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Learn to juggle

Disclaimer: I just recently learned the 3 ball shuffle, and it was hard to find satisfactory information from the web quickly. There were some good sites, but this is a collection of my experiences of how anyone can learn to juggle in a couple of hours. So far, it concentrates on the methods I know of.

What is juggling?
Juggling is a physical human skill involving the movement of objects, usually through the air, for entertainment.

Forms of juggling
There are different forms of juggling. You can juggle different kind of objects (balls, pins, knives, torches, anything else, mixtures). You can have different number of objects and move them in different patterns. You can toss, bounce or otherwise move the objects. You can juggle alone or together with others.

Required equipment
In the beginning, balls or beanbags may be the easiest to juggle. They are symmetric and you can catch them with an open palm of your hand regardless of the ball's orientation or spin. Similarly, you can just toss them from an open palm and no specific spin is required. Basically any balls should do, such as tennis balls, but it might be easier to learn if you follow these guidelines. That said, any balls will do. The best ones could be small round bags filled with beans, seeds or rice. They are moderately heavy, do not bounce and provide a good grip.
 * They should not be too large, or you will have trouble holding more than one in your hand
 * They should not be too small, or you will have trouble tossing them accurately
 * They should not be too heavy, or you will tire easily while practicing and may hurt yourself or break something
 * They should not be too light, or you will not be able control the height of the toss
 * They should not be too bouncy, or they will bounce away if you drop them
 * They should not be too different from each other, or else each has to be tossed differently

Your environment should be clear enough that dropped balls do not roll under things and can be easily picked up. If you want to reduce fetching bouncing balls and picking up dropped balls, you can stand next to your bed and have the balls drop to your mattress instead of the floor.

Learn to to juggle
The following tasks are ordered roughly from the easiest to more complicated. It might help if you learn them in order and one at a time in the beginning.

Note: These instructions are mostly for right-handed people. If you are left-handed exchanging left and right should make things slightly easier. Or you can just learn to use your lefthanded by reversing the directions given.

The Scoop
An important technique in juggling is the scoop. It allows you to quickly catch and throw the balls without and also makes the juggling look better visually. The scoop looks literally like a scoop. There are two kinds of scoop, inwards and outwards.

 * Inwards: move your hand in a semicircular motion, 'inwards', or towards the other hand. If it is your right hand it will be a clockwise semicircle and if it is the left it will be an anticlockwise semicircle. The ball is caught on the outside and thrown on the inside.
 * Outwards: move your hand in a semicircular motion, 'outwards', or away from the other hand. If it is your right hand it will be an anti-clockwise semicircle and if it is the left it will be an clockwise semicircle. The ball is caught on the inside and thrown on the inside.

1-ball shuffle with two hands
This may barely count as juggling (it may not be entertaining) but it is the basis for most of the toss styles and you can always use this exercise to polish your style in the future.

The task is to toss the ball from one hand to the other.

 * 1) First, take the stance (ball in the right hand)
 * 2) * Stand up in good balance (feet moderately apart)
 * 3) * Take one ball in the right hand
 * 4) * Keep your elbows relatively close to your body, low arms horizontal point directly forward, the ball on the palm of the hand facing up
 * 5) * Look directly forward, do not look at your hands or at the ball
 * 6) Ball moves from the right hand to the left hand
 * 7) * Toss the ball in an arc from your right hand upwards and slightly leftwards
 * 8) * It should reach at least your eye level, slightly left from your center (nose)
 * 9) * Catch it with your left hand without reaching upwards
 * 10) Return to the stance (with the ball in the left hand)
 * 11) * Return your left hand to the original position
 * 12) Ball moves from the left hand to the right hand
 * 13) * Toss the ball in an arc from your left hand upwards and slightly rightwards
 * 14) * It should reach at least your eye level, slightly right from your center (nose)
 * 15) * Catch it with your right hand without reaching upwards
 * 16) Return to the stance
 * 17) * Return your right hand to the original position
 * 18) Repeat from 2)

Practice this until you can toss the ball between your hands sufficiently many times without dropping the ball. For example, when you can make three consecutive series of at least 50 tosses. The numbers are arbitrary, you can always proceed further and return later to polish your basic technique.

These are important things to concentrate on
 * The tosses should be similar (arc, height, landing position) from one to the next (repetition and catching are easier)
 * Occasionally stop see how much you have to move you catching hand when you move it back to the original position (are the tosses similar and accurate? if you are throwing many balls, you must have economical movement and accurate tosses as time is limited)
 * Do not reach upwards with your hand when you catch the ball (economical movement)
 * Look directly forwards or even a little upwards, not the hands or the ball (in the future you will be managing many balls and you cannot concentrate on just one)
 * Do not let the ball roll of your fingers, toss it up from your palm (it will be easier to control)

Variations:
 * Eyes closed (if the tosses are accurate and consistent, you do no have to move your hands)
 * Different heights for the toss (from vertical tosses to ones that touch the ceiling)
 * Different style balls (size, weight, can you do it with a balloon?)
 * One ball with one hand (toss directly upwards)

3-ball shuffle
There is only one style toss that is being repeated alternately with the right hand and the left, and the throw is the basic toss from one hand to the other. The heights of the arcs should be equal. At this point do not care if the balls collide in the air. Just start again.

The starting position is the standard stance with two balls in the right hand and one ball in the left hand.

These are the tosses. Next is how you should learn them.
 * 1) Starting toss, toss the 1st ball from the right hand (not part of the pattern)
 * 2) Toss the 2nd ball from the left hand when the 1st ball reaches the highest point of its arc
 * 3) Catch the 1st ball with the left hand
 * 4) Toss the 3rd ball from the right hand when the 2nd ball reaches the highest point of its arc
 * 5) Catch the 2nd ball with the right hand
 * 6) Toss the 1st ball from the left hand when the 3nd ball reaches the highest point of its arc
 * 7) Catch the 3nd ball with the left hand
 * 8) Toss the 2nd ball from the right hand when the 1st ball reaches the highest point of its arc
 * 9) Catch the 1st ball with the right hand

Can you detect the pattern after the starting toss in the instructions? The pattern (left:toss-catch, right:toss-catch, left:toss-catch, ...) is easier to learn in steps.

Starting from the starting position and doing first only the first toss-catch (practice steps 1,2 and 3). In the beginning, you can let the 2nd ball just drop. If you can, you can catch it as well (it is step 5, but its easier when you don't have to pick up the balls from the floor). Practice, starting from the starting position, until you can do the first three steps almost always.

Then, try to add a second toss-catch (practice steps 1,2,3,4,5) and repeat until you can do it almost always. You can also do step 7 if you can make it.

The process is similar to the rest of tosses: add one more toss-and-catch, repeat until you can almost always do it. Remember to stop after you have made the correct number of tosses-and-catches, then start again from the starting position.

If you add the extra catch as soon as you can, you won't have to pick up dropped balls all the time. At some point you might notice that you are back in the starting position with two balls in the right hand and one ball in the left hand. After how many steps does this happen? After adding a few more tosses, you might notice that you are doing the repeated motion of "pumping" or doing little circles with your hands. Now you should get some sense of the rhythm of juggling and you no longer have to concentrate on individual movements.

Colliding balls can be avoided easily when the throws are accurate and have similar heights. There is no need to have every second throw be higher or keep one hand closer to your body. There are two main patterns in the 3-ball shuffle to avoid collisions. The perhaps easiest one is to always toss under the ball you are going to catch next with the same hand. It is possible because the arc of the ball is not quite symmetric, the highest point of the arc should be closer to the catching hand than to the tossing hand (perhaps of wind resistance) Or if the tosses are not that accurate, make the toss a little longer so that you have to move your catching hand outwards (left if the toss if from right to left). Remember to move your hand back to the original stance before tossing again. The other main pattern is to always throw the ball over the incoming ball. If you want to exaggerate the movement in this case, you can make your tosses little short so that you catch the balls closes to the center line of your body.

Note: It can be claimed that the 3-ball shuffle is easier than most 2-ball patterns, for instance, juggling two balls with one hand.

title: Learn to juggle


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Biochemistry for novices

What will I get by learning Biochemistry?
You are going to enjoy learning biochemistry more if you have background in organic chemistry or cell biology. We will have parts that cater to these groups, marked ADVANCED. If you had only high school chemistry, you're also welcome! Just recall your lessons, and we'll ride comfortably. There are parts marked REVIEW on this page, so you can re-study these parts if you forgot already. But I'm warning you, these parts are not intended to give a comprehensive review. If you need more review on the topics than these parts give, you may get some help from books and other chemistry texts.

The (bio)chemical basis of life
What is the basis of life as biochemists view it? One word: MOLECULES! Biochemists see life as a magnum opus, played by a giant orchestra composed of lots and lots of molecules. These molecules act on each other, or act together, so that all living matter can exercise the processes that define LIFE: the eating and drinking, the reproducing, the feeling, the loving, the moving, the breathing, every activity we consider as part of LIVING. So, be ready to encounter lots of molecules, big and small. We will be dealing with five classes of molecules in biochemistry.

Small molecules
Small molecules are called "small" because they are small - comparing them to the next four molecule classes. They have molecular masses below 600 atomic mass units, which means that one mole of molecules of this kind would weigh less than 600 grams! REVIEW Atomic mass units are the units of the weight of a molecule. An atomic mass unit is about 10-27 kilogram. A mole of molecules is 6.022 x 1023 molecules. The molar mass of a compound is the mass of one mole of molecules of that compound. Almost all molecules enountered in high school chemistry are therefore considered small. Benzene (MM = 72 g/mol) is small. Sucrose (table sugar; MM = 180 g/mol) is small. Even coronene (C24H12) is small. Can you calculate the molar mass of coronene?

Proteins
Proteins are made by joining together small molecules called amino acids. Glycine is one such amino acid. Glycine has two main parts, an -NH2 part and a -COOH part. The -NH2 part of one molecule of glycine can react with the -COOH part of another molecule of glycine to form glycylglycine. And glycylglycine, which has an -NH2 part and a -COOH part as well, can react with another molecule of glycylglycine in the same way! This way, long molecules called polypeptides are formed. And proteins are polypeptides that have been formed by the body naturally. Some famous proteins are:
 * Papain: used in whitening soaps. It's called a protein  hydrolase because it's a protein that hydrolyzes other proteins. Another term for protein hydrolase is protease.
 * Collagen: it forms the basis for connective tissues in animals' bodies (for example, the skin). It's a protein made up of mostly glycine and proline, another amino acid. Of course it has other amino acids in it.

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

title: Biochemistry for novices


text: File:Parallel axis theo.PNG

title: File:Parallel axis theo.PNG


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Social psychology (psychology)/Guides/Week 10

Week 10

 * 1) Complete the Week 07 guide.
 * 2) The lecture this week is on relationships
 * 3) If you haven't already done so, participate in the cross-cultural training tutorial.


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User:Eml4500.f08.group/Homework1/LectureNotes

=Chapter 4: Trusses, Beams, and Frames=

Trusses
The image below is a two-truss system with elastic (deformable) bars.

Global node numbers are denoted by circles while element numbers are denoted by triangles. The elements are separable members of a system which can be individually analyzed. Nodes are points on a system on which forces appear. The x-y coordinate system can be located in any rotation about the z-axis, however in this case the y-axis is chosen to be parallel to the applied force P. Furthermore, the triangles beneath nodes 1 and 2 denote that the node displacement is fixed to zero in both the x- and y-direction.

Forces acting on the system are shown below with a systematic example of how to name the forces on the member. The names given to each of forces are arbitrary, but must be systematic and consistent.

This picture is another example of how to name the forces, nodes and elements in a systematic manner, rather than a rigid, consistent manner. The element number is denoted by a triangle again. The local nodes (nodes relative only to the specific FBD section) are denoted by squares as opposed to the global nodes (nodes relative to the entire structure) which are circled. The subscript of the forces are the degrees of freedom (dof) or internal force of the element. Each force is a separate degree of freedom and should have their own number. Also they are numbered in a consistent order: x-, y-, x-, y-axis. The superscript represent the element number in parentheses. For this global system, the dof's are 1, 2, 3, and 4 and the element numbers are 1, 2, which come to eight total numbers in both elements.

Bar element 2:

Force Displacement (FD) Relation

The diagram above shows the relationship between force and displacement in a spring with one end fixed. It both ends are free, the case becomes as follows.

Because there are now two independent variables, the spring must be analyzed with a matrix form of F = kd. Matrices are names by "row x column"

\begin{pmatrix} f_1 \\ f_2 \end{pmatrix}_{2x1} = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix}_{2x2} \begin{pmatrix} d_1 \\ d_2 \end{pmatrix}_{2x1} $$

There are 2 cases which can be examined:

Case 1 is the result with respect to node 1:

$$f_2 = k (d_2 - d_1)$$

Case 2 is the result with respect to node 2:

$$f_1 + f_2 = 0$$

→ $$f_1 = -f_2 = -k (d_2 - d_1) = k (d_1 - d_2)$$

Steps to solve simple truss system
Steps to solve simple truss system described on Wed, 3 Sep 08. (Recipe)

Global picture
(description) at structure level
 * Global degrees of freedom (disp coeffs)
 * Global Forces

Actually the displacement degrees of freedom are partitioned into Simularly for the global forces
 * A known part, eq, fixed defs, constraints
 * an unknown part, solved using FEA
 * a know part: applied force
 * an unknown part: reactions

Element picture
either in global coord system on in local coord system
 * element degrees of freedom
 * element forces

Global Force Displacement relationship

 * Elemenet Stiffness matrix in global coord
 * element force matrices in global coord.
 * Assembly of element stiffness matrix and element force matrix into a global Force Displacement relationship

Elimination of known degrees of freedom
Elimination of known degrees of freedom to reduce the global Force Displacement Relationship (stiffness matrix non-singular => invertible)
 * m = number of unknown displacement degrees of freedom
 * n = number of known and unknown and displacement degrees of freedom

Kbar non singular => Kbar^-1   (Kbar invertible)

Compute Element forces
Compute Element forces from now known dbar => element stresses

Compute Reactions
Compute reactions (unknown forces)

Example
Take a specific example to see how the recipe works (no justification yet)

Element Length - L(1) = 4, L(2) = 2

Youngs Modulus - E(1) = 3, E(2) = 5

Cross Section Area - A(1) = 1, A(2) = 2

Inclination Angle - θ(1) = 30°, θ(2) = -45°

=Mon, 8 Sep 08= 1.) Global Picture 2.) Element Picture

3.) Global Force Displacement at element level

k is the element stiffness matrix for element e

d is the the element displacement matrix

f is the element force matrix

Matrix p 225 book, 2nd equation form the bottom

axial stiffness of bar elem "e". e = 1,2

l^(e), m^(e) = director cosines of ~x axis (goes from [1] to [2]) with respect to global (x,y) coords


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User:Eml4500.f08.delta 6.guzman/MATLAB Sections 10 & 11

Section 10: Command Line Editing and Recall
MATLAB's line editing and recall are extremely helpful when the related commands are needed repeatedly. These commands are activated by using the arrow keys. Therefore an earlier command can be recalled, edited, and then executed as follows:

Rather than creating an M-file, this convenient feature allows recalling the same command and modifying it before it is executed. In this case the command was modified to generate 9 random numbers instead of the original 3.

Section 11: Submatrices and Colon Notation
In order to manipulate fairly complex data, vectors and submatrices are often use. The effective use of these diminishes the time length, the number of loops and the quantity of code.

Colon notation can be used to retrieve submatrices from the main matrix. By using x(1:7,1) the first 7 elements of the matrix will be obtained.


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User:Eml4500.f08.group/Homework1/MATLAB

Link to MATLAB Primer: http://apollo.mae.ufl.edu/~vql/courses/feem/matlab_primer_3rd.pdf

MATLAB PRIMER IS OFFLINE - FOUND SECONDARY LOCATION! http://www.uib.no/med/avd/miapr/arvid/MOD3/Matlab/matlab_primer_3rd.pdf Eas4200c.f08.carbon.clausen 04:01, 12 September 2008 (UTC)

To enter source code use:

Eas4200c.f08.carbon.clausen 18:31, 12 September 2008 (UTC)

=MATLAB PRIMER NOTES=

1. Accessing MALAB.
MATLAB can be accessed in a variety of ways. For example, most systems can access MATLAB by using the “matlab” system command. Similarly, MATLAB can be shut down using the internal MATLAB command of “quit” or “exit”. Since a MS Windows operation system is being used, MATLAB is accessed through the Windows start menu and closed via the red “X” icon in the upper right corner of the open MATLAB window.

Again, since Windows is being used, it will be convenient to keep both MATLAB and a text editor open in separate windows. Reasons for this will be discussed in section 14 (Managing M-files).

2. Entering matrices.
MATLAB functions via the use of rectangular matrices. Numbers, vectors, and variables can all be represented as a matrix with numbers coming in the form of a 1x1 matrix, vectors as 1xN / Nx1, and variable as an NxN matrix (N = 1,2,3…). Matrices can be generation by built in MATLAB functions, loaded from document files or external data files, and be entered directly as a list of elements. For example, typing into MATLAB generates a 3x3 matrix of numbers assigned to variable A.

MATLAB also accommodates the use of imaginary numbers in calculations. Complex matrices can be entering in one of two ways. Entering will generate the same complex matrix. The variable i and j denote complex values in MATLAB. New complex variable may be generated by entering into MATLAB. This will assign the variable ii equal to an imaginary number thus allowing for its use as a complex variable.

Loading large arrays into MATLAB is best accomplished by utilizing as ASCII file from a text editor. This file should consist of a rectangular array of matrix values. This file can then be loaded into MATLAB with the command “load file.ext” where file is the name of the file and .ext if the file extension.

Three built in MATALB commands can be used to generate matrices. Entering rand(n) or rand (m,n) will create a (n x n) or (m x n) matrix of random numbers. The magic(n) command will create a (n xn ) matrix in which all columns, rows, and diagonals have the same sum. Finally hilb(n) will create a (n x n) Hilbert matrix.

Individual matrix numbers can be accessed by entering A(2,3)). This command will find the numerical value assigned to the second row of column three in matrix A.

3. Matrix operations, array operations.
There are several matrix operations that are available for use in MATALB and they apply to both matrix and scalar operations. Should there be incompatibilities in the size of the matrices to be operated on, an error message will result. The following are available matrix operators.

Some of the above matrix operations do not operate entry-wise. The ones that do not (*, ^, /, and \) can be made to operate entry-wise by preceding them with a period.

4. Statements, expressions, and variable; saving a session.
MATLAB is an expression based language where expressions are composed of operators, and variable names. These expression usually take the form of “variable = expression.” The resulting matrix is then displayed on the screen and assigned a variable. Should no variable be assigned by the used, a standard variable of ans (answer) is assigned to the matrix.

Statements usually terminate at the end of a single line. In the case that a longer statement is necessary, three periods (…) followed by pressing return will continue the statement onto the next line. Conversely, several statements can be placed on a single line of code provided that each expression is separated by commas or semicolons. By placing a semicolon at the end of a line, the expression is evaluated but not printed to the screen.

It is important to note that MATLAB is case sensitive i.e. the variable “A” is not the same as variable “a”. Should it be necessary, typing the command “who” will list are current variable in the workspace and any variable may be removed by typing “clear” followed by the variable name that the user wishes to clear. Infinite loops can be broken by pressing “ctrl-break” on the keyboard

All data and variable names are lost whenever MATLAB is exited unless the save command is used. Using this command will restore the workspace to its previous version the next time MATALB is opened.

For Loop
For loop syntax example, to produce a certain n-vector for a given n:

or can be written:

 * Note: x = [] creates an empty matrix

The for statement permits any matrix to be used instead of 1:n. The variable consecutively assumes the value of each column of the matrix. The following example shows this syntax as it sums all the entries of matrix A by adding its column sums.

While loop
General form:

The statements will be repeated continuously while the relation is true. Once the relation is no longer true the loop ends.

An example for a given number a, will compute and display the smallest nonnegative integer n, such that 2^n >= a:

 * Note: The semicolon after the statement keeps the values of n from printing on screen until after the loop is over and then the program calls to print n on the screen.

If statement
General form:

The statements are executed only if the relation is true.

Multiple branching is also possible as shown in the following example:

Relations
The relational operators in MATLAB are:

 * Note: = is used in an assignment statement while == is used in a relation

Relations may be connected or qualified by the following logical operators:

When applied to scalars a relation is just the scalar 1 (if true) or 0 (if false). Similarly for matrices of the same size a relation is a matrix of 1’s or 0’s. When a relation between matrices is interpreted by a while or if to be true each entry of the relation matrix must be nonzero.

Examples:
To execute statement 	when matrices A and B are equal:

To execute statement when A and B are not equal:

Or

 * Note: The above example does not work with the syntax:

This will not give the intended result because statement would only execute if each of the corresponding entries of A and B differ.

7. Scalar functions
A scalar function is one whose range is one dimensional and is made up of one or more variables. In MATLAB scalar functions generally operate on scalars, but will operate element-wise on matrices. some of the more common functions are:

8. Vector functions
A vector function is one whose range is three or n-dimensional. In MATLAB these functions generally operate on a vector. When applied to matrices in a column-by-column fashion to give you a row vector. If row-by-row action is desired it can be generated using the transpose. Some of the more common of these functions are:

10. Command line editing and recall
Editing the command line is very simple in MATLAB. Positioning of the cursor can be done with the left and right arrows. Backspace and Delete can be used to remove characters to the left and right of the cursor respectively. Edits can also be made to previous commands by using the up and down arrows to scroll through the commands to find the one you wish to edit and execute again. These edits work best for small projects and revisions.

11. Submatrices and colon notation
Colon notation is used to produce vectors and to reference submatrices. Colon Notation in addition to subscripting by key vectors are necessary to effectively manipulate matrices. They can also be used in place of loops to make the code more readable and simple. The formula for the colons is initial value: increment: terminating number. An example of this is: The Incrementing value can also be left out, causing the default of one o be used, such as: Colon Notation can be used to obtain submatrices of a matrix. An example of this is:

12 M-files: script files, function files
“M-files” are an integral part of MATLAB as they are files which can execute a sequence of statements stored in diskfiles – they are designated by the extension “.m” appended to the end of the filename. “M-files” can be broken down into two distinct forms: script files and function files.

Script files:

Script files are “M-files” which can contain a number MATLAB commands which can be executed by calling the script file by name. The advantage of a script file is evident; a single MATLAB command will actually execute a series of previously developed commands. Script files can be used to easily input data into a function, such as a matrix, which can help reduce entry errors. It is important to note that script file variables are global; and, the effect of the global variables contained in script files is that variables of the same name in the current MATLAB window will be altered by running the script command.

Function files:

Function files are another type of “M-file” similar to a script file, but allowing for (in certain ways) more control over how MATLAB handles an occurrence of this file type. Function files allow users to create their own functions in MATLAB, which can be used as any ordinary MATLAB function would (such as how “sqrt” is a MATLAB function that takes the square root of a number or of the value of a variable). A key difference between function files and script files is that variables in function files can be made either global or local – meaning that local variables can be created that pertain only to the function file (and thus a variable with the same name in a current MATLAB session would be unaffected by a call to the function file).

13 Text Strings, Error Messages and Input
Text strings are variables that contain a series, string, of characters. They are created by assigning text in single quotes to a variable name as exemplified below.

The function disp displays text strings.

However, the function error is better suited for displaying error messages as it also aborts execution of an M-file.

The user can be prompted to enter data during runtime of an M-file with the function input. This function assigns user input to it's return value. The return key denotes end of user input.

14 Managing M-Files
MATLAB has the ability to execute system commands without exiting MATLAB with the !-feature. Precede any system command with '!', for example.

This allows you to edit M-files without exiting MATLAB. For modern computers using a graphical user interface, simply opening the text editor in a new window is much simpler.

MATLAB also has a variety of debugging tools for M-files. Type help dbtype or see the last section for more details.

In MATLAB, the pwd command returns the name of the present working directory. dir or ls will list it's contents. If you only wish to see the M-files in the working directory use the what command. delete can be used to delete a file and type shows a M-file on the screen. These duplicate many common system commands while avoiding the use of the '!'.

Use the why command for entertainment

To use M-Files they must be in a directory accessible to MATLAB. M-Files in the current working directory are always accessible, as are M-file in a subdirectory of the home directory on most network installations. To see a list of directories currently in MATLAB's search path, or add and delete directories from the search path, use the path command.

15 Comparing efficiency of algorithms: Flops, tic and toc
MATLAB provides two method for measuring the efficiency of an algorithm. These are the number of floating point calculations (flops) and the elapse time it took to preform the algorithm. Note however that modern computers are time sharing machines and so elapse time may not be a reliable measure of efficiency. The computer may be busy performing another task in the middle of yours, adding to your task's time.

To measure elapse time, use the tic and toc commands. tic starts the time while toc returns the elapse time since tic was called. For example:

To measure the number of floating point calculations, use the flops and flops(0) commands. flops(0) resets the flop counter to zero and flops returns the current flop count. For example:

16 Output format
The precision of the displayed output can be controlled with the format command. This doesn't effect the computations themselves, which are always done in double precision. The format chosen from the table below remains in effect until changed.

additionally the command format compact will suppress blank lines while format loose will unsurpress blank lines. These do not effect which of the other format commands are active.

17 Hard copy
Command:  filename

This command pulls up the hardcopy and copies the content excluding graphics to that file, and will go to a default file if no filename is given. It continues to copy to the file until you command it to  or.

18 Graphics
MATLAB plots curves, 3D plots of curves , #D mesh surface plots , and 3D faceted surface plots. allows you to preview these features.

Planar plots
The command  creates an x-y plot as shown below the format for the range is 'lower rang:meshsize: upper range'.

Figures can be suppressed or exposed as the "current" figure by the command according to its figure number, and command   will give you the number of the current figure.

To make graphs, first put a function in an M-file then command it open with a domain:

Parametric graphs can also be defined as shown:

To enter text in the graph, follow these commands:

title of graph x-axis label y-axis label text on graph (placement defined by mouse) text position at specific coordinates places grid on the graph axis limits makes standard axis for all graphs auto-scaling shows current scaling for vector v puts same scale on both axes puts same scale and tic marks on both axes turns off both axis scaling and tic marks turns on both axis scaling and tic marks

Put the  commands after the   command.

To make multiple plots on a graph the format may be one of the two as follows

You can also command  to freeze the current plot and from there superimpose another plot on the same graph. This can be turned off with. To make line types use the codes: solid(-), dashed (--), dotted, dashdot (-.). To make marktypes, use the codes: point(.), plus (+), star (*), circle (o), x-mark (x). To make colors, use the codes: yellow(y), magent(m), cyan(c), red(r), green(g), blue(b), white(w), black(k). An example is a green dotted line:.

Several other features are:

Graphics hardcopy
To send the graphics to the printer, command, while   filename saves the figure to the specified format or to a default format if not specified.

3-D line plots
An example format of a 3-D line plot is as follows:

A 3-D graph has all the same commands plus.

3-D mesh and surface plots
3-D mesh plots shows a plots of matrix z with the command. A surface plot is controlled by. To obtain a mesh graph of a function over a rectangle, an example is as follows:

The shading can be  and are entered after the   command. Color profiles may be:.

Handle Graphics
To see a more complete list of the options to control a graph enter  and.

19 Sparse matrix computations
MATLAB generally assumes a matrix is dense, meaning that there is a possibility that any entry in the matrix could be non-zero. When a matrix has enough zero entries calculation time could be reduced and less memory will be needed to store the non-zero entries of the matrix. For this reason solutions to lager problems could be possible. MATLAB takes advantage of this by having two storage modes with the following functions: full and sparse.

The Following commands are used to create sparse matrices:

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User:Eml4500.f08.group/Homework2/Contributors

=HW 2 Contributing Team Members=

Jeffrey Wright Eml4500.f08.group.w 20:13, 25 September 2008 (UTC)

Julia Setlak Eml4500.f08.group.setlak 22:12, 25 September 2008 (UTC)

Brian Taylor Eml4500.f08.group.taylor 23:09, 25 September 2008 (UTC)

Tom Cady Eml4500.f08.group.cady 03:06, 26 September 2008 (UTC)

Aaron Clausen Eas4200c.f08.carbon.clausen 17:25, 26 September 2008 (UTC)

Melisa Gaar Eml4500.f08.group.gaar 18:51, 26 September 2008 (UTC)


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User:Eas4200c.f08.gator.edwards/Week Two Class Notes

 Class Notes (current) Archived

 Problem 1.1 continued (current) Archived

 Additional Questions (current) Archived

Contributing Team Members:

The following members contributed to this report:

Greg Edwards Eas4200c.f08.gator.edwards 03:32, 26 September 2008 (UTC)

William Mueller Eas4200c.f08.gator.mueller 04:57, 26 September 2008 (UTC)

Justin Dickinson Eas4200c.f08.gator.dickinson 05:41, 26 September 2008 (UTC)

Taylor Lusk Eas4200c.f08.gator.lusk 17:54, 26 September 2008 (UTC)


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User:Eml4500.f08.group/AAnotes

Mon, 25 Aug 08
Class Website http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/

FEM wide spread use in all reas of engineering applied math, heologival esplorations of oil reseriors etc

Password for mediawiki: fall08.wiki

Wed, 27 Aug 08
Deadline for the team structure: - Wed, 3 Sep 08 Each team, (5 or 6 students) Submit a paper:

 * Team Name: 4 Char preferred, max 6 char.
 * member names + UFID Numbers
 * Wikipedia Class usernames

Put your contribution is collapsible table

Vision - Professor smoking the Wiki Pipe

 * Big Picture re Wikipedia
 * textbooks, Wikiversity
 * MIT's OpenCourseWare

 * MediaWiki
 * Wikiversity - MIT's OpenCourseWare
 * Collaboration between students in a team, collaboration of whole class, collaboration of prof and class

Old approach
10% HW, 30% + 30% + 30% Exams

Trusses, Matrix Method (Chapter 4)
Also MIT's Open Course Ware

Class Wiki Pages Truss with 2 elastic (deformable) bars

Both "Bases" Constrained, no displacement

Global FBD
Whole Structure

4 unknown reactions, 3 equations of equilibria, statically indeterminate

The look at Free body diagrams for each element,

(n) Global Node number n [n] Local Node number n fi = ith internal force of element e: 1 = 1,2,3,4 : e = 1, 2

=Fri, 5 Sep 08=

Reading
READ: Chap 4 (Trusses, beams, frames)
 * Sections
 * 1.1.1 Plane Truss Elem
 * 1.2 Assembly of elem eqs
 * ex 1.4 Five-bar truss
 * Elem Solutions and model ralidity
 * 1.4.1 Plane Truss element

Steps to solve simple truss system
Steps to solve simple truss system described on Wed, 3 Sep 08. (Recipe)

Global picture
(description) at structure level
 * Global degrees of freedom (disp coeffs)
 * Global Forces

Actually the displacement degrees of freedom are partitioned into Simularly for the global forces
 * A known part, eq, fixed defs, constraints
 * an unknown part, solved using FEA
 * a know part: applied force
 * an unknown part: reactions

Element picture
either in global coord system on in local coord system
 * element degrees of freedom
 * element forces

Global Force Displacement relationship

 * Elemenet Stiffness matrix in global coord
 * element force matrices in global coord.
 * Assembly of element stiffness matrix and element force matrix into a global Force Displacement relationship

Elimination of known degrees of freedom
Elimination of known degrees of freedom to reduce the global Force Displacement Relationship (stiffness matrix non-singular => invertible)

 * m = number of unknown displacement degrees of freedom
 * n = number of known and unknown and displacement degrees of freedom

Kbar non singular => Kbar^-1   (Kbar invertible)

Compute Element forces
Compute Element forces from now known dbar => element stresses

Compute Reactions
Compute reactions (unknown forces)

Example
Take a specific example to see how the recipe works (no justification yet)

Element Length - L(1) = 4, L(2) = 2

Youngs Modulus - E(1) = 3, E(2) = 5

Cross Section Area - A(1) = 1, A(2) = 2

=Mon, 8 Sep 08= 2.) Element Picture

3.) Global Force Displacement at element level

k is the element stiffness matrix for element e

d is the the element displacement matrix

f is the element force matrix

Matrix p 225 book, 2nd equation form the bottom

axial stiffness of bar elem "e". e = 1,2

l^(e), m^(e) = director cosines of ~x axis (goes from [1] to [2]) with respect to global (x,y) coords

=Wed, 10 Sep 08= [7] Model 2-bar truss system (cont'd)

Note: The director cosines are the components of i^~-> (unit vector along x axis, with respect to basics (i^->,j^->)

Element 1

Observations

 * 1) You only need to compute 3 numbers. Other coefficients have the same absolute values, just differ by (+) or (-).
 * 2) Symmetry [[Image:FEAeq-7-3.JPG]]

transpose results in (note the dangle symmetry

Element 2

k(1,1) = 2.5

Observation

 * 1) The absolute values of all coefficients are the same. You only need to compute 1 coefficient
 * 2) Symmetry [[Image:FEAeq-7-3.JPG]]

Element force displacement relationship:

Global Force Displacement relationship

n = 6, this is because there are 6 outside forces acting on the system. 2 in each of the X and Y directions at the 3 points of the truss.

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Summary
Image for HW2, Problem 1.7 Cross Sectional Areas Second Moment of Inertia Calculations

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User:Eas4200c.f08.aeris.guan/919lecture

Torque, Shear Flow, and Shear Panels
Recall that from the problem 1.1 that the torque of the thin-walled cross section equaled to twice the shear stress multiplied by the product of the dimensions given by the equation $$ \tau=\frac{T}{2abt}$$. This expression for the torque's relationship with the cross sectional dimensions is only for a particular case where the cross section of the thin-walled beam or member being analyzed is rectangular. For a more general case of relating the torque applied on an object, it is necessary to introduce the concept of shear flow. Shear flow is the product of the shear stress and the thickness of the cross section. Usually, this thickness is very small compared to the contour length. For more general case of relating torque with dimensions and shear flow, a formula is introduced in section 3.5 of "Mechanics of Aircraft Structures" by C.T. Sun. $$\displaystyle T= \oint{\rho qds}=\int\int_{A_{bar}}{2qdA}=2qA_{bar}$$ This formula describes the torque of a thin-walled bar with an arbitrary cross section of thickness t. Here, q is the shear flow, t is the thickness, A bar is the average area of the cross section, and T is the torque. In summary, for a nonuniform thin-walled cross section,$$T=2qA_{bar}:$$  and for a thin-walled rectangular cross section$$T=2 \tau ab$$:  Deriving the torque for the thin-walled rectangular cross section was an ad hoc method because of the assumptions used to determine the equation. However, it is logical. The derivation of the non-uniform cross section torque is a method based on elasticity. Upon scrutinization of the stringers of an aircraft that is part of a monocoque structure, the open thin-walled box structure was previously determined to be more practical that the closed structure because of ease of riveting and installation. However, as the stringers were examined more carefully, it is realized that the vertical sections are slightly slanted. Why is this? That is because this structure is practical for stockpiling and storage and at the same time, not losing much moment of inertia due to the slanting of the vertical sections.  Starting chapter two, the structure of a shear panel is introduced. It is a thin sheet of material, usually metal aluminum, that can withstand shear load. There is a relationship that relates the shear modulus of a material and the shear strain with the shear stress that is applied the structure. $$\displaystyle \tau =G \gamma$$, where G is the shear modulus, and gamma is the shear strain. Gamma describes the change in a right angle between two fictitious lines in the material due to deformation. It can be expressed by displacements in an (x,y) coordinate system relative to the structure. $$\displaystyle \gamma = \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=\frac{\partial u_{x}}{\partial y}+\frac{\partial u_y}{\partial y}$$ where $$u=u_x$$= the displacement along the x direction and similar for v. This shear strain is called the engineering shear strain and not the tensorial shear strain. These two quantities are different in the fact that $$\displaystyle \epsilon_{xy}=$$ (tensorial shear strain)$$\displaystyle = .5 \gamma_{xy}$$. The following picture describes the deformation of an infinitesimally small element (i.e. part of a shear panel) subject to shear forces. The blue shape is the deformed infinitesimally small rectangle.

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User:Eml4500.f08.bottle.barnes/HW2

 * $$ \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} $$

 * $$ f^{(e)} = \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix}$$
 * $$ d^{(e)} = \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}$$

 * $$ \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} d_1 \\ d_2 \\ d_3 \\d_4 \\d_5 \\d_6 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6  \end{Bmatrix}$$

 * $$ \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_{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)} \\ & & 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} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\d_4 \\d_5 \\d_6 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} $$

 * [[Image:K_Matrix.JPG]]

 * $$ \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} $$

 * $$ \begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix} \begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix} $$

title: User:Eml4500.f08.bottle.barnes/HW2


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User:Eas4200c.f08.gator.edwards/Problem 1.1cont.

This is a continuation of Problem 1.1 from the first homework.

Part of 1.1 from week one

Problem 1.1 is taken from the textbook Mechanics of Aircraft Structures by C.T.Sun, copyright 2006.

Find the optimum ratio of $$b/a$$ to maximize the load carrying ability of the beam. Assume $$M=T$$ and $$\displaystyle \sigma_{allowable}=2\displaystyle \tau_{allowable}$$.

The question is asking for the optimal cross section carrying the maximum bending moment and maximum torque.

Since the cross section walls are very thin, assume shear stress distribution in the thin walls of the beam are distributed uniformly. This results in the equation $$\displaystyle \tau=\frac{T}{2abt}$$.

Case 1: Assume $$\displaystyle \sigma$$ (bending moment stress) reaches $$\displaystyle \sigma_{allowable}$$ first. Verify $$\displaystyle \tau \le 2\displaystyle \tau$$.

From Mechanics of Materials, the normal stress caused by a bending moment is $$\displaystyle \sigma=\frac{Mz}{I}$$

Case 1
Assume that $$\displaystyle \sigma_{allow}$$ is reached first. Recall from mechanics of materials the equation below.

Next calculate the second area moment of inertia of the box beam. The details are shown below.

To find the maximum bending moment the $$\displaystyle \frac{I}{b}$$ term must be maximized to find the ratio of $$\displaystyle \frac{b}{a}$$. The details are shown below.

Now plug the maximum value of I/b into the bending moment equation to get an equation of Tmax, L, and t. Then compare the value to the shear stress max and compare.

Case 2
Assume $$\tau _{max} = \tau _{allow}$$ first.

$$\tau = \frac{T}{2abt} = \tau _{max}$$ because shear stress is uniformly distributed.

$$T=(2 t \tau _{all})(ab)$$ where $$ab$$ is the only variable, so therefore maximizing $$T$$ = maximizing $$ab$$

By substitution and from the assumption M=T:

$$T_{max}^{(2)} = (2t\tau_{allow})\left( \frac{L}{4}\right)^{2} = \frac{1}{8}tL^{2}\tau _{allow} = M_{max}^{(2)}$$

Now from the assumption $$\sigma _{allow} = 2\tau _{allow}$$ :

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

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

Recall: $$f(b) = \frac{I(b)}{b}$$ and $$b^{(2)} = \frac{L}{4}$$

$$f(b^{(2)}) = \frac{tb^{(2)}}{6}\left[ 3a^{(2)} + b^{(2)}\right]= \frac{t\frac{L}{4}}{6}\left[ 3\left( \frac{L}{4}\right) + \frac{L}{4}\right]$$

$$f(b^{(2)}) = \frac{tL^{2}}{24}$$

Now plugging this into the equation for  $$\displaystyle \sigma_{max} = \frac{Mb}{2I}$$,

we get $$\sigma_{max}^{(2)} = \frac{M_{max}^{(2)}}{2f(b)} = \frac{24M_{max}^{(2)}}{2tL^{2}} = \frac{12M_{max}^{(2)}}{tL^{2}}$$

Recalling that $$\sigma _{allow} = \frac{16}{tL^{2}}M_{max}^{(2)}$$

We find that $$\sigma_{max}^{(2)} = \left( \frac{3}{4}\right) \sigma_{allow}^{(2)}$$

Since $$\sigma_{max}^{(2)}$$ is not greater than $$\sigma_{allow}^{(2)}$$, we find that Case 2 works.

Conclusion
Since we found that Case 1 was not acceptable since the maximum stress was greater that the allowable stress and Case 2 is acceptable because the max stress was less than the allowable stress, we can now conclude that Case 2 has the optimum cross-section. Since in Case 2, $$a = b = \frac{L}{4}$$ we conclude that the optimum cross-section is a square.


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User:Eml4500.f08.qwiki.bishop/Lecture 8

=Lecture 8 - Friday, September 12, 2008=
 * Global Force-Displacement relation:
 * $$\mathbf{K_{(nxn)}d_{(nx1)} = F_{(nx1)}}$$
 * For this example, n=6

$$\begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}\\ k_{21} & k_{22} & k_{23} & k_{24} & k_{25} & k_{26}\\ k_{31} & k_{32} & k_{33} & k_{34} & k_{35} & k_{36}\\ k_{41} & k_{42} & k_{43} & k_{44} & k_{45} & k_{46}\\ k_{51} & k_{52} & k_{53} & k_{54} & k_{55} & k_{56}\\ k_{61} & k_{62} & k_{63} & k_{64} & k_{65} & k_{66}\\ \end{bmatrix}\begin{pmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\\ \end{pmatrix}=\begin{pmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{pmatrix}$$

 * In compact notation, this matrix can be written in the form:

$$\begin{bmatrix} k_{ij}\\ \end{bmatrix}_{(6x6)}\begin{pmatrix} d_j\\ \end{pmatrix}_{(6x1)}=\begin{pmatrix} F_i\\ \end{pmatrix}_{(6x1)}$$

 * (More generally, nxn)

$$\sum_{j=1}^{6}{K_{ij}d_j}=F_i, i=1,...,6$$

The above global matrices can be define as follows:

 * $$K_{(nxn)}=[K_{ij}]_{(nxn)}$$ = Global Stiffness Matrix
 * $$d_{(nx1)}=(d_j)_{(nxn)}$$ = Global Displacement Matrix
 * $$F_{(nx1)}=(f_i)_{(nxn)}$$ = Global Force Matrix

When dealing with a single element, the matrices can be written as:

 * $$K_{(4x4)}^{(e)}=[K_{ij}^{(e)}]_{(4x4)}$$ = Element Stiffness Matrix
 * $$d_{(4x1)}^{(e)}=(d_j^{(e)})_{(4x4)}$$ = Element Displacement Matrix
 * $$F_{(4x1)}^{(e)}=(f_i^{(e)})_{(4x4)}$$ = Element Force Matrix

An assembly process is used to go form an element matrix to a global matrix. Identify the correspondence between element displacement degrees of freedom and global displacement degrees of freedom. The following examples show how to go from the global level to the element level for each node.

 * Examples for node 1:
 * $$d_1=d_1^{(1)}$$
 * $$d_2=d_2^{(1)}$$

 * Examples for node 2:
 * $$d_3=d_3^{(1)}=d_1^{(2)}$$
 * $$d_4=d_4^{(1)}=d_2^{(2)}$$

 * Examples for node 3:
 * $$d_5=d_3^{(2)}$$
 * $$d_6=d_4^{(2)}$$

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User:Eml4500.f08.jamama.justin/HW2

The Global Forces are determined by the following matrix

$$\begin{Bmatrix} f1\\ f2\\ f3\\ f4\\ f5\\ f6\\ \end{Bmatrix} = \begin{bmatrix} & &  &  &  & \\  &  &  &  &  & \\  &  &  K&  &  & \\ & &  &  &  & \\  &  &  &  &  & \\  &  &  &  &  & \end{bmatrix} \times \begin{Bmatrix} d1\\ d2\\ d3\\ d4\\ d5\\ d6\\ \end{Bmatrix} $$ 6X1      6X6           6X1

Global Force      Global Stiffness           Global Displacement

Column Matrix      Matrix           Column Matrix

Before the global matrix can be constructed, you first need to find the displacements in the elements themselves. To do this for each element, the following equation may be used:

$$K^{\left(e \right) } \times D^{\left(e \right)} = F^{\left(e \right)} $$

4X4      4X1           4X1

where: $$K^{\left(e \right) }$$=Element stiffness matrix for element e

$$D^{\left(e \right)}$$=Element displacement matrix of element e

$$F^{\left(e \right)}$$=Element force matrix

Element Stiffness Matrix:

The element stiffness matrix $$K^{\left(e \right)}$$ for the equation above is calculated with the matrix shown (below).

$$k^{\left(e \right)}\times \begin{bmatrix} \left(l^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} & -\left(l^{\left(e \right)} \right)^2 & -l^{\left(e \right)}m^{\left(e \right)}\\ l^{\left(e \right)}m^{\left(e \right)} & \left(m^{\left(e \right)} \right)^2 & -l^{\left(e \right)}m^{\left(e \right)} & -\left(m^{\left(e \right)} \right)^2 \\ -\left(l^{\left(e \right)} \right)^2& -l^{\left(e \right)}m^{\left(e \right)} & \left(l^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} \\ -l^{\left(e \right)}m^{\left(e \right)}& -\left(m^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} & \left(m^{\left(e \right)} \right)^2 \end{bmatrix} $$

where: $$k^{\left(e \right)}= \frac{E^{\left(e \right)}A^{\left(e \right)}}{L^{\left(e \right)}}$$ Axial stiffness of bar element "e"

and $$l^{\left(e \right)}, m^{\left(e \right)}$$=Director cosines of the $$\tilde{x}$$ axis with respect to the (x,y) coordinates for element e.

Director Cosines

Director cosines relate the angle at which the element is on to the traditional x,y axis. These values are needed to calculate the stiffness matrix k.

$$l^{\left(e \right)}=\vec{\tilde{i}}\bullet \vec{i}=\cos \theta^{\left(e \right)}$$

$$m^{\left(e \right)}=\vec{\tilde{i}}\bullet \vec{j}=\cos \left(\frac{\pi }{2} - \theta^{\left(e \right)}\right)=\sin \theta ^{\left(e \right)}$$

$$\vec{\tilde{i}}=\cos \theta ^{\left(e \right)} \vec{i} + \sin \theta ^{\left(e \right)}\vec{j}$$

$$\vec{\tilde{i}}\bullet \vec{i}=\cos \theta ^{\left(e \right)} \vec{i}\bullet \vec{i} + \sin \theta ^{\left(e \right)}\vec{j}\bullet \vec{i}$$

$$\vec{\tilde{i}}\bullet \vec{j}=\cos \theta ^{\left(e \right)} \vec{i}\bullet \vec{j} + \sin \theta ^{\left(e \right)}\vec{j}\bullet \vec{j}$$

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User:Eml4500.f08.RAMROD.A/HW2

Statically Indeterminate 3 Truss System
A look at the system with global nodes identifited:

Element Picture
We want to look at the elemental picture with $$\displaystyle {\color{Green} Global }$$ and $$\displaystyle {\color{Blue} Local }$$ coordinate systems:

While the $$\displaystyle {\color{Blue} Local }$$ Coordinates are denoated by $$\displaystyle {\color{Blue} (\tilde{x}, \tilde{y}) } $$.

 * $$\displaystyle \tilde{x} $$. is defined as along the element from local node 1 to 2.

Vector Notation

 * A common occurrence when dealing with forces such as P or the lengths of elements is to use vector notation: $$\displaystyle ({i}, {j})$$ a long with the angle between $$\displaystyle x $$ and $$\displaystyle \tilde{x}$$; denoted as $$\displaystyle \theta $$.

Next Step of Finite Element Method
It is important to go back to the equation described in the recipe from [earlier articles] on Finite Element Method:

K(e) * d(e) = f(e)

Where:

 * K is the element stiffness matrix

 * d is the element displacement matrix

 * f is the element force matrix

It is now possible to determine ke using the aforementioned $$\displaystyle \theta $$ and what we call the direction cosines

Direction Cosines
$$\displaystyle l^e $$ and $$\displaystyle m^e $$.

 * $$\displaystyle l^e = \cos (\theta) $$

 * $$\displaystyle m^e = \cos (90-\theta) = \sin (\theta) $$

Element Stiffness Matrix, k
$$\displaystyle \mathbf{k^e}=\frac{E^e*A^e}{L^e}\left[Direction Cosine Matrix \right] $$

Where: "$\displaystyle L = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} $"
 * E = Young's Modulus of element
 * A = Cross sectional Area of element
 * L = Length of element

When the cosine matrix is added in, we get: $$\displaystyle \mathbf{k^e}=\frac{E^e*A^e}{L^e} \begin{bmatrix} l^2 & (l\cdot m) & -l^2  & -(l\cdot m) \\ \\(l\cdot m) & m^2 & -(l\cdot m) & -m^2 \\ \\-l^2 & -(l\cdot m) & l^2 & (l\cdot m) \\ \\-(l\cdot m) & -m^2 & (l\cdot m) & m^2 \end{bmatrix}$$

= Element One Stiffness Analysis =

Taking all of the knowledge we have gained into consideration, we can now begin the analysis of the stiffness matrix for element one. Because we know that the element stiffness matrix is derived with the use of only three variables,

$$\displaystyle \ l^{e}$$     $$\displaystyle \  m^{e}$$     $$\displaystyle \ k^{e}$$

it is imperative to the completion of this problem to compute these values so that the element stiffness matrix for element one can be constructed. And if you have already forgotten, $$\displaystyle \ k^{e}$$ is the axial stiffness of the bar element $$\displaystyle \ e $$ and is given by the following formula,

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

where $$\displaystyle \ E^{e} $$, $$\displaystyle \ A^{e} $$, and $$\displaystyle \ L^{e} $$ are the modulus of elasticity, cross sectional area, and length of an element $$\displaystyle \ e $$ respectively. If you remember from previous work, the equations for $$\displaystyle \ l^{e}$$ and $$\displaystyle \ m^{e} $$ have already been computed and are as follows;

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

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

where θ is the angle at which the truss element is tilted. If you recall from the problem statement, the following values for the length, angle, modulus, and cross sectional area were given for element one.

With this information it is possible to begin to calculate the three variables needed to construct the element stiffness matrix.

$$\displaystyle \ l^{1} = cos (30) = \left ( \frac{\sqrt{3}}{2} \right ) $$

$$\displaystyle \ m^{1} = sin (30) = \left ( \frac{1}{2} \right ) $$

$$\displaystyle \ k^{1} = \left ( \frac{(3)(1)}{4} \right ) = \left ( \frac{3}{4} \right ) $$

Having calculated these values, it is now possible to construct the stiffness matrix for element one. Before we defined the element stiffness matrix as the following;

$$\displaystyle \ \mathit{k}^{(e)} = k^{e} \begin{bmatrix} (l^{e})^2 & l^em^e & -(l^e)^2 & -l^em^e\\ l^em^e & (m^e)^2 & -l^em^e & -(m^e)^2\\ -(l^{e})^2 & -l^em^e & (l^e)^2 & l^em^e\\ -l^em^e & -(m^e)^2 & l^em^e & (m^e)^2 \end{bmatrix}$$

To make things a little more condensed and easier to understand, we will multiply the axial stiffness value through and represent the element stiffness matrix as follows;

$$\displaystyle \ k^{(1)} = $$ $$\displaystyle \begin{bmatrix} k_{1,1}^1 & k_{1,2}^1 & k_{1,3}^1 & k_{1,4}^1 \\ \\k_{2,1}^1 & k_{2,2}^1 & k_{2,3}^1 & k_{2,4}^1 \\ \\ k_{3,1}^1 & k_{3,2}^1 & k_{3,3}^1 & k_{3,4}^1 \\ \\ k_{4,1}^1 & k_{4,2}^1 & k_{4,3}^1 & k_{4,4}^1\end{bmatrix}$$,

where   $$\displaystyle \ k_{1,1}^1 = k^{1}* l^{1^{\,\!2}} $$ and    $$\displaystyle \ k_{1,2}^1 = k^{1}*{l^{1}}*{m^{1}} $$ and so on until $$\displaystyle\ k^{e} $$ (axial stiffness) has been multiplied through the director cosines matrix. A short hand form of the same matrix can be constructed in which the entire four by four matrix is represented by a one by one matrix. This form is as follows,

$$\displaystyle \begin{bmatrix} k_{i,j}^{(e)} \end{bmatrix}$$  $$\displaystyle \ 0 \le i \le 4 $$     $$\displaystyle \ 0 \le j \le 4 $$

where $$\displaystyle \ 0 \le i \le 4 $$ and $$\displaystyle \ 0 \le j \le 4 $$. This matrix along with the equivalent matrix above is the element stiffness matrix. All of the required numerical values are at our disposal so it is now time to compute the values that will be placed into the matrix. As stated before,

$$\displaystyle \ k_{1,1}^1 = k^{1}*l^{1^{\,\!2}} $$ and $$\displaystyle \ k_{1,2}^1 = k^{1}*{l^{1}}*{m^{1}} $$

If a closer look is taken at the director cosine matrix, one will notice two important aspects that characterize the values within the matrix. The first is that there are only three numbers need to be calculated in order to completely fill the matrix, two of which are displayed in the two equations above. The second is that this matrix is symmetrical, meaning that if a line was drawn diagonally across the matrix as to cut it into two pieces, the upper portion would be a mirror image of the bottom portion. In general this means

$$\displaystyle \ k_{i,j}^e = k_{j,i}^e $$ or  $$\displaystyle \ K^e = K^{{e}^T} $$

where $$\displaystyle \ T$$ represents the transpose of the element stiffness matrix of an element $$\displaystyle \ e $$.

As was previously stated, the element stiffness matrix is composed of only three values, of which are computed below.

$$\displaystyle \ k_{1,1}^1 = k^{1}*l^{1^{\,\!2}} = \left ( \frac{9}{16} \right )$$ $$\displaystyle \ k_{1,2}^1 = k^{1}*{l^{1}}*{m^{1}} = \left ( \frac{3\sqrt{3}}{16} \right )$$ $$\displaystyle \ k_{2,2}^1 = k^{1}*m^{1^{\,\!2}} = - \left ( \frac {3}{16} \right )$$

It is now possible to construct the stiffness matrix $$\displaystyle \ k $$ for element one.

$$\displaystyle \ k^1 = $$ $$\displaystyle \begin{bmatrix} \left ( \frac{9}{16} \right ) & \left ( \frac{3\sqrt{3}}{16} \right ) & - \left ( \frac{9}{16} \right ) & -\left ( \frac{3\sqrt{3}}{16} \right ) \\ \\ -\left ( \frac{9}{16} \right ) & -\left ( \frac{3\sqrt{3}}{16} \right ) & \left ( \frac{9}{16} \right ) & \left ( \frac{3\sqrt{3}}{16} \right ) \\ \\ -\left ( \frac{3\sqrt{3}}{16} \right ) & \left ( \frac {3}{16} \right ) & \left ( \frac{3\sqrt{3}}{16} \right ) & - \left ( \frac {3}{16} \right ) \\ \\ -\left ( \frac{3\sqrt{3}}{16} \right ) & \left ( \frac {3}{16} \right ) & \left ( \frac{3\sqrt{3}}{16} \right ) & - \left ( \frac {3}{16} \right )\end{bmatrix}$$

=Element Two Stiffness Analysis=

Now that the stiffness matrix for element one is completed, construction of the stiffness matrix for element two can begin. To do so requires us to perform the same method as used for element one. To begin, lets remind ourselves of the parameters for element two given in the problem statement:

With this data we can compute the axial stiffness as well as the director cosine values $$\displaystyle \ l^2, m^2 $$.

$$\displaystyle \ k^{2} = \left ( \frac{E^{2} A^{2}}{L^{2}} \right ) = \left ( \frac{5*2}{2} = 5\right )$$ $$\displaystyle \ l^{1} = cos (-45) = \left ( \frac{\sqrt{2}}{2} \right ) $$ $$\displaystyle \ m^{1} = sin (-45) = -\left ( \frac{\sqrt{2}}{2} \right ) $$

And just as before, the stiffness matrix can be represented as the director cosine matrix multiplied through by the axial stiffness matrix.

$$\displaystyle \ k^2 = $$ $$\displaystyle \begin{bmatrix} k_{1,1}^2 & k_{1,2}^2 & k_{1,3}^2 & k_{1,4}^2 \\ \\k_{2,1}^2 & k_{2,2}^2 & k_{2,3}^2 & k_{2,4}^2 \\ \\ k_{3,1}^2 & k_{3,2}^2 & k_{3,3}^2 & k_{3,4}^2 \\ \\ k_{4,1}^2 & k_{4,2}^2 & k_{4,3}^2 & k_{4,4}^2\end{bmatrix}$$

where

$$\displaystyle \ k_{1,1}^{2} = k^2 * (l^1)^2 $$.

All values that are needed to complete the stiffness matrix are now known, but if a further observation is made the process of constructing this matrix becomes very simple. If you notice that the absolute value of the director cosine values of $$\displaystyle \ l^2, m^2 $$ are exactly the same, it is only necessary to compute one value of the stiffness matrix. This is possible because the director cosine matrix is a combination of three different processes:

$$\displaystyle \ (l^1)^2, (m^1)^2, m^2 * l^2 $$

Since all of these calculations yield the same values, only one calculation is needed to complete the matrix. Then the value can be placed and the sign of the values can appropriately be assigned as necessary. Below is the once calculation needed in order to carry out this process.

$$\displaystyle \ k_{1,1}^2 = k^2 * (l^1)^2 = \left ( \frac{5}{2} \right )$$

This then makes the element two stiffness matrix to be the following,

$$\displaystyle \ k^1 = $$ $$\displaystyle \begin{bmatrix} \left ( \frac{5{2}}{2} \right ) & -\left ( \frac{5}{2} \right ) & - \left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) \\ \\ -\left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) & -\left ( \frac{5}{2} \right ) \\ \\ -\left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) & -\left ( \frac{5}{2} \right ) \\ \\ \left ( \frac{5}{2} \right ) & -\left ( \frac{5}{2} \right ) & -\left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right )\end{bmatrix}$$

Just as the element stiffness matrix for element one, this stiffness matrix is symmetric. In other words,

$$\displaystyle \ k_{i,j}^2 = k_{j,i}^2 $$ or  $$\displaystyle \ K^{(2)} = K^{{2}^T} $$

where $$\displaystyle \ T$$ once again represents the transpose of the element stiffness matrix of an element $$\displaystyle \ 2 $$.

=Element Force-Displacement Relationship=

Now that we have found the stiffness matrices for both elements, we must now find the relationship between the force and displacement. In order to do so we must analysis the following equation:

$$\displaystyle \ k^e * d^e = f^e $$

where $$\displaystyle \ k, d, $$ and $$\displaystyle \ f $$ are the stiffness matrix, displacement matrix , and force displacement of an element $$\displaystyle \ e $$ respectively. The math works as the stiffness matrix is a $$\displaystyle \ 4 X 4 $$ matrix and when multiplied by a $$\displaystyle \ 4 X 1 $$ matrix (element displacment matrix), the product becomes a $$\displaystyle \ 4 X 1 $$ matrix that is the element force matrix.

$$\displaystyle \ k^{(e)} = $$ $$\displaystyle \begin{bmatrix} k_{1,1}^{(e)} & k_{1,2}^{(e)} & k_{1,3}^{(e)} & k_{1,4}^{(e)} \\ \\k_{2,1}^{(e)} & k_{2,2}^{(e)} & k_{2,3}^{(e)} & k_{2,4}^{(e)} \\ \\ k_{3,1}^{(e)} & k_{3,2}^{(e)} & k_{3,3}^{(e)} & k_{3,4}^{(e)} \\ \\ k_{4,1}^{(e)} & k_{4,2}^{(e)} & k_{4,3}^{(e)} & k_{4,4}^{(e)}\end{bmatrix}$$ $$\displaystyle \ d^{(e)} = $$ $$\displaystyle \begin{bmatrix} d_{1}^{(e)} \\ \\ d_{2}^{(e)} \\ \\ d_{3}^{(e)} \\ \\ d_{4}^{(e)}\end{bmatrix}$$ $$\displaystyle \ f^{(e)} = $$ $$\displaystyle \begin{bmatrix} f_{1}^{(e)} \\ \\ f_{2}^{(e)} \\ \\ f_{3}^{(e)} \\ \\ f_{4}^{(e)}\end{bmatrix}$$

In Compact Notation

 * [kij]6x6{dj}6x1 = {Fi}6x1
 * (more generally, the matrices are nxn)

$$\sum_{j=1}^{6}{ K_{ij}d_j }\! = F_i$$,    i = 1, 2, 3, ...., 6

The three matrices used are shown below:

 * Knxn = [Kij]nxn = global stiffness matrix

 * dnxn = {dj}nxn = global displacement matrix

 * Fnxn = {Fi}nxn = global force matrix

Recall the element force-displacement relation
 * k(e)4x4d(e)4x1=f(e)4x1

 * kn4x4 = [kij(e)] = element stiffness matrix

 * dnxn = {dj(e)} = element displacement matrix

 * Fnxn = {fi(e)} = elementforce matrix

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

The appropriate numbers will be inputted into matrix K which is a 6x6 matrix.

$$\mathbf{K}= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}+k_{11}^{(2)}) & (k_{34}^{(1)}+k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)}\\k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}+k_{21}^{(2)}) & (k_{44}^{(1)}+k_{22}^{(2)}) & k_{23}^{(2)} & k_{24}^{(2)}\\0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\end{bmatrix}$$

$$k_{11}^{(1)}=\frac{9}{16}=0.5625$$

$$k_{12}^{(1)}=\frac{3\sqrt{3}}{16}=0.3248$$

$$k_{13}^{(1)}=-k_{11}^{(1)}=-0.5625$$

$$k_{14}^{(1)}=-k_{12}^{(1)}=-0.3248$$

$$k_{22}^{(1)}=\frac{3}{16}=0.1875$$

$$k_{24}^{(1)}=-k_{22}^{(1)}=-0.1875$$

$$k_{13}^{(2)}=-k_{11}^{(2)}=\frac{-5}{2}=-2.5$$

$$k_{14}^{(2)}=-k_{12}^{2}=2.5$$

$$k_{24}^{(2)}=\frac{-5}{2}=-2.5$$

$$k_{33}^{(2)}=\frac{5}{2}=2.5$$

$$k_{34}^{(2)}=\frac{-5}{2}=-2.5$$

$$k_{44}^{(2)}=\frac{5}{2}=2.5$$

$$K_{15}^{}=K_{16}=K_{25}=K_{26}=K_{51}=K_{52}=K_{61}=K_{62}=0$$

$$\mathbf{}= \begin{bmatrix} 0.5625 & 0.32475 & -0.5625 & -0.32475 & 0 & 0\\0.32475 & 0.1875 & -0.3247 & -0.1875 & 0 & 0\\-0.5625 & -0.32475 & (0.5625+2.5) & (0.32475-2.5) & -2.5 & 2.5\\-0.32475 & -0.1875 & (0.32475-2.5) & (0.1875+2.5) & 2.5 & -2.5\\0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\0 & 0 & 2.5 & -2.5 & -2.5 & 2.5\end{bmatrix}$$

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

Elimination of known degrees of freedom in order to reduce the global Force-Displacement relationship
For this problem, the reduction is due to the fact that global nodes 1 and 3 are fixed and therefore have zero displacement, or:

$$d_{1} = d_{2} = d_{5} = d_{6} = 0$$

Setting these displacements to zero is the equivalent of eliminating the corresponding columns (1,2,5 & 6) of the global stiffness matrix, K. This elimination is shown below.

$$K = \begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix}$$

In addition to removing these columns because of the fixed boundary conditions, rows 1, 2, 5, & 6 can also be eliminated due to the principle of virtual work (PVW). After deletion of these rows, the resulting K becomes:

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

And, the resulting global Force-Displacement relationship is:

$$\begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44}\\ \end{bmatrix} * \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \begin{Bmatrix} F_{3}\\ F_{4} \end{Bmatrix}$$

We have already solved for each element in K previously such that

$$K = \begin{bmatrix} 3.0625 & -2.175\\ -2.175 & 2.6875\\ \end{bmatrix}$$

Looking at the problem drawing, we can see that F3 is zero and that F4 is equal to -7. Plugging these values and the values of K into the above global F-D relationship gives the relationship:

$$ \begin{bmatrix} 3.0625 & -2.175\\ -2.175 & 2.6875\\ \end{bmatrix} * \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \begin{Bmatrix} 0\\ -7 \end{Bmatrix}$$

Now, to solve for the unknown displacements, we can multiply both sides by K-1, where:

$$K^{-1} = \frac{1}{det(K)}\begin{bmatrix} K_{33} & -K_{34}\\ -K_{43} & K_{44} \end{bmatrix} = \begin{bmatrix} 0.875 & 0.621\\ 0.621 & 0.768 \end{bmatrix}$$

It is important to not that K-1 is not the same as the transpose of K(KT).

Therefore, the equation to solve for the unknown displacements are:

$$\begin{bmatrix} 0.875 & 0.621\\ 0.621 & 0.768 \end{bmatrix}*\begin{Bmatrix} 0\\ -7 \end{Bmatrix} = \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix}$$

Solving for these displacements gives

$$d_{3} = 4.352 $$ $$d_{4} = 6.1271$$

Finding Reactions
The final step for this problem is to solve for the unknown reactions at global nodes 1 and 3. For this we use the element F-D relationships:

$$K^{(1)}* d^{(1)} = F^{(1)}$$ $$K^{(2)}* d^{(2)} = F^{(2)}$$

In the above equation, K(e), d(e), and  half of f(e) are known for each element, e. For this situation, the F-D relationships with all known values substituted is:

$$\begin{Bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16}\ & \frac{3}{16} \end{Bmatrix} \begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix}= \begin{Bmatrix} f^{(1)}_1\\ f^{(1)}_2\end{Bmatrix}$$

$$\begin{Bmatrix} \frac{5}{2} & \frac{-5}{2}\\ \frac{-5}{2} & \frac{5}{2} \end{Bmatrix} \begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix}= \begin{Bmatrix} f^{(2)}_3\\ f^{(2)}_4 \end{Bmatrix}$$

Using this equation, we can now solve for the f(1) matrix and thus we find the values for the reactions: $$f^{(1)}_{1} = F_{1} = -4.4378$$ $$ f^{(1)}_{2} = F_{2} = -2.5622$$

Using this same procedure applied to element 2, we can solve for the values of the other two unknown reactions giving us: $$f^{(2)}_{3} = F_{5} = 4.4378$$ $$f^{(2)}_{4} = F_{6} = -4.4378$$

Explaining Resulting Forces
Element 1:

K(1) · d(1) = F(1)

$$= \begin{bmatrix} 0.5625 & 0.032476 & -0.5625 & -0.032476 \\ 0.032476 &

0.1875 & -0.032476 & -0.1875 \\ -0.5625 & -0.032476 & 0.5625 & 0.032476 \\ -0.032476 & -0.1875 & 0.032476 & 0.1875 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

You may simplify the matrix because of the boundary conditions e.g. the first two numbers in the displacement matrix are zero.

$$= \begin{bmatrix} -0.5625 & -0.032476 \\ -0.032476 & -0.1875 \\ 0.5625 & 0.032476 \\ 0.032476 & 0.1875 \end{bmatrix} \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

$$=\begin{Bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{Bmatrix} = \begin{Bmatrix} F_1^1 \\ F_2^1 \\ F_3^1 \\ F_4^1 \end{Bmatrix}$$

Reactions forces are actually $$F_1^1$$ & $$F_2^1$$

Internal forces are actually $$F_3^1$$ & $$F_4^1$$

Element 1:

$$P_1^1 = \sqrt((F_1^1)^2 + (F_2^1)^2)$$

$$P_1^1 = \sqrt((-4.4378)^2 + (-2.5622)^2$$

$$ P_1^1 = 5.69064                     $$

Element 2:

$$P_1^2 = \sqrt((F_1^2)^2 + (F_2^2)^2)$$

$$P_1^2 = \sqrt((-4.4378)^2 + (-2.5622)^2$$

$$P_1^2 = 5.69064  $$

Method Two: Statics
Solving the two bar truss system using Euler Cut Principle.

Back to FE Solution: How to bring P back into the big picture?

Ans: Using the Euler cut method to isolate node two, equilibrium can be established at global node 2. Statics can be applied to the system by using summation of the forces in the x and y directions. This provides two equations with two unknowns that can be solved by using a system of equations.

$$\sum{F_x}= P_1cos30 - P_2cos45=0$$

$$\sum{F_y}=7- P_1cos30 -P_2cos45=0$$

$$\sum{F_x}= P_1 \frac{\sqrt{3}}{2} - P_2\frac{\sqrt{2}}{2}=0$$

$$\sum{F_y}=7- \frac{P_1}{2}-P_2\frac{\sqrt{2}}{2}=0$$

$$P_1=5.122436\;$$

$$P_2 = 6.27603\;$$

Two Bar Truss System MATLAB Code
This section goes over the finite element MATLAB code for a two bar truss system. All of the resulting matrices are below the code. The MATLAB example code can also be found on Dr. Vu-Quoc's website:

http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt

% Two bar truss example clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4; nodes = [0, 0; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta)]; dof=2*length(nodes); conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1); %load vector R = zeros(dof,1); R(4) = P; % Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e, A, ...            nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results k_local = 0.75       -0.75         -0.75         0.75 k = 0.5625     0.32476      -0.5625     -0.32476       0.32476       0.1875     -0.32476      -0.1875       -0.5625     -0.32476       0.5625      0.32476      -0.32476      -0.1875      0.32476       0.1875 k_local = 5   -5     -5     5 k = 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 K = 0.5625     0.32476      -0.5625     -0.32476            0            0       0.32476       0.1875     -0.32476      -0.1875            0            0       -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5      -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5             0            0         -2.5          2.5          2.5         -2.5             0            0          2.5         -2.5         -2.5          2.5 R = 0     0      0      7      0      0 d = 0            0         4.352        6.1271             0             0 reactions = -4.4378      -2.5622        4.4378       -4.4378 results = 1.7081      5.1244       8.5406       5.1244       17.081        0.6276       1.8828        3.138       1.8828        6.276

Comparison of Methods
The Matlab code produced the exact same results for the reaction forces located at global nodes 1 and 3, as did the Finite Element Method learned in class:

Matlab reactions:

reactions = -4.4378      -2.5622        4.4378       -4.4378

Finite Element Method:

Using this equation, we can now solve for the f(1) matrix and thus we find the values for the reactions: $$f^{(1)}_{1} = F_{1} = -4.4378$$ $$ f^{(1)}_{2} = F_{2} = -2.5622$$

Using this same procedure applied to element 2, we can solve for the values of the other two unknown reactions giving us: $$f^{(2)}_{3} = F_{5} = 4.4378$$ $$f^{(2)}_{4} = F_{6} = -4.4378$$

Statics

Using the Euler cut method, you find the following components from the resultant P forces.

$$P_x^1 = -4.4378$$ $$P_y^1=-2.5622$$ $$P_x^2=4.4378$$ $$P_y^2=-4.4378$$

Contributing Team Members
Eml4500.f08.ramrod.D 19:19, 25 September 2008 (UTC)

Eml4500.f08.ramrod.c 04:24, 26 September 2008 (UTC)

Eml4500.f08.RAMROD.A 13:30, 26 September 2008 (UTC)

EML4500.f08.RAMROD.E 16:56, 26 September 2008 (UTC)

Eml4500.f08.RAMROD.F 19:38, 26 September 2008 (UTC)

EML4500.F08.RAMROD.B 19:38, 26 September 2008 (UTC)

title: User:Eml4500.f08.RAMROD.A/HW2


text:
User:Eml4500.f08.qwiki.bishop/HW2

 Comment: The pictures that were previously used in this homework submission have been changed. The previous images ( and ) were not originally authored by Team Qwiki, however I did not intentionally plagiarize Team Echo's work. While working on Homework 2, both Team Qwiki and Team Echo created two images, with the same name: and . Due to the nature of MediaWiki, if two different teams create an image with the same name, only the latest version is displayed when called upon in an article. Therefore, even though Team Qwiki uploaded two of our own images (Element1.jpg and Element2.jpg) during homework 2, Team Echo's images were displayed because their upload was the most recent. In the future, we will make sure to upload files that have a unique name. The correct images that were authored by Team Qwiki are included in this updated version of Homework 2.

A comparison between the current version of Homework 2 and our prior submission can be found here. Eric Gatch

=Lecture 6 - Monday, September 8, 2008=

Continued from Lecture 5
2. Element Picture

3. Global FD at element level

Element Stiffness Matrix
Element Stiffness Matrix:

$$\mathbf{k^{(e)}}=\begin{bmatrix} \mathbf{k}_{11} & \mathbf{k}_{12} & \mathbf{k}_{13} & \mathbf{k}_{14}\\ \mathbf{k}_{21} & \mathbf{k}_{22} & \mathbf{k}_{23} & \mathbf{k}_{24}\\ \mathbf{k}_{31} & \mathbf{k}_{32} & \mathbf{k}_{33} & \mathbf{k}_{34}\\ \mathbf{k}_{41} & \mathbf{k}_{42} & \mathbf{k}_{43} & \mathbf{k}_{44}\\ \end{bmatrix} $$

Knowing the above element stiffness matrix, is the result of:

$$\mathbf{k^{(e)}}=k^{(e)}*\begin{bmatrix} (l^{(e)})^{2} & l^{(e)}m^{(e)} & -(l^{(e)})^{2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & (m^{(e)})^{2} & -l^{(e)}m^{(e)} & -(m^{(e)})^{2}\\ -(l^{(e)})^{2} & -l^{(e)}m^{(e)} & (l^{(e)})^{2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -(m^{(e)})^{2} & l^{(e)}m^{(e)} & (m^{(e)})^{2} \end{bmatrix}$$

Where (e) is the element number.

$$k^{e}=\frac{E^{e}A^{e}}{L^{e}}$$ This is the axial stiffness of bar element "e"

$$l^{e},m^{e}$$ are director cosines of $$\tilde{x}$$ axis (from local node 1 to local node 2) with respect to the global (x,y) coordinate system.

$$l^{e}=\vec{\tilde{i}}\cdot \vec{i}=cos\theta^{e}$$

$$m^{e}=\vec{\tilde{i}}\cdot \vec{j}=sin\theta^{e}$$

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

Distributing by unit vector "i" it is easily seen that the equation reduces to $$l^{e}$$ because a unit vector dotted with itself equals 1, whereas a unit vector dotted with its perpendicular counter part equals 0.

Note: the director cosines are the components of $$\vec{\tilde{i}}$$

$$\vec{\tilde{i}}=cos\theta ^{e}\vec{i}+sin\theta ^{e}\vec{j}$$

=Lecture 7 - Wednesday, September 10, 2008= To continue with modeling the 2-bar truss system, element 1 is further examined to find the stiffness matrix k (1). Element 1 is located at an angle $$\theta ^{(1)}={\color{red}30^{\circ}}$$

Therefore inputting this angle into the following equations of the director cosines yields:

$$l^{(1)}=cos\theta ^{(1)}=cos30^{\circ} =\frac{\sqrt{3}}{2}=0.866$$

$$m^{(1)}=sin\theta ^{(1)}=sin30^{\circ} =\frac{1}{2}=0.50$$

Next, to find the stiffness coefficient the below equation is used.

$$k^{(1)}=\frac{E^{(1)} A^{(1)}}{L^{(1)}}=\frac{3*1}{4}=0.75$$

An example of finding out what each element of the stiffness matrix is shown below.

$$\mathbf{k_{11}}=k^{(1)}*(l^{(1)})^{2}=\frac{3}{4}*(\frac{\sqrt{3}}{2})^{2}=\frac{9}{16}=0.5625 $$

To compute element 2, the same steps are taken. Element 2 is at an angle $$\theta ^{(2)}={\color{red}45^{\circ}}$$ Again inputting this angle into the following equations of the director cosines yields:

$$l^{(2)}=cos\theta ^{(2)}=cos45^{\circ} =\frac{\sqrt{2}}{2}=0.707$$

$$m^{(2)}=sin\theta ^{(2)}=sin45^{\circ} =\frac{\sqrt{2}}{2}=0.707$$

Next, to find the stiffness coefficient the below equation is used.

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

An example of finding out what each element of the stiffness matrix is shown below.

$$\mathbf{k_{11}}=k^{(2)}*(l^{(2)})^{2}=5*(\frac{\sqrt{2}}{2})^{2}=\frac{5}{2}=2.5 $$

=Lecture 8 - Friday, September 12, 2008=
 * Global Force-Displacement relation:
 * $$\mathbf{K_{(nxn)}d_{(nx1)} = F_{(nx1)}}$$
 * For this example, n=6

$$\begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}\\ k_{21} & k_{22} & k_{23} & k_{24} & k_{25} & k_{26}\\ k_{31} & k_{32} & k_{33} & k_{34} & k_{35} & k_{36}\\ k_{41} & k_{42} & k_{43} & k_{44} & k_{45} & k_{46}\\ k_{51} & k_{52} & k_{53} & k_{54} & k_{55} & k_{56}\\ k_{61} & k_{62} & k_{63} & k_{64} & k_{65} & k_{66}\\ \end{bmatrix}\begin{pmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\\ \end{pmatrix}=\begin{pmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{pmatrix}$$

 * In compact notation, this matrix can be written in the form:

$$\begin{bmatrix} k_{ij}\\ \end{bmatrix}_{(6x6)}\begin{pmatrix} d_j\\ \end{pmatrix}_{(6x1)}=\begin{pmatrix} F_i\\ \end{pmatrix}_{(6x1)}$$

 * (More generally, nxn)

$$\sum_{j=1}^{6}{K_{ij}d_j}=F_i, i=1,...,6$$

The above global matrices can be define as follows:

 * $$K_{(nxn)}=[K_{ij}]_{(nxn)}$$ = Global Stiffness Matrix
 * $$d_{(nx1)}=(d_j)_{(nxn)}$$ = Global Displacement Matrix
 * $$F_{(nx1)}=(f_i)_{(nxn)}$$ = Global Force Matrix

When dealing with a single element, the matrices can be written as:

 * $$K_{(4x4)}^{(e)}=[K_{ij}^{(e)}]_{(4x4)}$$ = Element Stiffness Matrix
 * $$d_{(4x1)}^{(e)}=(d_j^{(e)})_{(4x4)}$$ = Element Displacement Matrix
 * $$F_{(4x1)}^{(e)}=(f_i^{(e)})_{(4x4)}$$ = Element Force Matrix

An assembly process is used to go form an element matrix to a global matrix.

=Lecture 9 - Monday, September 15, 2008= $$\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} & k_{13}^{(2)} &          k_{14}^{(2)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} & k_{23}^{(2)} &   k_{24}^{(2)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\\ \end{bmatrix}\begin{pmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\\ \end{pmatrix}=\begin{pmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{pmatrix}$$
 * Global Force/Displacement Relationship
 * (Global stiffness matrix, K) X (global displacement, d) = (global force, F)
 * $$\mathbf{K_{(6x6)}d_{(6x1)} = F_{(6x1)}}$$
 * The global stiffness matrix, K, is found by combining the element stiffness matrices $$k^{(1)}$$ & $$k^{(2)}$$ as seen below.
 * NOTE: There is an overlap for displacements $$d_1$$ & $$d_1$$, which results from Element 1 and Element 2 sharing a node; and their k values are then summed together.

=Lecture 10 - Wednesday, September 17, 2008=

Elimination of Known Degree's of Freedom
If d1 = d2 = d5 = d6 = 0, then rows 1, 2, 5 and 6 can be eliminated from the global stiffness matrix. This is possible because when the stiffness matrix is multiplied by a displacement matrix containing those zero values, the respective values for the force components will equal zero. So. ..

$$	\mathbf{D} = \begin{Bmatrix} d_1=0\\d_2=0\\d_3\\d_4\\d_5=0\\d_6=0 \end{Bmatrix}$$

Respective Global Stiffness Matrix
Now. ..

$$	\mathbf{K} = \begin{vmatrix} k_{13} & k_{14} \\ k_{23} & k_{24}\\k_{33} & k_{34}\\k_{43} & k_{44}\\k_{53} & k_{54}\\k_{63} & k_{64}\end{vmatrix}$$

Then. . . by the Principle of Virtual Work (PVW), we delete the Global Stiffness Matrix rows corresponding to those rows that equal zero in the Global Stiffness Matrix.

Therefore, the Global Stiffness Matrix, K, reduces to

$$	\mathbf{K} = \begin{vmatrix} k_{33} & k_{34}\\k_{43} & k_{44}\end{vmatrix}$$

Resulting Overall Force Displacement Relationship
$$\begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \end{bmatrix} \begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}=\begin{Bmatrix} 0\\P\end{Bmatrix}$$

Now we go back to the 2 bar truss problem, for which the solution is:

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}=\mathbf{K}^{-1}\begin{Bmatrix} 0 \\ P \end{Bmatrix}=\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Computing the Reactions
There are two different methods for finding the reactions:

Method 1

Use the Element Force Displacement Relationship

For this case, the element stiffness matrix is known and the element displacements are known. Some of the element forces are known.

Element 1

$$\mathbf{k}^{(1)}\mathbf{d}^{(1)}=\mathbf{f}^{(1)}$$

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

Element 2

$$\mathbf{k}^{(2)}\mathbf{d}^{(2)}=\mathbf{f}^{(2)}$$

$$d^{(2)}=\begin{Bmatrix} 4.352 \\ 6.1271 \\ 0 \\ 0 \end{Bmatrix}=\begin{Bmatrix} d_1^{(2)} \\ d_2^{(2)} \\ d_3^{(2)} \\ d_4^{(2)} \end{Bmatrix}$$

=Lecture 11 - Friday, September 19, 2008=

Element 1 Equilibrium
$$\mathbf{f}^{(1)}=\mathbf{k}^{(1)}\mathbf{d}^{(1)}= \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248\\ 0.3248 & 0.1875 & -0.3248 & 0.1875\\ -0.5625 & -0.3248 & 0.5625 & 0.3248\\ -0.3248 & -0.1875 & 0.3248 & 0.1875\\ \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 4.352\\ 6.127\\ \end{bmatrix}= \begin{bmatrix} 0.5625 & 0.3248\\ 0.3248 & 0.1875\\ \end{bmatrix} \begin{bmatrix} 4.325\\ 6.127\\ \end{bmatrix}= \begin{bmatrix} 4.438\\ 2.562\\ \end{bmatrix} $$ $$\mathbf{f}^{(1)}= \begin{bmatrix} f^{(1)}_1\\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4\\ \end{bmatrix}= \begin{bmatrix} -4.438\\ -2.562\\ 4.438\\ 2.562\\ \end{bmatrix} $$ Note that:
 * Recall that for Element 1, the local force relationship is given by:
 * To be correct, we must include both the internal and reaction forces of element 1. Since the internal and reaction forces are equal and opposite, the local force matrix for element 1 is given by:
 * $$ f^{(1)}_3 $$,and $$ f^{(1)}_4 $$ are the internal forces of Element 1
 * $$ f^{(1)}_1 $$ and $$ f^{(1)}_2 $$ are reaction forces of Element 1

Observation: Element 1 in Equilibrium We know that Element 1 is in equilibrium because the sum of forces in the x- and y- directions equals zero and the force moment about any point in the element also equals zero. $$\sum F_x = f_1^{(1)} + f_3^{(1)} = -4.438 + 4.438 =0$$ $$\sum F_y = f_2^{(1)} + f_4^{(1)} = -2.562 + 2.562 =0$$ $$\sum M_{any pt} = \sum M_{node1} = -L^{(1)}sin(30)f_3^{(1)} + L^{(1)}cos(30)f_4^{(1)}= -4cos(30)(2.5622) + 4sin(30)4.4378 = -8.867 + 8.867 = 0$$

Element 2 Equilibrium
$$\mathbf{f}^{(2)}=\mathbf{k}^{(2)}\mathbf{d}^{(2)}= \begin{bmatrix} 2.5 & -2.5 & -2.5 & 2.5\\ -2.5 & 2.5 & 2.5 & -2.5\\ -2.5 & 2.5 & 2.5 & -2.5\\ 2.5 & -2.5 & -2.5 & 2.5\\ \end{bmatrix} \begin{bmatrix} 4.352\\ 6.127\\ 0\\ 0\\ \end{bmatrix}= \begin{bmatrix} 2.5 & -2.5\\ -2.5 & 2.5\\ \end{bmatrix} \begin{bmatrix} 4.352\\ 6.127\\ \end{bmatrix}= \begin{bmatrix} f_1^{(2)}\\ f_2^{(2)}\\ \end{bmatrix}= \begin{bmatrix} -4.438\\ 4.438\\ \end{bmatrix} $$
 * Recall for Element 1 and 2 that the x- and y- nodal displacements are equal at their intersecting node. Simply put, $$ d_3^{(1)} = d_1^{(2)} $$ and $$ d_2^{(2)} = d_4^{(1)}$$. As a result, we can express the force-displacement relationship of Element 2 as:

$$\mathbf{f}^{(2)}= \begin{bmatrix} f^{(2)}_1\\ f^{(2)}_2\\ f^{(2)}_3\\ f^{(2)}_4\\ \end{bmatrix}= \begin{bmatrix} -4.438\\ 4.438\\ 4.438\\ -4.438\\ \end{bmatrix} $$
 * To be correct, we must include both the internal and reaction forces of element 2. Since the internal and reaction forces are equal and opposite, the local force matrix for Element 2 is given by:
 * $$ f^{(2)}_1 $$,and $$ f^{(2)}_2 $$ are the internal forces of Element 2
 * $$ f^{(2)}_3 $$ and $$ f^{(2)}_4 $$ are reaction forces of Element 2

Observation: Element 2 in Equilibrium We know that Element 1 is in equilibrium because the sum of forces in the x- and y- directions equals zero and the force moment about any point in the element also equals zero. $$\sum F_x = f_1^{(2)} + f_3^{(2)} = -4.438 + 4.438 =0$$ $$\sum F_y = f_2^{(2)} + f_2^{(2)} = 4.438 - 4.438 =0$$ $$\sum M_{node2} = -L^{(2)}sin(45)f_2^{(2)} + L^{(2)}cos(45)f_1^{(2)}= -2sin(45)(-4.4378) + -2sin(45)(4.4378) = 6.276 - 6.276 = 0$$

Analysis Methods
In general, there are two separate methods that can be used to solve for a 2-bar truss system using statics:

Method 1: FBD Analysis
In this analysis method:
 * Isolate the supports
 * Assign the applied force to only one FBD
 * Examine FBD 1 and FBD 2 Seaparately

Method 2: Nodal Finite Element Analysis
In this analysis method:
 * Isolate the supports AND the nodes
 * Assign the applied force to its respective node
 * Examine Element 1 and Element 2
 * Examine the node independently

=HW 2 Assignment: Matlab Code=

The following is an M-file created in MATLAB.

function two_bar_truss e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4; nodes = [0, 0; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta), ... L(1)*sin(pi/2-alpha)-L(2)*cos(beta)]; dof=2*length(nodes); conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1); %load vector R = zeros(dof,1); R(4) = P; % Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e, A, ...          nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function k = PlaneTrussElement(e, A, coord) % PlaneTrussElement(e, A, coord) % Generates stiffness matrix for a plane truss element % e = modulus of elasticity % A = area of cross-section % coord = coordinates at the element ends x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L; k = e*A/L*[ls^2, ls*ms,-ls^2,-ls*ms; ls*ms, ms^2,-ls*ms,-ms^2; -ls^2,-ls*ms,ls^2,ls*ms; -ls*ms,-ms^2,ls*ms,ms^2]; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function [d, rf] = NodalSoln(K, R, debc, ebcVals) % [nd, rf] = NodalSoln(K, R, debc, ebcVals) % Computes nodal solution and reactions % K = global coefficient matrix % R = global right hand side vector % debc = list of degrees of freedom with specified values % ebcVals = specified values dof = length(R); df = setdiff(1:dof, debc); Kf = K(df, df); Rf = R(df) - K(df, debc)*ebcVals; dfVals = Kf\Rf; d = zeros(dof,1); d(debc) = ebcVals; d(df) = dfVals; rf = K(debc,:)*d - R(debc); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function [eps, sigma, force] = PlaneTrussResults(e, A, coord, disps) % [eps, sigma, force] = PlaneTrussResults(e, A, coord, disps) % Compute plane truss element results % e = modulus of elasticity % A = Area of cross-section % coord = coordinates at the element ends % disps = displacements at element ends % The output quantities are eps = axial strain % sigma = axial stress and force = axial force. x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L; T=[ls,ms,0,0; 0,0,ls,ms]; d = T*disps; eps= (d(2)-d(1))/L; sigma = e.*eps; force = sigma.*A;

The following are results from the above M-file.

k_local = 0.75       -0.75       -0.75         0.75

k = 0.5625     0.32476      -0.5625     -0.32476     0.32476       0.1875     -0.32476      -0.1875     -0.5625     -0.32476       0.5625      0.32476    -0.32476      -0.1875      0.32476       0.1875

k_local = 5   -5   -5     5

k = 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

K = 0.5625     0.32476      -0.5625     -0.32476            0            0     0.32476       0.1875     -0.32476      -0.1875            0            0     -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5    -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5           0            0         -2.5          2.5          2.5         -2.5           0            0          2.5         -2.5         -2.5          2.5

R = 0   0    0    7    0    0

d = 0          0       4.352      6.1271           0           0

reactions = -4.4378    -2.5622      4.4378     -4.4378

results = 1.7081      5.1244       8.5406       5.1244       17.081   0.6276       1.8828        3.138       1.8828        6.276

=Contributing Team Members= The following students contributed to this report

 * Chris Bishop Eml4500.f08.qwiki.bishop 19:19, 25 September 2008 (UTC)
 * Michael Berry Eml4500.f08.qwiki.berry 19:41, 25 September 2008 (UTC)
 * Andrea Booher Eml4500.f08.qwiki.booher 20:21, 25 September 2008 (UTC)
 * David Nobles Eml4500.f08.qwiki.nobles 00:35, 26 September 2008 (UTC)
 * Eric Gatch Eml4500.f08.qwiki.gatch 00:01, 26 September 2008 (UTC)
 * Lee HarrisEml4500.f08.qwiki.harris 18:59, 26 September 2008 (UTC)

title: User:Eml4500.f08.qwiki.bishop/HW2


text: File:Wiki Fig 7.JPG

title: File:Wiki Fig 7.JPG