both of these methods are used to easily manipulate complicated vectors and submatricies. Creative use of the colon notation discussed here can often eliminate the need for loops thus making the execution of MATLAB code faster.

Vectors typically seen as [a b c d e f....n] can take on the notation of [a:n] in Matlab. Using this colon notation makes it easier and simpler to read and execute. One example of the use of this expression 1:5. This expands out to the row matrix [1 2 3 4 5]. When using this notation, the vector does not need to increment by just one either. Adding a middle term defines what level the vector should increment. For example, the expression 0.2:0.2:1.2 is the vector [0.2 0.4 0.6 0.8 1.0 1.2]. Another example is the expression 5:-1:1 gives [5 4 3 2 1].

The colon notation makes it easy to access submatricies as well as alter a given matrix. Below are some examples.

If A is a predefined matrix of arbitrary size n by n, the notation A(1:4,3) is the column vector consisting of the first four entries of the third column of A.

Using the colon by itself takes an entire row or column from a given matrix, in this case the matrix A. For example, the expression A(:,3) is the entire third column of A. The expression A(1:4,:) is the first four rows of A.

Creating new matricies using arbitrary integral vectors also proves useful in Matlab. The expression A(:,[2 4]) is a matrix whose two columns consist of column 2 and 4 of A. This method of manipulation can also be used in an assignment statement. One example of this is the expression A(:,[2 4 5]) = B(:,1:3) replaces columns 2, 4, and 5 of A with the first three columnds of B. Running this command prints and assigns the entire altered matrix A.

This notation also makes multiplying matricies easier. The expression A(:,[2,4]) = A(:,[2,4])*[1 2;3 4] multiplies the columns 2 and 4 in matrix A by the 2x2 matrix [1 2;3 4]. Running this command once again prints the entire altered matrix.

By comparison to other coding languages, using this notation makes adds to Matlab's usefulness when dealing with vectors and matricies.

Section 12 - M-files
M-files are MATLAB files with the ".m" file extension. These are usually created with a local editor, and contain a sequence of statements that can be executed. There are two types of M-files, script files and function files.

Script Files
A script file consists of a sequence of normal MATLAB statements. For example, if you create a script file called, then typing   into the command window will execute the statements that are in the script.

Note: Variables used in scripts are global and will interfere with any variables of the same name within the current MATLAB session.

Function Files
Function files allow you to create your own functions specific to what you are trying to do, and these will have the status as any other function within MATLAB. Function file variables are local by default, but can be set as global.

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

You would save this code as a diskfile with the name. Notice that the first line of code gives the name of the function, the input arguments, and output arguments; this line is the difference between a script file and a function file.

You would execute this function file by entering. This sends  and   to be sent as   and   respectively with the output result set as. Again, these variable are, by default, local variables.

The function are flexible in their use of inputs and outputs. The use of, ("number of input arguments") allows you to set a default value of an omitted input variable. A function may also have multiple output arguments.

To make comments, use '%' and the rest of the line is ignored by MATLAB. It is considered correct to include some beginning comment lines to describe the use of the function, and can displayed for example if  is entered. You should also try to write in efficient code, using vectorized operations and avoiding loops. If  loops must be used, you can make them faster by pre-allocating any vectors or matrices to which the output is stored.

Some MATLAB functions are built in while others are distributed as M-files. A listing of non-built-in functions can be seen with the MATLAB command  functionname.

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 Sections 13-17

Section 13 Text Strings, Error Messages, Input
Text strings can be entered into MATLAB using single quotation marks as follows:

string='text string'

Using the disp function, text strings can be displayed in the following way:

disp('displays text string')

Error messages can be displayed with the error function:

error('error message')

which is useful with M-files, as an M-file execution will be aborted.

Within M-files, the user may be prompted to input data using the input function:

variable=input('Enter number: ')

The variable is assigned the input value and execution continues.

Section 14 Managing M-files
Creating or editing an M-file from within MATLAB can be accomplished using the !-feature. This is useful because all variables would be reset upon exiting MATLAB. System sommands such as copying, editing, and printing can be executed using this feature. As an example, !ed can be used to access the system editor:

>> !ed fileforediting.m

After editing the file, the user will return to MATLAB as it was before entering the editor.

In addition, the editor and MATLAB may be active simultaneously on systems permitting multiple processes, suspending one while working in another.

Debugging tools can be seen using the help dbtype command.

The present working directory can be viewed with the pwd command and cd can be used to change directories. The dir and ls commands list the contents of the directory and what lists the M-files within the directory only. The delete command deletes a diskfile. The type command prints a M-file to the screen.

M-files must be stored in a MATLAB-accesible directory, such as the present working directory. Typically, these files are stored in a subdirectory of the home directory titled matlab.

Section 15 Comparing Efficiency of Algorithms: Flops, Tic and Toc
Efficiency can be measured by counting the floating point operations (flops)performed or the elapsed time.

The flops count is reset using flops(0). Using the command flops again immediately after the operation gives the flops count.

flops(0), var=2+2; flops

The time elapsed for any given command can be measured using the tic command immediately before the command and toc just after it.

tic, command, toc

The processing speed available for the operation will also affect these measurements.

Section 16 Output Format
Computations within MATLAB are performed using double precision, though the format with which the results are shown can be changed with the following commands:

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

Changes to format remain until changed again.

Section 17 Hardcopy
Hardcopy is obtained using the diary command. The command below causes text that appears subsequently on the screen to be written to the named diskfile (default: diary):

diary named diskfile

The command diary on will write text to the file until the command diary off is called.
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 Sections 18-19

Section 18: MATLAB Graphics
One of the functions of MATLAB is the ability to produce several different types of curves and plots. *These include the following, followed by their respective commands:

planar plots of curves		                       plot 3-D plots of curves				       plot3 3-D mesh surface plots				       mesh 3-D faceted surface plots				surf

Planar Plots
For example, to plot a cosine function over the interval -5 to 5 on an x-y plot of the elements of x versus the elements of y, the command is as follows:

x = -5 : .01 : 5 y = cos(x) plot(x,y) Another example of a sample function would be to plot the function y=e^(x^3 )over the interval of -3.5 to 3.5 on an x-y plot of the elements of x versus the elements of y, the command is as follows: x = -3.5 : .01 : 3.5 y = exp(x.^3) plot(x,y)

“fplots”
The function “fplot” allows the user to graph a function with great ease and efficiency. To plot the above function, the function must first be defined in an M-file (e.g. “expnormal.m”). The M-file for the above function would read as follows: function y = expnormal(x) y = exp(x.^3) followed by the command fplot(‘expnormal’, [-3.5,3.5])

Parametric Curves
Parametric curves can be created in MATLAB. The commands for an example parametric function are as follows: t = 0: 0.01: 2*pi x = sin(5*t) y = cos(3*t) plot(x,y) MATLAB graphs can be assigned with titles, labeled axes, as well as text placed within the graph. These options with their respective commands are as follows: graph title					title x-axis label					x label y-axis label					y label place text on graph using mouse		       gtext position text on coordinates			text For example, to assign the label “Force (N)” to the x-axis, the appropriate command would be as follows: xlabel(‘Force (N)’)

Axis Reformatting
MATLAB automatically formats the axes of graphs with an auto-scale. The command “axis” following the “plot” command can override this auto-scale. Some additional options with their respective commands are as follows: set axis scale to prescribed limits			axis([xmin, xmax, ymin, ymax]) freeze scale for subsequent graphs			axis(axis) return to auto-scale					axis auto returns vector v showing current scale		       v = axis same scale on both axes				       axis square same scale and tic marks on both axes		       axis equal turn off axis scaling and tic marks			axis off turn on axis scaling and tic marks			axis on

Superimposing Plots on Top of Each Other
In MATLAB it is possible to place multiple plots on a single graph. One example of such a case is listed in the following commands: x = 0: .01: 2*pi y1 = cos(x) y2 = cos(3*x) y3 = cos(5*x) plot(x,y1,x,y2,x,y3) The same plots can be generated using a matrix Y, as follows: x = 0: .01: 2*pi Y = [cos(x)’, cos(3*x)’, cos(5*x)’] plot(x,Y) Another way to superimpose plots on each other is with the command “hold”. “hold on” will freeze the current plot and will superimpose subsequent plots on it. “hold off” will release the specified hold.

Overriding Default Linetypes, Pointtypes, and Colors
One example of overriding defaults for a dotted line and a line with circles placed at each node is as follows: x = 0: .01: 2*pi y1 = cos(x) y2 = sin(3*x) plot(x,y1, ‘:’,x,y2, ‘o’) Options for linetypes includes the following: solid (-) dashed (--) dotted dashdot (-.) Options for marktypes include the following: point (.) plus (+) star(*) circle (o) x-mark (x) Options for colors include the following: yellow (y) magenta (m) cyan (c) red (r) green (g) blue (b) white (w) black (k) For example, to plot a blue dotted line, the commands are as follows: plot(x,y,’b:’)

Graphics Hardcopy
With the command “print”, MATLAB will send a high-resolution copy of the current graphics figure to the computer’s default printer. The command “printopt” specifies the default setting used the by the “print” command. The command “print filename” saves the current graphics figure to the specified filename. 3-D Line Plots MATLAB can also create curves in three dimensions with the command “plot3”. For example, a parametric graph in three dimensions can be created with the following commands: t = .01: .01: 2*pi x = sin(2*t) y = cos(t) z = t.^2 plot3(x,y,z)

3-D Mesh and Surface Plots
The command “mesh” creates a three-dimensional wire mesh surface plot.

The command “surf” creates a three-dimensional faceted surface plot.

For example, the graph of z=e^(y^3+x^2 )over the square [-3,3] x [-3,3] can be created with the following commands: xx = -3: .1 :3 yy = xx [x,y] = meshgrid(xx,yy) z = exp(y.^3 + x.^2) mesh(z)

Color Shading and Color Profile
The command “shading” edits the color shading of different surfaces in MATLAB. The three settings and their respective commands are as follows: faceted (default)		shading faceted interpolated			shading interp flat				shading flat Color profiles in MATLAB can also be edited using the command “colormap”. Possible colormap options include the following: hsv (default) hot cool jet pink copper flag gray bone For example, to set the color profile of a figure to hot, the command is as follows: colormap(hot)

Section 19: Sparse Matrix Computations
MATLAB has the capability of significantly reducing the computation time in regards to solving matrices. MATLAB has two storage modes, full (default) and sparse.

For a matrix A, the command “nnz(A)” would return the number of nonzero elements in matrix A.

To create a sparse matrix, the necessary commands are as follows: m = 5 n = 5 e = ones(n,1) d = -2*e T = spdiags([e,d,e], [-2,0,2], m, n) The sparse analogs of eye, zeros, ones, and randn are as follows: speye sparse spones sprandn The command “sparse” allows for the generation of a sparse matrix from the listing of its nonzero entries. i = [1 2 3 5 5 5] j = [1 2 3 3 2 1] s = [2 3 4 5 6 7] S = sparse(i, j, s, 4, 3), full(S) The command “sparse(i, j, s, m, n) is made up of several components. i and j list the row and column indices and s lists the nonzero entries of S. m and n are the dimensions of the matrix S.
 * }

Authors
Michael Aguilar Eml4500.f08.wiki1.aguilar 00:44, 23 September 2008 (UTC) Bryan Ambrosio Eml4500.f08.wiki1.ambrosio 14:36, 18 September 2008 (UTC) Clayton Brannon Eml4500.f08.wiki1.brannon 01:06, 18 September 2008 (UTC)

Benjamin Oatley Eml4500.f08.wiki1.oatley 21:22, 17 September 2008 (UTC)

Lyndsey Schaet Eml4500.f08.wiki1.schaet 17:49, 19 September 2008 (UTC)

title: User:Eml4500.f08.wiki1.aguilar/TeamHW1


text:
User:Eml4500.f08.wiki1.aguilar/TeamHW2

EML4500 Fall 2008 Notes
These class notes cover lecture material from Monday, Sept. 8th, 2008 up to Friday, Sept. 19th 2008.

HW1 left off with the description of elements as springs and explained the 6 steps of the FEM process.

In order to continue on with the FEM process and develop the element stiffness matrix, an understanding of director cosines must exits. The director cosines of the $$\tilde{x}$$ axis with respect to the global ($$ \ x $$,$$ \ y $$) coordinates are derived below from the figure below.

\ l^{(e)} =  \cos \theta^{(e)} $$

\ m^{(e)} =  \cos (90^{o} - \theta^{(e)}) $$

\ m^{(e)} =  \sin \theta^{(e)} $$

A further proof is presented below to show the relationship of the director cosines.

 * $$ \bar{\tilde{i}} = \cos \theta^{(e)} \bar{i} + \sin \theta^{(e)} \bar{j}  $$

\begin{array}{lcl} \begin{align} \bar{\tilde{i}} \cdot \bar{i} & = (\cos \theta^{(e)} \bar{i} + \sin \theta^{(e)} \bar{j}) \cdot \bar{i} \\ & = \cos \theta^{(e)} \bar{i} \cdot \bar{i} + \sin \theta^{(e)} \bar{j} \cdot \bar{i} \\ & = \cos \theta^{(e)} (1) + \sin \theta^{(e)} (0)\\ & = \cos \theta^{(e)} \end{align} \\ l^{(e)} =  \cos \theta^{(e)} \end{array} $$

\begin{array}{lcl} \begin{align} \bar{\tilde{i}} \cdot \bar{j} & = (\cos \theta^{(e)} \bar{i} + \sin \theta^{(e)} \bar{j}) \cdot \bar{j} \\ & = \cos \theta^{(e)} \bar{i} \cdot \bar{j} + \sin \theta^{(e)} \bar{j} \cdot \bar{j} \\ & = \cos \theta^{(e)} (0) + \sin \theta^{(e)} (1)\\ & = \sin \theta^{(e)} \end{align} \\ m^{(e)} =  \sin \theta^{(e)} \end{array} $$

In order to describe the global force displacement at the element level, the k(e) matrix must first be determined. The k(e) matrix for a 4x4 matrix is defined below.

\mathbf{k} ^{(e)} = k^{(e)} \begin{bmatrix} (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{bmatrix} $$

Where

 * $$ k^{(e)} =

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

And

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

Note: The director cosines are the components of ĩ (unit vector along the x-axis) with respect to basis (î,ĵ) î=cosθ(e)î + sinθ(e)ĵ Model 2-bar truss system (continued from previous lecture) Element 1:

\ \theta^{(1)} = 30^{o} $$

\ l^{(1)} = cos\theta^{(1)}=cos30^{o}=\frac\sqrt{3}{2} $$

\ m^{(1)} = sin\theta^{(1)}= sin30^{o}=\frac{1}{2} $$

\ k^{(1)} = \frac{E^1 A^1}{L^1}=\frac{(3)(1)}{(4)} = \frac{3}{4} $$

\mathbf{k} ^{(1)} = \begin{bmatrix} (k_{11}^{(1)}) & (k_{12}^{(1)})    & (k_{13}^{(1)})    & (k_{14}^{(1)})  \\ (k_{21}^{(1)}) & (k_{22}^{(1)})    & (k_{23}^{(1)})    & (k_{24}^{(1)})  \\

(k_{31}^{(1)}) & (k_{32}^{(1)})    & (k_{33}^{(1)})    & (k_{34}^{(1)})  \\ (k_{41}^{(1)}) & (k_{42}^{(1)})    & (k_{43}^{(1)})    & (k_{44}^{(1)}) \end{bmatrix} $$

where

\mathbf{k^{(e)}} = \left[k_{ij}^{(e)}\right] $$ where i = row, j = column, and e = element number. Therefore,

k_{11}^{(1)} = k^{(1)}\left (l^{(1)}\right)^2 = \left (\frac{3}{4} \right)\left(\frac\sqrt{3}{2}\right)=\frac{9}{16}$$
 * $$k_{12}^{(1)} = k^{(1)}\left (l^{(1)}m^{(1)}\right) = \frac{3\sqrt{3}}{16}$$
 * $$k_{42}^{(1)} = -k^{(1)}\left (m^{(1)}\right)^2 = -\frac{3}{16}$$

Two observations: 1. Only need to compute three values, l2, m2, and l*m. The other coefficients have the same absolute value. They just differ by a (+) or (-). 2. The matrix k(e) is symmetric. i.e. kij(e) = kji(e) or k(e)T = '''k(e)

Element 2:

\ \theta^{(2)}= \frac{-\pi}{4} $$

\ l^{(2)} = cos\theta^{(2)}=cos\frac{-\pi}{4}=\frac\sqrt{2}{2} $$

\ m^{(2)} = sin\theta^{(2)}= sin\frac{-\pi}{4}=-\frac\sqrt{2}{2} $$

\ k^{(2)} = \frac{E^1 A^1}{L^1}=\frac{(5)(2)}{(2)} = 5 $$

\mathbf{k} ^{(2)} = \begin{bmatrix} (k_{11}^{(2)}) & (k_{12}^{(2)})    & (k_{13}^{(2)})    & (k_{14}^{(2)})  \\ (k_{21}^{(2)}) & (k_{22}^{(2)})    & (k_{23}^{(2)})    & (k_{24}^{(2)})  \\

(k_{31}^{(2)}) & (k_{32}^{(2)})    & (k_{33}^{(2)})    & (k_{34}^{(2)})  \\ (k_{41}^{(2)}) & (k_{42}^{(2)})    & (k_{43}^{(2)})    & (k_{44}^{(2)}) \end{bmatrix} $$

Therefore, the three values needed to compute the matrix k(2) are as follows:
 * $$ k_{11}^{(2)} = k^{(2)}\left (l^{(2)}\right)^2 = \left (5 \right)\left(\frac\sqrt{2}{2}\right)^2=2.5$$
 * $$k_{12}^{(2)} = k^{(2)}\left (l^{(2)}m^{(2)}\right) = -2.5$$
 * $$k_{42}^{(2)} = -k^{(2)}\left (m^{(2)}\right)^2 = -2.5$$

The given homework was to compute both k(1) and k(2). They are as follows:

\mathbf{k} ^{(1)} = \begin{bmatrix} (\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{bmatrix} $$

\mathbf{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} $$  Element FD relationship: k(e)d(e)=f(e) k is a 4x4 matrix. d is a 4x1 matrix. $$\mathbf{d} = \begin{Bmatrix} d_1^{(e)} \\ \vdots \\ d_4^{(e)} \end{Bmatrix}$$ f therefore is a 4x1 matrix. $$\mathbf{f} = \begin{Bmatrix} f_1^{(e)} \\ \vdots \\ f_4^{(e)} \end{Bmatrix}$$

Global FD relationship: Kd=F K is a nxn matrix. d is a nx1 matrix. F is a nx1 matrix. Here, n is the number of dof's (degrees of freedom). For this problem, n=6.

$$ \begin{bmatrix} K_{11} & K_{12} & \cdots & K_{16}\\ \vdots & & & \vdots \\ K_{61} & \cdots & \cdots & K_{66} \end{bmatrix} \begin{Bmatrix} d_1 \\ \vdots \\ d_2 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ \vdots \\ F_2 \end{Bmatrix} $$

[Kij]6x6{dj}6x1 = {Fi}6x1

Knxn = [Kij]nxn = global stiffness matrix

dnx1 = {dj}nx1 = global displacement matrix

Fnx1 = {Fi}nx1 = global force matrix

k(e)4x4d(e)4x1 = f(e)4x1

k(e)4x4 = [k(e)ij]4x4 = elemental stiffness matrix

d(e)4x1 = [d(e)j]4x1 = elemental displacement matrix

f(e)4x1 = [f(e)i]4x1 = elemental force matrix

To get from  global matrices  to  elemental matrices : Use an  assembly process

Global level:

{d1, d2, d3, d4, d5, d6}

Elemental level:

Elem. 1: {d1(1), d2(1), d3(1), d4(1)}

Elem. 2: {d1(2), d2(2), d3(2), d4(2)}

Identification of Global-Local (Elemental) Degrees of Freedom (DOF's):

Global Node 1:

d1 = d1(1)

d2 = d2(1)

Global Node 2:

d3 = d3(1) = d1(2)

d4 = d4(1) = d2(2)

Global Node 3:

d5 = d3(2)

d6 = d4(2)

\begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16}\\ K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26}\\ K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36}\\ K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46}\\ K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56}\\ K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66} \end{bmatrix} = \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} & k_{13}^{(2)} & k_{14}^{(2)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} & k_{23}^{(2)} & k_{24}^{(2)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \end{bmatrix} $$

K_{11}=k_{11}^{(1)}= {9 \over 16}$$

K_{12}=k_{12}^{(1)}= {3 \sqrt{3} \over 16}$$

K_{13}=k_{13}^{(1)}= -{9 \over 16}$$

K_{14}=k_{14}^{(1)}= -{3 \sqrt{3} \over 16}$$

K_{21}=k_{21}^{(1)}= {3 \sqrt{3} \over 16}$$

K_{22}=k_{22}^{(1)}= {3 \over 16}$$

K_{23}=k_{23}^{(1)}= -{3 \sqrt{3} \over 16}$$

K_{24}=k_{24}^{(1)}= -{3 \over 16}$$

K_{31}=k_{31}^{(1)}= -{9 \over 16}$$

K_{32}=k_{32}^{(1)}= -{3 \sqrt{3} \over 16}$$

K_{33}=k_{33}^{(1)}+k_{11}^{(2)}= {9 \over 16}+{5 \over 2}=3.0625$$

K_{34}=k_{34}^{(1)}+k_{12}^{(2)}= {3 \sqrt{3} \over 16}+-{5 \over 2}=-2.1752$$

K_{43}=k_{43}^{(1)}+k_{21}^{(2)}= {3 \sqrt{3} \over 16}+-{5 \over 2}=-2.1752$$

K_{44}=k_{44}^{(1)}+k_{22}^{(2)}= {3 \over 16}+{5 \over 2}=2.6875$$ ...

The final global stiffness matrix, K, becomes:

\begin{bmatrix} {9 \over 16} & {3 \sqrt{3} \over 16} & -{9 \over 16} & -{3 \sqrt{3} \over 16} & 0 & 0\\ {3 \sqrt{3} \over 16} & {3 \over 16} & -{3 \sqrt{3} \over 16} & -{3 \over 16} & 0 & 0\\ -{9 \over 16} & -{3 \sqrt{3} \over 16} & 3.0625 & -2.1752 & -{5 \over 2} & {5 \over 2}\\ -{3 \sqrt{3} \over 16} & -{3 \over 16} & -2.1752 & 2.6875 & {5 \over 2} & -{5 \over 2}\\ 0 & 0 & -{5 \over 2} & {5 \over 2} & {5 \over 2} & -{5 \over 2}\\ 0 & 0 & {5 \over 2} & -{5 \over 2} & -{5 \over 2} & {5 \over 2} \end{bmatrix} $$

At this point the global force displacement relationship can now be shown by

\begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16}\\ K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26}\\ K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36}\\ K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46}\\ K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56}\\ K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66} \end{bmatrix}

\begin{Bmatrix} 0 \\  0  \\

d_{3} \\ d_{4} \\

0 \\

0 \\

\end{Bmatrix} = \mathbf{F} $$

By applying the boundary conditions we can delete the corresponding first, second, fifth and sixth columns in the global stiffness matrix K. Using the principal of Virtual Work we can also delete the corresponding first, second, fifth and sixth rows in the global stiffness matrix K.

The resulting force displacement relationship is

\begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix}

\begin{Bmatrix}

d_{3} \\ d_{4}

\end{Bmatrix} = \begin{Bmatrix}

F_{3} \\ F_{4}

\end{Bmatrix} $$

Because our only force, P, is in the X direction and equal to seven

\mathbf{\bar{F}} = \begin{Bmatrix} 0\\ 7 \end{Bmatrix} $$

Now we can solve for the displacements. However, first we must find the inverse of K. In order to do this without a calculator we use the determinate method.

det\mathbf{\bar{K}} = K_{33}K_{44}-K_{34}K_{43} $$

\mathbf{K^{-1}} = \frac{1}{det\mathbf{\bar{K}}}\begin{bmatrix} K_{44} & -K_{34}\\ -K_{43} & K_{33} \end{bmatrix} $$

In order to verify that we do indeed have the inverse matrix we can check

\mathbf{\bar{K}}\mathbf{\bar{K}^{-1}}=\mathbf{\bar{K}^{-1}}\mathbf{\bar{K}}=\mathbf{\bar{I}}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} $$

Now that we have the inverse K matrix we can plug it into

\mathbf{\bar{d}}=\mathbf{\bar{K}^{-1}}\mathbf{\bar{F}} $$

The result is then

\mathbf{\bar{d}}=\begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix}= \begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix} $$

Now that we have solved for the displacements, we can solve for the reactions using the element force displacement relationship

\mathbf{k^{e}}\mathbf{d^{e}}=\mathbf{f^{e}} $$

where

\mathbf{d^{1}}=\begin{Bmatrix} 0\\ 0\\ 4.352\\ 6.1271 \end{Bmatrix} $$

and

\mathbf{d^{2}}=\begin{Bmatrix} 4.352\\ 6.1271\\ 0\\ 0 \end{Bmatrix} $$

k1,k2,d1, and d2 are known, so we can solve for f1 and f2

Element 1: k(e)d(e)=f(e)

\mathbf{k} ^{(1)} = \begin{bmatrix} (\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{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix} = \begin{bmatrix} (-\frac{9}{16})   & (-\frac{3\sqrt{3}}{16})  \\ (-\frac{3\sqrt{3}}{16})   & (-\frac{3}{16})  \\ (\frac{9}{16})   & (\frac{3\sqrt{3}}{16})  \\ (\frac{3\sqrt{3}}{16})   & (\frac{3}{16}) \end{bmatrix}

\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix} = \begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \end{Bmatrix} = \begin{Bmatrix} f_1^{(1)} \\ f_2^{(1)} \\ f_3^{(1)} \\ f_4^{(1)} \end{Bmatrix} $$ where f1 and f2 are the reaction forces and f3 and f4 are the internal forces for element 1. Element 1 is in equilibrium so ΣFx=f1(1) + f3(1) = 0     (1) ΣFy=f2(1) + f4(1) = 0     (2)

Method 2: Statics for a 2-bar truss Use Euler Cut Method Back to FE solution. The question arises, how do you bring P back into the picture. Find the equilibrium at node 2. This is part of the homework.

P^{(1)} = [(f_1^{(1)})^2 + (f_2^{(1)})^2]^{\frac{1}{2}}$$

P^{(2)} = [(f_1^{(2)})^2 + (f_2^{(2)})^2]^{\frac{1}{2}}$$ Therefore,

P = [((f_1^{(1)})+ (f_1^{(2)}))^2 + ((f_2^{(1)}) + (f_2^{(2)}))^2]^{\frac{1}{2}} $$ Which equals our value for P. Therefore, node 2 is in equilibrium.

Authors
Eml4500.f08.wiki1.aguilar 20:23, 26 September 2008 (UTC) Eml4500.f08.wiki1.oatley 19:51, 26 September 2008 (UTC) Eml4500.f08.wiki1.schaet 20:45, 26 September 2008 (UTC) Eml4500.f08.wiki1.brannon 20:44, 26 September 2008 (UTC) Eml4500.f08.wiki1.handy 19:58, 26 September 2008 (UTC) Eml4500.f08.wiki1.ambrosio 14:36, 29 September 2008 (UTC)

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

title: User:Eml4500.f08.wiki1.aguilar/TeamHW2


text:
User:Eml4500.f08.wiki1.aguilar/hw2

= Meeting 7 - 10 September 2008= Note: The director cosines are the components of ĩ (unit vector along the x-axis) with respect to basis (î,ĵ) î=cosθ(e)î + sinθ(e)ĵ Model 2-bar truss system (continued from previous lecture) Element 1:

\ \theta^{(1)} = 30^{o} $$

\ l^{(1)} = cos\theta^{(1)}=cos30^{o}=\frac\sqrt{3}{2} $$

\ m^{(1)} = sin\theta^{(1)}= sin30^{o}=\frac{1}{2} $$

\ k^{(1)} = \frac{E^1 A^1}{L^1}=\frac{(3)(1)}{(4)} = \frac{3}{4} $$

\mathbf{k} ^{(1)} = \begin{bmatrix} (k_{11}^{(1)}) & (k_{12}^{(1)})    & (k_{13}^{(1)})    & (k_{14}^{(1)})  \\ (k_{21}^{(1)}) & (k_{22}^{(1)})    & (k_{23}^{(1)})    & (k_{24}^{(1)})  \\

(k_{31}^{(1)}) & (k_{32}^{(1)})    & (k_{33}^{(1)})    & (k_{34}^{(1)})  \\ (k_{41}^{(1)}) & (k_{42}^{(1)})    & (k_{43}^{(1)})    & (k_{44}^{(1)}) \end{bmatrix} $$

where

\mathbf{k^{(e)}} = \left[k_{ij}^{(e)}\right] $$ where i = row, j = column, and e = element number. Therefore,

k_{11}^{(1)} = k^{(1)}\left (l^{(1)}\right)^2 = \left (\frac{3}{4} \right)\left(\frac\sqrt{3}{2}\right)=\frac{9}{16}$$
 * $$k_{12}^{(1)} = k^{(1)}\left (l^{(1)}m^{(1)}\right) = \frac{3\sqrt{3}}{16}$$
 * $$k_{42}^{(1)} = -k^{(1)}\left (m^{(1)}\right)^2 = -\frac{3}{16}$$

Two observations: 1. Only need to compute three values, l2, m2, and l*m. The other coefficients have the same absolute value. They just differ by a (+) or (-). 2. The matrix k(e) is symmetric. i.e. kij(e) = kji(e) or k(e)T = '''k(e)

Element 2:

\ \theta^{(2)}= \frac{-\pi}{4} $$

\ l^{(2)} = cos\theta^{(2)}=cos\frac{-\pi}{4}=\frac\sqrt{2}{2} $$

\ m^{(2)} = sin\theta^{(2)}= sin\frac{-\pi}{4}=\frac\sqrt{2}{2} $$

\ k^{(2)} = \frac{E^1 A^1}{L^1}=\frac{(5)(2)}{(2)} = 5 $$

\mathbf{k} ^{(2)} = \begin{bmatrix} (k_{11}^{(2)}) & (k_{12}^{(2)})    & (k_{13}^{(2)})    & (k_{14}^{(2)})  \\ (k_{21}^{(2)}) & (k_{22}^{(2)})    & (k_{23}^{(2)})    & (k_{24}^{(2)})  \\

(k_{31}^{(2)}) & (k_{32}^{(2)})    & (k_{33}^{(2)})    & (k_{34}^{(2)})  \\ (k_{41}^{(2)}) & (k_{42}^{(2)})    & (k_{43}^{(2)})    & (k_{44}^{(2)}) \end{bmatrix} $$

Therefore, the three values needed to compute the matrix k(2) are as follows:
 * $$ k_{11}^{(2)} = k^{(2)}\left (l^{(2)}\right)^2 = \left (5 \right)\left(\frac\sqrt{2}{2}\right)^2=2.5$$
 * $$k_{12}^{(2)} = k^{(2)}\left (l^{(2)}m^{(2)}\right) = 2.5$$
 * $$k_{42}^{(2)} = -k^{(2)}\left (m^{(2)}\right)^2 = -2.5$$

The given homework was to compute both k(1) and k(2). They are as follows:

\mathbf{k} ^{(1)} = \begin{bmatrix} (\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{bmatrix} $$

\mathbf{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} $$  Element FD relationship: k(e)d(e)=f(e) k is a 4x4 matrix. d is a 4x1 matrix. $$\mathbf{d} = \begin{Bmatrix} d_1^{(e)} \\ \vdots \\ d_4^{(e)} \end{Bmatrix}$$ f therefore is a 4x1 matrix. $$\mathbf{f} = \begin{Bmatrix} f_1^{(e)} \\ \vdots \\ f_4^{(e)} \end{Bmatrix}$$

Global FD relationship: Kd=F K is a nxn matrix. d is a nx1 matrix. F is a nx1 matrix. Here, n is the number of dof's (degrees of freedom). For this problem, n=6.

$$ \begin{bmatrix} K_{11} & K_{12} & \cdots & K_{16}\\ \vdots & & & \vdots \\ K_{61} & \cdots & \cdots & K_{66} \end{bmatrix} \begin{Bmatrix} d_1 \\ \vdots \\ d_2 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ \vdots \\ F_2 \end{Bmatrix} $$

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User talk:Eml4500.f08.group

hey

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Lunar Boom Town/External links

Useful open/free design tools

 * Dia is free software for creating diagrams. It may be useful for process flow and instrumentation diagrams.
 * Art of Illusion is a free 3D Modeling and Rendering environment incredibly useful for design visualization.
 * Blender is also a 3D Modeling and Rendering environment augmented with exellent import/export data formats involving game development tools and environments.

U.S.G.S. Planetary Data System - Check licensing constraints before using. Summarize here for others. http://pdsmaps.wr.usgs.gov/maps.html

Background information

 * Lunar and Martian dust is a potential problem.
 * Possible solution for Lunar and Martian dust.
 * TANSTAFL Lunar vacuum will be compromised for chip making, requiring vacuum pumping.
 * http://www.kathryncramer.com/lunar_development/2004/11/maximizing_food.html
 * NASA Asteroid Surveys I feel safer already.  Might be a motherlode providing supplies to those asteroid miners!
 * http://en.wikiversity.org/wiki/Topic:Energy_storage
 * http://en.wikiversity.org/wiki/Networked_learning
 * http://en.wikiversity.org/wiki/Marketing_Strategy
 * Popular Mechanics Series on NASA's Planned Lunar Base

Applicable Google discussion groups

 * sci.space.policy
 * sci.space.history
 * sci.space.tech

Software

 * http://freerails.sourceforge.net/

Related news

 * April 2008 Minerals mapped on Lunar Surface

title: Lunar Boom Town/External links


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User talk:Eml4500.f08.team.Bengtson

Homework 1
Graphics Hardcopy

In MATLAB, there is often a need to obtain a copy of the graphic produced. The ‘Print’ option is what is used to do this. By itself this sends a high resolution copy directly to an attached printer. Adding commands such as a filename afterwards writes the current graphic to a file.

Print Eml4500

Writes current graphic to file Eml4500.

However, if the filename already exists then it will be overwritten. To attach several graphics to one file ‘-append’ should be put after the print command and before the filename that is being added to.

Print -append Eml4500

Adds current figure to eml4500 file.

Also, if one wants to get a figure besides the current one to print one just uses –fN with the print command with N being the number of the figure one wishes to write.

Print -f2 Eml4500

Prints the figure 2 to eml4500 file.

3-D Line Plots

A 3-D line plot can be with the plot3(x,y,z) command. This means that x, y, and z already have to be defined which usually is done parametrically.

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

This would show a helix that compresses toward the x-y plan. To add a label to the z-axis there is a command ‘zlabel’. As well as the axis command can be used to set the scaling to prescribed limits.

3-D mesh and surface plots

There a couple commands to create mesh and surface plots. ‘Mesh(z)’ creates a three-dimensional perspective plot of the elements of matrix z. The mesh surfaced is defined by the z-coordinates above the x-y plane. Surface plots are plotted with the command ‘surf(z)’, where z is a matrix. To draw a mesh within a set of perimeter one can define a rectangle with the xx and yy vectors	 [x,y] = meshgrid(xx,yy) where [x,y] is defined through ‘z=’

xx=-3:.3:3;

yy=xx;

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

z = exp(-x.^2 - y.^2);

mesh(z);

To impose shading on the plots and grids there is three options: shading faceted, shading interp, and shading flat. This command should be entered after the surf command. There are several predefined color profiles to choose from. To change it, use the command ‘colormap(hsv)’ which is default color profile to another selection such as ‘colormap(hot)’. The different predefined color values are hsv, hot, cool, jet, pink, copper, flag, gray, and bone.

The command ‘view’ can be used to specify from what location in coordinates the 3D object should be viewed from. It can be in either cartesian or spherical coordinates.

Sparse Matrix Calculations

MATLAB when performing matrix calculations by default assumes that any of the entries may be nonzero. However, sometimes this is not the case and to avoid lengthy calculations that become zero, there is a way to increase the efficiency so the processing power can find solutions to significantly larger problems.

The commands ‘full’ or ‘sparse’ change between the two settings, where a sparse matrix is stored as a linear array of the nonzero entries along with their row and column numbers. A full matrix records every single entry.

The sparse equivalents of eye, ones, and randn are simply done by adding sp before the command, for example ‘sprandn’. The command ‘sparse(m,n)’ creates a sparse zero matrix.

The sparse command can also create a sparse matrix by listing its nonzero components.

An example,

i = [1 2 3 4 4 4]; j = [1 2 3 1 2 3]; s = [5 6 7 8 9 10]

S= sparse(i, j, s, 4, 3);

The nonzero entries come from ‘s’ and their location by row and column number is i and j respectively. The matrix size is given by the last two numbers which makes it an m x n matrix or in this case 4x3.

Most of MATLAB operations can be used regardless of storage mode (either full or sparse). Most operation just involving a sparse matrix will continue to be in sparse mode such as S*S, S+S, inv(S), or S^n, while certain ones will be recorded in full mode such as S+F, S*F, S/F or F/S.

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User:Eas4200c.f08.carbon.orear/HW1

=Homework 1=

Lecture Notes (current version) Lecture Notes (archived version)

HW 2 Contributions (current version) HW 2 contributors (archived version)

Updated Pb1.1 (current version) Updated Pb1.1 (archived version)

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User:Eas4200c.f08.carbon.orear/HW1/Lecture Notes

When designing Aerospace Structures it is important that the structures are light weight, but also strong and stiff. Some commonly used aircraft materials are aluminum, steel, titanium, and carbon fiber composites.

Material Characteristics
Stiffness is defined by the slope of the plotted curve representing the stress and strain relationship. Stress is equal to the product of this slope (designated the elastic modulus, E) and strain. $$\displaystyle \sigma = E \epsilon$$

Strength can be characterized in many ways. Two common measurements important in engineering applications are the yield stress and the ultimate stress. Yield stress is the stress level that a material begins to deform and stretch. Ultimate stress is the stress level that the material ruptures.

Toughness is the ability to resist fracture. A material is said to possess "high fracture toughness" if it resists fracture well and "low fracture toughness" if it is less capable of resisting fracture.

The graph (drawn by Jeff Wright) depicts the general graph of the relationship between stresses and strains in metals.

Material Examples
We are interested in aerospace materials beyond just the cost and manufacurability of aircraft components. The nature of the materials' engineering properties (like strength, stiffness, and toughness) help us choose materials appropriate to a specific application.

The table (drawn by Jeff Wright) shows the mechanical properties of metals at room temperature in aircraft structures, and depicts the correlations between density, tensile ultimate stress, tensile yield stress, and stiffness.

Some material examples are given below:

HIGH stiffness and HIGH strength:
 * Steel Alloy
 * Titanium Alloy

Steel alloys have very high stiffness and strength. Titanium alloys have less high stiffness and strength, and Aluminum alloys still less. Modern aluminum lithium alloys are 10% stiffer and 10% lighter than conventional aluminum alloys.

-Aluminum alloys are resistant to fatigue and cyclic tensile stress, and thus are used on the fuselage and the lower wing skins. Aluminum has historically played a major roll as material used in aircraft structures because it is cheap to buy, manufacture and is abundant, as opposed to Titanium which is expensive.

-Titanium is heavier than aluminum and lighter than steel, and double the yield and ultimate stress of aluminum. It is popular for its corrosion resistance and ability to withstand extreme temperatures up to 1000°F.

-Steel Alloys have extremely high stiffness and strength, but is expensive and has poor resistance to corrosion. Thus, despite its unique material property, is general reserved for landing gear and fittings with high loads.

Reference Sun [2006], pg. 14-15

HIGH stiffness and LOW toughness:
 * Glass

Because of the very low toughness of glass, it has a very limited use on aircraft. Bulk glass is avoided on aircraft, and other forms of glass are preferred such as glass fiber for its improved mechanical properties. Because of the extreme forces acting on the fuselage in-flight, windows are designed to be very small and sturdy. Recent discoveries of new glass-like materials allow for larger windows. Also, the study of metallic glass has become increasingly significant and could prove to have major importance on aircraft in the future. The article below describes some of the issues and discoveries of metallic glass:

Metallic Glass

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

!CORRECTION to HW1 submitted error: Properties with low stiffness and high toughness, from team Carbon
 * LOW stiffness and HIGH toughness:
 * Nylon
 * Plastic
 * Aluminum
 * Aluminum

Aluminum has good fracture toughness and fatigue life, and is therefore used on the fuselage skin.
 * }

This table (drawn by Jeff Wright) shows the material properties of aluminum and steel alloys

HIGH stiffness and HIGH toughness:
 * Composites

Fiber-reinforced composites allow for complex loads by introducing fibers embedded in a matrix (polymers, metals, ceramics, etc.). They are stiff, strong, and light, and are ideal to withstand high loads experienced during flight while still minimizing weight to increase the aerodynamic properties. The composite laminates "have excellent fatigue life, damage tolerance, and corrosion resistance". Fiber composites are becoming very popular for their unique material properties and compose 50% of the new Boeing 787. More facts on the 787 can be found at:

Boeing 787 Fact Sheet

Reference Sun [2006], pg. 15-16

Aspects of Structure
It is key to look at both geometry and material in aerospace structure design. Some geometry examples are mentioned below.

Monocoque structure is that of a single shell design or thin-shell structures. The word has the French base 'coque' meaning, literally, shell. Semi-monocoque, or stiffened shell structures can also be worth exploring. It is important to note that stiffness and brittleness in aerospace structures can be either highly valuable or highly problematic. Normally, shell structures are chosen for their appealing strength for a very limited weight.

Goals with geometry and material
In considering geometry and material in an aerospace structure, one should try to optimize and integrate the two to minimize cost and weight. Sometimes geometry is limited by design goals and aerodynamic properties such as lift and drag. Careful consideration must be taken, therefore, not only to provide sufficient use of the materials for one purpose but to look at the integration of the entire system and how to maximize the utility of the available materials and possible geometric configurations.

The following website contains content not created by Team Carbon. The link was obtained in a google image search. It shows the material breakdown of an F/A-18 Hornet. F/A-18 Materials Full credit is due to http://www.military.cz/usa/air/in_service/aircraft/f18/f18pics/f18materials.jpg


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User:Eas4200c.f08.carbon.orear/HW1/Contributors

HW 1 Contributing Team Members
Jeff O'Rear Eas4200c.f08.carbon.orear 14:45, 18 September 2008 (UTC)

Jeff Wright Eas4200c.f08.carbon.w 16:19, 18 September 2008 (UTC)

Beau Guidry Eas4200c.f08.carbon.guidry 19:59, 18 September 2008 (UTC)

Aaron Clausen Eas4200c.f08.carbon.clausen 14:12, 19 September 2008 (UTC)

Abby Booth Eas4200c.f08.carbon.booth 16:29, 19 September 2008 (UTC)


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User talk:Eas4200c.f08.radsam/Homework

Nice work on Problem 1.1, z. :D Eas4200c.f08.radsam.a. 02:57, 23 September 2008 (UTC)

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User:EAS4200C.F08.WIKI.A/Homework

=Step by Step Procedure To Finding A Solution=

There are two cases for which need to be solved in order to compute the optimal ratio $$\displaystyle \ \frac{b}{a}$$. The first case which we will be looking at is the case in which the maximum stress $$\displaystyle \sigma _{max} $$ equals the allowable stress $$\displaystyle \sigma_{all} $$. The second case that needs to be concidered is the case in which the maximum shearing stress, $$\displaystyle \tau_{max} $$, equals the allowable shearing stress, $$\displaystyle \tau_{all} $$. Case one variables will be represented by a superscript of one and case two variables will be represented by a superscript two.

Case 1
To begin this analysis we have to first look at the equation for stress which is the following equation.

$$\displaystyle \sigma^{(1)} = \frac {M^{(1)}z}{I^{(1)}}$$

where $$\displaystyle \ M^{(1)} $$ is th bending moment, $$\displaystyle \ z $$ is the distance from the neutral axis, and $$\displaystyle \ I^{(1)} $$ is the second area moment of inertia. In this particular problem, $$\displaystyle \ z = \frac {b^{(1)}}{2} $$ and the stress has reached its allowable limit so therefore the stress can be expressed as

$$\displaystyle \sigma_{all}^{(1)} = \frac {M^{(1)}b^{(1)}}{2I^{(1)}}$$

For this case the bending moment must be maximized. The equation for the bending moment can be obtained from the stress equation.

$$\displaystyle \ M_{max}^{(1)} = 2\sigma_{all}^{(1)}(\frac {I^{(1)}}{b^{(1)}})$$

A closer look at the equation will reveal that there is only one variable that can be manipulated in order to maximize the moment equation and it is the fraction of $$\displaystyle \ \frac{I^{(1)}}{b^{(1)}}$$. the equation can therefore be writen as

$$\displaystyle \ M_{max}^{(1)} = 2\sigma_{all}^{(1)}(\frac {I^{(1)}}{b^{(1)}})_{max}$$

As stated above the ratio $$\displaystyle \ (\frac {I^{(1)}}{b^{(1)}})_{max}$$ must then be calculated, expressed in terms of $$\displaystyle \ b^{(1)} $$ using the relationship $$\displaystyle \ a = \frac{L}{2} - b^{(1)}$$ and then manipulated (take derivative and set equal to zero; solve)to obtain the maxium values of $$\displaystyle \ b^{(1)} $$ $$\displaystyle \ a^{(1)} $$.

The next step is to place the calculated values of $$\displaystyle \ b^{(1)} $$ and $$\displaystyle \ a^{(1)} $$ back into the equation representing the ratio of $$\displaystyle \ (\frac {I^{(1)}}{b^{(1)}})_{max}$$ and then putting this ratio back into the equation representing the bending moment.

$$\displaystyle \ M_{max}^{(1)} = 2\sigma_{all}^{(1)}(\frac {I^{(1)}}{b^{(1)}})_{max}$$

Now that the equation for the bending moment is complete and the ratio of $$\displaystyle \ \frac{b^{(1)}}{a^{(1)}} $$ is optimized, the correctness of this ratio must be check. To do so, the maximum shearing stress has to be compared to the allowable shearing stress and this can be completed by recognizing that the maximum bending moment is equal to the maximum torque as stated in the problem statement.

$$\displaystyle \ M_{max}^{(1)} = T_{max}^{(1)}$$

This expression will now be enetered into the expression for the maximum shearing stress given by the following equation;

$$\displaystyle \tau_{max}^{(1)} = \frac{T_{max}^{(1)}}{2a^{(1)}b^{(1)}t}$$

where t is the thickness of the cross-section. Reducing this equation down and replacing $$\displaystyle \sigma_{all}^{(1)}$$ with $$\displaystyle \tau_{all}^{(1)}$$, (stated in the problem), it is now possible to compare $$\displaystyle \tau_{max}^{(1)}$$ with $$\displaystyle \tau_{all}^{(1)}$$.

Case 2
In this case the same approach is taken, but this time it is assumed that $$\displaystyle \tau_{max}^{(2)} = \tau_{all}^{(2)}$$. Also, the torque will be maximized instead of the bending moment. That being said the following is true;

 $$\displaystyle \ T_{max}^{(2)} = 2ta^{(2)}b^{(2)}\tau $$

To maximize this equation, the values of $$\displaystyle \ a^{(2)} $$ and $$\displaystyle \ b^{(2)} $$ must be optimized. To do so, we again put $$\displaystyle \ a^{(2)} $$ in terms of $$\displaystyle \ b^{(2)} $$  ( $$\displaystyle \ a^{(2)} = \frac{L}{2} - b^{(2)}$$), plug this value into the previous equation, and solve this equation in order to find the maximum value of $$\displaystyle \ b^{(2)} $$. This is done by taking the derivative and setting the equation equal to zero and solving. Once the value of $$\displaystyle \ b^{(2)} $$ has been solved for, the value of $$\displaystyle \ a^{(2)} $$ can be solved for from the equation $$\displaystyle \ a^{(2)} = \frac{L}{2} - b^{(2)}$$.

After this has been completed and the values of $$\displaystyle \ a^{(2)} $$ and $$\displaystyle \ b^{(2)} $$ have been placed into the maximum torque equation this value can be equated to the maximum bending moment $$\displaystyle \ M_{max}^{(2)} $$. From this point we are interested in discovering the relationship between the maximum stress and the allowable stress. We proceed forward by looking at the maximum shearing stress equation:

 $$\displaystyle \sigma_{max}^{(2)} = \frac {M_{max}^{(2)}b^{(2)}}{I^{(2)}}$$

The expression for the maximum bending moment and the inverse of the ratio $$\displaystyle \ (\frac {I^{(2)}}{b^{(2)}})_{max}$$ (the corresponding values of $$\displaystyle \ a^{(2)} $$ and $$\displaystyle \ b^{(2)} $$ are input into the ratio) is placed in this equation. The resulting expression will be of the following format,

$$\displaystyle \sigma_{max}^{(2)} = C \tau_{all}^{(2)}$$

where $$\displaystyle C $$ is a constant that comes from reduction. Since we are interested in finding a relationship bewteen the maximum stress and the allowable stress, we have to relace the allowable shearing stress term with an equal value. to do so we must remember the following relationship,

$$\displaystyle \sigma_{all}^{(2)} = 2 \tau_{all}^{(2)}$$

It is now possible to determin whether the values of $$\displaystyle \ a^{(2)} $$ and $$\displaystyle \ b^{(2)} $$ are acceptable.

=Solving Case 1=

Now that the steps have been laid out, we can now proceed to solve Problem 1.1.

Recall:
 * $$\displaystyle \sigma={Mz \over I}$$
 * $$\displaystyle \sigma_{max}^{(1)} = \sigma_{allow}, z= {b \over 2} $$

Thus,
 * $$\displaystyle M^{(1)} = {2I\sigma_{max}^{(1)} \over b} $$
 * or $$\displaystyle M^{(1)}= (2\sigma_{allow})({I \over b})$$

Where $$\displaystyle (I/b)$$ is a function of $$\displaystyle a$$ and $$\displaystyle b$$. Recall from the first assumption:
 * $$\displaystyle L=2(a+b)=constant$$

Thus,
 * $$\displaystyle a={L \over 2}-b$$

Maximizing $$\displaystyle M^{(1)} $$:
 * $$\displaystyle M_{max}^{(1)} = (2\sigma_{allow})({I \over b})$$
 * $$\displaystyle I = \sum_{i=1}^4 [{b_i(h_i)^3 \over 12} + A_i(d_i)^2] $$

 * $$\displaystyle = 2{tb^3 \over 12} + 2[{at^3 \over 12} + (at)({b \over 2})^2] $$

Let
 * $$\displaystyle \alpha = {at^3 \over 12} + (at)({b \over 2})^2 ={at \over 12}[t^2+3b^2]$$

Since $$\displaystyle t $$ is much smaller than $$\displaystyle b $$ then one can conclude that $$\displaystyle t^2<<b^2, t^2<<3b^2 $$ Hence,
 * $$\displaystyle \alpha \approx {3ab^2t \over 12}$$

And therefore,
 * $$\displaystyle I \approx {2tb^3 \over 12} + {ab^2t \over 2} = {tb^2 \over 6}(3a+b)$$
 * $$\displaystyle f(b) := {I \over b} = {tb \over 6}(3a+b) $$
 * $$\displaystyle = {tb(3L-4b) \over 12} = \beta_0 + \beta_1 b^1 + \beta_2 b^2 $$

Where $$\displaystyle \beta_0 = 0, \beta_1 = {3Lt \over 12}, \beta_2 = -t/3 $$
 * $$\displaystyle {d^2f(b) \over db^2} = 2 \beta_2 = {-2t \over 3} < 0 $$
 * $$\displaystyle b = \begin{cases} 0 & \\ {3L \over 4} \end{cases} $$ [[Image:HW_2_9-10_2.JPG|thumb|400px|]]
 * $$\displaystyle {df(b) \over db} = \beta_1 + 2 \beta_2 b = 0 \Rightarrow b^{(1)} = {-\beta_1 \over 2 \beta_2} = {3L \over 8} $$
 * $$\displaystyle a^{(1)} = {L \over 2} - b^{(1)} = {L \over 8} $$

Thus,
 * $$\displaystyle {b^{(1)} \over a^{(1)}} = 3 $$
 * $$\displaystyle ({I \over b})_max = ({I^{(1)} \over b^{(1)} })$$
 * $$\displaystyle M_{max}^{(1)} = 2 \sigma_{allow}({I^{(1)} \over b^{(1)} }) = {3+L^2 \over 32} \sigma_{allow}$$

= Motivation and Solution Continued =

A box beam is a rectangular structure that is important to study due to the fact that many aerospace structures follow this same basic design. A box beam is a form of shell beam, and structures such as an airplane fuselage are also shell beams, the only difference being that their cross sections are circular rather than a rectangle. Likewise, the wings on an airplane are also shell beams with a slightly more complex cross section.

Now that there is a motivation for solving the problem, Case 1 from last lecture is continued as follows:

$$\tau ^{(1)}_{max}$$ is the shear stress corresponding to the maximum torque $$T^{(1)}_{max}$$

Continuing the solution, replace $$T^{(1)}_{max}$$ in the shear equation by $$M^{(1)}_{max}$$. This is done due to the 1st assumption in the problem statement. Secondly, assumption 2 allows for the substitution of $$\tau ^{(1)}_{max}$$ with $$\sigma _{allow}$$

$$\tau ^{(1)}_{max} = \frac{T^{(1)}_{max}}{2a^{(1)}b^{(1)}t}$$

$$\tau ^{(1)}_{max} = \frac{M^{(1)}_{max}}{2(\frac{L}{8})\frac{3L}{8}t} = \sigma _{allow}$$

$$\tau _{max}^{(1)} = \sigma _{allow} = 2\tau _{allow} > \tau _{allow}$$

After these substitutions, it is easy to see that $$\tau ^{(1)}_{max}$$ is equal to $$\sigma _{allow}$$, which is twice as big as $$\tau _{allow}$$. Therefore this answer is not acceptable, and the solver must move on to case 2.

Pictured to the right is a graphical representation of normal stress. The stress is distributed through the length of the bar with and is equal to the force, F, divided by the cross sectional area A.

A graphical representation of this result can be shown using a Mohr's circle. As shown on the figure to the right, as long as the shear and normal stress stay enclosed within the circle, the structure will not fail. However, since in this case the maximum shear encountered is greater than half the allowable normal stress, this result is unacceptable

Case 2

Assume that $$\tau _{max}$$ reaches $$\tau _{allow}$$ first.

Therefore:

$$\tau = \frac{T}{2abt}$$

$$T = (2(t)\tau_{allow})(ab)$$

$$T_{max} = (2(t)\tau_{allow})(ab)_{max}$$

Everything on the right hand side of the above equation is constant, therefore to maximize the torque, the cross sectional area must be maximized. With the constraint of keeping the perimeter constant, it is easy to see that the largest cross sectional area will be achieved by making a equal to b, and thus having a square cross section as follows:

$$a^{(2)} = b^{(2)} = \frac{L}{4}$$

$$T_{max}^{(2)} = (2t\tau _{all})(\frac{L}{4})$$

$$T_{max}^{(2)}=\frac{1}{8}tL^2\tau {all} = M_{max}^{(2)}$$

$$M_{max}^{(2)}=\frac{tl^2}{16}\sigma _{all}$$

$$\Rightarrow \sigma _{all}=\frac{16M_{max}^{(2)}}{tL^2} $$

$$Recall: F(b)= \frac{I(b)}{b}$$

$$F(b^{(2)})= \frac{I^{(2)}}{b^2}=tb^{(2)}\frac{(3L-4b^{(2)})}{12}$$

$$M_{max}^{(2)}=\frac{tl^2}{24}$$ $$Given \ B^{(2)}=\frac{L}{4}$$

$$ \sigma _{max} = \frac{M_{max}^{(2)}b^{(2)}}{2I^{(2)}}=M_{max}^{(2)}\frac{12}{tL^2}$$

$$\sigma _{max} = \frac{3}{4}\sigma _{all}\Rightarrow \frac{3}{4}\sigma _{all}\leq \sigma _{all}$$

Acceptable \ $$a^{(2)}=B^{(2)}= \frac{L}{4}$$

=Wed September 17 2008=

 Axial Members Continued

=Stringers, Shear Stress/Strain, and Elasticity=

Stringers Continued from Earlier
Notes

Longeron - The term stringer is interchangeable with the term longeron. Longerons have the same purpose as stringers - the only differences are that longerons are usually larger, there are less of them, and they are spaced farther apart. (Reference - )

Shear Panels - Also known as brace panels. Together, shear panels make up a shear wall to support lateral loads on a structure. (Reference - )

Shear Stress and Strain
Shear Strain - γ - Strain that acts parallel to the face it's acting on. (Reference - )

τ = Gγ
 * (G = shear modulus)
 * (γ = shear strain - See above definition)

γ = du/dy + dv/dx = dux/dy + duy/dx
 * $$u \equiv u_x $$ - displacement along the x-direction
 * $$v \equiv u_y $$ - displacement along the y-direction

Shear Panels, Shear Stress and Strain, and Elasticity
Engineering (not tensorial) shear strain

γ = du/dy + dv/dx = dux/dy + duy/dx
 * Remember:
 * $$u \equiv u_x $$

 * $$v \equiv u_y $$

duy/dx = dv/dx
 * (for a small angle, $$ \alpha \cong \tan \alpha $$)

γ = change in the right angle (90 degrees) due to the shear deformation

εxy = tensorial shear strain

εxy = (1/2)γxy

=Individual Contributions=

Braden Barnes Eas4200c.f08.wiki.b 17:51, 25 September 2008 (UTC)

Eric Viale Eas4200c.f08.wiki.c 14:55, 26 September 2008 (UTC)

Christopher Pergola EAS4200C.F08.WIKI.A 18:35, 26 September 2008 (UTC)

Jeff Huenink Eas4200c.f08.wiki.f 20:17, 26 September 2008 (UTC)

Michael Rolle Eas4200c.f08.wiki.d 19:50, 26 September 2008 (UTC)

Chris Fontana Eas4200c.f08.WIKI.E 20:33, 26 September 2008 (UTC)

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User:Eas4200c.f08.blue.a/Lecture 10

Homework Assignments

The first homework assigned in class consists of four individual problems.

The first Problem is to define the moment of inertia $$I_y$$ for a solid circular cross section through integration using polar coordinates. The integral being $$I_y=\int\int_A(Z^2dA)$$.

Also the derivation of $$I_y$$ for a square cross section is required which would be derived by the use of the integral $$I_y=\int\int_A(Z^2dYdZ)$$

The second problem assigned required comparing the moment of inertia defined for the circular cross section (case:1) in the previous problem to a redistributed equivalent area (case:2)

Where $$a=b$$, $$ A_2=3(a)(t)=\pi(r^2)$$,  and $$ t=a/10$$

The moment of inertia will be calculated assuming the y-axis passes through the centroid of case:1 and case:2 for fair comparison. The moment of inertia for case:2 will be calculated by using the fact that the moment of inertia for a rectagular cross section is $$I_y=1/12(b)(h^3)$$ along with the parallel axis theorem which is $$I_y=I+Ad^2$$

The third problem assigned was a modified version of problem 1.7. The modification was that all rectangular surfaces be changed to circular surfaces. The problem is to compare the moment of inertia $$I_y$$ of a circular cross section (case:1) to a modified cricular cross section (case:2) calculating the moment of inertia $$I_y$$ for both cases through the centroid of the cross section.

Where $$r_0=10cm$$, $$t=r_0/10$$, and $$A_1=A_2$$

The moment of inertia will be calculated for the circle from the answer to problem number one since it will be the same the moment of inertia for case 2 will be from a combination of rectangle moment of inertia and circular moment of inertia with use of the parallel axis theorem.

The fourth problem assigned was to show why a open rectangular cross section would be preferable to a hollow recangular cross section by comparing moment of inertia, manufacturing, and assembly techniques.

title: User:Eas4200c.f08.blue.a/Lecture 10


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User:Eas4200c.f08.wiki.f/9- -08

{| class="toccolours collapsible collapsed" style="width:100%" ! Metal Matrix Composite vs Fiber Metal Composite
 * align=center |
 * align=center |

A metal matrix composite is material made up of at least two parts - one part is a metal and the other can be a different metal or some other type of material. A hybrid composite is composed of at least three materials. A metal matrix composite is formed by inserting a reinforcing material into a metal matrix. Sometimes the reinforcing material will have to be coated to prevent a chemical reaction that could cause the metal to break down. Usually the matrix is made up of a lighter metal, such as aluminum or titanium. The reinforcement material is not always added just to add structural strength, but also to change the properties of the composite material.1 Unlike metal matrix composites, fiber metal composites make use of organic materials and plastic. Natural fibers are refined, mixed, and merged to form high strength composite material in a polymer matrix. An advantage of this material is that it can be recycled up to 20 times.2

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User:Eas4200c.f08.wiki.f/9-19-08

=9/19 Notes=

Stringers Continued from Earlier
Notes

Longeron - The term stringer is interchangeable with the term longeron. Longerons have the same purpose as stringers - the only differences are that longerons are usually larger, there are less of them, and they are spaced farther apart. (Reference - )

Shear Panels - Also known as brace panels. Together, shear panels make up a shear wall to support lateral loads on a structure. (Reference - )

Shear Stress and Strain
Shear Strain - γ - Strain that acts parallel to the face it's acting on. (Reference - )

τ = Gγ
 * (G = shear modulus)
 * (γ = shear strain - See above definition)

γ = du/dy + dv/dx = dux/dy + duy/dx
 * $$u \equiv u_x $$ - displacement along the x-direction
 * $$v \equiv u_y $$ - displacement along the y-direction

Shear Panels, Shear Stress and Strain, and Elasticity
Engineering (not tensorial) shear strain

γ = du/dy + dv/dx = dux/dy + duy/dx
 * Remember:
 * $$u \equiv u_x $$

 * $$v \equiv u_y $$

duy/dx = dv/dx
 * (for a small angle, $$ \alpha \cong \tan \alpha $$)

γ = change in the right angle (90 degrees) due to the shear deformation

εxy = tensorial shear strain

εxy = (1/2)γxy

title: User:Eas4200c.f08.wiki.f/9-19-08


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User:AminSanaei

glk:User:AminSanaei

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User:Daanschr/art

Moved to my Wikipedia user page.

title: User:Daanschr/art


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Template:Bcpm/aurco/7

July

This template makes a link to the July key for Aurora, Colorado, and adds plant pages to Category:BCP/AURCO/7 and Category:BCP/Flora of Colorado (it automatically adds ), for plants that have been recorded as flowering in May in Aurora, Colorado.

This template should only be used in the "Aurora, Colorado = " field of Bloom clock plant middle

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Template:Bcpm/aurco/8

August

This template makes a link to the May key for Aurora, Colorado, and adds plant pages to Category:BCP/AURCO/8 and Category:BCP/Flora of Colorado (it automatically adds ), for plants that have been recorded as flowering in May in Aurora, Colorado.

This template should only be used in the "Aurora, Colorado = " field of Bloom clock plant middle

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Template:Bcpm/aurco/9

September

This template makes a link to the September key for Aurora, Colorado, and adds plant pages to Category:BCP/AURCO/9 and Category:BCP/Flora of Colorado (it automatically adds ), for plants that have been recorded as flowering in May in Aurora, Colorado.

This template should only be used in the "Aurora, Colorado = " field of Bloom clock plant middle

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Template:Bcpm/aurco/10

October

This template makes a link to the October key for Aurora, Colorado, and adds plant pages to Category:BCP/AURCO/10 and Category:BCP/Flora of Colorado (it automatically adds ), for plants that have been recorded as flowering in May in Aurora, Colorado.

This template should only be used in the "Aurora, Colorado = " field of Bloom clock plant middle

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User talk:Mk~enwikiversity/ipc

Groups
Although I know that the "come together" will be kind of more efficient if the instructors put the students into groups, I think that a free and independent group finding process and the competences involved (ability to contact others, team playing etc.) should be taken into consideration as well.--Mk 10:30, 23 September 2008 (UTC)

degree of instruction
We should also be aware of too much instruction as far as the topics are concerned.--Mk 10:30, 23 September 2008 (UTC)

title: User talk:Mk~enwikiversity/ipc


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User:ESanchez013

Hello!

I'm a "new student" here at Wikiversity. I have experience at the English Wikipedia here and may be able to figure out some basic things by myself, but I'd love it if you could "tutor" me or "show me the ropes to campus life" here. Thanks!

title: User:ESanchez013


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User:Eas4200c.f08.blue.a/Lecture 9

Pb. 1.1 Continued
Problem Statement The problem is to find the optimal ratio of b/a in a thin walled beam to maximize the load-bearing capacity of a beam while assuming that the moment (M) is equal to the torsion (T).This is done by optimizing the cross section that carries the maximum bending moment and the maximum torsion.

Aircraft Structure
Aircraft structures are assembled from various basic structural elements to help withstand loading conditions like axial, bending, and torsion. Some basic elements of an aircraft are the axial members(truss and bar elements), sheer panels, bending members(like beams), and torsional members(bar and beam torsion).

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Syphilis

Syphilis is often referred to as the great imitator because the symptoms that it presents are often mistaken for another disease. Syphilis is most common in women between the ages of 20-24 but also is seen in men between the ages of 35-39 years old. The incidence of syphilis doubled in both sexes between the years of 2000-2006.

Syphilis is typically spread through direct contact with the genitals, anus, or mouth of an infected person. It can also be transmitted from mother to child during birth. The primary stage of syphilis is designated by the formation of a single or multiple chancres. Typically this is 21 days after infection but can take up to 90 days. The secondary stage is characterized as a subtle to severe rash appearing typically on the hands and feet. Associated symptoms include: fever, swollen lymph glands, sore throat, patchy hair loss, headaches, weight loss, muscle aches, and fatigue. The Late or Latency stage does not have the symptoms associated with the first two stages. Therefore, many people incorrectly believe that they no longer have the infection. The latency stage can take 10-20 years to appear after the onset of the disease. Due to severe internal damage this stage includes symptoms such as: difficulty coordinating muscle movements, paralysis, numbness, gradual blindness, dementia, or death.

Fortunately, syphilis is easily detected through a blood test and can be treated with strong antibiotics. Syphilis can cause extreme developmental disorders, seizures, or even death if infected newborns are untreated.

title: Syphilis


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User:Marius63

Marius Amado-Alves is a composer, computer scientist, linguist from Portugal.

title: User:Marius63


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Wikiversity:Request custodian action/Moulton's talk page

User:Moulton, given his interest in wiki editors' identities, and as part of the "ethics" project, has published personal information about wiki editors on wikis (including Wikiversity), blogs, and forums - a practice known as "outing". This is to request community comments on Moulton's posting of personal information to Wikiversity. Please read (and develop) the activity under review, and the actions taken (so far), and add comments in the discussion section. Cormaggio talk 19:16, 23 September 2008 (UTC)

Context
Cary Bass, WMF volunteer coordinator, sent an email to Foundation-l mailing list which spelled out his concerns - some of which related to Moulton's activity - as general project-wide questions. Wikiversity is a Wikimedia project, and therefore conforms to fundamental Wikimedia principles and policies (such as freedom of content); however Wikiversity, like all projects, has a certain autonomy in setting its own policies that support Wikiversity's scope and mission, provided they don't usurp fundamental WMF principles.

See:
 * Wikimedia's privacy policy - discusses what kinds of information you may make available to others by editing WMF sites, and what level of privacy you may expect with respect to the Wikimedia Foundation.
 * Wikiversity's (proposed) privacy policy - discusses the adding of personal information on Wikiversity
 * Wikipedia's policy on 'outing' and harassment - discusses the adding of others' personal information on Wikipedia

See also Request custodian action/Requests relating to Moulton for previous discussions around Moulton's activity.

Activity for community review
As stated, Moulton has been adding personal information to Wikiversity, or links from Wikiversity to other websites where personal information is presented. Moulton's co-founder of the ethics project, User:WAS_4.250, showed awareness and concern about the problems in posting personal info (eg here); and User:SB_Johnny warned Moulton that links to his blog that 'outed' people were not allowed (see). However Moulton has continued to post information linking names and/or email addresses to wiki user names, despite being asked several times on-wiki (across several WMF projects), and on IRC not to do so. The latest of these was today (12:06 (UTC), 23 September 2008), in which Moulton posted a private email he had sent to several Wikipedians, leaving names and email addresses intact. This latest edit has now been oversighted - i.e. deleted from the page's history - though Moulton has re-posted the content of that edit on the "media ethics" blog he contributes to. Several other requests for oversighting such material have been made to Wikimedia stewards over the past few weeks.

Additions
(If you know of other instances of personal information being added, or discussed, please add them here.)

Actions taken

 * Moulton's talk page has been protected from editing for now. This is intended to last until the outcomes of this discussion become clear.
 * Oversighting done (as mentioned).
 * The User:Moulton account has been blocked, but he has continued to edit as an anonymous ip. The following ip addresses that he has been using have also been blocked:

 * Special:Contributions/68.160.173.170
 * Special:Contributions/68.163.103.219
 * Special:Contributions/141.154.15.138
 * Special:Contributions/141.154.67.172
 * Special:Contributions/141.154.15.252
 * Special:Contributions/141.154.72.189
 * Special:Contributions/68.162.213.45
 * Special:Contributions/141.154.69.44
 * Special:Contributions/141.154.74.168
 * Special:Contributions/68.163.105.83
 * Special:Contributions/68.163.99.18
 * Special:Contributions/68.162.212.132
 * Special:Contributions/68.160.161.115
 * Special:Contributions/141.154.16.250
 * Special:Contributions/68.163.99.170
 * Special:Contributions/68.163.99.108
 * Special:Contributions/141.154.81.122
 * Special:Contributions/68.163.104.69

See block logs for more information

Discussion

 * I feel that Moulton's talk page should not be unprotected until he 1. wants to participate at Wikiversity and not continue his crusade and 2. is willing to conform to the community wide standards here. Ottava Rima (talk) 21:44, 23 September 2008 (UTC)

 * Personal information should not be posted carelessly or in a non-journalistic manner. It's not the mere act of someone's name being posted, but the effect that happens related to that event. For example, "X's real name is Y," followed with something like "Y is a Z." The fact that X is really Y may not be a big deal, but X's reputation is now Z. In order for X to show where X was "outed," X would have to point out the edit that stated "Y is a Z." Some will (and have) claim that X "outed" himself/herself just by pointing out that event, but obviously X was just trying to show that someone else made an earlier earlier statement that changes X's reputation. This may sound completely hypothetical, but this actually did happen (didn't involve Moulton). If it was said in any journalistic quality that X is Y and Y is a Z, then others may have not have seen it as X outing himself/herself. In this regards, I feel that Wales statement about the ethics project to only discuss notable people is a step in the right direction. With X being not notable, that leaves X to defend for himself/herself about Z, rather good or bad, because there is obviously no other (hence notable) sources to prove otherwise.
 * There are limits that WMF should recognize more, not that they haven't been recognized at all. One being able information posted on other websites. There are many websites linked to by Wikipedia and such, but they don't get the axe if they mention real identifiable information. If one does and it gets the axe for it, then maybe all should. Actually, that starts making WMF judge other websites by content, and that really can be avoided if WMF simply sets accord to ignore content of other websites. As long as the effect of "outing" (i.e. Y is a Z) is not being done on-wiki, then that covers the main concern.
 * There are too many "good" websites that will say "X's real name is Y." In fact, if WMF was to strictly say you can't link to anything that "outs" people, then the technical use of IP addresses does exactly that. There is lots of personal information that can be gathered by IP address. It appears by the stated benefits of having a regular account that this is clearly known. I've seen many log messages that state X and some IP address by admins. For example, they say, "this IP user is user X." *BAM* everybody can now find out Y about X. Moving on...
 * It appears Moulton has been given the chance to take a more careful and journalistic approach to his interests. He could easily take it all off-wiki and source himself here on Wikiversity in a way that would be more appropriate. He appears to know how to do it, and if he is sincere to journalistic quality on Wikiversity then he would actually do it. I'm wondering why he hasn't.
 * (If I need to expand this more, please note on my talk page.) Dzonatas 21:57, 23 September 2008 (UTC)
 * This is useful - thanks Dzonatas. I should point out however, that some of Moulton's 'outing' has been much more than the ability to link IP addresses and usernames (which, I think, is relevant to figuring out what's happening on the wiki), but rather links to real names, real jobs, real email addresses. You're right that his approach is journalistic - this is all possible to do - but the question is whether it's appropriate to do so on Wikiversity. Wikipedia very clearly frowns upon this (see policy above), but we haven't yet firmed up our own standards (pending further discussion of our own privacy policy). Also, I'll note that we clearly don't disallow pointing to external sites that contain personal information about living people (how would WP then cite sources for their biographies?), but rather this discussion concerns only information about WMF contributors. (Not sure if that was what you were referring to in your second paragraph.) Cormaggio talk 08:44, 24 September 2008 (UTC)

 * I'm in agreement with Ottava. Unless I missed it, Moulton has yet to recognize what he did wrong.  I find it unlikely that he's going to make any changes until he does. Sχeptomaniacχαιρετε 18:00, 24 September 2008 (UTC)

 * But those IPs are only posted because he abused them.... --Enric Naval 03:15, 29 September 2008 (UTC)

If Moulton's IP addresses will be allowed to freely cavort around without reprecussion he should be unblocked so he can be more easily ignored. An alternative solution would be to block anonymous IP addresses from verzion dsl in boston from editing wikiveristy or making accounts - this would force those IP addresses to request accounts. Salmon of Doubt 17:14, 30 September 2008 (UTC)
 * Possible support blocking all anon ip address from boston. Emesee 05:28, 1 October 2008 (UTC)


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Wikiversity talk:Request custodian action/Moulton's talk page

If you have an opinion on the content of the review, please leave a comment in the discussion section. This talk page is for overall comments about the review - eg. whether it should exist, how it should be conducted, etc. Cormaggio talk 08:48, 24 September 2008 (UTC)