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Microsoft Office/Word

Microsoft Word is a word processor created by Microsoft. Basically, it allows you to create eye-catching text and documents. Today, these documents are generally saved as a docx file, although they can be saved in other formats.

Create and manage documents

 * /Create a document/
 * /Navigate through a document/
 * /Format a document/
 * /Customize options and views for documents/
 * /Configure documents to print or save/
 * Use a template
 * Creating templates in Word 2010
 * Creating templates in Word 2019

Page breaks and section breaks
Scenario: How to create a single landscape page within a portrait-style document.

Answer: Put section breaks above and below the content and whilst on the target content go to page setup and change to landscape format.

Shortcuts
External link: Keyboard shortcuts for Word
 * Ctrl+A: Select entire document
 * Ctrl+B: Toggle Bold
 * Ctrl+C: Copy
 * Ctrl+D: Open font dialogue
 * Ctrl+E: Centralize Selection
 * Ctrl+F: Find in the open document
 * Ctrl+G: Open the 'Go To' dialogue box
 * Ctrl+H: Open the 'Find and Replace' dialogue box
 * Ctrl+I: Toggle Italic
 * Ctrl+J: Justify selection
 * Ctrl+K: Insert hyperlink
 * Ctrl+L: Align Left
 * Ctrl+M: Indent
 * Ctrl+N: Open New document
 * Ctrl+O: Open file
 * Ctrl+P: Print document
 * Ctrl+R: Align Right
 * Ctrl+S: Save file
 * Ctrl+U: Toggle Underlined
 * Ctrl+V: Paste
 * Ctrl+W: Close
 * Ctrl+X: Cut
 * Ctrl+Y: Redo last action
 * Ctrl+Z: Undo last action
 * Ctrl+8 (alphanumeric pad): Lower font size
 * Ctrl+9 (alphanumeric pad): Raise font size
 * Ctrl+=: Make subscript
 * Ctrl+Shift-+: Make superscript
 * Ctrl+Shift+A: Turn all caps
 * Ctrl+Shift+B: Switch to Symbol font
 * Ctrl+Shift+
 * Ctrl+Shift+Q: Turn into small caps
 * Ctrl-Alt-M: Add Comment
 * Alt+X: Convert to/from Unicode codepoint
 * Shift+F3: Cycle casing (ALL CAPS/all lowercase/Every First Letter Capitalized)
 * Ctrl+>: Raise font size
 * Ctrl+<: Lower font size
 * Ctrl+Shift+C: Format Painter (Copy)
 * Ctrl+Shift+V: Format Painter (Paste)
 * Ctrl+Space: Removes all manual character formatting from a selection

Styles

 * Understanding styles in Microsoft Word

Formats
Formats are one of the most important features in Microsoft Word. This is because they let you chose what size, colour, font, or weather your text is bold, in italics, or underlined. (See figure 4.1)

Format painter copies the format of text, not the actual text. (See figure 4.2 - 4.4)

Table of contents

 * How to create a table of contents in Microsoft Word (Shauna Kelly)

title: Microsoft Office/Word


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One-way ANOVA

Normality

 * Kruskal-Wallis one-way analysis of variance

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

title: One-way ANOVA


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One-way repeated measures ANOVA

A one-way repeated measures ANOVA can be understood as an extension of a paired-samples t-test to situations involving three or more related measures.

Non-parametric

 * Friedman test
 * See also Wilcoxon signed-rank test for pairwise follow-up tests.

title: One-way repeated measures ANOVA


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Kurtosis

Kurtosis (from the Greek word κυρτός, kyrtos or kurtos, meaning bulging) is a measure of the "tailedness" of the distribution of sample data. Positive kurtosis (leptokurtic) indicates a distribution more outlier-prone than predicted by a normal distribution, and negative kurtosis (platykurtic) indicates a distribution less outlier-prone than a normal distribution. 0 would indicate that the outlier-producing potential is similar to that of a normal distribution.

title: Kurtosis


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Skewness

 refers to asymmetry (or "tapering") in the distribution of sample data:
 * 1) negative skew: The left tail is longer; the mass of the distribution is concentrated on the right of the figure. It has a few relatively high values. The distribution is said to be left-skewed. In such a distribution, usually (but not always) the mean is lower than the median, and the median is lower than the mode; in which case the skewness is lower than zero.
 * 2) positive skew: The right tail is longer; the mass of the distribution is concentrated on the left of the figure. It has a few relatively low values. The distribution is said to be right-skewed. In such a distribution, usually (but not always) the mean is greater than the median, or equivalently, the mean is greater than the mode; in which case the skewness is greater than zero.

The relationship between mean and median is not sufficient to make a statement about the skewness of the distribution, but it can serve as an indicator.

In a skewed (unbalanced, lopsided) distribution, the mean is farther out in the long tail than is the median. If there is no skewness or the distribution is symmetric like the bell-shaped normal curve then the mean = median = mode. Usually there are three ways in which a set of data can be analyzed for its distribution: 1.By the range: This gives a rough idea of the spread of data but it is affected by extreme values. It is generally only used with small data groups together with either median or mode. 2.The interquartile range(IQR): It is not affected by extreme values and tells you how spread out the middle 50% of the observations are. The IQR is often used together with the median when data are skewed. 3.The mean with standard deviation: It is generally used when the data are symmetrical and the data size is not small.

title: Skewness


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Theoretical psychology

Theoretical psychology is concerned with theoretical and philosophical aspects of the discipline of psychology. It is an interdisciplinary field involving psychologists specialising in, amongst others, cognitive, social, developmental, personality, clinical, perceptual, neurological, biological, evolutionary, historical, political and critical psychology. In the United States, one group focused on theoretical and philosophical issues in Psychology is The Society for Theoretical and Philosophical Psychology (formerly known as The American Psychological Association, Division for Theory and Philosophy of Psychology or APA Division 24). This society regularly hosts symposia, lectures, roundtable discussion, and poster sessions at the annual APA conference.

title: Theoretical psychology


text: File:Fea schematic.jpg

title: File:Fea schematic.jpg


text: File:K concept.jpg

title: File:K concept.jpg


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User:Eml4500.f08.delta 6.lopez/Homework 2: Lecture Notes (Contribution)

 To custodian: This team appeared to have blanked out their wiki page. Please give them some time before deleting this page in case they need to recover anything from this page. Thank you. Eml4500.f08 09:28, 25 September 2008 (UTC)


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User talk:Mu301/Archive 2008

featured content
I just wanted to make sure that you know that Observational astronomy is featured content at Portal:Physical Sciences. Please make sure that the description there is coherent. --JWSchmidt 15:43, 26 January 2007 (UTC)

You could be more helpful if...
...you were a Custodian :). I'd be happy to serve as your mentor if you like: seems you're doing a lot of RC patrolling these days, and the tools really do come in handy from time to time. --SB_Johnny | talk 19:45, 1 January 2008 (UTC)
 * I agree. --McCormack 05:31, 7 January 2008 (UTC)

Deletions and clean-up
I think it's great you are looking at all these weird pages and marking them for deletion. As part of the deletion process, we need to check what links to these pages and tidy up the links. The links need tidying first, because once the page is deleted, you can't use the "what links here" feature any more. I'm happy to help you with this if you like, but I'm waiting for you to get probationary status - I don't want to take away your chance of practising with those tools! --McCormack 05:31, 7 January 2008 (UTC)

Custodianship
I've promoted you to probationary custodian. Good luck with tools. sebmol ? 12:51, 7 January 2008 (UTC)

New Courses?
I wanted to know where I could get information on designing a course on wikiversity, I am a consumer-student-activist who is working on my doctoral thesis in clinical psychology; my rather radical endeavor is to create an alternative behavior-social science called Recovery Psychology; the syllabus for the course that I want to place on wikiversity would be: although mostly just links to other sites and internet resources that I would say constitute a textbook or course materials with explanations of how it relates to the subject at hand...for example an outline like this:

 * What is mental illness? [Sociological]
 * 1) What is social deviance?
 * 2) http://www.geocities.com/tdeddins/deviance.htm
 * 3) http://en.wikipedia.org/wiki/Deviant_behavior
 * 4) http://www.umsl.edu/~keelr/200/intrdev.html
 * 5) What is conformity?
 * 6) 	Stanley Milgram Obedience to Authority about how easy the human mind bends to the will of perceived authority
 * 7) 	Solomon Asch about how people conform to group pressure and change their perceptions and behaviors
 * 8) 	Philip Zimbardo The Stanford Prison about how people fall in to assigned roles
 * 9) 	Jane Elliot: In the Eye of the Storm or A Class Divided about how people are quick to change their perception of others
 * 10) 	See Rosenhan below for a psychosocial lesson on the nature of psychiatric conditions
 * 11) 	Learned Helplessness
 * What is mental illness [Psychological]
 * 1) http://www.mental-health-matters.com/articles/article.php?artID=160
 * Etc, Etc, Etc...

Would that be allowed? Since the subject does not exist as a subject, although my arguementum would say it does exist as a thing with in itself although not yet releized by academics. Of course I would place a disclaimer statement that this an academic pursuit of an activist, but pretty much all the things I point at will be "considered to be authorities" I understand that your subject in astronomy not psychology, but no less you are a qualified academic for advisement (particularly here) --Recovery Psychology 03:03, 11 January 2008 (UTC) I had some of my work tagged as spam by a bot, do you think I should delete it?--Recovery Psychology 00:17, 21 January 2008 (UTC)

Machine sound
I am having better results in the Wikimedia media player with this new file: Image:Punchcardsound2.ogg

Venus and Mercury
If you are the one who created the pages Venus and Mercury, I added a ton more information to the pages. --km 21:38, 15 January 2008 (UTC)

Meteorology
hi i am an very keen on meteorology and weather forecasting...i have no proffessional degree nor professional knowledge about it; so i am just going to follow lessons...but i have a little bit experience about forecasting using models like gfs... i can share the web sites and some basic details of wether models.

thanks for the current lessons and waiting for other lessons(including forecasting meteorology)...

Image licensing
Hi, Mike. I saw you removed the public domain licence info on an image as a copyright violation. But now we're left with an image which has no licensing at all. So I'm wondering what the next step is ;-) (Delete; contact uploader, etc...)  --McCormack 07:00, 16 January 2008 (UTC)

Hi Mike. Aah - your secret weapon is the User:Mu301/todo page! I felt you were up to something that had a system to it somehow, but from the edits I could see, it looked like the "unlicenced images" were just going to vanish into wiki-oblivion. Great stuff, Mike, and thank you for your excellent efforts to make WV a better place. --McCormack 07:02, 17 January 2008 (UTC)

Observational astronomy lessons
I can't do it… I don't have the digital camera and a telescope. --km 01:41, 17 January 2008 (UTC)

Liquid water on Europa
I could do it if I have the chance… --km 01:42, 17 January 2008 (UTC)

Test edit
User:Daanschr says he can't post to your page. This is a test edit to see if I can. --McCormack 12:50, 17 January 2008 (UTC)

Commons images
Hi Mike. The images from your to-do list that are already on commons should just be deleted... you should probably also resize on the page any that were uploaded for that reason (I'm still trying to find a help file on image sizing, but may end up just having to import one). I'll pick off a few if you don't have time. --SB_Johnny | talk 10:51, 18 January 2008 (UTC)

 * Oh, here it is: Help:Media Files. --SB_Johnny | talk 10:59, 18 January 2008 (UTC)

Sandbox Server
Hi mikeu,

The sandbox server now has shell accounts. I'll setup a subdomain for you -- would mu301.sandboxserver.org do? Drop me a line at draicone at gmail dot com with some basic details of what you intend to do on the server, technical-wise (e.g. setup some PHP scripts, build a bot to upload captured images via SCP etc.).

Also let me know what password you want and what username you were hoping for. Finally, put a unique subject line in and some way for me to identify that it's you; a blank edit to your user page with that subject line will do, I just don't want to start randomly handing out shell accounts.

You'll get your own subdomain running Apache loaded with PHP, Perl and (I think) Ruby, maybe even Python (don't ask me, I don't run the server, I just stand there, look pretty and pay the bills). If you'd like to build a Jabber bot, I can hook you up with an account on the jabber server as well. I haven't really gone through your project much, but if you have any special technical requirements let me know and I'll see what I can do.

Draicone (talk) 11:46, 18 January 2008 (UTC)

Europa
I edited the liquid water on Europa page. Thanks for the advise ☺. --km 02:41, 23 January 2008 (UTC)

Introduction to computers
I finally got your notes, sorry it took me so long to notice. I am teaching a class, and I am trying to get them to check and document the copyright state of materials. I will pint them towards Commons. Andrew 04:40, 29 January 2008 (UTC)

Astrophysics
Sorry, I've been caught up with offline work that I had forgotten to check these pages. I'm hoping to check back periodically. I agree with you about not separating Astronomy into both a School and a Topic. I belive it should be under the School of Physics (joint school). However, I still maintain that Astronomy and Astrophysics should probably be in the same topic page.

I agree that many people may be interested in Astronomy but I'm not sure if wikiversity is supposed to have the scope to cater for high school level and university level. Although, given that astronomy isn't taught at any high school (that I know of), I can see good reason for starting with very basic lessons.

Too many sub-topics on the current Astronomy topic page are 1) redundant and 2) heavily physics based. Could we just merge Astronomy with Astrophysics, kill the redundant place markers (courses/topics) and include a set of pre-university topics/lessons? Well, I know we could but I'm wondering if you or anyone else would agree with that line of logic. :-) Augustus 00:59, 31 January 2008 (UTC)

Living on Earth
Hi! Can you please drop me a note, when the story about Ladd Obs. will be made and available online? I want to hear it, but as I know myself, I forget to check for it at radio's website. :) Thx! --Gbaor 06:32, 31 January 2008 (UTC)

Wikimedian Demographics update
Hello Mu301. I'm writing because you requested a reminder about progress on Wikimedian Demographics. For the time being these will be about once per month.

Speaking of reminders, it turns out the template question isn't the most efficient way of coming up with a mailing list. If you want to recieve this message on another of your talk pages (or don't want them at all), please edit the list on Wikimedian Demographics/Reminders.

To update on progress so far, there have been somewhere around 50 respondents over the past month, with more trickling in slowly but steadily. There are 7 major surveys (or 12, depending on how you count them):
 * 1) The basic survey (about your participation in Wikimedia)
 * 2) The "geography" surveys (actually 6 of those, for each inhabited continent)
 * 3) The Operating System survey (still stubby)
 * 4) The Wikipedia survey
 * 5) The Commons survey
 * 6) The US presidential elections survey
 * 7) The sexuality and relationships survey

Only the geography surveys have had any major modifications since being originally posted (more countries were added to those).

Several new surveys have been discussed at Talk:Wikimedian Demographics, as well as other places within Wikiversity. A few that might be ready sometime in February include:
 * A survey about Women in Wikimedia
 * A survey about Wikibooks
 * A survey about Wikiversity
 * A survey about administrative issues (including cross-wiki topics such as CheckUser and "global blocking")
 * A survey about environmental issues
 * A survey about religion
 * A survey about international trade

If you have ideas for more surveys, please share them on Talk:Wikimedian Demographics. Thanks for your participation thus far, and we hope to see you soon! --SB_Johnny | talk 11:59, 5 February 2008 (UTC)

us currency
I don't know why, but I felt like adding pages about the currency of the United States of America. Take a look @ it.

 * 1¢
 * 5¢
 * 10¢
 * 25¢
 * 50¢
 * 100¢
 * $1 bill
 * $2 bill
 * $5 bill
 * $10 bill
 * $20 bill
 * $50 bill
 * $100 bill

Since I am a seventh grader, I go to school five times a week and another school on Saturday's so I have a lot of homework everyday. I don't have time for anything unless I don't have the homework. So that means I wouldn't be able to be doing the experiments on the Observational astronomy page. All I can do for now is to read the page.

-km 12:10, 10 February 2008 (EST)

Wikiversity:Candidates for Custodianship/Mu301 (full custodian)|custodianship
custodianshipThe 5 days period is over now. Did one of the b'cats contact you already about making you full custodian ? Erkan Yilmaz Wikiversity:Chat  18:44, 23 February 2008 (UTC)

Welcome to the Bloom Clock!
Bloom Clock Hello, Mu301, and thanks for signing up as a contributor to the Bloom Clock!

To get started, all that you need to do is keep your eyes open for flowers on plants growing outdoors. Then, depending on how well you know plants, you can either try to find the plant using the global keys, or just check the master list where plants are listed by their scientific names.

Assuming you find a log page for the plant you saw, all that you need to do is click [edit] above the list of signatures, and add *~ on a new line below the last signature and above the line that begins with, which will make your signature appear on the profile page.

If you don't find a log page, just add an entry for the plant to the Master List using. Someone can then show you how to create a profile and log page for the plant.

If you don't know the name of the plant you saw, try asking on Bloom Clock/Unknown Plants using the template there. Photographs are very helpful, but if you don't have a digital camera, just give the best description you can and hopefully someone can identify it for you.

After you've logged 10 or more plants as flowering for your location, another bloom clock contributor will help you set up the categories and templates used to construct a key for your region, and try to match up your region to the global keys.

If you have any questions, ideas, or need something explained, please feel free to leave a message on my talk page, and I'll get back to you as soon as I can!

(Apologies for the long-belated welcome message... finally got around to writing one!) --SB_Johnny | talk 09:26, 13 March 2008 (UTC)

Re:Teaching with Technology
Thanks very much for these links, Mike! I had a very quick look (more deadlines to catch up with) at some of the work of Mark Tribe - some very interesting stuff there. I'd like to talk with him - perhaps a group voice chat session might be in order? Also, I taught a session a few weeks ago on wikis in education, in which I wanted to give an up-close look at Wikipedia, the result of which is at Inside Wikipedia (my slides are on LeMill for the moment). There are some readings in that page that might be relevant to your upcoming sessions - including a nice article on using Wikipedia to develop students' media literacies. In fact, this is something we could easily develop a pretty good course on... Cormaggio talk 13:51, 14 March 2008 (UTC)

Re:
I've removed the deletion tag, however though is there a chance if you could tell me what I need to improve the content, or change. Terra Welcome to my talkpage 07:53, 17 March 2008 (UTC)

Congratulations
Congratulations on being successfully elected "bureaucrat". --McCormack 08:28, 22 March 2008 (UTC)

Username Change
Hi, Is it possible for you to change my username from User:Reedy Boy to User:Reedy, as i am having it changed on the english wikipedia, and want to get the rest of my accounts inline for the single user login

Thanks

Reedy Boy 14:03, 23 March 2008 (UTC)
 * Hello, if you put your request at: Changing username also other bureaucrats see this faster. Erkan Yilmaz uses the Wikiversity:Chat (try) 14:25, 23 March 2008 (UTC)
 * Done, thanks - Changing_username Reedy Boy 14:41, 23 March 2008 (UTC)

 * Rename is done. --mikeu talk 16:46, 23 March 2008 (UTC)
 * Thanks! Reedy 17:20, 23 March 2008 (UTC)

Licensing
If you'd like some input regarding Wikiversity talk:Licensing policy, I'm probably not the person you want to do it as I'm not intimately familiar with Wikiversity. As for the bot, I can probably be more helpful there. The state of your images (as of when my bot last finished operating) is that the untagged ones are categorized in Category:Images with unknown copyright status, and all uploaders have been notified. If there's something more you'd like done to them, just let me know. Best of luck. – Mike.lifeguard | @en.wb 02:44, 1 April 2008 (UTC)

Image:Test file.gif
This is tagged for deletion as Copyrighted material, in the edit summery you've stated that it shouldn't be deleted, as with copyright and other copyright status all images must provide the copyright status so that we know whether it is allowed on this site, if it isn't copyrighted please say so and use the correct tag. Terra 18:57, 2 April 2008 (UTC)

How to use R/Tutorials/Introduction
How to use R/Tutorials/Introduction Looks like I made a bit of mess in terms of losing the trail with that import, sorry. I was semi-conscious at the time of leaving a gap in the editing history, but not quite sure what to do. The content has subsequently been split into different tutorials (loading and plotting), so maybe on this occasion we can live with it. Had I thought through it more, it probably should have been imported to How to use R/Tutorials rather than How to use R/Tutorials/Introduction, then the sub-directory structure would lead to the content. Not sure if its possible to somehow redirect the import for wikibooks history to How to use R/Tutorials. But thankyou, because this learning experience will definitely help for next time. If you have any other thoughts on how this could/should have been done, please let me know. -- Jtneill - Talk 23:57, 11 April 2008 (UTC)

 * Thanks, Mike. I've added a note to explain at the top of each of the subpages which can be shifted to the talk page once the content evolves. Yes, WBs wanted it left there, so they might expand. -- Jtneill - Talk 00:27, 12 April 2008 (UTC)

Copyvios in Design for the Environment course
Hi Mike, I'm going through the many copyvios in the Design for the Environment course, and I noticed you added and then a copyvio flag on one image - how come? Cormaggio talk 16:32, 14 April 2008 (UTC)
 * Yes, thanks - I thought as much. :-) I've created Template:Db-unknowncopyvio - do you think it's redundant with Template:No license?. Cormaggio talk 10:35, 15 April 2008 (UTC)
 * Right - we might do well to draft a page for face to face groups intending to use Wikiversity as a workspace - with copyright issues at the top in bold, blinking text. (Well, ok, we don't need blinking text.. :-)) Cormaggio talk 12:45, 15 April 2008 (UTC)

Thanks for help with Design for the Environment
Mike, I appreciate your help and that of other custodians with the labeling of the reports with the DFE category. It's the first year that I'm doing this and hope that with time and experience everything will go better. --SRego 18:52, 14 April 2008 (UTC)

Flowers?
Hi Mike,

Just wondering if you had blooms to log, and if you plan to keep logging generally. I ask because I'm going to have Mike.lifeguard do a template update within the next day or two, and need to know whether we need an entry field for Rhode Island :). --SB_Johnny | talk 15:58, 21 April 2008 (UTC)

I thought of you today :-)
Was wondering if you saw this show ? Erkan Yilmaz uses the Wikiversity:Chat (try) PS: Tag a learning project with completion status !! 22:46, 1 June 2008 (UTC)

Custodial flag for Remi
ping, Erkan Yilmaz uses the Chat (try) 16:50, 23 June 2008 (UTC)

subst
Hmm...I subst'ed the welcome template, but it had the unintended side effect of crediting me for the initial welcome. Maybe go back and sign your name over mine? Cheers, --HappyCamper 16:13, 7 September 2008 (UTC)

Welcome
Hi Mike, thanks for joining the Learning from conflict and incivility project - looking forward to reading more on your narrative. Cormaggio talk 07:41, 12 September 2008 (UTC)

A random mascot
 Hello , we are the Baah and Bahh twins and we'd both like to welcome you to Wikiversity. While you sit back and explore Wikiversity, enjoy some of our fresh goat milk. You can help us learn by creating new learning resources or help us by improving the guided tours and other exploration tools. We look forward to seeing you around. Baaaahbye for now.

-- Jtneill - Talk - c 07:29, 13 September 2008 (UTC)

CheckUser access
Thank you for the heads up. Emesee 01:00, 22 September 2008 (UTC)

Thank you
. . . for the welcome! :) Xdenizen 01:12, 29 September 2008 (UTC)

CheckUser on Vandal?
Mike, have you been able to run (or request) a CheckUser on this morning's vandal?

Moulton 15:35, 3 October 2008 (UTC)

 * blocked indef and I believe that all the diffs have been rolled back or reverted. We currently have no local CU, so I will contact SBJ on how to proceed.  --mikeu talk 15:45, 3 October 2008 (UTC)

 * You might want to ask Alison if she thinks this is Grawp, one of the more notorious Uber-Trolls who has been marauding WMF-sponsored projects. —Moulton  16:51, 3 October 2008 (UTC)

American History
Can you delete Topic:History of the Americas and move Topic:American History? Geo.plrd 18:08, 8 October 2008 (UTC)

At commons now
That Florida class is showing up at commons:Commons:Administrators%27_noticeboard now, any help would be appreciated. MBisanz 00:51, 10 October 2008 (UTC)

inappropriate page locks
There's this one, too:   The Jade Knight (d'viser) 09:14, 13 October 2008 (UTC)

Sockpuppets
There's this one, too:   The Jade Knight (d'viser) 06:46, 19 October 2008 (UTC)

Your revert
I disagree with your revert:. It is obvious that the added words do not address consensus specifically and only confuse what consensus is not by not being easily precise, which it can be. I rather you discuss such change more than to revert like you did. Dzonatas

 * The section is titled "What consensus isn't" and not "What consensus isn't and may be what it is," which obviously diverts from what it simply is not. That kind of 'diversion' is 'amusing', synonymously. Dzonatas 17:42, 25 October 2008 (UTC)

 * Acknowledge that you don't understand the intention to keep it simple and move everything else that quickly falls out of scope to the amusing section. Maybe it shouldn't be called the amusing section, but you could have made a more productive change than reverting me because you don't understand. Dzonatas 18:37, 25 October 2008 (UTC)

 * You made a lot of criticism towards me:, which was not justified at all. You stated you did not understand, so you could have asked me on my talk page to clarify. I have no problem to admit a mistake if something was not understood and taken the wrong way. You didn't give any chance for that before you reverted. It takes discussion to understand that first. If you still think I have some intent to be uncivil especially without any of your prior attempt at discussion, then may I suggest you take a wiki break. Dzonatas

Poor job
The way you worded your comments took me off guard, as I felt there was no need to use words like "you did a poor job," "you should...," and "border on uncivil." Today I read, "You should take your time and make edit summaries and edits that a reasonable person can understand without confusion." I felt you didn't criticize edit there at all. I hope you didn't mean to make me feel stupid over that comment. A reasonable person would have first talked about it to understand and suggest alternative. You expressed you don't understand. I tried to talk here with you, but you just tell me "you should." Dzonatas 18:11, 26 October 2008 (UTC)

A Unified Account
A unified account would really be nice! I intend to spend a good deal of time on WB getting that book right. Please let me know what information you might need to do this (if any).

Thanks again, for all your help, you've been very helpful.--Jolie 13:05, 27 October 2008 (UTC)

 * what I minute, I think I'm talking to the wrong 'mike'. Its mike.lifeguard I think that can get my setup with a unified account.  my bad. sorry.--Jolie 19:57, 28 October 2008 (UTC)

SUL
Don't worry, all accounts are using the same password and email address. I just decided to wait a while for the bugs to come out. Right now I am working to make sure that I have all accounts renamed. Geo.plrd 15:23, 28 October 2008 (UTC)

Discussion on Mike.lifeguard's candidacy
Mu301, I have reverted your closure on the community review on Mike.lifeguard's candidacy, which is very premature, your comment "with near unanimous support" hides the fact that the discussion is clearly unfinished, the last comment being written even in less than 24 hours, with questions not yet answered. You can expect more wikiversians to voice their supports or oppositions as the discussion moves on. As has been said not so long ago, requests for custodial tools are not votes. Hillgentleman | //\\ |Talk 20:59, 16 November 2008 (UTC)

thanks for the astronomical advice, BTW I've done some astronomy too
thank you for your helpful entry on writing to our justification

'mayans forecasting the end of the world based on studying asteroids' is part of the sometimes ridicilous content of the WB book "US History"

by the way, I'm into amateur astronomy, as well. I have a 10" dob and we explore the relatively dark skies of NW Pennsylvania for the deepspace objects; galaxies, nebulas, clusters. I've also become rather interested in some lunar work and have been known to pick through the craters for the interesting stuff; rimae, rupes, and the like.

Is there observational astronomy content of wikiversity; have you ever done a course or a project?

I've spent a good bit of the summer fishing- but the season is passing. I will be ramping up for my favorite yearly ritual; the messier marathon (which like a rainy parade has been ruined year after year; by weather, work, you name it). I like the think tank idea and will lurk around waiting for a good opportunity to contribute.--Jolie 14:25, 23 October 2008 (UTC)

 * No imaging is a little beyond me. I am a pretty manual guy.  thought about a barn-door mount before.  didn't pursue it.  Observations is free and film and mistakes with film are not.  I'll look into your links.  I've played around with variable stars, before. I've also thought it might be cool to look into supernova.  I check out comets when I get the chance.  Their interesting objects and there seems to normally be a few around viewable with a 10" mirror.  We have two right now in the north, correct?  I've also done a little variable star work.  fine discrimination of reference stars is HARD!

 * I've been at times and places a sucker for challenges and have chased some of the tough stuff. stephens quartet is a favorite target.  the crescent nebula in the summer, some of the tougher planetaries, a few palomar globular (though most are beyond me).  the coma cluster and the one in perseus.  there's a long list.

 * Part of what hurt the excitement a little was moving to town. I went from 6+ skies to 5, with a little drive for good skies that still seems far with a telescope in the backseat.  But I'm definitely open for ideas.  I think a course that better explains things up there would be great!
 * --Jolie 15:25, 23 October 2008 (UTC)

 * I've added some content; the sky is awash in comets right now!

todays sky was that beautiful deep blue that has you anticipating some observing. Awesome! we should do so stuff together you and I. Do you like to oberve comets, there's some real neat ones around 17P is back and scwassman is in the late sky with the possiblity of outbursts.--Jolie 16:47, 23 October 2008 (UTC)

 * Hurray! I've accomplished one of my big achievements; to record a globular (or two) in a careful sketch of the Andromeda galaxy.  I've written a page about it here; Globulars_in_M31

The two comets near casseopia are tough under smalltown skies (5th magnitude)! nevertheless I nabbed them and will post observations soon.--Jolie 13:45, 24 October 2008 (UTC)

Supernovae
I have stumbled into your supernovae page and I am a little confused; what would one do? do you have a reguarly imaging routine of galaxies. Is it possible to get raw images of current supernovae, make photometric measurements than report the values back? I had heard someone say the manipulating images are actually quite time consuming for many amateurs but I'm not sure I understand how I would make perform that analysis. is photometric software availabe as freeware??

--Jolie 17:48, 23 October 2008 (UTC)
 * hold up, I see some sort of procedure using aladdin in another page. I'm typing faster than I'm searching.

you know If I like doing this, I'm going to want more images :) --Jolie 18:01, 23 October 2008 (UTC)your jan 5 photo of ngc 105 loaded. I tried 'matching stars' and a calibration pane was created.  How do I find out what the value of the marked supernova is??

 * Its time consuming and a little confusing. I'm uncertain whether technique 1 or 2 is for photometric measurements of a local image.--Jolie 18:53, 23 October 2008 (UTC)

 * Yep I'm doing something wrong. Aladdin has got the 2mass and dss image for Ngc 105. I have

Requested course;

Astronomical imaging analysis and reduction (including utilizing online information) I have downloaded aladin. Astronomical calibration is definitely beyond me. --Jolie 21:44, 23 October 2008 (UTC)

 * I love it! Have you performed the calibration and are you planning to report it?
 * I've been remodeling the backyard astronomy page and thinking about new horizons in astronomy; I've been feeling that I need to give variable star work a second chance.--Jolie 12:36, 28 October 2008 (UTC)

Backyard Astronomy
Lots of changes in backyard astronomy. I thought it wasn't very good advice for someone interested in doing some astronomy. Intend it to progress from initial experiences, too equiptment advice, to colloration and networking with other amateurs (such as you and I). Drop by and tell me what you think. As in the case of all my writing, I'm not as suncinct as I would like to be. hopefully over time, I'll improve much of my wording.--Jolie 20:00, 28 October 2008 (UTC)
 * aha, so you've undoubtably seen my fitful attempt to improve 'backyard astronomy'. Well, Hopefully I will have time to press on with it.  the last few days haven't seen much time to do anything.
 * But now what shall we do with 'stargazing'?? I'm glossing over with synonyms; when we 'stargaze' we are likely doing backyard astronomy (particularly if we are using a telescope).  While stargazing does seem to be an introductory peice on many different observing experiences outdoors.  it seems to short to do any of its subject justice and too indirect, to help a new or aspiring observer.
 * My efforts at backyard astronomy was to give (what I think) is good advice for beginning and aspiring observers. but somepeople might charge that my peice about 'backyard astronomy' omits that the topic backyard astronomy is about telescopic observation, yet my page advises the new observer to NOT buy a telescope.
 * This is from personal experience. People ruin the hobby, IMHO, by poorly purchased, junk scopes before they know what they are doing.  People are particularly fond of ruining the hobby for kids whom might not have the patience and temperiment for it at all.  the kids would be better visiting a planetarium or science museum.
 * It would be nice to organize the astronomy resources of this website, generally. I like to work on making sure all our resources are 'connected'.  --Jolie 20:12, 4 November 2008 (UTC)

Images
really like M15c2. M15 is the one with the planetary, pease1 in it, Isn't it? Your exposure is just about perfect here! Last night was the first Lake effect snow storm of the year. 'Lake effect' is a almost untolerable curse to observing in NW Pennsylvania.--Jolie 11:54, 29 October 2008 (UTC)

Chemistry through astronomy
Following up on yesterday, I talked to my wife about it (and my mother-in-law), and they both really liked the idea, both saying chemistry was a really hard subject to stay interested in, but astrophysics is much more interesting. Like I said, hydrogen is probably the hardest one, but it's simple enough to explain and easy to remember in any case (atomic number 1, valence 1).

I got the idea from this show: http://www.astronomycast.com/astronomy/ep-107-nucleosynthesis-elements-from-stars/

I guess the idea would be to go through how each element is made in the star (avoiding too much of the actual math part of the physics), and then explain why the element behaves as it does according to how it was made and it's structure. A one element per week course would be nice to give people an "intimate feel" for each element, and start picking up the patterns of how they form compound molecules. The basic story could be used for a Wikijunior book, while the wikiversity project could focus more on a question-and-answer approach (updating the wikijunior book when a question/answer seems like it might make a young person even more excited about the element).

One thing I was a bit unsure about (from the show) was how neutrons become protons (or is it the other way around?), but that seems to be a central part of it. There's another ephisode I'm downloading now that's about that.

Thanks again for recommending the podcast... I'm hooked! :-). --SB_Johnny talk 12:36, 25 October 2008 (UTC)

Oh by the way, let me know if you can think of a better name for this... I'd like to develop it through the think tank. --SB_Johnny talk 13:36, 25 October 2008 (UTC)

I'll be interested to see what comes out of this. Nuclear chemistry isn't my specialty but as a chemist and an amateur astronomy, I have a little background.--Jolie 12:39, 28 October 2008 (UTC)

Image license
Hi Mikeu! I am tiding up the image ns and I have found some of your pictures, such as this: Image:06rsMU C200701022326.jpg. Could you double licence them GFDLxcc (e.g. CC-SA-BY 3.0)? GFDL is not a free licence for media.--Juan 00:01, 28 October 2008 (UTC)

Watching with interest...
Hiya Mike. I've been watching those projects with interest (the NEO one too), but still busy getting things done before the freeze. As far as what I'd like to learn, it's really more about what I can tell 'lil SB about what we see in the sky in the evening or early morning, so mostly planets in relation to the moon.

BTW, the computer I had been using for IRC is in the shop (RAM chip blown). Taking the opportunity (excuse) to live IRC-free for a while :-). --SB_Johnny talk 22:53, 5 November 2008 (UTC)

Busy organizing
Here's what I was thinking. Could you enter some of your information into internet astronomy? I can see that in the past in an effort to drum up projects and activity that you and others looked to astronomical projects that could be entirely done on the internet.

I'm going to generally struggle to define much around imaging/ CCD work, etc. yet this has always been a little more interest here. talk to you later.--Jolie 16:56, 6 November 2008 (UTC)

 * I've been working on some beggining skygazing content. I intend to make a beginners astronomy course out of it. the geist of it, is if, by taking some of my advice- a beginner can learn the constellations and become interested in astronomy.

 * I've watched a lot of beginners waste money on cheap scopes ;)--JoliePA 20:55, 13 November 2008 (UTC)
 * This is now developed enough for an adventerous wikiversity participant to attempt Initial_experiences. feel free to tidy things up, I now things are a little roughly said.--JoliePA 20:48, 14 November 2008 (UTC)

Happy holidays
Have a happy holiday and take care. I hope everything goes well. :) Ottava Rima (talk) 06:14, 25 December 2008 (UTC)

import
Thank you i have already done so. The talk page import is unnecessary70.24.79.253 14:43, 30 December 2008 (UTC)

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User:Eas4200c.f08.blue.a/Lecture 11

How Ahead We Have Been
While the pace of the course seems to be slow, we have actually been covering material several chapters ahead of where we are currently at. An example is found when comparing problems 1.1 and 3.5. As it turns out, problem 1.1 is just a specific case of problem 3.5.

For the first case, the box, $$tau = T/2abt$$. In the second case, T = 2qA, where A is the average area and $$q = tau*t$$. This is a case where the first example, the box is a specific case of the second figure. The box problem is solved using the ad-hoc method and the generic shape is solved using a method based on the elasticity of the object.

Getting back to the airplane stringer problem, the question was posed last time as to why the walls of the stringer are sloped outward rather than being parallel as shown below. Also the question of why an open cross section is used rather than a closed cross section.

First off, there are 2 reasons why an open cross section is used. 1) Manufacturing - sheet metal is used to make the stringer. It is much easier to stamp an open cross section than to extrude a closed cross section. 2) Airplane construction - stringers are rivited to the outside skin of the plane and it is next to impossible to rivit a closed cross section. To use a closed cross section, spot wels would be needed.

Next, why are the walls of the stringer not parallel. It was suggested that the reason why they are not parallel is for storage purposes. Stringers can be stacked vertically when being stored prior to construction which saves a lot of space in a factory.

Shear Panels, Shear Stress/Strain
Engineering Shear Strain (not torsional shear strain)

The angle between the original and deformed box is gamma which is defined as dv/dx. Using small angle approximations, alpha = tan(alpha).

Tensorial shear strain is defined as (1/2)gammaxy.

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Wikiversity:Publications

Journals

 * WikiJournal User Group
 * Portal:Wiki Scholar

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Architecture/Bibliography

Master books
This is a list of very important books that everyone interested in learning architecture should know. They are usually part of architecture courses in the whole world.

 * Lessons for Students in Architecture, by Herman Hertzberger - Hertzberger is a world-famous dutch architecture teacher, and this book, full of techniques and methods for making architecture projects, is considered a must-read by many architecture schools, if not all of them. Hertzberger is considered a master at studying the function of architecture in an urban environment.
 * Architecture: Form, Space, & Order, by Francis D. K. Ching - is another bible for architects, it explains a lot of formal rules, order and mechanisms used when doing architecture, and the interaction between formal rules and function in architecture.
 * Towards a new architecture, by Le Corbusier - This is a manifest that Le Corbusier, one of the fathers of the Modernism wrote. It is a key to understand our post-modernism current time, and what modernism changed in the paradigms used by architects.
 * The art of projecting architecture by Ernst Neufert - A very practical book about designing with human proportions in mind, full of examples and precise measurements.

Advanced books
This is a list of more secific books that usually are considered important as someone advances on the path of architecture.

 * The image of the city, by Kevin Lynch - A reference in analyzing the city and what makes distinctive points "urban marks" for its inhabitants
 * Learning from Las Vegas, by Robert Venturi- This is a very important book about how architecture can cope, deviate or stand out against rules and classical orthodoxy.
 * Urbanisme, utopie et réalités, by Francoise Choay
 * The death and life of great american cities, by Jane Jacobs

Architecture magazines
This is a collection of world-recognized architecture publications.

 * El Croquis (Spain) - http://www.elcroquis.es/
 * Architectural Record (USA) - http://archrecord.construction.com/
 * Architectural Review (UK) - http://www.arplus.com/home.htm
 * L'Architecture aujourd'hui (France) - http://www.jeanmichelplace.com/fr/revues/fiche.cfm?ProduitCode=13
 * Techniques et Architecture (France) - http://www.techniquesarchitecture.com/
 * Architecture Week (USA) - http://www.architectureweek.com/ - This one can be read online freely
 * GA Houses (Japan) - http://www.ga-ada.co.jp/english/ga_houses
 * GA Document (Japan) - http://www.ga-ada.co.jp/english/ga_document/
 * A+U (Japan) - http://www.japan-architect.co.jp/english/1all/top_frame.html
 * L'Arca (Italy) - http://www.arcaedizioni.it/web_pages_IT/main/index.htm
 * Detail (Germany) - http://www.detail.de/thema_architecture-magazine-construction_57_En.htm
 * 2G (Spain) - http://www.ggili.com/2G.cfm?IDIDIOMA=EN
 * Arquitectura viva (Spain) - http://www.arquitecturaviva.com/avProyectos/Default.asp
 * A10 (Netherlands) - http://www.a10.eu/

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I am not familiar with wiki way of writing book references... Can someone adapt if needed? --Yorik 21:21, 21 September 2008 (UTC)

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User:EvilWendyMan

I am EvilWendyMan, the evil mascot for Wendy's!

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User:Liquider

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Category:Wikiversity votes

Pages describing proposed policies, guidelines and processes currently being voted on. All pages in this category should use.

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User:Eml4500.f08.ateam.mcnally/Week 4

Step 4 Cont. The Theory behind the Elimination of known degrees of freedom (dofs)
Meeting 10 on Wednesday September the 17th of 2008 was a continuation of elimination of the known dofs to reduce the Force Displacement relationship. By applying the fixed boundary conditions it becomes apparent that d1=d2=d5=d6=0. With elements d1=d2=d5=d6 of the global displacement matrix d =0 computation time can reduced by realizing that columns 1,2,5, and 6 can be removed from the global displacement matrix K. See the example below for further explanation.

\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}

\begin{bmatrix} d_{3}\\ d_{4}\\ \end{bmatrix} =

\begin{bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{bmatrix}

$$

From the principle of virtual work the corresponding rows 1, 2, 5, and 6 of the global stiffness and global force matrix K and F may also be deleted. This allows the force displacement relation to reduced further. The resulting force displacement relation is shown below.

\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} $$

F3 =0 and F4=P by inspection of Global Force pictures shown below.

Global displacements d₃ and d₄ can then solved for by inverting K and multiplying by F.

\begin{bmatrix} d_{3}=4.352\\ d_{4}=6.127\\ \end{bmatrix} =K^{-1}

\begin{bmatrix} F_{3}\\ F_{4}\\ \end{bmatrix} $$

Displacements d3 and d4 correspond to X and Y displacement of Node 2 in the global picture. See the following example for a review on inverting matrices.

K^{-1}=1/det(K) \begin{bmatrix} K_{44} & K_{-34} \\ K_{-43} & K_{33} \\ \end{bmatrix} $$ With

det(K)=(K_{33}*K_{44})-(K_{34}*K_{43}) $$

Please see the MATLAB solution at the end of this test for the full solution.

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Talk:Spanish 2

It would really be a good idea to use articles when introducing nouns. Although most students will at this point recognize that libro, ending in -o-, is masculine, they may not know that resumen is masculine or that carnet is feminine.

Completion
Why are the 100% completion boxes being used here? The Jade Knight (d'viser) 09:33, 20 November 2008 (UTC)

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User:Eml4500.f08.ateam.nobrega/Homework Report Two

A 2-Beam Truss System Solved Statically

$$\sum F_{x} = 0 = R_{1}cos(30) - R_{2}cos(-45)$$

$$\sum F_{y} = 0 = R_{1}sin(30) + R_{2}sin(-45) + P $$

$$ R_{1} = \frac{R_{2}cos(-45)}{cos(30)} = \frac{\sqrt{2}}{\sqrt{3}}R_{2} $$

$$ 0 = \frac{\sqrt{2}}{2\sqrt{3}}R_{2} - \frac{\sqrt{2}}{2}R_{2} + P $$

$$ 0 =\frac{\sqrt{2}(1 - \sqrt{3})}{2\sqrt{3}}R_{2} + P $$

$$ R_{2} = -\frac{2\sqrt{3}}{\sqrt{2}(1 - \sqrt{3})}P $$

$$ R_{1} = -\frac{2}{1 - \sqrt{3}}P $$

An Example of a Statically Indeterminate Problem
An example of a statically indeterminate problem is given below:

This system is statically indeterminate because the beam is not a two-force member, with more unknowns than equations relating to the system. The two equations available are the sum of forces in the x- and y-direction. An equation relating moments to the system is not applicable in this situation.

The Definition of a Moment
A force moment is a force applied to a system at a radial distance from the origin of that system. The SI units for a moment is the Newton-meter(N·m), the US Customary Units is the foot-pound(ft·lbf). The equation for a moment is:

$$ \overrightarrow{M_o} = \overrightarrow{F}\times\overrightarrow{r} $$

This is the definition of a moment about point O. A moment can also be referred to as the torque of a system.

Solving A Truss System Using Matrices
The Force Equation as seen in matrix form is as follows:

\begin{bmatrix} f_1\\ .\\     .\\      .\\      .\\      f_6 \end{bmatrix} =     \begin{bmatrix} &  &   &   &   \\          &   &   &   &   \\          &   & K &   &   \\ &  &   &   &   \\          &   &   &   &      \end{bmatrix} \cdot \begin{bmatrix} d_1\\ .\\     .\\      .\\      .\\      d_6 \end{bmatrix} $$

Where F is the global force column matrix, K is the global stiffness matrix, and d is the global displacement matrix.

It was pointed out that the previous example, a truss system with two elastic bars, was in fact a solvable system correcting that it was originally stated to be indeterminate. The class was then told that the system was to be solved and turned in with the next homework report. As a reminder it is important that when the system is broken up into it's individual elements to place any forces shared by two or more elements in the global force diagram on only one of the individual elements.

To begin solving the above problem one must compute the the k matrix for each element, a depiction of what is to go in to each k matrix is given below:

K^{(e)}= k^{(e)} \cdot \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} $$ Here l(e) and m(e) are director cosines for the x'-axis and the equation for k(e) is:

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

The director cosines are based on the $${\hat{x}}$$ coordinate system of the individual elements where,

l(e) = î • i = cosθ(e)

m(e) = î • j = cos(π/2 - θ(e)) = sinθ(e)

Here,

î = cosθ(e) i + sin θ(e) j

î • i = (cosθ(e) i + sinθ(e) j) • i

î • i = cosθ(e) i • i + sinθ(e) j • i

î • i = cosθ(e)

Therefore, î • j also equals sinθ(e)

It is important to note that the matrix, k(1), is symmetric along the diagonal. This means that values with interchangeable row and column indices are equal; for example, k(13) is equal to k(31) and k(12) is equivalent to k(21). Thus, only three values need to be computed. The remaining coefficients have the same absolute value and only differ by a positive or negative sign. In general, the transpose of matrix k(ij)(e) is equal to the original matrix.

The yellow highlighted portion of matrix k(1) is referred to as the upper triangular portion; whereas the lower triangular portion is not highlighted. The blue highlighted portion provides the line of symmetry.

Element 1:
L(1) = 4 E(1) = 3 Ac(1) = 1 θ(1) = 30° - l(1) = cosθ(1) l(1) = cos30° = √ 3 /2 m(1) = sinθ(1) m(1) = sin30° = 1/2

Element 2:
L(2) = 2 E(2) = 5 Ac(2) = 2 θ(2) = -45° - l(2) = cosθ(1) l(2) = cos(-45°) = √ 2 /2 m(2) = sinθ(1) m(2) = sin(-45°) = √ 2 /2

The Stiffness Matrix:

K^{(e)}= k^{(e)} \cdot \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} $$

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

k^{(1)} = \frac{(3)\cdot(1)}{4} = \frac{3}{4} $$

k^{(2)} = \frac{(5)\cdot(2)}{2} = 5 $$

k_{11}^{(1)} = (l^{(1)})^2 = (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} $$

k_{12}^{(1)} = l^{(1)}m^{(1)} = (\frac{\sqrt{3}}{2})\cdot(\frac{1}{2}) = \frac{\sqrt{3}}{4} $$

k_{22}^{(1)} = (m^{(1)})^2 = (\frac{1}{2})^2 = \frac{1}{4} $$

Due to symmetry, where k12(1) = k21(1),

K^{(1)} = \frac{3}{4}\cdot \begin{bmatrix} \frac{3}{4} & \frac{\sqrt{3}}{4} & -\frac{3}{4} & -\frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4} & \frac{1}{4} & -\frac{\sqrt{3}}{4} & -\frac{1}{4}\\ -\frac{3}{4} & -\frac{\sqrt{3}}{4} & \frac{3}{4} & \frac{\sqrt{3}}{4}\\ -\frac{\sqrt{3}}{4} & -\frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{1}{4} \end{bmatrix} =    \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} $$

For Element 2,

k_{11}^{(2)} = (\frac{\sqrt{2}}{2})^2 = \frac{2}{4} = \frac{1}{2} $$

k_{12}^{(2)} = \frac{1}{2} $$

k_{22}^{(2)} = \frac{1}{2} $$

K^{(2)} = 5\cdot \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} =    \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} $$

Observations:

1) The absolute values of all coefficients k(ij)(e), e=2 (I,j) =1,…,4 are the same =>  compute1 coefficient For other coefficients the corresponding sign (positive or negative).

2) (k(2))^T = k(2) . Thus k(2) is symmetric.

Elem. FD rel: [k(e)][d(e)]= [F(e)]

$$ d^{(e)} = \begin{bmatrix} d_{(1)}^{(e)}\\ d_{(2)}^{(e)}\\ d_{(3)}^{(e)}\\ d_{(4)}^{(e)}\\ \end{bmatrix}

F^{(e)} = \begin{bmatrix} F_{(1)}^{(e)}\\ F_{(2)}^{(e)}\\ F_{(3)}^{(e)}\\ F_{(4)}^{(e)}\\ \end{bmatrix} $$

The column matrices d and F are both 4x1.

The global FD relation is [K][d] = [F], where K is an nxn matrix, d is an nx1 matrix and F is a nx1. In the example provided n is equal to six. Professor Vu-Quoc also mentioned a book titled "The Dragons of Eden" by [| Carl Sagan].

Global Force Displacement relationship

Global Force Displacement Relationship
The FD Relation:

\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}

$$

In compact notation the above equation simplifies to:

$$[ K_{ij} ] \cdot { d_{j} } = { F_{i} }$$

The stiffness coefficient, K, is a 6x6 matrix; d and F are 6x1 matrices.

 * $$\sum_{j=1}^6 F_{i} = K_{ij} \cdot d_j $$

 * K = global stiffness matrix.
 * d = global displacement matrix.
 * F = global force matrix.
 * k = elemental stiffness matrix.
 * f = elemental force matrix.

To go from an elemental matrix to a global matrix the matrix has to go through an assembly process.

First identify the correspondence between elemental displacement dof's and global displacement dof's.

At the Global level:

{d1, d2, ............, d6}

At the Elemental level:

* Element 1:{d(1)1, d(1)2, d(1)3, d(1)4} * Element 2:{d(2)1, d(2)2, d(2)3, d(2)4}

Identification of the Global-Local Degrees of Freedom:

Node 2 d2 = d2(1) d3 = d3(1) = d1(2) d4 = d4(1) = d2(3)

Node 3 d5 = d3(2) d6 = d4(2)

Conceptual step of assembly: (topology of K)

The highlighted portion in the figure above represents the overlap between the two elemental matrices of K.

The Global Stiffness Matrix can be seen below,

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

Below are sample calculations when solving for the global stiffness matrix. One can see how both element stiffness matrices come together to form the global stiffness matrix,

$$ K_{11} = k_{11}^{(1)} = \frac{9}{16} = 0.5625 $$ $$ K_{22} = k_{22}^{(1)} = \frac{3\sqrt{3}}{16} = 0.3248 $$ $$ K_{33} = k_{33}^{(1)} + k_{11}^{(2)} = \frac{9}{16} + \frac{5}{2} = 3.0625 $$ $$ K_{34} = k_{34}^{(1)} + k_{21}^{(2)} = \frac{3\sqrt{3}}{16} + -\frac{5}{2} = -2.1752 $$ $$ K_{44} = k_{44}^{(1)} + k_{22}^{(2)} = \frac{3}{16} + \frac{5}{2} = 2.6875 $$

Once all the values have been computed the following matrix will be the result of combining the element stiffness matrix 1 with the element stiffness matrix 2.

$$ \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0.00 & 0.00\\     0.3248 & 0.1875 & -0.3248 & -0.1875 & 0.00 & 0.00\\      -0.5625 & -0.3248 & 3.0625 & -2.1752 & 2.5 & -2.5\\      -0.3248 & -0.1875 & -1.9375 & 2.6875 & -2.5 & 2.5\\      0.00 & 0.00 & 2.5 & -2.5 & 2.5 & -2.5\\      0.00 & 0.00 & -2.5 & 2.5 & -2.5 & 2.5\\     \end{bmatrix} $$

Elimination of Known Degrees of Freedom (dofs)
In this step one will apply known degrees of freedom. This will allow us to reduce the global force displacement relation. For example in the problem being described it is known that global node 1 and global node 3 are both fixed. A fixed constraint is the equivalent of stating that displacements 1, 2, 5, 6 are all equal to 0. Knowing this information allows us to reduce the 6x1 global displacement column matrix to a 2x1. Which then allows one to reduce the global stiffness matrix from a 6x6 to a 6x2, as it is shown below.

First Apply Boundary Conditions:

 * Global Node 1: Fixed
 * Results in:
 * d1=0
 * d2=0

 * Global Node 3: Fixed
 * Results in:
 * d5=0
 * d6=0

By applying the fixed boundary conditions it becomes apparent that d1=d2=d5=d6=0. With elements d1=d2=d5=d6 of the global displacement matrix d =0 computation time can reduced by realizing that columns 1,2,5, and 6 can be removed from the global displacement matrix K. See the example below for further explanation.

\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}

\begin{bmatrix} d_{3}\\ d_{4}\\ \end{bmatrix} =

\begin{bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{bmatrix}

$$

From the principle of virtual work the corresponding rows 1, 2, 5, and 6 of the global stiffness and global force matrix K and F may also be deleted. This allows the force displacement relation to reduced further. The resulting force displacement relation is shown below.

\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} $$

F3 =0 and F4=P by inspection of Global Truss System shown below.

Global displacements d₃ and d₄ can then solved for by inverting K and multiplying by F.

\begin{bmatrix} d_{3}=4.352\\ d_{4}=6.127\\ \end{bmatrix} =K^{-1}

\begin{bmatrix} F_{3}\\ F_{4}\\ \end{bmatrix} $$

Displacements d3 and d4 correspond to X and Y displacement of Node 2 in the global picture. See the following example for a review on inverting matrices.

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

\det{(K)}=(K_{33}\cdot K_{44})-(K_{34}\cdot K_{43}) $$

Please see the MATLAB solution at the end of this test for the full solution.

Computing the Element Forces from New Known Displacements
Since displacements of Element 1 are now known, the element forces can be determined.

Recall Element 1:

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

Where k and d have previously been defined:

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} $$

d^{(1)} =

\begin{bmatrix} 0\\    0\\     4.352\\     6.1271\\     \end{bmatrix} $$

To compute the element forces in Element 1, simply multiply the k and d matrix of Element 1. Note that the first two columns of matrix k will be multiplied to zero and can be ignored if a manual calculation is performed.

\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}\cdot

\begin{bmatrix} 0\\    0\\     4.352\\     6.1271\\     \end{bmatrix}

= \begin{bmatrix} f_{1}^{(1)}\\ f_{2}^{(1)}\\ f_{3}^{(1)}\\ f_{4}^{(1)} \end{bmatrix}

$$

Executing matrix multiplication to solve for the element and reaction forces, F:

\begin{bmatrix} -4.4378\\ -2.5622\\ 4.4378\\ 2.5622\\ \end{bmatrix}= \begin{bmatrix} f_{1}^{(1)}\\ f_{2}^{(1)}\\ f_{3}^{(1)}\\ f_{4}^{(1)} \end{bmatrix}

$$

Observation: Note Element 1 is in equilibrium. If the sum of the forces in the x and y direction are computed, one will note that they sum to zero.

 * $$\sum f_{1}^{(1)} + f_{3}^{(1)}=0$$
 * $$\sum f_{2}^{(1)} + f_{4}^{(1)}=0$$

Verifying Equilibrium:

To prove that the magnitude and direction of the forces acting on Element 1 are equal and opposite, first compute the magnitude of the reaction. This is done by the root-sum-square method:

 * $$P_1^{(1)}=[(f_1^{(1)})^2 + (f_2^{(1)})^2]^{1/2}$$

 * $$P_1^{(1)}=[(-4.4374)^{2} + (-2.5622)^{2}]^{1/2} $$

 * $$P_1^{(1)}=5.124 $$

The magnitude of P2 is computed the same way:

 * $$P_2^{(1)}=[(f_3^{(1)})^2 + (f_4^{(1)})^2]^{1/2}$$

 * $$P_2^{(1)}=[(4.4374)^{2} + (2.5622)^{2}]^{1/2} $$

 * $$P_2^{(1)}=5.124 $$

Thus, the forces are equal in magnitude but opposite in direction. The element is in equilibrium.

Element 2 is also in equilibrium and this can be proven the same way as above.

Method 2: To solve for 2 bar truss system using statics method:

Two different cut methods can be used as demonstrated above. The red outlines the different applications of the Euler cut principle. Individual free body diagrams can be drawn for each element circled in red and the problem can be solved as statically determinate.

Verification of the equilibrium of node 2 statically:

 * $$\sum F_{x}=0=-P_{1}^{(1)}cos(30) + P_{2}^{(1)}cos(45)$$
 * $$\sum F_{y}=0=P - P_{1}^{(1)}sin(30) - P_{2}^{(1)}sin(45)$$

P, P1, and P2 are known from previous steps:

 * $$P_{applied}^{} = 7 $$
 * $$P_{1}^{(1)}=5.124$$
 * $$P_{2}^{(2)}=6.276$$

Plugging these known values into the force equations:

 * $$\sum F_{x}=-5.124 \cdot cos(30) + 6.276 \cdot cos(45)$$
 * $$\sum F_{y}=7 - 5.124 \cdot sin(30) - 6.276 \cdot sin(45)$$

 * $$\sum F_{x}=0$$
 * $$\sum F_{y}=0$$

Global node 2 is in equilibrium.

MATLAB

% Two bar truss example clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4;

nodes = [0, 0; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta)];

dof=2*length(nodes);

conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1);

%load vector R = zeros(dof,1); R(4) = P;

% Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e, A, ... nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results

k_local =

0.75       -0.75        -0.75         0.75

k =

0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local =

5   -5    -5     5

k =

2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5

K =

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

R =

0    0     0     7     0     0

d =

0           0        4.352       6.1271            0            0

reactions =

-4.4378     -2.5622       4.4378      -4.4378

results =

1.7081      5.1244       8.5406       5.1244       17.081       0.6276       1.8828        3.138       1.8828        6.276

Team Members contributing to this report
Tatiana Nobrega Eml4500.f08.ateam.nobrega 14:50, 26 September 2008 (UTC) Taylor Boggs   Eml4500.f08.ateam.boggs.t 03:20, 26 September 2008 (UTC) Akash Shah     Eml4500.f08.ateam.shah 14:59, 26 September 2008 (UTC) Pedro Rivero   Eml4500.f08.ateam.rivero 21:42, 25 September 2008 (UTC) Josh McNally   Eml500.f08.ateam.mcnally 20:57, 25 September 2008 (UTC) Michael Carr   Eml4500.f08.ateam.carr 14:36, 25 September 2008 (UTC)


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User:Eml4500.f08.group/Homework2

=Homework 2=

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Homework Problems (Tom's Homework Problems)

title: User:Eml4500.f08.group/Homework2


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User:Eml4500.f08.group/Homework2/Lecture Notes

= Solving 2 Element 2D Truss=

Global Picture

As a standard, mark local load numbers in a manner in which they increase along the x axis. As shown above, the global system can be separated into two separate local systems, where P can only be applied only to one arbitrary local system. On the bottom right local node system, taking the moment about either node shows that Fx(1) = Fy(2) = 0.

Solution:

ΣFx = F1cosθ1 + Pcosθ2 + F2cosθ3

ΣFy = F1sinθ1 + Psinθ2 + F2sinθ3

ΣM2 = 0 (not useful)

ΣM1 = -Pcosθ2L1sinθ1 + Psinθ2L1cosθ1 + F2cosθ3(L1sinθ1 - L2sinθ3) + F2sinθ3(L1cosθ1 + L2cosθ3)

Here are three variables and three equations which can be solved by arithmetic.

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

!Contribution Statically Indeterminate Systems, from team Carbon
 * A statically indeterminate system cannot be solved by the three classical equations of statics:
 * A statically indeterminate system cannot be solved by the three classical equations of statics:

ΣFx=0

ΣFy=0

ΣM=0 (take the sum of moments about any arbitrary node)

If the number of unknown forces exceed the number of useful static equilibrium equations, then the system is statically indeterminate. In these cases, mechanics equations such as that of deformation may be used to solve the system. Some examples are below:

Contribution by group.w
 * }

Global Force Displacement at element level

k is the element stiffness matrix for element e (4x4 in this case)

d is the the element displacement matrix (4x1 in this case)

f is the element force matrix (4x1 in this case)

Matrix p 225 book, 2nd equation form the bottom

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

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

→ l(e), m(e) = director cosines of x~ axis

l(e) = i~ • i = cosθ(e)

m(e) = i~ • j = cos(π/2 - θ(e)) = sinθ(e)

Note: Physically, the director cosines represent the components of $$\hat{\tilde{i}}$$ with respect to the basis $$(\hat{i},\hat{j})$$.

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

!Contribution Director cosine derivation, from team Carbon
 * i~ = cosθ(e)i + sinθ(e)j
 * i~ = cosθ(e)i + sinθ(e)j

→ l(e) = i~ • i = (cosθ(e)i + sinθ(e)j) • i = cosθ(e)(i • i) + sinθ(e)(j • i), where i • i = 1 and j • i = 0, = cosθ(e)

→ m(e) = i~ • j = (cosθ(e)i + sinθ(e)j) • j = cosθ(e)(i • j) + sinθ(e)(j • j), where i • j = 0 and j • j = 1, = sinθ(e)

Contribution by group.w
 * }

Note in the first matrix that there are only three numbers to compute in each matrix below, then change the respective sign. The first matrix is also symmetric, Kij(e) = Kji(e).

Note in the second matrix that the absolute value of all numbers are 5/2, and the transpose of the matrix is equal to the matrix (the matrix is symmetric).

The Force-Distance relationship: K(e)d(e) = F(e)

In compact notation:

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

Σkijdj=Fi, i=1,...6

Knxn=[Kij]nxn = global stiffness matrix

dnx1={dj}nx1 = global displacement matrix

Fnx1={Fi}nx1 = global force matrix

Recall the element FD relationship:

[kij]4x4{dj}4x1={fi}4x1

knxn=[kij]nxn = element stiffness matrix

dnx1={dj}nx1 = element displacement matrix

fnx1={fi}nx1 = element force matrix

Lower case designates the element matrix while upper case designates the global matrix.

How do you go from element matrices to global matrices? Through an assembly process:

Identify the correspondence between element displacement dof(s) (element1: {d1(1), d2(1), d3(1), d4(1)} element2: {d1(2), d2(2), d3(2), d4(2)})and global displacement dof(s)({d1, d2, d3, d4})

Global - Local dof identification:

 * Node 1:
 * d1 = d1(1)
 * d2 = d2(1)
 * Node 2:
 * d3 = d3(1) = d1(2)
 * d4 = d4(1) = d2(2)
 * Node 3:
 * d5 = d3(2)
 * d6 = d4(2)

Global conceptual step of assembly (topology of K): $$\begin{bmatrix} & &  |&  &k^{(2)}  &|\\ & &  |-&  -&  -& -  | \end{bmatrix} \begin{Bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6} \end{Bmatrix} = \begin{Bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}
 * & &k^{(1)}  &|  &  & \\
 * & &k^{(1)}  &|  &  & \\

\end{Bmatrix}$$

Global stiffness matrix K:

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

K11=k11(1)=$$\frac{9}{16}$$

K12=k12(1)=$$\frac{3\sqrt{3}}{16}$$

K33=k33(1)+k11(2)=$$\frac{9}{16} + \frac{5}{2}$$

K34=$$\frac{3\sqrt{3}}{16} + \frac{-5}{2}$$ = -2.1752

K44=$$\frac{3}{16} + \frac{5}{2}$$ = 2.6875

...etc

A general rule because of symmetry: K43=K34, K33 is not equal to K44

Elimination of known dofs by reducing the global FD relation matrices
Applying the fixed boundary equations, d1=d2=d5=d6=0, therefore d3 and d4 are the only displacement unknowns. Then by principle of virtual work (PVW) the rows(1,2,5,6) corresponding to the F supplied by the reactions are also deleted. Resulting in the FD relationship:

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

\end{Bmatrix}$$

Applying P=7, Then F3=0, F4=7 Then solving for the displacement:

$$\begin{Bmatrix} d_{3}\\ d_{4}

\end{Bmatrix} \begin{bmatrix} K \end{bmatrix}^{-1} = \begin{Bmatrix} 0\\ p

\end{Bmatrix}$$

d3 = 4.354

d4 = 6.1271

Inverting Stiffness Matrix
What is $$\displaystyle \overline{\underline K}^{-1}$$, the inverse of $$\displaystyle \overline{\underline K}$$? How do you find it by hand (without calculator)?

We begin with the determinant of $$\displaystyle \overline{\underline K}$$

$$\displaystyle det(\overline{\underline K})=K_{33}K_{44}-K_{34}K_{43}$$

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

$$\overline{\underline K}^{-1} \overline{\underline K} =\overline{\underline I}= \begin{bmatrix} 1 & 0\\   0 & 1 \end{bmatrix}$$

$$=\frac{1}{det(\overline{\underline K})} \begin{bmatrix} det(\overline{\underline K}) & 0                           \\ 0                           & det(\overline{\underline K}) \end{bmatrix}$$

NOTE:
Do Not confuse the above with the transpose of K

$$\overline{\underline K}^{T} = \begin{bmatrix} K_{33} &-K_{43} \\ -K_{34}  & K_{44} \end{bmatrix}$$

$$\overline{\underline K}^{-1} \neq \frac{1}{det(\overline{\underline K})}\overline{\underline K}^{T}$$

Solving 2 Bar Truss Reactions
Back to the 2-bar truss system:

$$\begin{Bmatrix} d_3 \\d_4 \end{Bmatrix} = k\begin{Bmatrix} 0 \\P\end{Bmatrix} = \begin{Bmatrix} 4.352 \\6.1271\end{Bmatrix}$$

Compute Reactions using two methods, FEA and statics

What is Known
Global or element FD relations can be used (2 methods)

$$\displaystyle k^{(e)}d^{(e)} = f^{(e)} e=1,2 $$

Use the element Force-Displacement (FD) relations:

Element 1 (e=1)

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

where: $$d^{(1)} = \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix} = \begin{Bmatrix} d_1^{(1)} \\ d_2^{(1)} \\ d_3^{(1)} \\ d_4^{(1)} \end{Bmatrix}$$ $$f^{(1)} = \begin{Bmatrix} f_1^{(1)} \\ f_2^{(1)} \\ f_3^{(1)} \\ f_4^{(1)} \end{Bmatrix}$$
 * k is known
 * d is known
 * f is the unknown reaction

Element 2 (e=2)

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

where: $$d^{(2)} = \begin{Bmatrix} 4.352 \\ 6.1271 \\ 0 \\ 0 \end{Bmatrix} = \begin{Bmatrix} d_1^{(2)} \\ d_2^{(2)} \\ d_3^{(2)} \\ d_4^{(2)} \end{Bmatrix}$$ $$f^{(2)} = \begin{Bmatrix} f_1^{(2)} \\ f_2^{(2)} \\ f_3^{(2)} \\ f_4^{(2)} \end{Bmatrix}$$
 * k is known
 * d is known
 * f is the unknown reaction

Using this information we can now solve for the reaction forces acting on each element

Element 1 Reaction Forces
From:

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

we find that the reaction forces are equal to the solution of:

$$\underline{f} ^{(1)} = \begin{bmatrix} 0.5625 & 0.32476 &-0.5625 & -0.32476    \\  0.32476 & 0.1875  &-0.32476 & -0.1875       \\  -0.5625 & -0.32476  &0.5625 & 0.32476       \\  -0.32476 & -0.1875  &0.32476 & 0.1875       \\ \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\  4.352 \\  6.1271 \\ \end{Bmatrix}$$

The zeros in the 1 x 4 matrix eliminate the first two columns in the 4x4 matrix resulting in:

$$ \underline{f} ^{(1)} = \begin{bmatrix} &-0.5625 & -0.32476    \\  & -0.32476 & -0.1875       \\  &0.5625 & 0.32476       \\  & 0.32476 & 0.1875       \\ \end{bmatrix} \begin{Bmatrix} 4.352 \\ 6.1271 \\ \end{Bmatrix} = \begin{Bmatrix} -4.4378 \\ -2.5622 \\  4.4378 \\  2.5622 \\ \end{Bmatrix} = \begin{Bmatrix} \underline{f}_1 ^{(1)} \\ \underline{f}_2 ^{(1)} \\ \underline{f}_3 ^{(1)} \\ \underline{f}_4 ^{(1)} \\ \end{Bmatrix} $$

$$\underline{f}_1 ^{(1)}$$ and $$\underline{f}_2 ^{(1)}$$ are Reaction forces where as $$\underline{f}_3 ^{(1)}$$ and $$\underline{f}_4 ^{(1)}$$ are internal forces within the element. This element is a two force member as Verified Here. And so the following summations should be true, and are.

$$\sum_{} F_x = \underline{f}_1 ^{(1)} + \underline{f}_3 ^{(1)} = 0$$

$$\sum_{} F_y = \underline{f}_2 ^{(1)} + \underline{f}_4 ^{(1)} = 0$$

$$\displaystyle \sum_{} M_{any point} = 0$$

Element 2 Reaction Forces
$$P_1^{(1)} = [(f_1^{(1)})^2 + (f_2^{(1)})^2]^{1/2}$$

Statics Method
There are two methods that can be used to break up a truss, both are based on the Euler Cut Principle. Method 2 breaks up the truss segments further such that is is made up of nodes and 2-Force members

 * [[Image:EulerCutPrinciple.JPG]]

Back to the Finite Element Solution: How to bring $$\underline P$$ back into the big picture

The Equilibrium of node (2) is Verified Here


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University of Florida/Eml4500/f08.qwiki/Lecture 6

The global force relationship is given by the equation $$F=KD$$, where $$F$$ is the resultant force matrix, $$K$$ is the global stiffness matrix, and $$D$$ is the global displacement matrix.

In matrix format,

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


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User:Eas4200c.f08.aeris.krammer/HW2

=Class Notes=

Problem 1.1 continued
The member being analyzed is a shell or box beam and it can resemble an airfoil. Analyzing a simple box beam will give insight into more completed components such as the airfoil. This member is described with the word "shell" because the cross sectional thickness is very small and "beam" because it resists beaming. To the bottom right is a picture of a simple shell or box beam. Now, the problem posed in problem 1.1 is to find the ratio of dimensions b to a that will maximize the load-bearing capacity of the beam. First, in case 1, we assumed that the normal stress $$\displaystyle \sigma$$ will reach the allowable normal stress first. However, we must also include the condition in which the shear stress $$\displaystyle \tau$$ will reach the allowable shear stress first. This is designates as case 2. Recall, that the assumptions are that (1) the contour length of the box beam is equal to twice the sum of the dimensions a and b, (2) the magnitudes of the moment and torque loads on the beam are equal, and (3) the allowable normal bending stress is equal to twice the allowable shear stress. $$\displaystyle L=2(a+b)$$ $$\displaystyle |M|=|T|$$ $$\displaystyle \sigma_{allowable}=2*\tau_{allowable}$$ Also, recall that since the thickness of the shell beam is very small compared to the length and width, the shear stress on the member can be assumed to be uniform. As seen in fig. A, the arrows are the shear forces acting on the member. All the arrows are the same same denoting equal shear forces. Fig. B is the actual distribution of the shear forces on the member laterally along the y direction. Fig. C is the distribution of the stresses that is not a real representation of the stresses but is statically the same. The loading conditions are also shown below.  Knowing this knowledge allows for the steps of case 2 to be described. Assume $$\displaystyle \tau$$ reaches the allowable shear stress first. The purpose is to verify that $$\displaystyle \sigma_{max}<\sigma_{allowable}$$ given that the shear is maxed already. $$\displaystyle \tau=\frac{T}{2abt}$$ => $$\displaystyle T_{max}=2t\tau_{allow}(ab)$$ In the equations above, $$\displaystyle \tau$$ is the shear stress, T is the torque, a and b are the width and length respectively, and t is the thickness of the beam. Given this, in order to maximize torque, the product (ab) must be maximized where $$\displaystyle a=\frac{L}{2}-b$$. This means that this will only be one independent variable and the max torque is a function of either only b or only a. For conventional purposes, torque will be a function of b however, the problem can also be solved with torque being a function of a. Maximizing the product (ab) keeping the contour length or circumference constant will yield a square cross section of the beam. The derivation of these lengths are shown in the collapsible box below.

Box Beam with Rectangular Cross Section
A box beam with a rectangular cross section will be studied as a model for Aerospace Structures such as a fuselage or wings. Let's begin by analyzing shear stresses due to maximum Torque. Case 1: $$\displaystyle \tau_{max}=(\frac{T_{max}}{2abt})$$ Then equate the above formula in terms of maximum bending moment to obtain the allowable stress. $$\displaystyle (\frac{T_{max}}{2abt})=(\frac{M_{max}}{2(\frac{L}{8})(\frac{3L}{8})t})=\sigma_{allow}$$ It can be seen that: $$\displaystyle \tau_{max}=\sigma_{allow}$$  Based on assumption 3, $$\displaystyle \sigma_{allow}=2\tau_{allow}$$  This case is not accepted because: $$\displaystyle 2\tau_{allow}>\tau_{allow}$$  Case 2: Assume $$\displaystyle \tau_{max}=\tau_{allow}$$ ($$\displaystyle \tau_{max}$$ reaches $$\displaystyle \tau_{allow}$$ first). $$\displaystyle \tau=\frac{T}{2abt}=\tau_{max}=\tau_{allow}$$ $$\displaystyle T=(2t\tau_{allow})(ab)$$  $$\displaystyle T_{max}=(2t\tau_{allow})(ab)_{max}$$

Knowing that the lengths will equal L/4,the resulting max torque is $$\displaystyle T_{max}=2t\tau_{allow}(\frac{L}{4})^2=\frac{1}{8}tL^2\tau_{allow}$$ This value in turn equals the maximum bending moment given from assumption 2 ($$\displaystyle T_{max}=M_{max}$$). Now, obtaining the maximum bending moment, we must check that for the resulting maximum bending normal stress and decide whether it exceeds the allowable normal stress. $$\displaystyle \sigma_{max}=\frac{M_{max}b^{(2)}}{2I^{(2)}}$$ If $$\displaystyle \sigma_{max}<\sigma_{allow}$$ then the ratio $$\displaystyle \frac{b^{(2)}}{a^{(2)}}=1$$ is acceptable. Upon acquiring the steps to solve this problem we can determine the ratios numerically. For case 1, the numerical process is as follows. "I" the second area moment of inertia given by the general formula : $$\displaystyle I=\int\int_{A_{cross}}{z^2dydz}$$ Here, "z" is the distance from the moment axis to the point of interest. Generally, in order to increase the moment of inertia of a structure, one would design the structure so that a majority of the material is at a far distance away from the moment axis. Since in the formula the distance from t he moment axis is squared, it has a profound effect on the moment of inertia. In this problem, the distance from the moment axis is b/2. For a rectangle about the conventional y axis (as displayed on all above diagrams), the second area moment of inertia is: $$\displaystyle I=\frac{1}{12}bh^3$$ In this problem, for both cases, "I" can be found by by summing the moment of inertia of 4 segments of the rectangle along with using the parallel axis theorem. This gives the following equation. The four segments can also be referenced to the below figure of the case one thin-walled rectangular cross section.  $$\displaystyle I=\sum_{i=1}^4[\frac{b_ih_i^3}{12}+A_id_i^2]=\frac{2tb^3}{12}+2[\frac{at^3}{12}+at(\frac{b}{2})^2]\approx{\frac{tb^2}{6}(3a+b)}$$ The approximation sign assumes that the thickness of the walls is very small compared to the length and width of the cross section. This also means that the thickness cubed will be even smaller thus, the term with a $$t^3$$ can be neglected. So: $$\displaystyle \frac{I}{b}=tb(3L-4b)/12$$ The second derivative here is negative on the whole domain thus, setting the derivative to zero and solving for b will yield the value of b in which will maximize the second moment of inertia.
$$f^'(b)=0, 0=\frac{tL}{4}-\frac{2tb}{3} , b^{(1)}=\frac{3L}{8}$$ Plugging this value into assumption one will yield the corresponding a value. The ratio of $$\frac{b^{(1)}}{a^{(1)}}=3, M_{max}=2\sigma_{allow}(\frac{I^{(1)}}{b^{(1)}})=\frac{3tL^2}{32}\sigma_{allow}$$. The max bending stress is greater than the allowable bending stress. This ratio is not acceptable.

Torque, Shear Flow, and Shear Panels
Recall that from the problem 1.1 that the torque of the thin-walled cross section equaled to twice the shear stress multiplied by the product of the dimensions given by the equation $$ \tau=\frac{T}{2abt}$$. This expression for the torque's relationship with the cross sectional dimensions is only for a particular case where the cross section of the thin-walled beam or member being analyzed is rectangular. For a more general case of relating the torque applied on an object, it is necessary to introduce the concept of shear flow. Shear flow is the product of the shear stress and the thickness of the cross section. Usually, this thickness is very small compared to the contour length. For more general case of relating torque with dimensions and shear flow, a formula is introduced in section 3.5 of "Mechanics of Aircraft Structures" by C.T. Sun.
$$\displaystyle T= \oint{\rho qds}=\int\int_{A_{bar}}{2qdA}=2qA_{bar}$$ This formula describes the torque of a thin-walled bar with an arbitrary cross section of thickness t. Here, q is the shear flow, t is the thickness, A bar is the average area of the cross section, and T is the torque. In summary, for a nonuniform thin-walled cross section,$$T=2qA_{bar}:$$
 and for a thin-walled rectangular cross section$$T=2 \tau ab$$:
 Deriving the torque for the thin-walled rectangular cross section was an ad hoc method because of the assumptions used to determine the equation. However, it is logical. The derivation of the non-uniform cross section torque is a method based on elasticity. Upon scrutinization of the stringers of an aircraft that is part of a monocoque structure, the open thin-walled box structure was previously determined to be more practical that the closed structure because of ease of riveting and installation. However, as the stringers were examined more carefully, it is realized that the vertical sections are slightly slanted. Why is this? That is because this structure is practical for stockpiling and storage and at the same time, not losing much moment of inertia due to the slanting of the vertical sections.

 Starting chapter two, the structure of a shear panel is introduced. It is a thin sheet of material, usually metal aluminum, that can withstand shear load. There is a relationship that relates the shear modulus of a material and the shear strain with the shear stress that is applied the structure. $$\displaystyle \tau =G \gamma$$, where G is the shear modulus, and gamma is the shear strain. Gamma describes the change in a right angle between two fictitious lines in the material due to deformation. It can be expressed by displacements in an (x,y) coordinate system relative to the structure.
$$\displaystyle \gamma = \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=\frac{\partial u_{x}}{\partial y}+\frac{\partial u_y}{\partial y}$$ where $$u=u_x$$= the displacement along the x direction and similar for v. This shear strain is called the engineering shear strain and not the tensorial shear strain. These two quantities are different in the fact that $$\displaystyle \epsilon_{xy}=$$ (tensorial shear strain)$$\displaystyle = .5 \gamma_{xy}$$. The following picture describes the deformation of an infinitesimally small element (i.e. part of a shear panel) subject to shear forces. The blue shape is the deformed infinitesimally small rectangle.

=Homework Problems=

Example Problem - Second moment of area
Problem:

 Actually, in class, I asked to do the problem with the channel cross section rotated by $$\displaystyle (-90^\circ)$$ relative to the orientation of your channel cross section. So you are solving a different problem. Eml4500.f08 20:50, 26 September 2008 (UTC)

Given two cross sections of the same area, determine the ratio of the second moment of area for the two cases shown.

To solve this problem we must find the second moment of area for each, then relate the two using the area constraint. For the first case, the circle, the second moment of area is $$I^{(1)}_{y} = \frac{\pi r^4}{4}$$.

Solution: Using the parallel-axis theorem and the known value for a rectangle, the second moment of area for case 2 can be calculated as follows:
$$I^{(2)}_{y} = \frac{ta^3}{12}+2[\frac{bt^3}{12}+tb(\frac{a}{2})^2]$$

Plugging in $$b=a$$ and $$t=a/10$$ gives:
$$I^{(2)}_{y} =\frac{a^4}{120}+\frac{a^4}{20}=\frac{7a^4}{120}$$

Given that the cross-sectional area of each case is the same, we are able to relate the radius of circle one to the dimensions of case 2. Specifically, we can relate $$r$$ to $$a$$.
 $$A^{(1)}_{}=A^{(2)}$$ $$\pi r^2=3at=\frac{3a^2}{10}$$ $$r=a\sqrt{\frac{3}{10\pi}}$$ We now have all the information needed to solve the ratio of second moment of area.
$$\frac{I^{(2)}_{y}}{I^{(1)}_{y}}= \frac{7a^4/120}{\pi r^4/4}=\frac{7a^4/120}{\pi (a\sqrt{\frac{3}{10\pi}})^4/4}=\frac{70\pi}{27}=8.145$$ This result shows that the second moment of area for case 2 is just over eight times larger than case one. The larger second moment of area allows for the element to withstand higher bending moments without yielding. This helps explain why stringers take on a shape closer to case 2 rather than a circular rod.

Problem 1.7 [C.T. Sun] with modifications
Problem Statement: Two load carrying beams are given with circular cross-sections, where $$\displaystyle R_{0}= 10 cm$$, $$t=\frac{1}{10}R_{0}=1 cm$$ and $$\displaystyle R_{1}$$ is such that the beam cross sectional areas are equal $$\displaystyle A^{(1)}= A^{(2)}$$

Question: Find the second moment of inertia of the two beams $$\displaystyle I_{y}^{(1)}$$ and  $$\displaystyle I_{y}^{(2)}$$ and compare them.

Solution: Case 1: The second moment of inertia for a circle is $$I_{x}=I_{y}=\frac{1}{4} \pi r^{4}$$ Thus, for this case: $$I_{y}^{(1)}=\frac{1}{4} \pi R_{0}^{4}$$. Also, $$A^{(1)} = \pi R_{0}^{2}$$

Case 2: The cross sectional area of the element in Case 2 is $$\displaystyle A^{(2)}=2R_{0}t+2\pi R_{1}^2=A^{(1)}$$. For the second moment of inertia calculation, for Case 2, the parallel axis theorem is to be used. Since the moment of inertia of a circular area about a centroidal axis is $$I_{x}=I_{y}=\frac{1}{4} \pi r^{4}$$, the moment of inertia for the circle of radius $$\displaystyle R_{1}$$ about its centrodial axis is $$I_{y}=\frac{1}{4} \pi R_{1}^{4}$$. Thus,
$$I_{y}^{(2)}= 2(\frac{\pi R_{1}^4}{4}+\pi R_{1}^2(R_{0}+R_{1}))+\frac{t(2R_{0})^3}{12}$$. Therefore,
 $$\displaystyle A^{(1)}=A^{(2)}=314.16 cm^{2}$$, $$\displaystyle R_{1}=6.84cm$$ The moments of inertia for Case 1 and 2 are
 $$\displaystyle I_{y}^{(1)}=7853.98 cm^4$$ and $$\displaystyle I_{y}^{(2)}=9055.29 cm^4$$ The moment of inertia in Case 2 is greater than the moment of inertia in Case 1.

Homework Statement from 9/10
Find the moment of inertia, with respect to y, of a circular cross section and a u-shaped cross section, as shown below.

CASE 1: Beam with circular cross section.

We begin with the definition of moment of inertia, I. Specifically here we are calculating moment of inertia with respece to the y-axis, Iy.

 * $$\ I_{y}=\int r^{2} dm $$

Solving for this integral gives us the moment of inertia of the circle in Case 1.

 * $$\ I_{y}=\int r^{2} dm$$

 * $$\ I_{y}=\int r^{2} \pi r dr$$

 * $$\ I_{y}=\int \pi r^{3} dr$$

 * $$\ I_{y}=\frac{\pi}{4} r^{2}$$

CASE 2: Beam with u-shaped cross section.

As can be seen in the diagram, we can divide the "u-shaped" beam into 3 separate parts, with the symmetrical side piece being equal. We can calculate the moments of inertia of these three parts separately, and then combine them using the Parallel Axis Theorem.

We know that the moment of inertia of a rectangle is equal to $$\frac{b^{3} h}{12}$$.

From here we can calculate the moment of inertia for each of the 3 pieces, centered about each centroid.

For the 2 symmetrical pieces.

 * $$\ I_{y}=\frac{b^{3} h}{12}$$

 * $$\ I_{y}=\frac{c^{3} b}{12}$$

Using the Parallel Axis Theorem, we can move the centroid of these moments to the centroid about the x-axis.

 * $$\ I_{y}= \frac{c^{3} b}{12} + \frac{(a + c)^{2} b c}{4}$$

Noting we have two of these pieces we can multiply this moment by 2.

 * $$\ I_{y}=2 \frac{c^{3} b}{12} + 2 \frac{a + c^{2} b c}{4}$$

For the bottom piece.

 * $$\ I_{y}=\frac{b^{3} h}{12}$$

 * $$\ I_{y}=\frac{a^{3} c}{12}$$

Because the centroid of this piece is already on the x-axis, we do not need to use the Parallel Axis Theorem here.

Combining the pieces.

We can combine the moments of the calculated pieces to obtain the total moment of inertia with respect to the y-axis, Iy.

 * $$\ I_{y,total}=\frac{c^{3} b}{6} + \frac{a^{3} c}{12} + \frac{(a + c)^{2} b c}{2}$$

= Contributions = Jin Yu Guan Eas4200c.f08.aeris.guan 19:50, 26 September 2008 (UTC) Anett Krammer Eas4200c.f08.aeris.krammer 19:54, 26 September 2008 (UTC) Joshua Holladay Eas4200c.f08.aeris.holladay 19:55, 26 September 2008 (UTC) Nelson Caceres Eas4200c.f08.aeris.caceres 19:56, 26 September 2008 (UTC) Jesse W. McRae Eas4200c.fo8.aeris.mcrae 20:36, 26 September 2008 (UTC)

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User:Ybit/Spanish/Pronunciación de la r

For the letter R/r there are in Spanish language two ways of the pronunciation:

ɾ
Lightly pronounce "r" (e.g. trabajo).

 * 1) Write down some examples:

formato, mora, tradición

r
Thrilling pronounced "r". It is placed on the beggining of the word (e.g. río), in the word after n (e.g. enriquecimiento) and s (e.g. Israel). Also "rr" (leter - some alphabets include it) is situated in the word between two wovels (e.g. correo).

 * 1) Write down some examples for all three cases:

beginning: rojo, rosado, rudeza

nr: enrollar, enredo, enraizado

sr: Sri Lanka

rr: narrador, porrazo, forro

title: User:Ybit/Spanish/Pronunciación de la r


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User:Ybit/Spanish/Flamenco

 * 1) Listen to this performance by Camarón de la Isla: http://www.youtube.com/watch?v=yBHuqvEw7FE
 * 2) Read this article: Flamenco and write down a shor sumary not exceeding 600 words:

title: User:Ybit/Spanish/Flamenco


text:
User:Ybit/Spanish/Galicia

 * 1) Go to www.youtube.com and place links two 3 excelent videos about Galicia:

title: User:Ybit/Spanish/Galicia


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User:Ybit/Spanish/Andalucía

Qué es Anadalucía para mí:
 * calor

title: User:Ybit/Spanish/Andalucía


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Wikiversity talk:Appropriate Venues

Point 2 seems confusing to me. Also: Something should be said about heeding names; ie, if you're editing under "Spanish" but actually writing a course on dialectal difference in Spanish, you should move your page to "Dialectal Differences in Spanish", etc.  The Jade Knight (d'viser) 12:20, 22 September 2008 (UTC)

 * Ah well, that's a one edit stub policy :-). What I was getting at is that the various Spanish courses (e.g.) could coordinate on the topic page, rather than randomly here and there... just makes it easier to have "department" pages for related projects, designated discussion areas for specific projects, and use fora like the colloquium for very new things or issues of interest to Wikiversity as a whole. --SB_Johnny talk 13:14, 22 September 2008 (UTC)

title: Wikiversity talk:Appropriate Venues