text:
Large Hadron Collider

About the LHC

The LHC is an international research project based at CERN in Geneva, Switzerland, where scientists, engineers and support staff from 111 nations are combining state-of-the-art science and engineering in one of the largest scientific experiments ever conducted.

The LHC is the latest and most powerful in a series of particle accelerators that, over the last 70 years, have allowed us to penetrate deeper and deeper into the heart of matter and further and further back in time. The next steps in the journey will bring new knowledge about the beginning of our Universe and how it works, as the LHC recreates, on a microscale, conditions that existed billionths of a second after the birth of our Universe.

What is the LHC?

The LHC is exactly what its name suggests - a large collider of hadrons. Strictly, LHC refers to the collider; a machine that deserves to be labelled ‘large’, it not only weighs more than 38,000 tonnes, but runs for 27km (16.5m) in a circular tunnel 100 metres beneath the Swiss/French border at Geneva.

However, the collider is only one of three essential parts of the LHC project. The other two are:

* the detectors, which sit in 4 huge chambers at points around the LHC tunnel and * the GRID, which is a global network of computers and software essential to processing the data recorded by LHC’s detectors.

The LHC’s 27km loop in a sense encircles the globe, because the LHC project is supported by an enormous international community of scientists and engineers. Working in multinational teams, at CERN and around the world, they are building and testing LHC equipment and software, participating in experiments and analysing data. The UK has a major role in leading the project and has scientists and engineers working on all the main experiments.

What will the LHC do?

The LHC will allow scientists to probe deeper into the heart of matter and further back in time than has been possible using previous colliders.

Researchers think that the Universe originated in the Big Bang (an unimaginably violent explosion) and since then the Universe has been cooling down and becoming less energetic. Very early in the cooling process the matter and forces that make up our world ‘condensed’ out of this ball of energy.

The LHC will produce tiny patches of very high energy by colliding together atomic particles that are travelling at very high speed. The more energy produced in the collisions the further back we can look towards the very high energies that existed early in the evolution of the Universe. Collisions in the LHC will have up to 7x the energy of those produced in previous machines; recreating energies and conditions that existed billionths of a second after the start of the Big Bang.

The results from the LHC are not completely predictable as the experiments are testing ideas that are at the frontiers of our knowledge and understanding. Researchers expect to confirm predictions made on the basis of what we know from previous experiments and theories. However, part of the excitement of the LHC project is that it may uncover new facts about matter and the origins of the Universe.

One of the most interesting theories the LHC will test was put forward by the UK physicist Professor Peter Higgs and others. The different types of fundamental particle that make up matter have very different masses, while the particles that make up light (photons) have no mass at all. Peter’s theory is one explanation of why this is so and the LHC will allow us to test the theory. More of the Big Questions about the universe that the LHC may help us answer can be found here.

How does the LHC work?

The LHC accelerates two beams of atomic particles in opposite directions around the 27km collider. When the particle beams reach their maximum speed the LHC allows them to ‘collide’ at 4 points on their circular journey.

Thousands of new particles are produced when particles collide and detectors, placed around the collision points, allow scientists to identify these new particles by tracking their behaviour.

The detectors are able to follow the millions of collisions and new particles produced every second and identify the distinctive behaviour of interesting new particles from among the many thousands that are of little interest.

As the energy produced in the collisions increases researchers are able to peer deeper into the fundamental structure of the Universe and further back in its history. In these extreme conditions unknown atomic particles may appear.

Who benefits?

There are two types of benefit that the LHC project produces for the UK. The less easily measured benefits are:

* new understanding of the physical world, * training of world class scientists and engineers, * maintenance of a vibrant, world class UK research base and, * a leading role in a major international project.

More easily appreciated are the knowledge, expertise and technology that is spun off from the LHC and can be directly applied to development of new medical, industrial and consumer technologies (more...)

The science of the LHC is far removed from everyday life, but the fact that the science is so extreme constantly pushes the boundaries of existing technical and engineering solutions. Simply building the LHC has generated new technology.

The LHC is not primarily about building a better world. Rather, it allows us to test theories and ideas about how the Universe works, its origins and evolution. The questions asked, and answers found, are so fundamental that the information from LHC experiments will only be applied many years in the future, if at all. However, this is an experiment and one of the surprises from the experiment may be new science that can be applied almost immediately.

For Students Exploring the outer limits of knowledge

The LHC is the most exciting science adventure of this decade; a huge international project exploring the boundaries of our knowledge and theories about the world we live in. The LHC experiments will help us understand the origins and evolution of our universe and possibly reshape how we think about the physical world. Screen shot of particle detectives home page

See a young physicist describing how the study of particle physics has affected our lives and our understanding of the universe we live in and hear how the LHC may be our window into a weird world of missing matter and quantum physics.

the fundamental building blocks of nature, the fundamental forces of nature, the quantum world and the hunt for the Higgs boson.

Drive our simulator at www.particledetectives.net and learn how LHC scientists will conduct their experiments. This website includes a collection of resources for 14-16 and 16+ year old students on 'how science works' and particle physics. Your own discoveries or listen to the video clips of students questionning scientists about the LHC project and download Powerpoint presentations on LHC science.

There are simulations and pages of information on particle physics at the Lancaster Particle Physics pages. These are suitable for AS/A2 16+ students. There is a section dedicated to the LHC.

A simple game that illustrates how scientists search for new particles can be found on the Science Museum's website.

UK scientists have major roles in this project, which involves thousands of scientists and engineers from around the world working in harmony, to build and run the biggest experiment in the history of science. The project will continue for at least 15 years, so anyone considering science as a GCSE option now could be working on the LHC ... and contributing to new discoveries … in the future.

The science of the LHC is directly relevant to areas of the A-level (and equivalent) curriculum, but the project is also an example of how science works and there are many role models and career stories among the diverse community of scientists, engineers and administration staff working on the LHC.

A lot of resources on the LHC and particle physics are available and more will be produced over the life of the project. Visit the resources pages on the STFC website to download, or order, a stunning A1 poster of the ATLAS detector and an A5 fold out booklet that describes some of the big questions that researchers expect the LHC to answer. You can also download a poster describing the CMS detector (jpg 69kb) and a booklet describing the evolution of the universe (pdf 9.1mb) from the CMS area of CERN website. The ATLAS experiment has two posters available to download (or order) from the ATLAS store. Booklets, DVDs and other materials are also available. --Ashimjgec08 15:23, 24 May 2009 (UTC)#REDIRECT LHC#REDIRECT lhc

title: Large Hadron Collider


text: File:Loadedtruss.JPG

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

title: File:Loadedtruss.JPG


text:
User:Eml4500.f08.delta 6.guzman

Finite Element Analysis

Diana Guzman

HW 1:
 * Lecture Notes 09/05
 * MATLAB Sections 10 & 11

HW 2:
 * Lecture Notes 09/12
 * Two Bar Truss Example
 * New Two Bar Truss Example

HW 3:
 * Lecture Notes 09/26
 * Computation of Eigenvalues of Matrix $K_{6x6}$
 * MATLAB Code (plot of undeformed two-bar truss structure)

HW 4:
 * Lecture Notes 10/10
 * Three Bar Truss Example and Plot

HW 5:
 * Composites with Moduli that Change with Length

HW 6:
 * Lecture Notes 11/07

HW 7:
 * Lecture Notes 11/19
 * MATLAB Code and Plot of Undeformed and Deformed Shapes for the Two Bar Truss System
 * Evaluation of Wikiversity vs E-learning

title: User:Eml4500.f08.delta 6.guzman


text:
User:EAS4200c.f08.Blue.E/Lecture 4

 Introduction to Aerospace structures

The goals of aerospace structures is to design aircraft that are very light yet also strong. this is accomplished through the use of materials that have a high stiffness, high strength, and lightweight.

The stiffness of a material is determined by  $$E$$  The young's modulus of the material which is relates the stress and strain on the material through hooke's law  $$\sigma=E*\epsilon$$  hooke's law is a linear equation and the Young's modulus is the slope of the equation. This means for the higher the Young's modulus the faster the stress will increase meaning little deformation will be able to occur in the material before the yield stress is reached.

The strength of a material is determined by  $$\sigma_y $$, $$ \sigma_u$$  which are the yield stress and the ultimate stress respectively.

The higher the stresses that the material is able to handle before yielding the stronger the material.

Toughness is also considered in materials because it is the ability of the material to resist fracturing. Which means that a material with a high toughness would be able to resist fracturing very well. An example of a very stiff material but has a very low toughness would be glass.

Two aspects of aircraft design are the geometry and materials used. The geometry for the aircraft body is constructed by a thin wall shell design instead of an internal truss design. Aerodynamics also play a larger role in the geometry of the aircraft. Materials that are used need to be used to their full capacity to avoid more weight than necessary in the design. Also when able replace aluminum and titanium with fiber reinforced composites which will reduce weight by approximately thirty to forty percent.

Book pg 1-3

title: User:EAS4200c.f08.Blue.E/Lecture 4


text:
User talk:Eml4500.f08.delta 6.guzman

 Hello Eml4500.f08.delta 6.guzman, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 22:15, 9 September 2008 (UTC)

title: User talk:Eml4500.f08.delta 6.guzman


text:
User:Eml4500.f08.lulz.strack/monobook.js

// install User:Cacycle/wikEd in-browser text editor document.write('');


text:
User:Eml4500.f08.lulz.strack

FEAD class page for Andrew Strack

Individual Homework Portions

 * Homework 1
 * Homework 2
 * Homework 3
 * Homework 4
 * Homework 5
 * Homework 6
 * Homework 7

Group Homework Reports

 * Homework 1
 * Homework 2
 * Homework 3
 * Homework 4
 * Homework 5
 * Homework 6
 * Homework 7

Grades

 * Homework 1: 86.4
 * Homework 2: 87.8
 * Homework 3: 83.5
 * Homework 4:
 * Homework 5:
 * Homework 6:
 * Homework 7:
 * Exam 1: 95 (raw), 90.9 (scaled)
 * Exam 2: 140 (raw), 89.9 (scaled)
 * Exam 3:

title: User:Eml4500.f08.lulz.strack


text:
User:Eml4500.f08.group.setlak

5. Matrix Building Functions:

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

6. Loops, if statements, and relations

For Loop

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

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

or can be written:

x = [];

for i= 1:n

x = [x, i^2]

end

 * 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. s = 0;

for c = A

s = s + sum(c);

end

While loop

General form:

while relation

statements

end

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:

n = 0;

while 2^n < a

n = n + 1;

end

n

 * 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:

if relation

statements

end

The statements are executed only if the relation is true.

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

if n < 0

parity = 0;

elseif rem(n,2) == 0

parity = 2;

else

parity = 1;

end

Relations

The relational operators in MATLAB are:

<	less than

>	greater than

<=	less than or equal

>=	greater than or equal

==	equal

~=	not equal

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

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

&	and

|	or

~	not

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:

if A == B

statement

end

To execute statement when A and B are not equal:

if any(any(A ~=B))

statement

end

Or

if A == B else

statement

end

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

if A ~= B, statement, end

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

title: User:Eml4500.f08.group.setlak


text:
MATLAB Tutorial: Part 2

MATLAB Tutorial: Part 2

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

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

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

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

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

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

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

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

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

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

 * Title: graph title

 * xlabel: x-axis label

 * ylabel: y-axis label

 * gtext: place text on graph using mouse

 * text: place text at specified coordinates

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

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

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

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

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

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

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

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

 * yy = xx;

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

 * z = exp(-x.^2 – y.^2);

 * mesh(z)

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

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

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

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

title: MATLAB Tutorial: Part 2


text:
User:Eas4200c.f08.aero6.behnke/nothing

The following information is from the EAS4200c lectures beginning on August 25, 2008 to September 5, 2008.

Leactures 1-4: 8/25/2008 - 9/1/2008 For this class we will be submitting homework through Wikiversity. The homework assignments will include lecture summaries, assigned homework problems, and extra group contributions. The advantage to submitting assignments this way is that the knowledge can be shared.

Lecture 5: 9/3/2008 When designing aircraft, we want it to be lightweight, and have high stiffness and strength. Stiffness can be defined as the resistance of a body to deformation. It is mathematically defined as (insert eqn here sigma = E*epsilon)

where E is Younge's Modulus, sigma is stress, and epsilon is strain. A large value of E corresponds to high stiffness. For brittle materials the stress vs. strain curve is similar to that of figure 1. For metals, the stress vs. strain curve is similar to that of figure 2.

title: User:Eas4200c.f08.aero6.behnke/nothing


text:
User talk:Eml4500.f08.wiki.dagulo.r

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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 00:17, 10 September 2008 (UTC)

title: User talk:Eml4500.f08.wiki.dagulo.r


text:
User talk:Eml4500.f08.group.setlak

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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 00:18, 10 September 2008 (UTC)

title: User talk:Eml4500.f08.group.setlak


text:
User talk:Eas4200c.f08.wiki.c

 Hello Eas4200c.f08.wiki.c, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 00:19, 10 September 2008 (UTC)

control on who edits
If your team does not want anyone else to edit a page, just put a message at the top of the page asking others not to edit. An example of a template that can be used for this is shown at Page protection templates. It is easy to look in the page history and see if someone tried to sneak in an edit.

Copying homework. If you wait until the last minute to post the homework, you destroy the whole point of using wiki (easy online collaboration). It might be possible to develop a feature that would allow a wiki page to be password protected and only visible to selected users. We could experiment with how to do this at the sandbox server. --JWSchmidt 00:31, 10 September 2008 (UTC)

title: User talk:Eas4200c.f08.wiki.c


text:
User:Eml4500.f08.team.figuerrez/hw1

 1. Accessing MATLAB

After logging in, you can enter MATLAB by using the system command matlab and exit MATLAB with the command exit. On systems that     permit multiple users, such as MS Windows and a Unix system, it is suitable to leave MATLAB active.

 2. Entering matrices

Inputing either statements: A = [1 2 3; 4 5 6; 7 8 9]

and/or

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

will construct a 3 x 3 matrix assigning it to a variable A. The elements within a row of a matrix can be separated by either commas or a     blank space. However, when listing a number in exponential form, blank spaces should be avoided.

Two ways to enter complex matrices:

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

When listing complex numbers in a matrix, blank spaces must also be avoided. Either i or j may be used as the imaginary unit. If i and j are     used as variables, a new imaginary unit may be generated. (i.e. ii = sqrt(-1).)

The built-in functions rand, magic, and hilb, provide an easy way to create matrices.

rand(n) will create an n x n matrix with randomly generated entries between 0 and 1.

rand(m,n) will create an m x n matrix.

magic(n) will create an integral n x n which is a magic square (rows, column, and diagonals have common sum).

hilb(n) will create the n x n Hilbert matrix.

 3. Matrix operations, array operations

Matrix operations used in MATLAB:

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

These matrix operations apply to scalars (1 x 1 matrices) as well.

 Array operations

These operations, *, ^, \, and /, can be made to operate entry-wise by placing a period prior to the operation used.

For example, either [1,2,3,4].*[1,2,3,4] or [1,2,3,4].^2 will yield [1,4,9,16].

 4. Statements, expressions, and variables; saving a session

MATLAB statements are usually of the form:

variable = expression, or simply expression

Expressions usually consist of operators, functions, and variable names. If the variable name and = sign are omitted, a variable ans (for answer) is automatically created. MATLAB is case-sensitive in the names of commands, functions, and variables. (i.e. solveUT is not     the same as solveut).

The command who (or whos) will list the variables currently in the workspace. A variable can be cleared from the workspace with the command clear variablename.

 Saving a session

The command save before exiting causes all variables to be written to a non-human-readable disk file named matlab.mat. The command load will restore the workspace to a previous state.

 5. Matrix building functions

Convenient matrix building functions:

eye identity matrix zeros matrix of zeros ones matrix of ones diag create or extract diagonals triu upper triangular part of a matrix tril lower triangular part of a matrix rand random generated matrix hilb Hilbert matrix magic magic square

 6. For, while, if – and relations

 For

Example:

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

or         x = [];

for i = 1:n

x = [x,i^2]

end

will produce the same vector in reverse order.

 While

Example: n = 0;

while 2n < a n = n + 1;

end n

will, for a given number a, compute and display the smallest nonnegative integer n such that 2n > or = to a

 If

Example:

if n < 0

parity = 0;

elseif rem(n,2) = = 0

parity = 2;

else

parity = 1;

end

In two-way branching the elseif portion would be omitted.

 Relations

Relationship operators in MATLAB: <     	less than

>     	greater than'''

<=     less than or equal

>=     greater than or equal

==     equal

~=     not equal

title: User:Eml4500.f08.team.figuerrez/hw1


text:
User:Eml4500.f08.Lulz.Layton.Eric

3-bar Truss System

For element 1:

$$E^{(1)}=2$$ $$A^{(1}=3$$ $$L^{(1)}=5$$ $$\theta^{(1)}=30$$(deg)

For element 2:

$$E^{(2)}=4$$ $$A^{(2}=1$$ $$L^{(2)}=5$$ $$\theta^{(2)}=30$$(deg)

For element 3:

$$E^{(3)}=3$$ $$A^{(3}=2$$ $$L^{(3)}=10$$ $$\theta^{(3)}=45$$(deg)

If we break down each element into its free-body diagram we get the following 3 pictures:

If we put these 3 pictures together we get:

We cannot solve this problem by statics because we only have 2 useful equations.

$$\Sigma F_x=0$$

$$\Sigma F_y=0$$

The moment equation $$\Sigma M_A=0$$ is trivial and does not help us solve the problem.

But what about taking the moment about an arbitrary point B?

Following the math for this we get:

$$\Sigma\textbf{M}_B=\textbf{BA}\times\textbf{F}$$

$$\Sigma\textbf{M}_B=\textbf{BA}^'\times\textbf{F}$$

$$\textbf{BA}^'=\textbf{BA}\times\textbf{AA}^'$$

$$\Sigma\textbf{M}_B=(\textbf{BA}+\textbf{AA}^')\times\textbf{F}$$

$$\Sigma\textbf{M}_B=\textbf{BA}\times\textbf{F} + \textbf{AA}^'\times\textbf{F}$$

Where: $$\textbf{AA}^'\times\textbf{F}=0$$ which is trivial and doesn't help us.

Now if we look back at the three-bar truss example we see that node A is in equilibrium.

$$\Sigma_{i=0}^3\textbf{F}_i=\overrightarrow{0}$$

So, where A' is any point on the line of action of Fi:

$$\Sigma\textbf{M}_{Bi}=\Sigma\textbf{BA}_i^' x \textbf{F}_i$$

Where $$\Sigma_i\textbf{F}_i=0$$ and is trivial.

Now we need to find the global stiffness matrix. This can be found in the following manner:

So, for example we see:

$$K_{33}=k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)}$$

$$K_{34}=k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)}$$

Axial displacement calculation for local node 2
During meeting 12, a calculation for the axial displacement of local node 1 for an arbitrary element e was considered. What follows is a similar calculation for local node 2 of the same element.

First, we define the displacement vector at local node 2 (Please note the [#] subscript was used in place of the "box" subscript used in class):

$$ \vec{\mathbf{d}^{(e)}_{[2]}} = d_{3}^{(e)} \vec{i} + d_{4}^{(e)} \vec{j} $$

Next, the axial displacement at node 2 is defined: $$ q_{2}^{(e)} = \vec{\mathbf{d}^{(e)}_{[2]}}\cdot\vec{\tilde{i}} $$

$$ q_{2}^{(e)} = (d_{3}^{(e)}\vec{i} + d_{4}^{(e)}\vec{j})\vec{\tilde{i}} $$

$$ q_{2}^{(e)} = d_{3}^{(e)}(\vec{i}\cdot\vec{\tilde{i}}) + d_{4}^{(e)}(\vec{j}\cdot\vec{\tilde{i}}) $$

From a previous homework question and the class notes, the following relations are valid:

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

Substituting,

$$ q_{2}^{(e)} = d_{3}^{(e)}\cos \theta^{(e)} + d_{4}^{(e)}\sin\theta^{(e)} $$

From the notes, the following relations are valid:

$$\cos\theta^{\left(e\right)} = l^{(e)}$$ $$\sin\theta^{\left(e\right)} = m^{(e)}$$

Substituting,

$$ q_{2}^{(e)} = d_{3}^{(e)}l^{(e)} + d_{4}^{(e)}m^{(e)} $$

Putting the above equation yields the following result, consistent with that of the axial displacement of local node 1:

$$ q_{2}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}\begin{Bmatrix} d_{3}^{(e)}\\ d_{4}^{(e)} \end{Bmatrix} $$

Proof of $$\mathbf{k}^{(e)} = \mathbf{T}^{(e)^{T}} \mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$
In this question we are asked to prove that the above matrix product generates the generic element stiffness matrix $$\mathbf{k}^{(e)}$$.

First, the three matrices used to generate the product:

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

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

$$\mathbf{\hat{k}}^{(e)} = \begin{bmatrix} 1&-1 \\ -1&1 \\ \end{bmatrix}$$

The product of the first two matrices:

$$\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} = \begin{bmatrix} 1&-1 \\ -1&1 \\ \end{bmatrix} \begin{bmatrix} l^{(e)}&m^{(e)}&0&0 \\ 0&0&l^{(e)}&m^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)}&m^{(e)}&-l^{(e)}&-m^{(e)} \\ -l^{(e)}&-m^{(e)}&l^{(e)}&m^{(e)} \\ \end{bmatrix}$$

The entire product multiplied out:

$$\mathbf{T}^{(e)^{T}}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} = \begin{bmatrix} l^{(e)}&0 \\ m^{(e)}&0 \\ 0&l^{(e)} \\ 0&m^{(e)} \\ \end{bmatrix} \begin{bmatrix} l^{(e)}&m^{(e)}&-l^{(e)}&-m^{(e)} \\ -l^{(e)}&-m^{(e)}&l^{(e)}&m^{(e)} \\ \end{bmatrix} $$ $$ = \begin{bmatrix} l^{(e)^{2}} & m^{(e)}l^{(e)} & -l^{(e)^{2}} & -m^{(e)}l^{(e)} \\ m^{(e)}l^{(e)} & m^{(e)^{2}} & -m^{(e)}l^{(e)} & -m^{(e)^{2}}\\ -l^{(e)^{2}} & -m^{(e)}l^{(e)} & l^{(e)^{2}} & m^{(e)}l^{(e)}\\ -m^{(e)}l^{(e)} & -m^{(e)^{2}} & m^{(e)}l^{(e)} & m^{(e)^{2}} \\ \end{bmatrix} $$

As shown in previous homework question, the element stiffness matrix:

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

Thus, it is easily shown that the following relation is indeed true:

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

John Saxon Eml4500.f08.Lulz.js 18:09, 8 October 2008 (UTC)

title: User:Eml4500.f08.Lulz.Layton.Eric


text:
User:Eml4500.f08.wiki1.brannon

HW1
Bar Element 1: Global Free-Body Diagram

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

string='text string'

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

disp('displays text string')

Error messages can be displayed with the error function:

error('error message')

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

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

variable=input('Enter number: ')

The variable is assigned the input value and execution continues.

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

>> !ed fileforediting.m

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

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

Debugging tools can be seen using the help dbtype command.

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

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

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

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

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

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

tic, command, toc

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

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

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

Changes to format remain until changed again.

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

diary named diskfile

The command diary on will write text to the file until the command diary off is called.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The final global stiffness matrix, K, becomes:

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

Eml4500.f08.wiki1.brannon 20:44, 26 September 2008 (UTC)

Infinitesimal displacement (related to virtual displacement)
Global Free-Body Diagram showing Infinitesimal Displacements

$$ \overline{AC}= {|P_2^{(1)}| \over k^{(1)}}={5.1243 \over 3/4}=6.8324 $$

$$ \overline{AB}= {|P_1^{(1)}| \over k^{(2)}}={6.2762 \over 5}=1.2552 $$

Find (x,y) coordinates of B and C:

Point B:

$$ x= \overline{AB} cos{135^\circ}=1.2552*-{\sqrt 2 \over 2}=-0.8876 $$

$$ y= \overline{AB} sin{135^\circ}=1.2552*{\sqrt 2 \over 2}=0.8876 $$

Point C:

$$ x= \overline{AC} cos{30^\circ}=6.8324*{\sqrt 3 \over 2}=5.917 $$

$$ y= \overline{AC} sin{30^\circ}=6.8324*{1 \over 2}=3.4162 $$

We now have two unknowns, (XD,YD)

We need the equations for line $$\overline{AB}$$ and $$\overline{BC}$$

Method for the Determination of the Slope between Two Arbitrary Points
PQ Line Segment Diagram

$$ \overline{PQ}=(PQ) \tilde{i}=(PQ)[cos \theta \vec i +sin \theta \vec j]=(x-x_P)\vec i +(y-y_P)\vec j $$

$$ x-x_P=(PQ)cos \theta$$

$$ y-y_P=(PQ)sin \theta$$

$$ {y-y_P \over x-x_P}=tan \theta$$

$$ y-y_P=tan \theta(x-x_P)$$

Equation for line perpendicular to PQ passing through P:

$$ y-y_P=tan( \theta + {\pi \over 2})(x-x_P)$$

Determination of (XD,YD)
Line Perpendicular to Point B:

$$y-0.8876=tan{225^\circ}(x+0.8876)$$

$$y=x+1.7752$$

Line Perpendicular to Point C:

$$y-3.4162=tan{120^\circ}(x-5.917)$$

$$y=-1.732x + 13.665$$

Intersection at Point (XD,YD):

$$y_D+1.732(y_D-1.7752)=13.665$$

$$y_D=6.127$$

$$(x_D+1.7752)=-1.732x_D + 13.665$$

$$x_D=4.352$$

$$\vec {AD} = (x_D-x_A) \vec i + (y_D-y_A) \vec j$$

Simplification: $$x_A=0, y_A=0$$

$$\vec {AD} = x_D \vec i + y_D \vec j$$

$$\vec {AD} = 4.352 \vec i + 6.127 \vec j$$

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

$$ d_3=4.352, d_4=6.127$$

3-bar Truss System
3-bar Truss System

$$ E^{(1)}=2, A^{(1)}=3, L^{(1)}=5$$

$$E^{(2)}=4, A^{(2)}=1, L^{(2)}=5$$

$$E^{(3)}=3, A^{(3)}=2, L^{(3)}=10$$

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

$$P=30$$

Convenient Local Node Numbering:

3-bar Truss System Local Node Numbering

Eml4500.f08.wiki1.brannon 01:29, 8 October 2008 (UTC)

Eigenvalues and Eigenvectors of K
For the global stiffness matrix K computed previously, the eigenvectors and eigenvalues are displayed below as matrix V and D, respectively. EDU>> k

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

EDU>> [V,D]=eig(k)

V = Mode Shapes =

0.1201  -0.5000   -0.0021    0.5951    0.6174   -0.0139    0.0693    0.8660    0.0036    0.3436    0.3565   -0.0080    0.4636    0.0000    0.0000    0.4796   -0.5409    0.5123   -0.5257    0.0000    0.0000    0.5438   -0.4330   -0.4904    0.4947    0.0030   -0.7071   -0.0321   -0.0765   -0.4984   -0.4947    0.0030   -0.7071    0.0321    0.0765    0.4984

D = Diag[Eigenvalues] =

-0.0000        0         0         0         0         0         0    0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    1.4706         0         0         0         0         0         0   10.0294

EDU>> frequency=sqrt(D)

frequency =

0 + 0.0044i       0                  0                  0                  0                  0 0            0.0000                  0                  0                  0                  0                  0                  0             0.0000                  0                  0                  0                  0                  0                  0             0.0046                  0                  0                  0                  0                  0                  0             1.2127                  0                  0                  0                  0                  0                  0             3.1669

EDU>> modeShapes=V

modeShapes =

0.1201  -0.5000   -0.0021    0.5951    0.6174   -0.0139    0.0693    0.8660    0.0036    0.3436    0.3565   -0.0080    0.4636    0.0000    0.0000    0.4796   -0.5409    0.5123   -0.5257    0.0000    0.0000    0.5438   -0.4330   -0.4904    0.4947    0.0030   -0.7071   -0.0321   -0.0765   -0.4984   -0.4947    0.0030   -0.7071    0.0321    0.0765    0.4984

  Truss System Eigenvectors Plot

These mode shapes (corresponding to the zero eigenvalues) may come out as a linear combination of the pure mode shapes. The mode shapes are in the form of pure rigid body motion or translation in the x, y, and z directions, as well as one representing the mechanism motion, the pin joint joining the two elements at the point of applied force, allowing independent motion of each element.

Eigenvectors and Eigenvalues of Modified K for Case A
$$\mathbf{\overline{k}v} = \lambda \mathbf{v}$$

$$\Rightarrow \mathbf{\overline{k}}$$ is stiffness matrix for constrained system

  Truss System for Case A

  Truss System for Case A (Local Node Numbering)

Matlab code: Modify 2-bar truss code

$$conn= \begin{bmatrix} 1&2\\ 2&3\\ 3&4 \end{bmatrix}$$

$$lmm= \begin{bmatrix} 1&2&3&4\\ 3&4&5&6\\ 5&6&7&8 \end{bmatrix}$$

$$ a = b = 1m$$

$$E = 2$$

$$A = 3$$

$$ k^{(1)}= \begin{bmatrix} 0&0&0&0\\ 0&6&0&-6\\ 0&0&0&0\\ 0&-6&0&6 \end{bmatrix},

k^{(2)}= \begin{bmatrix} 6&0&-6&0\\ 0&0&0&0\\ -6&0&6&0\\ 0&0&0&0 \end{bmatrix},

k^{(3)}= \begin{bmatrix} 0&0&0&0\\ 0&-6&0&6\\ 0&0&0&0\\ 0&6&0&-6\\ \end{bmatrix} $$

K= \begin{bmatrix} 0&0&0&0&0&0&0&0\\ 0&6&0&-6&0&0&0&0\\ 0&0&6&0&-6&0&0&0\\ 0&-6&0&6&0&0&0&0\\ 0&0&-6&0&6&0&0&0\\ 0&0&0&0&0&-6&0&6\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&6&0&-6\\ \end{bmatrix} $$

ka =

0    0     0     0     0     0     0     0     0     6     0    -6     0     0     0     0     0     0     6     0    -6     0     0     0     0    -6     0     6     0     0     0     0     0     0    -6     0     6     0     0     0     0     0     0     0     0    -6     0     6     0     0     0     0     0     0     0     0     0     0     0     0     0     6     0    -6

EDU>> [Va,Da]=eig(ka)

Va = Mode Shapes =

0        0         0         0    1.0000         0         0         0         0   -0.7071         0         0         0         0   -0.7071         0         0         0   -0.7071         0         0         0         0   -0.7071         0   -0.7071         0         0         0         0    0.7071         0         0         0   -0.7071         0         0         0         0    0.7071   -0.7071         0         0         0         0    0.7071         0         0         0         0         0   -1.0000         0         0         0         0    0.7071         0         0         0         0    0.7071         0         0

Da = Diag[Eigenvalues]

-12    0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0    12     0     0     0     0     0     0     0     0    12

  Case A Truss System Eigenvectors Plot

The global stiffness matrix K for case A shows 5 zero eigenvalues, representing 3 rigid body motions and 2 mechanism motions for the two pin joints in the truss system

Eml4500.f08.wiki1.brannon 01:56, 24 October 2008 (UTC)

Justification for Eliminating Rows 1,2,5, and 6 to Obtain $$\overline{K}$$ in 2-bar Truss
Force-Displacement Relationship: K*d=F K*d-F=0 $$\rightarrow Equation(1)$$ By the Principle of Virtual Work:W•(Kd-F)=0 for all W $$\rightarrow Equation(2)$$ $$Equation(1)\Leftrightarrow Equation(2)$$ Proof: $$(1) \rightarrow (2)$$: Trivial $$(2) \rightarrow (1)$$: Not Trivial We want to show $$(2) \rightarrow (1)$$. Choice 1: Select W such that W1=1, W2=0, W3=0, W4=0, W5=0, W6=0 W=$$ \begin{Bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ 0\end{Bmatrix}$$ W•(Kd-F)= 1.) $$[\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot [\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot [\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{1j}d_j=F_1$$   $$\rightarrow (1st Equation)$$ Choice 2: Select W such that W1=0, W2=1, W3=0, W4=0, W5=0, W6=0  W=$$ \begin{Bmatrix} 0\\1\\0\\0\\0\\0\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot [\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{2j}d_j=F_2 \rightarrow (2nd Equation)$$ Choice 3: Select W such that W1=0, W2=0, W3=1, W4=0, W5=0, W6=0 W=$$ \begin{Bmatrix} 0\\0\\1\\0\\0\\0\end{Bmatrix}$$ 2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{3j}d_j=F_3 \rightarrow (3rd Equation)$$ Choice 4: Select W such that W1=0, W2=0, W3=0, W4=1, W5=0, W6=0 W=$$ \begin{Bmatrix} 0\\0\\0\\1\\0\\0\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+[\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{4j}d_j=F_4 \rightarrow (4th Equation)$$ Choice 5: Select W such that W1=0, W2=0, W3=0, W4=0, W5=1, W6=0 W=$$ \begin{Bmatrix} 0\\0\\0\\0\\1\\0\end{Bmatrix}$$ 2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+[\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{5j}d_j=F_5 \rightarrow (5th Equation)$$ Choice 6: Select W such that W1=0, W2=0, W3=0, W4=0, W5=0, W6=1 W=$$ \begin{Bmatrix} 0\\0\\0\\0\\0\\1\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+[\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{6j}d_j=F_6 \rightarrow (6th Equation)$$ $$\begin{bmatrix} K_{11}&\cdots&K_{16}\\ \vdots&\ddots&\vdots\\ K_{61}&\cdots&K_{66}\end{bmatrix}* \begin{Bmatrix} d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{Bmatrix}= \begin{Bmatrix} F_1\\F_2\\F_3\\F_4\\F_5\\F_6\end{Bmatrix}$$ Hence, Kd=F (or Equation (1) from above)

Principle of Virtual Work, Relating to W
The Principle of Virtual Work accounts for the boundary conditions.

For the two-bar truss system, d1=d2=d5=d6=0 Weighting coefficients must be "kinematically admissible". In other words, the weighting coefficients cannot violate the boundary conditions.

Weighting Coefficients$$\equiv$$Virtual Displacements $$W_1=W_2=W_5=W_6=0$$ W•(Kd-F) $$\begin{Bmatrix}W_3\\W_4\end{Bmatrix}\cdot (\bar{\mathbf{K}} \bar{\mathbf{d}} -\bar{\mathbf{F}})=0$$ for all $$\begin{Bmatrix}W_3\\W_4\end{Bmatrix}$$ (3) $$\bar{\mathbf{K}}=\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix}\qquad \bar{\mathbf{d}}=\begin{Bmatrix}d_3\\d_4\end{Bmatrix}\qquad \bar{\mathbf{F}}=\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$

$$\begin{Bmatrix}W_3=n\\W_4=n\end{Bmatrix}\cdot (\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix} -\begin{Bmatrix}F_3\\F_4\end{Bmatrix})=0$$

Because the $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}$$ term is equal to $$\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$, $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix} -\begin{Bmatrix}F_3\\F_4\end{Bmatrix}=\begin{Bmatrix}0\\0\end{Bmatrix}$$ $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}=\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$(4)

Part 4
The conversion from axial forces and element forces in global coordinates is as follows: $$\mathbf{f}^{(e)}=\begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix}$$ in global coordinates $$\mathbf{\hat{k}^{(e)}q^{(e)}=p^{(e)}}={EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{pmatrix} q_1^{(e)}\\q_2^{(e)} \end{pmatrix}= \begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$ in axial coordinates $$\mathbf{q^{(e)}=T^{(e)}d^{(e)}}$$ $$\mathbf{\hat{k}^{(e)}T^{(e)}d^{(e)}=p^{(e)}}$$ $${EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)}  & m^{(e)} & n^{(e)} \end{pmatrix}\begin{pmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)} \end{pmatrix}=\begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$ Multiplying both sides by T(e)T as completed in the 2D case, $$\mathbf{T^{(e)T}\hat{k}^{(e)}T^{(e)}d^{(e)}=T^{(e)T}p^{(e)}}$$ $$\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix} {EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix} \begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)}  & m^{(e)} & n^{(e)} \end{pmatrix} \begin{pmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)} \end{pmatrix}=\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix} \begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix} $$ and with $$\mathbf{k^{(e)}=T^{(e)T}\hat{k}^{(e)}T^{(e)}},\qquad\mathbf{f^{(e)}=T^{(e)T}p^{(e)}}$$ We have $$\mathbf{(T^{(e)T}\hat{k}^{(e)}T^{(e)})d^{(e)}=(T^{(e)T}p^{(e)})}$$ or $$\mathbf{k^{(e)}d^{(e)}=f^{(e)}}$$ $$\begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix}=\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix}\begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$

Eml4500.f08.wiki1.brannon 18:40, 7 November 2008 (UTC)

HW6
Bar Broken Up Into Elements

Single Element Function Interpolation

Assume $$u(x) \,$$ for $$x_i \le x \le x_{i+1} \,$$

Motivation for Linear Interpolation of $$u(x) \,$$
2-bar Truss

2-Bar Truss Deformation

Rube Goldberg Machine
[Rube Goldberg Machine]

Honesty, Imagination, and Ethics
[Honesty],[Imagination], and [Ethics]

[The University of Florida Honor Code] We, the members of the University of Florida community, pledge to hold ourselves and our peers to the highest standards of honesty and integrity.

The University of Florida Honor Pledge

"On my honor, I have neither given nor received unauthorized aid in doing this assignment.”

Continuous Principle of Virtual Work to Discrete Case (continued)
Lagrangian Interpolation

Motivation for Form of $$N_i(x) \,$$ and $$N_{i+1}(x) \,$$:

$$N_i(x) \,$$ and $$N_{i+1}(x) \,$$ are linear (straight lines), thus any linear combination of $$N_i(x) \,$$ and $$N_{i+1} \,$$ is also linear, and in particular, the expression for $$u(x) \,$$.

$$N_i(x)=\alpha_i + \beta_i x \,$$, with $$\alpha_i \,$$, $$\beta_i \,$$ numbers

$$N_{i+1}(x)=\alpha_{i+1}+\beta_{i+1}x \,$$, with $$\alpha_{i+1} \,$$, $$\beta_{i+1} \,$$ numbers

Linear Combination of $$N_i \,$$ and $$N_{i+1} \,$$:

$$N_id_i+N_{i+1}d_{i+1}=(\alpha_i+\beta_ix)d_i+(\alpha_{i+1}+\beta_{i+1}x)d_{i+1} \,$$ $$\quad=(\alpha_id_i+\alpha_{i+1}d_{i+1})+(\beta_id_i+\beta_{i+1}d_{i+1})x \,$$ is clearly linear.

Recall equation for $$u(x) \,$$ (interpolation of $$u(x) \,$$):

$$u(x_i)=N_i(x_i)d_i+N_{i+1}(x_i)d_{i+1} \,$$ $$u(x_i)={x_{i+1}-x_i \over x_{i+1}-x_i}d_i+{x_i-x_i \over x_{i+1}-x_i}d_{i+1} \,$$ $$N_i(x_i)=1 \,$$ $$N_{i+1}(x_i)=0 \,$$ $$u(x_i)=d_i \,$$

Using the same logic:

$$u(x_{i+1})=d_{i+1} \,$$

Dimensional Analysis
$$\left[\varepsilon \right]={[du] \over [dx]}={L \over L}=1$$

$$[ \sigma ]=[E]={F \over L^2} \,$$

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

$$\left[ {EA \over L} \right]=[\tilde{k_{11}}]={{F \over L^2}L^2 \over L}={F \over L}$$

$$[\tilde{k}_{11} \tilde{d}_1]=[\tilde{k}_{11}][\tilde{d}_1]=F$$

$$[\tilde{k}_{14} \tilde{d}_4]=[\tilde{k}_{14}][\tilde{d}_4]=\left[ {-EA \over L} \right][L]={-{F \over L^2}L^2 \over L}L=-F$$

$$[\tilde{k}_{22} \tilde{d}_2]=[\tilde{k}_{22}][\tilde{d}_2]={12EI \over L^3}L={{F \over L^2}L^4 \over L^3}L=F$$

$$[\tilde{k}_{23} \tilde{d}_3]=[\tilde{k}_{23}][\tilde{d}_3]={[6][E][I] \over [L^2]}={1 {F \over L^2}L^4 \over L^2}=F$$

$$[\tilde{k}_{25} \tilde{d}_5]=[\tilde{k}_{25}][\tilde{d}_5]=-[\tilde{k}_{22} \tilde{d}_2]=-F$$

$$[\tilde{k}_{26} \tilde{d}_6]=[\tilde{k}_{26}][\tilde{d}_6]=[\tilde{k}_{23} \tilde{d}_3]=F$$

$$[\tilde{k}_{33} \tilde{d}_3]=[\tilde{k}_{33}][\tilde{d}_3]={[4][E][I] \over [L]}={1{F \over L^2}L^4 \over L}=F*L$$

$$[\tilde{k}_{35} \tilde{d}_5]=[\tilde{k}_{35}][\tilde{d}_5]=-[\tilde{k}_{23} \tilde{d}_3]=-F$$

$$[\tilde{k}_{36} \tilde{d}_6]=[\tilde{k}_{36}][\tilde{d}_6]={[2][E][I] \over [L]}={1{F \over L^2}L^4 \over L}=F*L$$

$$[\tilde{k}_{44} \tilde{d}_4]=[\tilde{k}_{44}][\tilde{d}_4]={[E][A] \over [L]}[L]={{F \over L^2}L^2 \over L}L=F$$

$$[\tilde{k}_{55} \tilde{d}_5]=[\tilde{k}_{55}][\tilde{d}_5]=[\tilde{k}_{22} \tilde{d}_2]=F$$

$$[\tilde{k}_{56} \tilde{d}_6]=[\tilde{k}_{56}][\tilde{d}_6]=-[\tilde{k}_{23} \tilde{d}_3]=-F$$

$$[\tilde{k}_{66} \tilde{d}_6]=[\tilde{k}_{66}][\tilde{d}_6]=[\tilde{k}_{33} \tilde{d}_3]=F*L$$

Element Force-Displacement Relationship in Global Coordinates from the Force-Displacement Relationship in Local Coordinates
$$\mathbf{k}^{(e)}_{6x6} \mathbf{d}^{(e)}_{6x1}=\mathbf{f}^{(e)}_{6x1}$$

$$\mathbf{k}^{(e)}=\tilde{\mathbf{T}}^{(e)T} \tilde{\mathbf{k}}^{(e)} \tilde{\mathbf{T}}^{(e)}$$

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

$$\begin{Bmatrix} \tilde{d}_1\\ \tilde{d}_2\\ \tilde{d}_3\\ \tilde{d}_4\\ \tilde{d}_5\\ \tilde{d}_6 \end{Bmatrix}= \begin{bmatrix} l^{(e)}&m^{(e)}&0 &0& 0& 0\\ -m^{(e)} &l^{(e)} &0 &0 &0 &0\\ 0 &0 &1 &0 &0 &0\\ 0 &0 &0 &l^{(e)}& m^{(e)} &0\\ 0 &0 &0 &-m^{(e)} &l^{(e)} &0\\ 0 &0 &0 &0 &0 &1 \end{bmatrix} \begin{Bmatrix} d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{Bmatrix}$$

$$\tilde{d}_3=d_3$$

$$\tilde{d}_6=d_6$$

Derivation of $$\tilde{\mathbf{k}}^{(e)}$$ from the Principle of Virtual Work, focusing only on the bending effect.

$$-{\partial^2 \over \partial x^2}((EI){\partial^2 v \over \partial x^2}) + f_t(x)=m(x)\ddot v$$

$${\partial \over \partial x}((EA){\partial u \over \partial x}) + f_a(x,t)=m(x) \ddot u$$

Motivation: deformed shape of truss element interpolation of transverse displacement $$v(s)=v(\tilde{x})$$

The Principle of Virtual Work for Beams
$$\int_0^L W(\tilde{x})[-{\partial^2 \over \partial x^2}(EI){\partial^2 v \over \partial x^2}+f_t-m \ddot v] dx=0$$ for all W(x)
Equation (1)

Integration by Parts of the 1st term:

$$\alpha =\int_0^L W(x) {\partial^2 \over \partial x^2}((EI){\partial^2 v \over \partial x^2})dx$$

$$W(x){\partial \over \partial x}({\partial \over \partial x}((EI){\partial^2 v \over \partial x^2}))$$

$$=[W{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})]_0^L-\int_0^L {dW \over dx}{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})dx$$

$$=[W{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})]_0^L-[{dW \over dx}(EI){\partial^2 v \over \partial x^2}]_0^L+\int_0^L{d^2W \over dx^2}(EI){\partial^2 v \over \partial x^2}dx$$

Thus, Equation (1) becomes:

$$[W{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})]_0^L-[{dW \over dx}(EI){\partial^2 v \over \partial x^2}]_0^L+\int_0^L{d^2W \over dx^2}(EI){\partial^2 v \over \partial x^2}dx+\int_0^L W f_t dx-\int_0^L W m \ddot v dx =0 for all possible W(x)$$

Derivation of the Stiffness Matrix and Shape Functions for Beams
Focus on stiffness term, $$\int_0^L{d^2W \over dx^2}(EI){\partial^2 v \over \partial x^2}dx$$, to derive the beam stiffness matrix and identify shape functions for beams

$$v(\tilde{x})=N_2(\tilde{x})\tilde{d}_2+N_3(\tilde{x})\tilde{d}_3+N_5(\tilde{x})\tilde{d}_5+N_6(\tilde{x})\tilde{d}_6$$
Equation (2)

Recall:$$u(\tilde{x})=N_1(\tilde{x})\tilde{d}_1+N_4(\tilde{x})\tilde{d}_4$$
Equation (1)

$$N_2(\tilde{x})=1-{3\tilde{x}^2 \over L^2}+{2\tilde{x}^3 \over L^3}; N_3(\tilde{x})=\tilde{x}-{2\tilde{x}^2 \over L} + {\tilde{x}^3 \over L^2}; N_5(\tilde{x})={3 \tilde{x}^2 \over L^2} - {2\tilde{x}^3 \over L^3}; N_6(\tilde{x})=-{\tilde{x}^2 \over L} + {\tilde{x}^3 \over L^2}$$

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

Computation of $$u(\tilde{x})$$ and $$v(\tilde{x})$$:

$$u(\tilde{x})=u(\tilde{x})\tilde{\vec{i}}+v(\tilde{x})\tilde{\vec{j}}$$

$$=u_x(\tilde{x})\vec{i}+u_y(\tilde{x})\vec{j}$$

Computation of $$u(\tilde{x})$$ and $$v(\tilde{x})$$ using Equations (1) and (2):

$$\begin{Bmatrix}u_x(\tilde{x})\\u_y(\tilde{x})\end{Bmatrix}=\mathbf{R}^T \begin{Bmatrix}u(\tilde{x})\\v(\tilde{x})\end{Bmatrix}$$

$$\begin{Bmatrix}u_x(\tilde{x})\\u_y(\tilde{x})\end{Bmatrix}=\begin{bmatrix}N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix}\begin{Bmatrix}{\tilde{d}_1^{(e)}}\\{\tilde{d}_2^{(e)}}\\{\tilde{d}_3^{(e)}}\\{\tilde{d}_4^{(e)}}\\{\tilde{d}_5^{(e)}}\\{\tilde{d}_6^{(e)}}\end{Bmatrix}$$

$$\mathbb{N}(\tilde{x})=\begin{bmatrix}N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix}$$

\begin{Bmatrix}u_x(\tilde{x})\\u_y(\tilde{x})\end{Bmatrix}=\mathbf{R}^T\mathbb{N}(\tilde{x})\tilde{\mathbf{T}}^{(e)}\mathbf{d}^{(e)}

title: User:Eml4500.f08.wiki1.brannon