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User:Eml4500.f08.gravy.jad/NotesRedo+Oldversion

 Summery of Matlab tutorial by Kermit Sigmon

Accessing MATLAB
After MATLAB is installed to the computer, it may be accessed either by clicking on a icon on the desktop, or if that icon was not created, by selecting the icon in the start menu. An alternate method to access MATLAB is to enter it into the system command which can also be found in the start menu

For convenience, MATLAB, as well as the local editor should both remain open in separate windows for any computer system that allows for multiple processes. Any questions in this area should be directed towards the computer manufacturer.

Entering Matrices
Matrices are the method by which MATLAB represents all variables. Single variables or interpreted as a 1 x 1 matrix, with no indicated direction. This is also known as a scalar quantity. A vector is represented by any matrix consisting of one column, or one row. All matrices are two dimensional and must consist of numerical values.

There are multiple methods by which to introduce a matrix into MATLAB including loading the matrix from a disk or file, inserting the matrix manually, or using a matrix that has been generated by another mathematical equation. The following two statements are examples of the correct syntax to generate a 3 x 3 matrix:

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

A= [

1 2 3 4 5 6 7 8 9]

The first example separates the values by spaces and the rows by semi-colons while the second separates the rows by lines.

To indicate an exponent, values must be entered in without spaces (i.e. 2.34e-5) for proper syntax. MATLAB also has the ability to interpret imaginary numbers using i and j. These values are also entered without spaces (i.e. 3+4i). If the values of i and j are assigned prior to their use as an imaginary value, a new value can be used by first setting it equal to the square root of negative one.

If a matrix is too long and tedious to enter manually it should be entered via an ASCII file from the local editor to avoid possible error from input. To load files in MATLAB, use the load command followed by the name of the file.

For example: load filename.ext If a user wishes to experiment with matrices of random numbers the rand, magic, and hilb functions provide a solution. The rand function creates a matrix of randomly generated numbers for a matrix with the same number of rows as columns. The magic function creates a magic square matrix. This is a particular type of matrix where the rows, columns, and diagonals have a common sum. The hilb function creates a Hilbert matrix.

Matrix and Array Operations
MATLAB has a variety of operations available including addition (+), subtraction (-), multiplication (*), power (^), conjugate transpose (‘), left division (\), and right division (/) all of which can be applied to both scalars and vectors. Order of operations applies.

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

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

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

Matrix Building Functions
Some commonly used functions in Matlab consists of:

For example, by saying zeros(m,n) produces an mxn matrix containing zeros. By saying zeros(n) produces an nxn matrix containing zeros. By saying zeros(A) while A is an existing matrix, it produces a matrix size A containing zeros.

For, if, while -and relations
For, if and while are different types of loops used in Matlab.

for:

-The "for" loop produces a vector of size "n" for a given function "x"

-saying, i=n:-1:1, will write the vector backwards

-There can be double nested "for" loops (two or more for loops in the same loop. This will produce an mxn matrix.

-A matrix can be used instead of "n" and it will assume to compute the matrix

while:

-The "while" function is a loop that will continue to loop until the condition statement becomes false

if:

-The"if" function is a single loop that will only enter the loop if the primary condition is met

-it is a single loop

Relations

Some common relational operators used in Matlab consists of:

logical operators are as follows:

In addition, any true/false relations appear as a scalar number 1/0.

Scalar Functions
There are some functions in MATLAB that were designed and created for scalar functions. These functions can still operate element-wise on matrices. Some common examples are sin, cos, tan, asin, acos, atan, exp, log, ln, rem (remainder), abs (absolute value), sqrt (square root)

Vector Functions
Unlike scalar functions, vector functions are, as their name hints, used on vectors (i.e. matrices or size m x n). These functions when used yield answers which are also in vector form. Some common examples of vector functions are max (max value), min (min value), sort (sort values), median, mean, std (standard deviation), sum, prod (product)

Matrix Functions
As mentioned above, matrices are used as a representation of variables in MATLAB. This is what makes the program such a useful tool. Some common examples of matrix functions are eig (eigenvalues), chol (choleski factorization), inv (inverse), hess (hessenberg form), det (determinant)

Command line editing and recall
- editing is easy -> backspace or delete to the right or left of the cursor

- commands: Ctl-a, Ctl-e, Ctl-d, Ctl-k for Unix or PC systems

- the cursor can be moved up or down to recall any previous line, therefore one can move the cursors up, down, left, right and recall and/or edit anything on the page

Submatrices and colon notation
-matlab uses vectors and submatrices, often generated by "colon notation"

- "colon notation" and subscripting is used to manipulate data sets

- ex: A(1:4,3) = matrix A, column 3, elements 1 through 4

A(:,3) = matrix A, column 3, all elements

A(:,[2 4]) = matrix A, column 2 and 4, all elements

M-Files
M-files M-files are are stored files usually ending with the “.m” extension. They allow MATLAB to run an entire preprogrammed series of commands. They effectivly make up the backbone of MATLAB, and most work is usually done to create or edit these files. There are two main types of m-files: script files and function files.

Script Files Script files are a type of m-file that contains a sequence or normal MATLAB commands. They are normally used for the creation and mataince of large variables, such as matrices. The varibable created in a script file are global. To call a script file, simply enter a command into MATLAB that is the name of the script file with the “.m” extension left off. For example, if the file is named “example1.m” than typing in the command “example1” will cause the script file to run.

Function Files Function files are the true brains behind MATLAB. They allow for the creation of any type of function programmable, which greatly extends the cabalities of MATLAB. Function files, unlike script files, default to creating local variables. However they can be declared global if the need arises. A function files starts with the line “function variable=execute(a,b)”. The word “function” tells MATLAb that the following is a function. “variable” is the output argument, and the name can be changed to be more descriptive. “execute(a,b)” is the name of the function (execute) and the variables that need to be passed into the function (a,b). The list of variables (a,b), called the input arguments, can be expanded or deleted to fit the need of the function. Function files, like all m-files, are able to use and call other files within them, allowing for a greater range of programming options.

An example of a function file is shown below. Remember that the “%” symbol allows for comments without affecting any code. MATLAB ignores all statements after the % symbol.

function a=randint(m,n) %RANDINT Randomly generated integral matrix. %       randint(m,n) returns an m-by-n matrix with entries

%      between 0 and 9 a=floor(10*rand(m,n))

Putting the command “z=randint(3,6)” into MATLAB will return a 3-by-6 matrix variable named “z”. The great advantage to function files is the ability to expand and create huge programs. This allows for the creation of many smaller function, similar to the randint(a,b) created before, to be combined into a large file that can be programmed to do thousands of different things. Also, there is the ability to combine self-made functions with some of MATLAB’s built-in functions, much like the built in function “rand(x,y)” was used above.

A useful command to find out a list of all the functions that MATLAB has access to, a combination of built-in and user programmed functions, can be seen by typing the command “type functionname”.

Text strings, error messages, input
To enter a text string in MATLAB, use single quotes: s = 'This is a test' This will assign the text string ‘This is a test’ to the variable s.

To display a text string, use the function disp: disp('this message is hereby displayed');

To display an error message, use the function error: error('Sorry, the matrix must be symmetric'); This also will exit the M-file upon execution.

To prompt the user for input data, use the function input: iter = input('Enter the number of iterations: '); This will display ‘Enter the number of iterations’ and pause execution while the user inputs the data which is then data assigned to the variable iter.

Managing M-files
To keep MATLAB active while editing a file in order to avoid losing all variables, use the !-feature by inserting a ! before any system command: >> !ed rotate.m This will let you edit rotate.m using the local editor without exiting MATLAB.

See help dbtype and references given there for more concerning debbugging tools.

The following commands will duplicate system commands without the use of a !:

M-files must be in an accessible directory. Storing personal M-files in a subdirectory named Matlab in one’s home directory should work. For further information see the discussion of MATLABPATH in the User's Guide.

Comparing efficiency of algorithms: flops, tic, toc
There are two measure of algorithm efficiency: floating point operations (flops) performed and the elapsed time (tic and toc).

The MATLAB function flops keeps a running total of the flops performed.

To reset the flops count to 0, use the command fops(0), not flops=0. Do this at the beginning of the code, and then use the command fops at the end to get the fops count of the algorithm: fops (0); x=A\b; fops

The MATLAB functions tic and toc give the elapsed time between the two commands: tic; x=A\b; toc

However, on timesharing machines, elapsed time is not as reliable as flops at determining algorithm efficiency because the rate of execution is dependent on how busy the computer is at the time.

Output format
To control the format of the displayed output, use the following commands:

Once invoked, the chosen format remains in effect until changed.

To suppress most blank lines, use the command format compact. To reverse the format compact command, use format loose. These are independent of the other format commands.

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

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

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

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

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

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

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

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

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

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

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

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

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

axis(axis), freezes scaling for subsequent graphs.

axis auto, resets the scale to the default.

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

axis square, makes the same scale for both axes.

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

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

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

'''

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

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

Sparse Matrix Computations
There are two storage modes in Matlab for a matrix: full and sparse. Full mode stores everything, while sparse only stores nonzero numbers. The default is to store elements of a matrix in full mode. In full mode, the zero elements are taken into account during calculations. Sparse mode will remove these zero elements during calculations, thus reducing time spent performing them. It saves time, but still outputs the same numbers.

Matrices can be converted between sparse and full fairly easily. The command “S=sparse(F)” will convert the full matrix, F, to a sparse matrix, S. It can then be converted back from sparse to full using the command "F=full(S)”. Usually though, if a sparse matrix is desired, it will be created directly rather than by converting from a full matrix. There are sparse analogs to many full matrices. The functions speye, sparse, spones, and sprandn are sparse analogs to the full functions eye, zeros, ones and randn respectively.

When performing calculations using both full and sparse matrices, the storage mode of the output will vary. If calculations are performed on only sparse functions, the output will be sparse. If calculations are performed only on full functions, the output will be full. However, if calculations are performed using both full and sparse functions, the output will always come out as full unless the user specifically states otherwise.

 New Lecture Notes
Notes

Online sources for free college resources.

Wikipeida, wikversity, MIT. Open course ware.

Labeling system
1.      Number in a triangle is the global label for an element

2.      Number in a circle is the global label for a node

3.      Number in a square is the local label for a node

Next big step: Force-Displacement relation (FD)

$$\begin{Bmatrix} f_1 \\ f_2 \end{Bmatrix}_{2X1}=\begin{bmatrix} k  &- k  \\-k &k    \end{bmatrix}_{2X2}*    \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix}_{2X1}$$

Case 1; observer sits on node 1

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

Case 2: observer sits on node 2

$$\ f_1=-f_2=-k(d_2-d_1)=k(d_1-d_2) \,$$

Ch1. : Big Picture
1.1 Discretization

1.1.1         Plane Truss element

1.2      Assembly of element

Ex. 1.4 Fire-bar truss

1.4  Element solution of model validity

1.4.1         Plane Truss element

Steps to solve simple truss systems

1.      Global picture (description) At structures level

·        Global degrees of freedoms (dofs) and displacement degrees of freedoms (disp. dofs)

·        Global forces

Actually the disp dofs are partitioned into:

·        A known part, e.g fixed dofs, constrains

·        An unknown part: Solved using FEM

Similarly for the global forces:

·        A known part: Applied forces

·        An unknown part : Reactions

2.      Element Picture:

·        Element  dofs

·        Element forces

·        Either  of the above in global coordinates systems or in local coordinates systems

3.      Global FD relation

·        Element stiffness matrices in global coordinates

·        Element force matrices in global coordinates

·        Assembly of element stiffness matrix and element force matrix into global FD relation

$$\ K_{nXn} * d_{nX1}=F _{nX1} \,$$

4.      Elimination of known dofs. To reduce the global FD relation (stiffness matrix non-singular) invertible

$$\ \overline{ K_{mXm}} * \overline{ d_{mX1} }= \overline{ F _{mX1}} \,$$

M= number of unknown disp dofs

N= number of both known and unknown disp dofs

k non singular so k-1 exists (k invertible)

$$\ \overline{ d_{mX1} }= \overline{ K^{-1}_{mXm}} * \overline{ F _{mX1}} \,$$

5.      Compute element forces from  now known d= element stresses

6.      Compute reactions. (unknown forces)


text:
User:Sciurus

w:de:Benutzer:Sciurus

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

title: User:Sciurus


text: User talk:Sciurus

title: User talk:Sciurus


text:
MediaWiki:Gadget-contribsrange.js

// _________________________________________________________________________________________ // |                                                                                | // |                    === WARNING: GLOBAL GADGET FILE ===                                  | // |                 Changes to this page affect many users. | // | Please discuss changes on the talk page or on Wikipedia_talk:Gadget before editing. | // |_________________________________________________________________________________________| // // Imported from version 312947638 as of September 10, 2009 from User:Splarka/contribsrange.js // See notes below

/* Special:Contributions Wildcard/CIDR lookup, version [0.2.7] Originally from: https://en.wikipedia.org/wiki/User:Splarka/contribsrange.js

Notes:
 * Uses the API which is faster than most CIDR contrib tools.
 * Needs 1.14 r42198+ for advanced continue.
 * Currently uses a GET json via to avoid ajax problems.
 * Only currently works if submitted (or called by URL parameter, eg Special:Contributions/User).
 * Checks namespace and date options in form.
 * Also utilizes &ucstart and &ucend date parameters if supplied manually (not supported in the UI atm).
 * These parameters override the form date options.
 * Uses: Submit any IP CIDR range /16 or from /24 to /32. Submit any string (at least 3 characters) with a suffixed asterisk.
 * eg: [123.123.123.0/24 ] or [123.123.123.* ] or [Willy* ].
 * Now keeps searching until it finds crMax (or forever with crshowall=true URI parameter).
 * Puts them in a nice collapsed div stack, like enhanced recent changes.
 * /25 and /26 ranges now disabled, since they are inaccurate (50 parameter limit), please use /24 (ucuserprefix) or /27 (32 parameters)

To do:
 * use ajax (let sysops/bots = 5000)?

en.wikiversity version forked from https://w.wiki/EWA (version 0.2.7) local version: 0.2.12 /*jshint scripturl:true*/ /*global jQuery, mediaWiki, prefixContribsToggleDiv */ ( function ( mw, $ ) {

function prefixContribsInit { const $show = $( '#contentSub' ) || $( '#topbar' ); if ( $show ) { $show.append(			' \u2022 Javascript-enhanced contributions lookup 0.2 enabled. You may enter a CIDR range or append an asterisk to do a prefix search. For IPv6, you must use capital letters for gadget matching, native contribution matching is not case-sensitive.'		); }	const ucfrm = $( 'form' )[0]; if ( !ucfrm.target ) { return; }

/** general optionlets independent of type of search. */	var options = ''; if ( queryString( 'ucstart' ) ) { options += '&ucstart=' + encodeURIComponent( queryString( 'ucstart' ) ); }	if ( queryString( 'ucend' ) ) { options += '&ucend=' + encodeURIComponent( queryString( 'ucend' ) ); }	if ( options === '' ) { if ( ucfrm.start.value ) { options += '&ucstart=' + ucfrm.start.value + 'T23:59:59Z'; }		if ( ucfrm.end.value ) { options += '&ucend=' + ucfrm.end.value + 'T23:59:59Z'; }	}	if ( typeof ucfrm.namespace != "undefined" ) { const chosenNamespace = ucfrm.namespace[ucfrm.namespace.selectedIndex].value; if ( parseInt( chosenNamespace ) > -1 ) { options += ( '&ucnamespace=' + chosenNamespace ); }	}	options += '&uclimit=max';

const patternCIDR = /(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\/(2[7-9]|3[0-2]|24|16)/i; const patternWild = /^.{3,}\*$/i;

if ( ucfrm.target.value.search( patternCIDR ) === 0 ) { var cidr = ucfrm.target.value.match( patternCIDR )[0]; const range = cidr.match( /[^/]\d{1,2}$/i )[0]; if ( range === '24' ) { options += '&ucuserprefix=' + cidr.match( /\d{1,3}\.\d{1,3}\.\d{1,3}\./ )[0]; } else if ( range === '16' ) { options += '&ucuserprefix=' + cidr.match( /\d{1,3}\.\d{1,3}\./ )[0]; } else { //complex CIDR, lets figure it out const oct3 = cidr.match( /\.\d{1,3}\//i )[0].replace( /(\.|\/)/g,'' ); cidr = cidr.match( /\d{1,3}\.\d{1,3}\.\d{1,3}\./ )[0]; const num = Math.pow( 2,32 - range ); const start = oct3 - oct3 % num; options += '&ucuser='; for ( var iii = start; iii <= start + num; iii++ ) { options += '' + cidr + iii; if ( iii !== start + num ) { options += '|'; }			}		}	} else if ( ucfrm.target.value.search( patternWild ) === 0 ) { //very simple wildcard, lets do-er var prefix = ucfrm.target.value.replace( /\*$/, '' ); prefix = prefix.substr( 0, 1 ).toUpperCase + prefix.substr( 1 ); options += '&ucuserprefix=' + prefix; } else { console.log( 'Contribs Range - not a valid range, closing' ); return; }	prefixContribsStartbox; const url = mw.config.get( 'wgServer' ) + mw.config.get( 'wgScriptPath' ) + '/api.php?action=query&format=json&callback=prefixContribs&rawcontinue=&list=usercontribs' + options; mw.loader.load( url ); crContURI = url; }

function prefixContribsStartbox { $( '.mw-userpage-userdoesnotexist, .mw-contributions-user-tools' ).hide; $( 'form:first' ).after(		$( ' ' )			.attr( 'id', 'results-from-CIDR' )			.append( $( '' ) .attr( 'href', 'javascript:prefixContribsToggleAll;' ) .text( 'toggle all' ) .attr( 'id', 'prefixcontribs-tog' ), $( ' ' )					.attr( 'id', 'prefixcontribs-prog' ) .text( 'Searching.' ), $.createSpinner( 'prefixcontribs-spin' ) )	); }

window.prefixContribs = function ( obj ) { if( !obj.query || !obj.query.usercontribs ) { return; }	const cidr = obj.query.usercontribs; if ( !cidr.length ) { $.removeSpinner( 'prefixcontribs-spin' ); $( '#prefixcontribs-prog' ).text( 'No changes were found for this wildcard/CIDR range.' ); return; }	cidr.forEach((edit) => {		const id = 'cr-' + encodeURIComponent( edit.user.replace( / /g,'_' ) ).replace( /%3A/g,':' ).replace( /%/g,'.' );		// group each result set based on the user name, create new div for new names		if ( $( '*[id="' + id + '"]' ).length === 0 ) {			$( '#results-from-CIDR' )				.append( $( ' ' )						.attr( 'id', 'd-' + id ) .append(							$( '' )								.attr( 'href', 'javascript:prefixContribsToggleDiv("' + id +'")' )								.text( edit.user )								.attr( 'id', 'm-' + id )								.addClass( 'cr-toggle-hidden' ),							'\n'						), '\n', $( '' ) .attr( 'id', id ) .addClass( 'cr-list' ) .hide, '\n' );		}		const $li = $( '' )			.append( edit.timestamp.replace( /T[\d:]*Z/,' ' ), $( '' ) .attr( 'href', mw.util.getUrl( 'Special:Contributions/' + edit.user ) ) .text( edit.user ), ' (',				$( '' )					.attr( 'href', mw.util.getUrl( 'User_talk:' + edit.user ) )					.text( 'talk' )					.addClass( 'mw-mightexist' ),				') edited (',				$( '' )					.attr( 'href', mw.util.getUrl( edit.title, { diff: 'prev', oldid: edit.revid } ) )					.text( 'diff' ),				') ', $( '' ) .attr( 'href', mw.util.getUrl( edit.title, { curid: edit.pageid } ) ) .text( edit.title ) );		if ( edit.comment ) {			$li.append( ' (' + edit.comment + ')' );		}		$( '*[id="' + id + '"]' ).append( $li );	}); mw.hook( 'wikipage.content' ).fire( $( '.cr-list' ) ); //continue? crMax = crMax - cidr.length; const $prog = $( '#prefixcontribs-prog' ); if( !obj['query-continue'] || !obj['query-continue'].usercontribs || !obj['query-continue'].usercontribs.uccontinue ) { $.removeSpinner( 'prefixcontribs-spin' ); $prog.remove; prefixContribsNumerate; return; }	if ( crMax <= 0 && queryString( 'crshowall' ) !== 'true' ) { var gaurl = document.location.href; if ( gaurl.indexOf( '#' ) !== -1 ) { gaurl = gaurl.substr( 0, gaurl.indexOf( '#' ) ); }		if ( gaurl.indexOf( '?' ) === -1 ) { gaurl += '?'; }		gaurl += '&crshowall=true';

$prog .append(				' Whoa! Finding a lot. To see them all click ',				$( '' )					.attr( 'href', gaurl )					.text( 'here' ),				'. (Warning: May bog down browser!)'			);

$.removeSpinner( 'prefixcontribs-spin' ); prefixContribsNumerate; } else { $prog.append( '.' ); const url = crContURI + '&uccontinue='+ obj['query-continue'].usercontribs.uccontinue; mw.loader.load( url ); } };

function prefixContribsNumerate { const $lists = $( 'ul.cr-list' ); for ( var iii = 0; iii < $lists.length; iii++ ) { $( '*[id="d-' + $lists[iii].id + '"]' ).append( ' ' + $lists[iii].childNodes.length + ' found' ); }	if ( $lists.length === 1 ) { prefixContribsToggleDiv( $lists[0].id ); } }

window.prefixContribsToggleAll = function { const $lists = $( 'ul.cr-list' ); if ( $lists.length === 0 ) { return; }	const togglefrom = $lists[0].style.display; for ( var iii = 0; iii < $lists.length; iii++ ) { if ( $lists[iii].style.display === togglefrom ) { prefixContribsToggleDiv( $lists[iii].id ); }	} };

window.prefixContribsToggleDiv = function ( id ) { const i = $( '*[id="' + id + '"]' )[0] || false; const m = $( '*[id="m-' + id + '"]' )[0] || false; if ( !i || !m ) { return; }	if ( i.style.display === 'none' ) { i.style.display = 'block'; m.className = 'cr-toggle-shown'; } else { i.style.display = 'none'; m.className = 'cr-toggle-hidden'; } };

function queryString( p ) { const re = RegExp( '[&?#]' + p + '=([^&#]*)' ); const matches = re.exec( document.location ); if ( matches ) { try { return decodeURI( matches[1] ); } catch ( e ) { console.log( 'Contribs range error: ', e ); }	}	return null; }

if ( mw.config.get( 'wgCanonicalSpecialPageName' ) === 'Contributions' ) { var crContURI = ''; //query URI for continuing later var crMax = 10000; //bypass with &crshowall=true URI parameter $( prefixContribsInit ); } }( mediaWiki, jQuery ) );

title: MediaWiki:Gadget-contribsrange.js


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MediaWiki:Gadget-contribsrange

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User:Eml4500.f08.gravy.jad/NotesRedo2+Oldversion

FEA Notes and Homework 2

Truss Problem description
Take a specific example to see how the recipe works.

Data:

Element Length

$$\ L^{(1)}=4 \,$$

$$\ L^{(2)}=2 \,$$

Young's Modules

$$\ E^{(1)}=3 \,$$

$$\ E^{(2)}=5 \,$$

Cross-sectional Area

$$\ A^{(1)}=1 \,$$

$$\ A^{(2)}=2 \,$$

Inclination angle

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

 * $$\ \theta^{(2)} = -45^o \,$$

Solving a Truss Problem with Statics

First take the sum of the moments about global node 3 and solve for $$R_3$$ which is the reaction at global node 3 in the y-direction

$$ \sum {M_1}= P(4cos(30)) +R_3(4cos(30) + 2sin(45))$$

$$R_3= \frac{-P(4cos(30))} {4cos(30) + 2cos(45)} = -4.4378$$

Then use this information to take the sum of the forces in the y-direction and solve for $$R_1$$ which is the reaction at global node 1 in the y-direction.

$$\sum{F_y}= -4.4378 + P+R_1$$

$$\ R_1= -2.5622 \,$$

By observation it can be determined that there are no forces in the x-direction so $$R_1$$ and $$R_3$$ are the only reaction forces

Solving a Truss Problem with Finite Element Analysis
1)Global Picture

$$\begin{Bmatrix} f_1 \\\vdots & \\ f_6 \end{Bmatrix}_{1X6}=\begin{bmatrix} & &  \\  & k & \\ & &   \end{bmatrix}_{6X6}*    \begin{Bmatrix} d_1 \\\vdots & \\ d_6 \end{Bmatrix}_{1X6}$$

$$\begin{matrix} global& &global& &global \\ force & &stiffness& &displacement \\  column &   &matrix  & &column \\ matrix & & & &matrix \end{matrix}$$

2) Element Picture

3) Global FD at element level:

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

$$\ k^{(e)}\,$$ = the element stiffness matrix

$$\ d^{(e)}\,$$= the element displacement matrix

$$\ f^{(e)}\,$$= the element force matrix

$$k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -(l^{(e)}m^{(e)})  \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(e)})^2  \\ -(l^{(e)})^2 & -(l^{(e)}m^{(e)}) & (l^{(e)})^2 & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2    \end{bmatrix}_{4X4} = \underline{k^{(e)}}$$

$$\ k^{(e)}= \,$$ $$\frac{E^{(e)}A^{(e)}}{L^{(e)}}$$ axial stiffness of bar element "e" where e=1,2

$$\ l^{(e)},m^{(e)} = \,$$ director cosines of x axis (goes from 1 to 2) with respect to global (x,y coordinates)

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

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

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

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

$$\ \vec \tilde{i} \cdot \vec j= HW \,$$

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

$$\ l^{(1)}=cos\theta^{(1)}= cos 30^o= \frac{\sqrt{3}}{2} \,$$

$$\ m^{(1)}=sin\theta^{(1)}= sin 30^o= \frac{1}{2} \,$$

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

$$\ \underline{k^{(1)}}=\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}  \\  k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}  \\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

i=row number =1,2....4

j=column number =1,2....4

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

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

$$\ k_{42}^{(1)}=k_{24}^{(1)}= -k^{(1)}\left( m^{(1)} \right)^2=\frac{-3}{4}*\frac{1}{2}= \frac{-3}{16} \,$$

Observation:

1) Only need to compute 3 numbers. Other coefficients have same absolute value, just differ by (+ -)

2) Matrix $$ \underline{k^{(1)}}$$ is synmetrix, $$k_{ij}^{(1)}=k_{ji}^{(1)}$$

$$k_{13}^{(e)}=k_{31}^{(e)}$$ (just interchange the row and column index)

$$\ \underline{k^{(1)}}=\begin{bmatrix} \frac{3}{4}(\frac{\sqrt{3}}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{\sqrt{3}}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2})  \\  \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{1}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{1}{2})^2  \\ -\frac{3}{4}(\frac{\sqrt{3}}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{\sqrt{3}}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2})\\-\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{1}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{1}{2})^2    \end{bmatrix}_{4X4}=  \underline{k^{(1)}} _{4X4} \,$$

$$\ \underline{k^{(1)}}=\begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}  \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}  \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16}    \end{bmatrix}_{4X4}=  \underline{k^{(1)}} _{4X4} \,$$

In general, $$k_{ij}^{(e)}=k_{ji}^{(e)} or k^{(e) T} =k^{(e)}$$

The transpose of $$ k^{(e)}$$ is equal to $$ k^{(e)}$$

$$\ \underline{k^{1}}=\begin{bmatrix} k_{11}^{(e)} & k_{12}^{(e)} & k_{13}^{(e)} & k_{14}^{(e)}  \\   & k_{22}^{(e)} & k_{23}^{(e)} & k_{24}^{(e)}  \\  &  & k_{33}^{(e)} & k_{34}^{(e)}\\ sym. & &  & k_{44}^{(e)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

Element 2:
$$\ k^{(2)}= \frac{E^{(2)}A^{(2)}}{L^{(2)}} \frac{\left(5\right ) \left(2\right )}{2}= 5\,$$

$$\ \theta^{(2)} = \frac{-\pi}{4}= -45^o \,$$

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

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

$$\ k^{(2)} = \left [k_{ij}^{(2)} \right ]_{4X4} \,$$

$$\ k_{11}^{(2)} = k^{(2)}(l^{(2)})^2 = 5\frac{2}{4} = 2.5 \,$$

Observation:

1) Absolute value of all coefficients $$\ k_{ij}^{(e)} \,$$, $$\ e=2 \,$$ , $$\ (i,j)=1, ..., 4 \,$$ are the same => comp 1 coefficient. For other coefficients, and (+) or (-)

2) $$\ k ^{(2)T} = k^{(2)} \,$$ ie. $$\ k^{(2)} \,$$ sym

$$\ \underline{k^{(2)}}=\begin{bmatrix} \frac{5}{2}& \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2}  \\  \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2}  \\ -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}\\ -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}    \end{bmatrix}_{4X4}=  \underline{k^{(2)}} _{4X4} \,$$

Elem FD rel: $$\ k^{(e)}d^{(e)}= f^{(e)} \,$$
 * e=1,2

$$\begin{Bmatrix} d_1 \\\vdots & \\ d_6 \end{Bmatrix}_{1X6}, \begin{Bmatrix} f_1 \\\vdots & \\ f_6 \end{Bmatrix}_{1X6} $$

Global FD relationship:

$$\ \underline{K}_{nxn} \underline{d}_{nx1} = \underline{F}_{nx1} \,$$

there, n=6

$$\ \begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}  \\ k_{21}& k_{22} & k_{23} & k_{24} & k_{25} & k_{26} \\ k_{31}& k_{32} & k_{33} & k_{34} & k_{35} & k_{36} \\ k_{41}& k_{42} & k_{43} & k_{44} & k_{45} & k_{46} \\ k_{51}& k_{52} & k_{53} & k_{54} & k_{55} & k_{56} \\ k_{61}& k_{62} & k_{63} & k_{64} & k_{65} & k_{66}  \end{bmatrix}  \begin{Bmatrix}  d_{1} \\ d_{2} \\ d_{3} \\ d_{4} \\ d_{5} \\ d_{6} \end{Bmatrix} = \begin{Bmatrix}  F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \\ F_{6} \end{Bmatrix}\,$$

Global F-D Relationship
Noting the fact that the equations for the elemental $$\underline{d^{(e)}}$$ and global $$\underline{d}$$ overlap, as seen in the below picture, the following equations can be derived.

$$d_1 = d_1^{(1)}$$ $$d_2 = d_2^{(1)}$$ $$d_3 = d_3^{(1)} = d_1^{(2)}$$ $$d_4 = d_4^{(1)} = d_2^{(2)}$$ $$d_5 = d_3^{(2)}$$ $$d_6 = d_4^{(2)}$$

Below is a representation of the global $$\underline{K}$$ matrix, which is the stiffness matrix for the system. The two colored boxes represent the contributions of the elemental $$\underline{k^{(e)}}$$. The $$\underline{k^{(1)}}$$ matrix is shown in blue, while the red box represents $$\underline{k^{(2)}}$$.

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

EX:

$$\ K_{11} = k_{11}^{(1)} \,$$

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

$$\ K_{43} = K_{34} = k_{34}^{(1)} + k_{12}^{(2)} = k_{43}^{(1)} + k_{21}^{(2)} \,$$

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

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

$$ K_{33} = \frac{9}{16} + \frac{5}{2} = 3.0625 $$

$$ K_{43} = \frac{3\sqrt{3}}{16} + \frac{-5}{2} = 2.1752 $$

$$ K_{44} = k_{44}^{(1)} + k_{22}^{(2)} = \frac{3}{16} + \frac{5}{2} = 2.6875 $$ Mbr> $$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & (\frac{9}{16} + \frac{5}{2}) & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & (\frac{3}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} \\ 0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$ $$\ \ = \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3428 & 0 & 0 \\  0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & (0.5625 + 2.5) & (0.3248 + 2.5) & -2.5 & -2.5 \\ -0.3248 & -0.1875 & (0.3248 + 2.5) & (0.1875 + 2.5) & -2.5 & -2.5 \\  0 & 0 & -2.5 & -2.5 & 2.5 & 2.5 \\ 0 & 0 & -2.5 & -2.5 & 2.5 & 2.5  \end{bmatrix}_{6X6}\,$$

Elimination of Known Degrees of Freedom
Knowing that global nodes 1 and 3 are actually fixed points, it is seen that there is no possibility of them moving. Therefore, the displacement at those points is zero.

$$\ d_1 = d_2 = d_5 = d_5 = d_6 = 0 \,$$

$$\ \underline{d}=\begin{Bmatrix} d_1 \\ d_2  \\  d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}= \begin{Bmatrix}  0 \\ 0 \\  d_3 \\ d_4 \\ 0 \\ 0 \end{Bmatrix}$$

Since 4 of the values for $$\underline{d}$$ are zero, we can reduce the system. Turning $$\underline{d}$$ into a 2x2 matrix, would mean that you would be able to reduce $$\underline{K}$$ to a 6x2 matrix. This is down by removing the columns that are associated with the rows within $$\underline{d}$$ that are zero. The new $$\underline{K}$$ is shown below.

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

This makes the F-D equation: $$\ \underline{K}_{6x2} \underline{d}_{2x1} = \underline{F}_{6x1} \,$$

The next step is to see that since global nodes 1 and 3 are fixed, we can able the concept of "Virtual Work" which will allow us to ignore their effects for the moment and let us turn $$\underline{F}$$ into a 2x1 by removing the components from those fixed nodes. However, by doing this, we need to adjust $$\underline{K}$$ to keep the equation true. We can further reduce $$\underline{K}$$ down to a 2x2, and the following equation results:

$$\ \underline{K}_{2x2} * \underline{d}_{2x1} = \underline{F}_{2x1} \,$$

Where $$\underline{K}$$ is equal to:

$$\ \underline{K}=\begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix}$$ Since $$\underline{F}$$ is now only made up of the forces from global node 2, we now know the value of $$\underline{F}$$. $$\ \underline{F} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}= \begin{Bmatrix}  0 \\ P \end{Bmatrix}$$ We also can solve for $$\underline{K}$$: $$\ \underline{K}=\begin{bmatrix} 3.0625 & 2.1752 \\ 2.1752 & 2.6875 \end{bmatrix}$$

At this point, the only term not known in the global F-D equation is $$\underline{d}$$. In order to solve the system, we must first solve for $$\underline{d}$$, which gives the following equation: $$\ \underline{d}_{2x1} = \underline{K}^{-1}_{2x2} * \underline{F}_{2x1} \,$$

Where $$\underline{K^{-1}}$$ is equal to: $$\ \underline{K^{-1}} = \frac{1}{det(\underline{K})}*\begin{bmatrix} K_{44} & -K_{43} \\ -K_{34} & K_{33} \end{bmatrix}$$ $$\ det(\underline{K}) = K_{33} * K_{34} - K_{34} * K_{43} $$

$$\ \underline{K^{-1}} = \begin{bmatrix} 0.7681 & -0.6217 \\ -0.6217 & 0.8753 \end{bmatrix}$$ Plugging everything in and solving for $$\underline{d}$$ gives:

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

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix} = \begin{bmatrix} 0.7681 & -0.6217 \\ -0.6217 & 0.8753 \end{bmatrix} * \begin{Bmatrix} 0 \\ P \end{Bmatrix} = \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix} $$

$$\ \underline{d} = \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Computing Reactions
$$ Elem 1: \underline{k}^{(1)} \underline{d}^{(1)} = \underline{f}^{(1)} $$

$$\ d^{(1)} = \begin{bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} 0.5625 & 0.32476 & ... & ... \\ & & & \\ & & & \\ & & & \end{bmatrix}_{4X4} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -0.5625 & -0.32476 \\ ... & ... \\ & \\ & \end{bmatrix}_{4X2} \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} f_{1}^{(1)} \\ f_{2}^{(1)} \\ f_{3}^{(1)} \\ f_{4}^{(1)} \end{bmatrix}_{4X1}\,$$

 * f1(1) and f2(1) = Reactions

 * f3(1) and f4(1) = Internal Forces

 * Observation: Element 1 is in equilibrium

 * ΣFx = f1(1) + f3(1)

 * ΣFy = f2(1) and f4(1)

HW Redo New Pictures
Element 1:

Element 2:

ΣMat any point = 0

$$P_{1}^{(1)} = [(f_{1}^{(1)})^2 + (f_{2}^{(1)})^2]^ \frac{1}{2}$$

$$ Elem 2: \underline{k}^{(2)} \underline{d}^{(2)} = \underline{f}^{(2)} $$

 * The same method to solve for element one can also be used to solve for element 2

HW: Verify equilibrium of element 1
Let $$P_{1}^{(1)} = f_{1}^{(1)} + f_{2}^{(1)} = -4.4378 i\hat{} -2.5622j\hat{}$$ and let $$P_{2}^{(1)} = f_{3}^{(1)} + f_{4}^{(1)} = 4.4378 i\hat{} + 2.5622j\hat{}$$

Note that $$P_{1}^{(1)} + P_{2}^{(1)} = 0$$ which illustrates that $$\sum{}F=0$$.

Also, note that $$P_{1}^{(1)} \times P_{2}^{(1)} = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) \times \left( 4.4378 i\hat{} +2.5622j\hat{} \right) = \left( \left( -4.4378 \times 2.5622 \right) - \left( 4.4378 \times -2.5622 \right) \right) k\hat{} = 0$$.

This shows that $$P_{1}^{(1)}$$ and $$P_{2}^{(1)}$$ are in line with each other. Since all forces are in line with each other, $$\sum{}M_{anypoint} = 0$$.

HW: Verify equilibrium of node 2
Note that $$\sum{}F = P_{1}^{(1)} + P_{2}^{(2)} + P = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) + \left( 4.4378 i\hat{} -4.4378j\hat{} \right) + 7j\hat{} = 0$$.

Also, since all forces are acting on the same point, $$\sum{}M_{2} = 0$$.

HW2 MATLab Redo HW problem
To solve the system above using an automated FEM process, we've turned to MATLab. Below is the code that was used:

In order to run this code, there are a few functions that are needed. All these functions were downloaded from the book's Website. They are listed below.

After running the preceding code, this was the output results: (Note: code was changed from MATLab laungue to a generic matrix in order to be easier to read.)

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 22:11, 25 September 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 00:52, 26 September 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 16:33, 26 September 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 17:32, 26 September 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 16:31, 26 September 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 17:20, 26 September 2008 (UTC)


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FEA Notes and Homework 3

Force Displacement
Derivation of element Force Displacement with respect to global coordinate system.

$$\ k^{(e)}_{4x4} d^{(e)}_{4x1}=f^{(e)}_{4x1} \,$$ equation 1

The figure is labeled with the correct conventions to aid with the solving method. The four forces and displacements are collapsed to two forces and displacements by combining the x and y directions into a single vector.

Force Displacement matrices calculations
The force displacement can be written as

$$ \ K^{(e)} * \begin{bmatrix} 1 & -1 \\ -1 &1 \end{bmatrix}_{\hat{K}_{2x2}^{(e)}}* \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}_{q_{2x1}^{(e)}} = \begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix}_{p_{2x1}^{(e)}} \,$$ equation 2

$$\ q_i^{(e)} = \,$$ axial disp.of element at local node i

$$\ p_i^{(e)} = \,$$ axial disp.of element at local node i

The force displacement formula simplifies to the above and the unknowns can be solved for.

Goal: Derive Equation 1 from Equation 2

Want the relationship between $$\underline {q^{(e)}}$$ and $$\underline {d^{(e)}}$$ also $$\underline {p^{(e)}}$$ and $$\underline {f^{(e)}}$$

These relationships can be expressed in the form:

$$\ \underline {q^{(e)}_{2x1}}= \underline { T^{(e)}_{2x4}} * \underline {d^{(e)}_{4x1}} \,$$

To relate the q and d matrices and the p and f matrices a new matrix must be introduced. This matrix is the T matrix.

Consider the displacement vector of local node 1 denoted by

$$ \vec{d^{(e)}_{1}} $$

$$ \vec{d^{(e)}_{1}}= \vec{d^{(e)}_{1}}* \vec{i} + \vec{d^{(e)}_{2}} *\vec{j} $$

$$\ q_i^{(e)} = \,$$ axial displacement of node 1 is the orthogonal projection of the displacement.

Vector $$ \vec{d^{(e)}_{1}} $$ of node 1 on the axis $$\tilde{x}$$ of element $$e^{(1)}$$

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

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

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

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

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

$$\ = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}_{1x2} * \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}_{2x1} \,$$

Similarly for node 2 : HW.
$$\ q_2^{(e)}=l^{(e)}*d_1^{(e)} + m^{(e)}* d_2^{(e)}= \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}_{1x2} * \begin{Bmatrix} d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} \,$$

$$ \begin{Bmatrix}q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}_{\underline{q_{2x1}^{(e)}}} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0& 0\\0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}_{\underline{T_{2x4}^{(e)}}}* \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}_{\underline{d_{4x1}^{(e)}}}$$

$$\ \underline{ q_{2x1}^{(e)} }= \underline{ T_{2x4}^{(e)} }* \underline{ d_{4x1}^{(e)} } \,$$

Similarly (same argument) :

$$\ \begin{Bmatrix} P_1^{(e)}\\P_2^{(e)} \end{Bmatrix}_{\underline{ P_{2x1}^{(e)} } } = \underline{ T_{2x4}^{(e)} } * \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} _{\underline{ f_{4x1}^{(e)} } } \,$$

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

Recall element axial Force Displacement relation:

$$\ \underline{\hat{k}_{2x2}^{(e)}} * \underline{q_{2x1}^{(e)}}= \underline{P_{2x1}^{(e)}} \,$$

$$\underline{\hat{k}^{(e)}} *( \underline{ T^{(e)} } * \underline{ d^{(e)} })_{\underline{q^{(e)}}} = ( \underline{ T^{(e)} } * \underline{ f^{(e)} })_{\underline{P^{(e)}}}$$

Goal: Want to have $$\ \underline{k^{(e)}} * \underline{d^{(e)}}= \underline{f^{(e)}} \,$$

So "move " $$\underline{ T^{(e)}}$$ from the right hand side to the left hand side by multiplying the equation by $$ \underline{ T^{(e) -1}}$$ (inverse of $$\underline{ T^{(e)}} $$)

Unfortunately, $$\underline{ T^{(e)}}$$ is a rectangular matrix (2x4) so you can not invert it.

Ans: $$[\underline{ T_{4x2}^{(e)T} } * \underline{\hat{k}_{2x2}^{(e)}} * \underline{ T_{2x4}^{(e)}} ]_{is the matrix we are looking for} * d_{4x1}^{(e)}= f_{4x1}^{(e)}$$

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

1) $$\ \underline{k^{(e)} } = \underline{T^{(e)T}} * \underline{\hat{k}^{(e)} }* \underline{T^{(e)} } \,$$

HW: verify above.
Verify the following equation:

$$\ \underline{k}^{(e)} = \underline{T}^{(e)T} * \underline{k}^{(e)} * \underline{T}^{(e)}$$

Where: $$\ \underline{T}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}$$ and $$\ \underline{k}^{e)} = k^{(e)} * \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

Since $$k^{(e)}$$ is a scaler, we can move it within the equation without changing the problem. So after moving the scaler and inserting in the matrices, the following equation is found:

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

Solving the first set of matrices gives:

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

Solving the second set of matrices gives:

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

Which is the definition of the stiffness matrix, therefore the equation is true.

Justification for 1) : PVW (later see 10-1 for fist approach of PVW Reduction of Global Force Displacement relation)

$$ \underline{K_{6x6}}* \underline{d_{6x1}}= \underline{F_{6x1}} which reduces to \underline{K_{2x2}}* \underline{d_{2x2}}= \underline{F_{2x2}}$$

Remark: Why not solve as follows

$$ \underline{d} = \underline{K^{-1}}* \underline{F} $$ ?

The reason is that K can't be unversed because it is singular

Singular matrix has a determinate of 0 so it is not invertible

Recall: need to compute $$\frac{1}{det(K)}$$ to find K-1

Why? For an unconstrained structure system there are three possible rigid body motion in 2-D (2 translation, and 1 rotation)

HW: Eigenvalues of the Global Stiffness Matrix K
For an unconstrained structure system, there are three possible rigid body motionsin 2-D (2 translation and 1 rotation).

Dynamic Evaluation Problem

 K v = λ Mv

$$\ \underline{k} = \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}_{6x6}\,$$

$$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{49}{16} & \frac{3\sqrt{3} - 40}{16} & -\frac{5}{2} & \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3} - 40}{16} & \frac{43}{16} & \frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} \\ 0 & 0 & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$

MATLAB Code

Values of the stiffness matrix are assigned to matrix K >> K= [9/16 (3*sqrt(3))/16 -9/16 -(3*sqrt(3))/16 0 0; (3*sqrt(3))/16 3/16 -(3*sqrt(3))/16 -3/16 0 0; -9/16 -(3*sqrt(3))/16 49/16 ((3*sqrt(3)-40)/16) -5/2 5/2; -(3*sqrt(3))/16 -3/16 ((3*sqrt(3)-40)/16) 43/16 5/2 -5/2; 0 0 -5/2 5/2 5/2 -5/2; 0 0 5/2 -5/2 -5/2 5/2]

Values of K in decimal form K = 0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

Matlab function to calculate eigenvalues >> d = eig(K)

List of the eigenvalues for global stiffness matrix K d = -0.0000  -0.0000    0.0000    0.0000    1.4705   10.0295

λ = 0, 0, 0, 0, 1.4705, 10.0295

It is important to note that four of the six eigenvalues are zero. These values represent zero stored elastic energy (rigid body at the modes).

//(end Remark)

HW: FD Rel. on p. 10-1
$$(\underline{k}_{6x2}) (\underline{d}_{2x1}) = \underline{F}_{6x1} $$

$$\ \begin{bmatrix} K_{13}^{(1)} & K_{14}^{(1)} \\ K_{23}^{(1)} & K_{24}^{(1)} \\ (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) \\ (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) \\ K_{31}^{(2)} & K_{32}^{(2)} \\ K_{41}^{(2)} & K_{42}^{(2)} \end{bmatrix}_{6x2}\,$$ $$\ \begin{bmatrix} d_3 \\ d_4 \end{bmatrix}_{2x1}\,$$ $$\ \ = \begin{bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{bmatrix}_{6x1}\,$$

$$\ \begin{bmatrix} -\frac{9}{16} & -\frac{3\sqrt{3}}{16} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} \\ \frac{49}{16} & \frac{3\sqrt{3} - 40}{16}\\ \frac{3\sqrt{3} - 40}{16} & \frac{43}{16} \\ -\frac{5}{2} & \frac{5}{2} \\ \frac{5}{2} & -\frac{5}{2} \end{bmatrix}_{6X2}\,$$ $$\ \begin{bmatrix} 4.3527 \\ 6.1271 \end{bmatrix}_{2x1}\,$$ $$\ \ = \begin{bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_3 \\ F_4 \end{bmatrix}_{6x1}\,$$

$$F_1 = (-\frac{9}{16})(4.3527) + (-\frac{3\sqrt{3}}{16})(6.1271) = -4.4382$$

$$F_2 = (-\frac{3\sqrt{3}}{16})(4.3527) + (-\frac{3}{16})(6.1271) = -2.5624$$

$$F_5 = (-\frac{5}{2})(4.3527) + (\frac{5}{2})(6.1271) = 4.436$$

$$F_5 = (\frac{5}{2})(4.3527) + (-\frac{5}{2})(6.1271) = -4.436$$

Note: F3 and F4 were calculated in previous section to be 0 and 7 respectively

Closing the loop between the FEM and statics: Virtual Displacement
There are two different ways on solving these types of problems: using FEM, or using statics.

Element 1:

Element 2:

These pictures are FBD's of 2 force bodies of element 1 & 2. By Statics reactions are known and therefore the member forces $$\ [P_{1}^{(1)}, P_{2}^{(2)}] \,$$ are equal and opposite in direction. We need to compute the axial displacement dofs ( amount of extension of bars):

$$\ q_{2}^{(1)}= P^{(1)}/k^{(1)}= \underline{AC} \,$$ . ( $$\ q_{1}^{(1)}=0 \,$$ because it's fixed at global node 1.

$$\ q_{1}^{(2)}= -P^{(2)}/k^{(2)}= -\underline{AC} \,$$ . ( $$\ q_{2}^{(2)}=0 \,$$ because it's fixed at global node 3.

Q : How to back out from above results of displacement of dofs of global node 2?

MATLAB code for truss displacement

The above picture is the output graph from MATLab after the above code was run. The solid line represents the deformed truss, while the dashed lines show the initial, or un-deformed, truss.

$$\ u_{y}=R*sin(\alpha) \simeq R*\alpha \,$$

$$\ u_{x}=R*(1-cos(\alpha)) \simeq 0 \,$$ $$\ (cos(\alpha) \simeq 1) \,$$

hence

Here we are assuming that alpha is very small. When we make that assumption sin(alpha) becomes alpha, and cos(alpha) becomes one. The picture shown above is exaggerated, and alpha is very small.

Infinitesimal Displacement (Related to Virtual Displacement)
We can solve for the displacement vectors AC and AB by first referencing all displacement from point A which will be (x,y)=(0,0) on the coordinate system. The actual displacement vector can be defined by the x and y displacement vectors in the form of d(3)(x-coordinates) and d(4)(y-coordinates). Points B and C are the axial displacement vectors in regards to element 1 and element 2.

$$\ AC= \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{3/4} = 6.8324 \,$$

HW: Solving AC and AB
From previous lecture

Element Length

$$\ L^{(1)}=4 \,$$

$$\ L^{(2)}=2 \,$$

Young's Modules

$$\ E^{(1)}=3 \,$$

$$\ E^{(2)}=5 \,$$

Cross-sectional Area

$$\ A^{(1)}=1 \,$$

$$\ A^{(2)}=2 \,$$

Inclination angle

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

 * $$\ \theta^{(2)} = -45^o \,$$

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

Since $$ \ P_{(e)} = T_{(e)} * f_{(e)}\,$$

And the T matrix is

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

we can solve for P

The f matrix was solved in a previous lecture and is:

$$\ F= \begin{bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{bmatrix}_{4X1}\,$$

l m and k were also solved in a previous lecture and are:

$$\ l^{(1)}=cos\theta^{(1)}= cos 30^o= \frac{\sqrt{3}}{2} \,$$

$$\ m^{(1)}=sin\theta^{(1)}= sin 30^o= \frac{1}{2} \,$$

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

$$\ l^{(2)}=cos\theta^{(2)}= cos -45^o= .707 \,$$

$$\ m^{(2)}=sin\theta^{(2)}= sin -45^o= -.707 \,$$

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

$$T^{(1)}$$ and $$T^{(2)}$$ are solved as the following:

$$\ T^{(1)} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix} \,$$

$$\ T^{(2)} = \begin{bmatrix} .707 & -.707 & 0 & 0 \\ 0 & 0 & .707 & -.707\end{bmatrix} \,$$

With T and f the P s were found as being,

$$P_{2}^{(1)}$$= $$P_{1}^{(2)}$$=5.1243

so

$$\ AC= \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{3/4} = 6.8324 \,$$

$$\ AC= \frac{|P_{1}^{(2)}|}{k^{(2)}} = \frac{5.1243}{5} = 1.0249 \,$$

HW sove (x,y) for points B and C
Knowing that the angle to find x_b and y_b is 30 degrees, and to find x_c and y_c is 135 degrees, we can use simple trigonometry to solve for the values:

$$\ x_i = L*cos(\theta), yi = L*sin(\theta)\,$$

$$\ x_b = AC*cos(30)= 6.8324 * cos(30) = 5.9170, y_b = AC*sin(30) = 6.8324 * sin(30) = 3.4162\,$$

$$\ x_c = AB*cos(135) = 1.0249 * cos(135) = -.7247, y_c = AB*sin(135) = 1.0249 * sin(135) = .7247 \,$$

(xb,yb) = (5.9170, 3.4162) (xc,yc) = (-0.7247, 0.7247)

- P is fixed, Q is "sliding" on the line

- the two lines are parallel $$\ \bar{PQ} =(PQ)*\vec {\tilde i} = (PQ)[cos\theta \vec {i} + sin\theta  \vec {j}] = (x-x_p)\vec i + (y-y_p)\vec j \,$$

$$\ \Rightarrow x-x_p=PQ cos\theta \,$$

$$\ \Rightarrow y-y_p=PQ sin\theta \,$$

$$\ \frac{y-y_p}{x-x_p}= tan\theta \,$$

$$\ (y-y_p)= (tan\theta)(x-x_p) \,$$

eq. for a line perpendicular to the line, passing P: $$\ (y-y_p)= \tan(\theta+\frac{\pi}{2})(x-x_p) \,$$

HW solve (x,y) for point D
To find x_d and y_d, the following equation is used:

$$\vec{AD} = x_d\vec{i}+ y_d\vec{j}$$

From HW2, we know that:

$$\vec{AD} = d_3\vec{i} + d_4\vec{j}$$

Also from HW2, we know that d3=4.352 and d4=6.1271. This means that:

$$\vec{AD} = 4.352\vec{i} + 6.1271\vec{j}$$

hence, (xd,yd) = (4.352, 6.1271).

since $$\ x_A=0 and y_A=0 \,$$ then $$\ \vec {AD} =(x_D-x_A)\vec i + (y_D - y_A)\vec j = x_D-\vec i + y_D\vec j \,$$

By definition: $$\ \vec{AD} = d_3 \vec i + d_4 \vec j \,$$ (these are displacement vectors of A from FEM

///end closing loop

3-Bar Truss Syst:

The FEA method of solving for a 3-Bar Truss system is done similarly to the 2-Bar Truss system except with an extra element. A 3-Bar Truss system cannot be solved though statics due to the fact that there are too many unknowns.

$$\ E^{(1)} =2, E^{(2)} =4, E^{(3)} =3 \,$$

$$\ A^{(1)} =3, A^{(2)} =1, A^{(3)} =2 \,$$

$$\ L^{(1)} =5, L^{(2)} =5, L^{(3)} =10 \,$$

$$\ P=30, \theta^{(1)}=30^\circ = -\theta^{(2)}, \theta^{(3)}=45^\circ \,$$

Local node numbering is equivalent to the convenience in assembly k.

Convenient local node numbering

$$ \sum{F_x} =0$$

$$ \sum{F_y} =0$$

$$ \sum{M_A} =0$$ trivial

The sum of the moments about A equals 0 so it is trivial. This means that there are three unknowns and only two equations. So the problem can not be solved using statics and must be solved using FEA.

Question: How about $$\sum{M_B}?$$ It still equals 0

$$ \sum{M_B} = \vec{BA'}\times \vec{F}$$

for A' on line of action of $$\vec{F}$$

$$\vec{BA'}= \vec{BA} + \vec{AA'}$$

$$ \sum{M_B}=(\vec{BA}+ \vec{AA'})\times \vec{F}$$ $$ = \vec{BA}\times F + \vec{AA'}\times \vec{F}= 0 $$

Back to 3-bar truss

Node A is in equilibrium : $$\sum_{i=0}{\vec{F_i}} =\vec{0}$$

$$ \sum_i{M_{B,i}}=\sum_i{\vec{BA'_i}}\times \vec{F_i}$$

A'i= any point on line of action of $$\vec{F_i}$$

$$ \sum_i{M_{B,i}}=\sum_i{\vec{BA}}\times \vec{F_i} \vec{BA} \times \sum_i{\vec{F_i}} = 0 $$

Converting from k matrices to K matrix

This figure is a representation of the k matrices forming the K matrix. The third element k matrix is split so that the ks corresponding to node 2 are added to the ks corresponding to node 2 for both element one and two.

Where $$K_{33}= k^{(1)}_{33}+ k^{(2)}_{11}+k^{(3)}_{11} $$ and

$$K_{34}= k^{(1)}_{34}+ k^{(2)}_{13}+k^{(3)}_{12} $$

HW: fill out the rest
$$ \begin{bmatrix} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14} & 0 & 0 & 0 & 0\\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24} & 0 & 0 & 0 & 0\\ k^{(1)}_{31} & k^{(1)}_{32} & k^{(1)}_{33}+k^{(2)}_{11}+k^{(3)}_{11} & k^{(1)}_{34}+k^{(2)}_{12}+k^{(3)}_{12} & k^{(2)}_{13} & k^{(2)}_{14} & k^{(3)}_{13} & k^{(3)}_{14}\\ k^{(1)}_{41} & k^{(1)}_{42} & k^{(1)}_{43}+k^{(2)}_{21}+k^{(3)}_{21} & k^{(1)}_{44}+k^{(2)}_{22}+k^{(3)}_{22} & k^{(2)}_{23} & k^{(2)}_{24} & k^{(3)}_{23} & k^{(3)}_{24}\\ 0 & 0 & k^{(2)}_{31} & k^{(2)}_{32} & k^{(2)}_{33} & k^{(2)}_{34} & 0 & 0\\ 0 & 0 & k^{(2)}_{41} & k^{(2)}_{42} & k^{(2)}_{43} & k^{(2)}_{44} & 0 & 0\\ 0 & 0 & k^{(3)}_{31} & k^{(3)}_{32} & 0 & 0 & k^{(3)}_{33} & k^{(3)}_{34}\\ 0 & 0 & k^{(3)}_{41} & k^{(3)}_{42} & 0 & 0 & k^{(3)}_{43} & k^{(3)}_{44} \end{bmatrix} $$

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 01:45, 8 October 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 15:37, 7 October 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 22:49, 7 October 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 01:01, 8 October 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 01:43, 8 October 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 01:47, 8 October 2008 (UTC)


title: File:VirileMail.m4a.OGG


text:
User:Eml4500.f08.team.figuerrez/hw5

Meeting 27 - The Principle of Virtual work - Part III
$$\tilde{\boldsymbol{W}}_{2x1}=$$ Virtual axial displacement. $$\tilde{\boldsymbol{W}}_{2x1}$$ corresponds to the axial displacement matrix, $${\boldsymbol{q}}^{(e)}_{2x1}$$.

$${\boldsymbol{w}}_{4x1}=$$ Virtual displacement in global coordinate system. $${\boldsymbol{w}}_{4x1}$$ corresponds to the global displacement matrix, $${\boldsymbol{d}}^{(e)}_{4x1}$$. When you combine equations  (3)  &  (4)  into equation  (2)  it yields:

 (5)  $$({\boldsymbol{T}}^{(e)}{\boldsymbol{w}})* \begin{bmatrix} \tilde{\boldsymbol{k}}^{(e)}({\boldsymbol{T}}^{(e)}{\boldsymbol{d}}^{(e)})-{\boldsymbol{p}}^{(e)} \end{bmatrix} =o $$  for all $${\boldsymbol{w}}_{4x1}$$

Recall:

 (6)  $$({\boldsymbol{A}}_{pxq}{\boldsymbol{B}}_{qxr})^{T}= {\boldsymbol{B}}^{T}{\boldsymbol{A}}^{T} $$ Equation (6) equates that the transpose of the multiplication of two matrices is equal to the transpose of matrix one multiplied by the transpose of matrix two. Equation (6) is verified below with given values for matrix A and B.

Recall:

 (7)  $${\boldsymbol{a}}_{nx1}{\boldsymbol{b}}_{nx1}= {\boldsymbol{a}}^{T}_{1xn}{\boldsymbol{b}}_{nx1} $$

 Note:  The right side of equation (7) is scalar: $${\boldsymbol{(1xn)}}{\boldsymbol{(nx1)}}={\boldsymbol{(1x1)}}=(Scalar)

$$

When you apply equations  (7)  &  (6)  into equation  (5)  it yields:

$$({\boldsymbol{T}}^{(e)}{\boldsymbol{w}})^{T}* \begin{bmatrix} \tilde{\boldsymbol{k}}^{(e)}({\boldsymbol{T}}^{(e)}{\boldsymbol{d}}^{(e)})-{\boldsymbol{p}}^{(e)} \end{bmatrix} =o $$  for all $${\boldsymbol{w}}_{4x1}$$

 Note:  $$({\boldsymbol{T}}^{(e)}{\boldsymbol{w}})^{T}$$ comes from equation (7). Applying the transpose rule for matrix multiplication from equation (6) yields:

$${\boldsymbol{w}}^{(T)}{\boldsymbol{T}})^{(e)T} \begin{bmatrix} \tilde{\boldsymbol{k}}^{(e)}({\boldsymbol{T}}^{(e)}{\boldsymbol{d}}^{(e)})-{\boldsymbol{p}}^{(e)} \end{bmatrix} =o $$

When you replace the transpose with the dot product it yields:

$${\boldsymbol{w}}* \begin{bmatrix} ({\boldsymbol{T}}^{(e)}\tilde{\boldsymbol{k}}^{(e)}{\boldsymbol{T}}^{(e)}){\boldsymbol{d}}^{(e)}-({\boldsymbol{T}}^{(e)T}{\boldsymbol{p}}^{(e)} \end{bmatrix} =o $$

Therefore, $${\boldsymbol{w}}* \begin{bmatrix} ({\boldsymbol{k}}^{(e)}{\boldsymbol{d}}^{(e)}-{\boldsymbol{f}}^{(e)} \end{bmatrix} =o $$  for all $${\boldsymbol{w}} $$

So far, only discrete cases have been given. Now, we have continuous cases, i.e. PDEs.

Motivational model problem:

We have an elastic bar (shown in Figure #1 below) with varying A(x), Young's Modulus E(x), which is subjected to a varying axial load (distributed), concentrated load, and an inertia force (dynamic force). Both the varying axial load and the concentrated load are both time dependent.

title: User:Eml4500.f08.team.figuerrez/hw5


text:
User:Eml4500.f08.a-team.melvin/Lecture 28

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

$$\Sigma$$ Fx = $$ \frac{\delta N}{\delta x} $$ dx + h.o.t. + f(x,t) dx - m(x) ,math> \ddot{u}, dx neglect h.o.t.

(h.o.t. = higher order terms)

Recall Taylor series expansion:

f(x+dx) = f(x) + $$ \frac{df(x)}{dx} $$ dx + $$ \underbrace{\frac{1}{2} \frac{d^2f(x)}{dx^2} dx^2 + ...}_{h.o.t.} $$

Eq. (1) $$ \Rightarrow \frac{\delta N}{\delta x} + f = m \ddot{u} $$ (2)

Equation of motion (EOM)

$$ N(x,t) = A(x) \underbrace{\sigma (x,t)}_{E(x) \underbrace{\epsilon (x,t)}_{\frac{\delta u}{\delta x} (x,t)}} $$ (3)

Constitutive relation

(3) in (2) yields

$$ \frac{\delta}{\delta x} [A(x)E(x) \frac{\delta u}{\delta x}] + f(x,t) = m(x) \underbrace{\ddot{u}}_{\frac{\delta^2 u}{\delta t^2}} $$

Need 2 b.c.s (2nd order derivative w.r.t. x)

2 initial conditions (2nd order derivative w.r.t. t)

(initial displacement, initial velocity)

u(0,t) = 0 = u(L,t)

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

1) u(0,t) = 0

2)

$$ \underbrace{N(L,t)}_{A(t)\underbrace{\sigma (L,t)}_{E(L) \underbrace{\epsilon (L,t)}_{\frac{\delta u}{\delta x} (L,t)}}} = F(t)$$

Initial condition at t=0, prescribe

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

$$ \frac{\delta u}{\delta t} $$ (x,t=0) = $$ \dot{u} $$ (x,t=0) = $$ \overline{\nu} $$ (x)

known function velocity


title: File:Woot1.jpg


text:
Wiki Campus Radio/Free Music

This page contains a list of sources of 'free' as in freedom music :

 * Wikimedia Commons
 * IMLSP
 * Mutopia
 * MusOpen

Using this site, we can get free scores to play in WCR. My idea is to create a radio station using Wiki Campus Radio for songs played by users of Wikiversity. I could help playing classical music, but all royalty-free or GFDL music scores could work. Davichito 01:31, 6 November 2008 (UTC)

 * http://imslp.org/index.php?title=Main_Page&

title: Wiki Campus Radio/Free Music


text: File:6bartruss with labels.GIF

title: File:6bartruss with labels.GIF


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text:
Music performances

Welcome to the Music performances audio project! This is the place where you can collaborate by performing free musical pieces, like those of Beethoven, Chopin, Mozart, etc. It will be an easy task; we need musicians. It will be done in steps.

 * 1) Get your royalty free score at http://imslp.org, for instance, Moonlight sonata score. You can find them in other websites too, as long as the license is appropriate for Wikiversity and Wiki Campus Radio.
 * 2) Play the song on your favorite instrument, including voice.
 * 3) Upload your file and request it for WCR.

Collaborators

 * David: I will perform classical music, specially Beethoven and Chopin sonatas and valses. See my performance of the Moonlight sonata at youtube
 * Gaidheal1 17:07, 2 March 2010 (UTC) I would love to contribute some recordings. I am currently without a computer, but expect this to be rectified shortly.
 * Add yourself

Requests (for recording/performance)

 * Skye Boat Song (Ideally on Pipes)
 * Amazing Grace (Ideally on Pipes)
 * Beethoven 3rd Symphony
 * Tchaikovsky 1st Piano Concerto

title: Music performances


title: File:Lulz5.1.svg


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User:Jasonallebush

TEST:

Goals and Objectives:

1.Outline a Civil Engineering curriculum <- 1 month

2.Start Mathematics Core (Pre-Calculus, Calculus, Differential Equations, Linear Algebra...) <-4 years to complete

3.General Engineering Core <- 3 years to complete

4.Civil Engineering Core <- 6 years to complete

5.Outline a Structural Engineering curriculum <- 1 month

6.Structural Engineering emphasis <- 4 years to complete

TOTAL TIME: Close to 20 years!

title: User:Jasonallebush


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Life satisfaction

Measuring life satisfaction

 * Satisfaction with Life Scale (SFWL, Diener, Emmons, Larsen & Griffin, 1985)

title: Life satisfaction


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User:Eml4500.f08.bottle.ranto/HW5

Debugging of Two Bar Truss Code
The code given in conjunction with the second homework report contained an error which was to be corrected. The previous code returned an error when attempting to generate a results matrix. This error was caused by the lines:

for i=1:elems results = [results; PlaneTrussResults(e, A, ...           nodes(conn(i,:),:), d(lmm(i,:)))]; end

This section of code attempted to pass two arrays as scalar values.

To fix this problem, the code was modified to read:

for i=1:elems results = [results; PlaneTrussResults(e(i), A(i), ...   nodes(conn(i,:),:), d(lmm(i,:)))]; end

The code in its entirety is as follows:

% Debugged Two-Bar Truss System Code clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4; nodes = [ 0, 0; % x,y position of node 1 L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); % x,y position of node 2 L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta) % x,y position of node 3 ]; 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(i), A(i), ...   nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results

This correction yields the following results:

k_local = 0.750  -0.750   -0.750    0.750 k = 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 k_local = 5   -5    -5     5 k = 2.5000  -2.5000   -2.5000    2.5000   -2.5000    2.5000    2.5000   -2.5000   -2.5000    2.5000    2.5000   -2.5000    2.5000   -2.5000   -2.5000    2.5000 K = 0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000 R = 0    0     0     7     0     0 d = 0        0    4.3520    6.1271         0         0 reactions = -4.4378  -2.5622    4.4378   -4.4378 results = 1.7081      5.1244       5.1244       0.6276        3.138        6.276

title: User:Eml4500.f08.bottle.ranto/HW5


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User talk:Jasonallebush

Welcome
Hello Jasonallebush, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Emesee 04:53, 6 November 2008 (UTC)

title: User talk:Jasonallebush


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User talk:Justinwhitaker~enwikiversity

Welcome
Hello Justinwhitaker, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * [ Ask] a question, discuss Wikiversity, etc. at the Colloquium;
 * Read and help develop our community policies, or
 * Chat with other Wikiversians on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Emesee 04:53, 6 November 2008 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:40, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:12, 19 April 2015 (UTC)

title: User talk:Justinwhitaker~enwikiversity


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User:WikimediaNotifier/translate

Machine translation : Service List | traduku

title: User:WikimediaNotifier/translate


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User talk:Wingamu

In Zibwabwe as we all know there is an economical crisis that needs to be eradicated if not minimized or wiped out. In this regard how can financial institutions (Banks, micro finance, discount houses,e.t.c) lend sustainably without harming the business community. Say a month on month inflation rate of 230 000% if prevailing. What interest rates can be charged if we are to isolate regulation as well as lending from the REserve Bank/ Federal Reserve/ Treasury of a country. A lot of factors come into play including the politica. How can we save beloved Zimbabwe or any other country for that matter. let us include all variables if a viable policy is to be envisaged and implemented.

title: User talk:Wingamu


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User:Dilos1

Hello! I have taught college level European History courses for years and I am preparing resources for similar courses at Wikiversity.

title: User:Dilos1


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Transporter Kinetics

The Transportome
The Human Genome Organisation (HUGO) recognises 1289 genes as transporters and channels, as a collective these are called the Transportome which comprises around 4% of the human genome. Of the 1289 genes 406 appear to be ion channels and the other 883 transporters, all these gens are classified into structurally related super-famillies and famillies. The solute carrier(SLC) gene super family includs 43 families and 298 transporter genes.

Important Characteristics of Transporters
 * solute flux >>> predicted by passive diffusion (avoids bilayer)
 * Substrate specific (hexoses,amino acids, lactate)
 * Saturable (theoretically)
 * Specific inhibitors/inactivators

Basic Transport Mechanisms
Uniporter-A uniporter is an integral membrane protein that is involved in facilitated diffusion. They can be either a channel or a carrier protein.

Symporter-A symporter is an integral membrane protein that is involved in movement of two or more different molecules or ions across a phospholipid membrane such as the plasma membrane in the same direction, and is therefore a type of cotransporter

Antiporter-An antiporter (also called exchanger or counter-transporter) is an integral membrane protein which is involved in secondary active transport of two or more different molecules or ions (i.e. solutes) across a phospholipid membrane such as the plasma membrane in opposite directions.

Uniporter
Uniporter carrier proteins work by binding to one molecule of solute at a time and transporting it with the solute gradient. Uniporter channels open in response to a stimulus and allow the free flow of specific molecules. Uniporters may not utilize energy other than the solute gradient. Thus they may only transport molecules with the solute gradient, and not against it.

There are several ways in which the opening of uniporter channels may be regulated:

 * 1) Voltage - Regulated by the difference in voltage across the membrane
 * 2) Stress - Regulated by physical pressure on the transporter (as in the cochlea of the ear)
 * 3) Ligand - Regulated by the binding of a ligand to either the intracellular or extracellular side of the cell

Uniporters are involved in many biological processes, including impulse transmission in neurons. Voltage-gated sodium channels are involved in the propagation of a nerve impulse across the neuron. During transmission of the signal from one neuron to the next, calcium is transported into the presynaptic neuron by voltage-gated calcium channels. Calcium released from the presynaptic neuron binds to a ligand-gated calcium channel in the postsynaptic neuron to stimulate an impulse in that neuron. Potassium leak channels, also regulated by voltage, then help to restore the resting membrane potential after impulse transmission.

Symporter
A symporter is an integral membrane protein that is involved in movement of two or more different molecules or ions across a phospholipid membrane such as the plasma membrane in the same direction, and is therefore a type of cotransporter. Typically, the ion(s) will move down the electrochemical gradient, allowing the other molecule(s) to move against the concentration gradient. The movement of the ion(s) across the membrane is facilitated diffusion, and is coupled with the active transport of the molecule(s). It should be noted that although two or more types of molecule are transported, there may be several molecules transported of each type.

Antiporter
An antiporter (also called exchanger or counter-transporter) is an integral membrane protein which is involved in secondary active transport of two or more different molecules or ions (i.e. solutes) across a phospholipid membrane such as the plasma membrane in opposite directions.

Uniport and Antiport
Transport mechanisms are "reversible", substrate loaded transporters change conformation quicker therfore antiport transport is quicker than uniport leading to trans-stimulation and co-transport

In primary active transport, one species of solute moves along its electrochemical gradient, allowing a different species to move against its own electrochemical gradient. This movement in contrast to primary active transport, in which all solutes are moved against their concentration gradients, fueled by ATP.

Transport may involve one or more of each type of solute. For example, the Na+/Ca2+ exchanger, used by many cells to remove cytoplasmic calcium, exchanges one calcium ion for three sodium ions.

Channels

 * Selective related to size and charge of hydrated ion
 * Confomational change? (but may be "gated")
 * Diffusion single-file through narrow point of channel(saturable)

Transporters

 * Selectivity related to molecular interaction between solutae "substrate" and binding-site on carrier protein ("recognition criteria")
 * Conformational change predicted
 * Carriage requires "adsorption of solute from bulk fluid phase
 * Requirement for co-substrates in many cases
 * Exhibit counter-transport

Kinetic Characteritics of Transporters
Cis-Effects
 * Each substrate binding site has characteristic properties, easiest to look at Uniport transporters
 * Saturability- substrate molecules competeing for transporter binding sites
 * Stereospecifity- Certain stereoisomers are better "fit" to binding site eg D-sugar, L-Amino acids; biologically important forms.

Trans-Effects
 * Exchange diffusion- trans-substrate accelerates exchange
 * Counter transport- trans-acceleration by a different substrate (hetero-exchange)

title: Transporter Kinetics


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Action potentials

Neuronal action potential
 This animated gif illustrates action potential propagation in an axon. Three types of ion channel are shown: potassium "leak" channels (blue), voltage-gate sodium channels (red) and voltage-gated potassium channels (green). The movement of positively-charged sodium and potassium ions through these ion channels controls the membrane potential of the axon. The negative-inside resting potential is mostly determined by potassium ions leaving the cell through leak channels. Action potentials are initiated in the axon initial segment after neurotransmitter activates excitatory receptors in the neuron's dendrites and cell body. This depolarizes the axon initial segment to the threshold voltage for opening of voltage-gated sodium channels. Sodium ions entering through the sodium channels shift the membrane potential to positive-inside, approaching the sodium equilibrium potential. The positive-inside voltage during the action potential in the initial segment causes the adjacent part of the axon membrane to reach threshold. When positive-inside membrane potentials are reached, voltage-gated potassium channels open and voltage-gate sodium channels close. Potassium ions leaving the axon through voltage-gated potassium channels return the membrane potential to negative-inside values near the potassium equilibrium potential. When the voltage-gated potassium channels gate shut, the membrane potential returns to the resting potential.

Videos

 * [[Image:Action potential video.ogv]]
 * smaller version

 * [[Image:Action potential video (larger).ogv]]
 * larger version

Estimations of the number of ions involved in generating the upstroke of the action potential
A question concerning a number of ions traversing the membrane during action potential and whether or not they can significantly alter the ionic gradients is addressed in this section. If K+ is leaving the cell, won't this deplete the cell of intracellular K+, thus decreasing the concentration gradient? The answer to the latter question is "not much", the concentration gradient is not dissipated very much. So how can this be? To answer this question, one must understand is that the number of potassium ions that must cross the membrane is actually quite small. This is because the membrane potential occurs only right next to the membrane. The membrane potential is stored in the membrane capacitance, which physically is only the charges in direct contact with the membrane surfaces. The voltage across the capacitor is the membrane potential (see action potential for an example of a membrane RC circuit). Knowing that, one needs only to calculate how much current needs to flow to discharge the membrane capacitance to calculate the number of charges that must flow across the membrane to change the capacitor voltage (membrane potential) by, say 100 mV.

The capacitative current that flows into or out of a capacitor to effect a change in membrane potential of amplitude dV, is given by the equation:

 * $$c_m A \frac{\mbox{d}V} {\mbox{d}t} =-I_c $$

To make a back-of-the-envelope calculation, let’s assume the normal conditions for cylindrical cardiomyocytes during the action potential upstroke:
 * dV: 100 mV
 * dt: 2 ms
 * cm: 0.01 to 0.02 farads·m-2, the latter for cardiomyocytes due to the membrane invaginations of the T tubules.
 * A: cell membrane surface area
 * Length L: 100 μm
 * radius r: 5 μm
 * Thus, since the cell is a cylinder,
 * Area, A: 3.3 m2
 * Volume, V: 7.85 liters
 * Intracellular sodium concentration, [Na+]i: 15 mM

 * $$I_c = \left(0.02 \ \mathrm{F\cdot m}^{-2} \right) \left(3.3\times 10^{-9} \ \mathrm{m}^2 \right) \frac{0.1 \ \mathrm{V}} {0.002 \ \mathrm{s}} = 3.3\times 10^{-9} \ \mathrm{C\cdot s}^{-1}$$

This equals 3.3 nanoamps. Now, we wanted to know the number of ions, n, this corresponds to. dt is 2 ms, every ion has 1 elementary charge, e, each, and one coulomb is 6.24 e.

 * $$n = \left(3.3 \times 10^{-9} \ \mathrm{C\cdot s}^{-1} \right) \left( 0.002\ \mathrm{s} \right) \left( 6.24 \times 10^{18} \ e \cdot \mathrm{C}^{-1} \right) = 4.12 \times 10^7 \ e$$

Therefore, it would take the movement of 41 million charges (i.e. sodium ions) across the membrane to change the voltage of the cardiomyocyte by 100 mV in 2 ms. We want to know what that will do to the intracellular sodium concentration, which in cardiomyocytes is about 15 millimoles per liter. Avogadro's number, which relates numbers to moles, is 6.022·1023 mol-1:

 * $$\Delta [\mbox{Na} ^+]_{ \mbox{i}} = \frac{4.12 \times 10^7 \ e} {\left(6.022 \times 10^{23} \ e \cdot \mathrm{mol}^{-1} \right) \left( 7.85 \times 10^{-12} \ \mathrm{L} \right)} = 8.71 \times 10^{-6} \ \mathrm{mol \cdot L}^{-1}$$

Since the normal inside concentration of sodium is some 15 mM, 8.7 μM entering the cell during the action potential upstroke is only about 0.058% extra sodium ions

The bottom line for those who don’t want to sift through the calculation is that only about 40,000,000 sodium ions must cross the membrane to move the membrane potential by 100 mV in 2 ms, and that this constitutes only some 0.06% of the sodium already present in the cell. This is at the action potential's most extreme and in a very large cell. In other cells, calculations may look different, depending on surface-to-volume ratio, max dV/dt etc.

title: Action potentials


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User:M.M.S./Test

{Vad heter Kirgizistans huvudstad - Dusjanbe + Bisjkek - Astana - Aschjabad
 * type=""}

title: User:M.M.S./Test