text:
File:Poke through example.jpg

Summary
I made this as an example for DAZ Studio D-Form tool.

title: File:Poke through example.jpg


text:
File:Horn examp.gif

Summary
I made this animated GIF as an example for DAZ Studio D-Form tool.

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

title: File:Horn examp.gif


text:
File:D-former one.jpg

Summary
Example for DAZ Studio D-Form tool.

title: File:D-former one.jpg


text:
File:Second D-Former.jpg

Summary
Example for DAZ Studio D-Form tool.

title: File:Second D-Former.jpg


text:
College Algebra/Real Numbers

Welcome! In this lesson we will review the concept of real numbers. After you are done, take the self-quiz to test your knowledges on this subject and then go on to the next lesson.

Real numbers
Materials to use:
 * Numbers and Points.
 * MIT Points and Vectors.
 * Algebra/Real Numbers.
 * The real number system (Book I have).

title: College Algebra/Real Numbers


text:
File:Frame 20 D-Forms.jpg

Summary
Example for DAZ Studio D-Form tool.

title: File:Frame 20 D-Forms.jpg


text:
File:D-form animation two.gif

Summary
Example for DAZ Studio D-Form tool.

title: File:D-form animation two.gif


text:
File:Dform.ogg

Summary
Example for DAZ Studio D-Form tool.

title: File:Dform.ogg


text:
Airy stress function with body force

Airy stress function with body force potential
If a body force exists, the Airy stress function ($$\varphi$$) has to be combined with a body force potential ($$V$$). Thus,

\sigma_{11} = \varphi_{,22} + V ~; \sigma_{22} = \varphi_{,11} + V ~; \sigma_{12} = -\varphi_{,12} \qquad \text{(3)} $$ or,

\sigma_{11} = \cfrac{\partial^2\varphi}{\partial x_2^2} + V ~; \sigma_{22} = \cfrac{\partial^2\varphi}{\partial x_1^2} + V ~; \sigma_{12} = -\cfrac{\partial^2\varphi}{\partial x_1 \partial x_2} \qquad \text{(4)} $$

Is equilibrium still satisfied ?
Recall the equilibrium equation in two dimensions:

\boldsymbol{\nabla}\bullet{\boldsymbol{\sigma}} + \mathbf{f} = 0 ~; \sigma_{\beta\alpha, \beta} + f_{\alpha} = 0 \qquad \text{(5)} $$ or,

\begin{align} \sigma_{11,1} + \sigma_{21,2} + f_1 & = 0 \qquad \text{(6)}\\ \sigma_{12,1} + \sigma_{22,2} + f_2 & = 0 \qquad \text{(7)} \end{align} $$ In terms of $$\varphi$$,
 * $$\begin{align}

\varphi_{,122} + V_{,1} - \varphi_{,212} + f_1 & = 0 \qquad \text{(8)}\\ -\varphi_{,112} + \varphi_{,211} + V_{,2} + f_2 & = 0 \qquad \text{(9)} \end{align}$$ or,
 * $$\begin{align}

V_{,1} + f_1 & = 0 \qquad \text{(10)}\\ V_{,2} + f_2 & = 0 \qquad \text{(11)} \end{align}$$ Therefore,

f_1 = - V_{,1} ~; f_2 = - V_{,2} \qquad \text{(12)} $$ or,

f_1 = - \cfrac{\partial V}{\partial x_1} ~; f_2 = - \cfrac{\partial V}{\partial x_2}\qquad \text{(13)} $$ Hence,

\mathbf{f} = -\boldsymbol{\nabla}{V}\qquad \text{(14)} $$

 * Equilibrium is satisfied only if the body force field can be expressed as the gradient of a scalar potential.
 * A force field that can be expressed as the gradient of a scalar potential is called conservative.

What condition is needed to satisfy compatibility ?
Recall that the compatibility condition in terms of the stresses can be written as $$ \nabla^2{\sigma_{\gamma\gamma}} = -\cfrac{1}{\alpha} f_{\gamma,\gamma} \qquad \text{(15)} $$ where, $$    \alpha = \begin{cases} 1-\nu & \rm{for~ plane~ strain} \\ \cfrac{1}{1+\nu} & \rm{for~ plane~ stress} \end{cases} \qquad \text{(16)} $$ or, $$   \nabla^2{(\sigma_{11}+\sigma_{22})} + \cfrac{1}{\alpha}(f_{1,1}+f_{2,2}) = 0 \qquad \text{(17)} $$ or, $$   \sigma_{11,11}+\sigma_{11,22}+\sigma_{22,11}+\sigma_{22,22} + \cfrac{1}{\alpha}(f_{1,1}+f_{2,2}) = 0 \qquad \text{(18)} $$ which is the same as, $$   \cfrac{\partial^2\sigma_{11}}{\partial x_1^2}+ \cfrac{\partial^2\sigma_{11}}{\partial x_2^2}+ \cfrac{\partial^2\sigma_{22}}{\partial x_1^2}+ \cfrac{\partial^2\sigma_{22}}{\partial x_2^2}+ \cfrac{1}{\alpha}(\cfrac{\partial f_1}{\partial x_1}+\cfrac{\partial f_2}{\partial x_2}) = 0 \qquad \text{(19)} $$ Plug in the stress potential and the body force potential in equation (18) to get $$   \varphi_{,2211}+V_{,11}+\varphi_{,2222}+V_{,22}+ \varphi_{,1111}+V_{,11}+\varphi_{,1122}+V_{,22}+ \cfrac{1}{\alpha}(-V_{,11}-V_{,22}) = 0 \qquad \text{(20)} $$ or, $$   2\varphi_{,2211}+2V_{,11}+\varphi_{,2222}+2V_{,22}+ \varphi_{,1111}+ \cfrac{1}{\alpha}(-V_{,11}-V_{,22}) = 0 \qquad \text{(21)} $$ Rearrange to get $$   \varphi_{,1111} + 2\varphi_{,2211} + \varphi_{,2222} + \left(2 - \cfrac{1}{\alpha}\right) (V_{,11} + V_{,22}) = 0 \qquad \text{(22)} $$ which is the same as, $$   \cfrac{\partial^4\varphi}{\partial x_1^4} + 2\cfrac{\partial^4 \varphi}{\partial x_1^2 \partial x_2^2} + \cfrac{\partial^4\varphi}{\partial x_2^4} + \left(2 - \cfrac{1}{\alpha}\right) \left(\cfrac{\partial^2 V}{\partial x_1^2} + \cfrac{\partial^2 V}{\partial x_2^2}\right) = 0 \qquad \text{(23)} $$ Therefore, $$   \nabla^4{\varphi} + \left(2 - \cfrac{1}{\alpha}\right) \nabla^2{V} = 0 \qquad \text{(24)} $$

Compatibility is satisfied only if equation (24) is satisfied.

Equations for Airy stress function with body force potential
The relation between the Cauchy stress and the Airy stress function is (in direct tensor notation)

\boldsymbol{\sigma} =  \boldsymbol{\nabla}\times{\boldsymbol{\nabla}\times{\varphi}} $$ The relation between the body force and the body force potential is

\mathbf{f} = -\boldsymbol{\nabla}{V} $$ We also have to satisfy the compatibility condition for the Airy stress function to be a true stress potential, i.e.,

\nabla^4{\varphi} + \left(2 - \cfrac{1}{\alpha}\right) \nabla^2{V} = 0 $$

In rectangular Cartesian coordinates
In rectangular Cartesian coordinates, the relation between the Cauchy stress components and the Airy stress function + body force potential can be written as

\sigma_{11} = \varphi_{,22} + V ~; \sigma_{22} = \varphi_{,11} + V ~; \sigma_{12} = -\varphi_{,12} $$ or,

\sigma_{xx} = \cfrac{\partial^2\varphi}{\partial y^2} + V ~; \sigma_{yy} = \cfrac{\partial^2\varphi}{\partial x^2} + V ~; \sigma_{xy} = -\cfrac{\partial^2\varphi}{\partial x \partial y} $$ The relation between the body force components and the body force potential are:

f_1 = - V_{,1} ~; f_2 = - V_{,2} $$ or,

f_x = - \cfrac{\partial V}{\partial x} ~; f_y = - \cfrac{\partial V}{\partial y} $$ The compatibility condition is written as

\varphi_{,1111} + 2\varphi_{,2211} + \varphi_{,2222} + \left(2 - \cfrac{1}{\alpha}\right) (V_{,11} + V_{,22}) = 0 $$ or,

\cfrac{\partial^4 \varphi}{\partial x^4} + 2\cfrac{\partial^4 \varphi}{\partial x^2 \partial y^2} + \cfrac{\partial^4 \varphi}{\partial y^4} + \left(2 - \cfrac{1}{\alpha}\right) \left(\cfrac{\partial^2 V}{\partial x^2} + \cfrac{\partial^2 V}{\partial y^2}\right) = 0 $$

In cylindrical coordinates
In cylindrical coordinates, the relation between the Cauchy stresses and the Airy stress function + body force potential can be written as

\sigma_{rr} = \cfrac{1}{r}\cfrac{\partial\varphi}{\partial r} + \cfrac{1}{r^2}\cfrac{\partial^2\varphi}{\partial \theta^2} + V ~; \sigma_{\theta\theta} = \cfrac{\partial^2\varphi}{\partial r^2} + V ~; \sigma_{r\theta} = -\cfrac{\partial}{\partial r}         \left(\cfrac{1}{r}\cfrac{\partial\varphi}{\partial \theta}\right) \qquad \text{(25)} $$

The components of the body force are related to the body force potential via

f_r = - \cfrac{\partial V}{\partial r} ~; f_{\theta} = - \cfrac{1}{r}\cfrac{\partial V}{\partial \theta} \qquad \text{(26)} $$

The compatibility condition can be expressed as
 * $$\begin{align}

\left(\cfrac{\partial^2}{\partial r^2} + \cfrac{1}{r}\cfrac{\partial}{\partial r} +         \cfrac{1}{r^2}\cfrac{\partial^2}{\partial \theta^2}\right) & \left(\cfrac{\partial^2\varphi}{\partial r^2} + \cfrac{1}{r}\cfrac{\partial\varphi}{\partial r} +         \cfrac{1}{r^2}\cfrac{\partial^2\varphi}{\partial \theta^2}\right) + \\ &\left(2 - \cfrac{1}{\alpha}\right) \left(\cfrac{\partial^2 V}{\partial r^2} + \cfrac{1}{r}\cfrac{\partial V}{\partial r} +         \cfrac{1}{r^2}\cfrac{\partial^2 V}{\partial \theta^2}\right) = 0 \qquad \text{(27)} \end{align}$$

Related Content
Introduction to Elasticity

title: Airy stress function with body force


text:
User:Damërung

es:User:Damërung

title: User:Damërung


text:
User:Alcuin~enwikiversity

"And gladly wolde he lerne and gladly teche." from Geoffrey Chaucer's Canterbury Tales, General Prologue, line 308

Re natura usoris
A medievalist by nature, I am a graduate student in historical Germanic linguistics, having also studied the literatures of medieval Spain and England (all at a brickiversity). I am at Wikiversity because of its commitment to the original structure and goals of the medieval university, something that has become quite rare in the more financially-oriented universities of our time.

I have been a member of the Wikiversity community since July 30th, 2007.

For the time being, I will be active on Wikiversity about once per week.

A few odd facts

 * I have been married for three years
 * I was born in California
 * I have no children
 * I enjoy Mexican food
 * I find winter nights south of 42º N too short
 * I love hiking
 * I have a secret penchant for Dungeons & Dragons (but let's just keep that between us).

Faculty of Humanities

 * Classical Greek
 * Classics
 * German
 * Dutch
 * Afrikaans
 * Chinese
 * Mythology
 * Intellectual property law
 * Constitutional law
 * Theology
 * Religious studies

Faculty of Mathematics

 * Algebra
 * Combinatorics
 * Pure mathematics
 * Philosophy of mathematics
 * History of mathematics
 * Cryptography

Faculty of Social Sciences

 * Archaeology
 * Economics
 * Political Science
 * History
 * American History
 * European History

Faculty of Practical Arts and Sciences

 * Agriculture
 * Food and drink
 * Firearms handling

Faculty of Engineering and Technology

 * Computer programming
 * Civil engineering

Learning Projects

 * Homseschooling
 * Collaborative play writing
 * Chain mail construction
 * War seminar

Wolde gladly teche

 * Linguistics
 * English language
 * Spanish language
 * English literature
 * Spanish literature
 * Writing and composition

title: User:Alcuin~enwikiversity


text:
User talk:John van v~enwikiversity

Hello John van v, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 03:26, 1 August 2007 (UTC)

middle school learning resources
"middle school science" <-- Many contributors at Wikiversity have been college students, so there is an existing bias towards making Wikiversity look like a college course catalog. However, the mission of Wikiversity calls for this project to be for learners of all ages. I think we should change the entry points for people who are browsing Wikiversity so as to include explicit entry points for educators working at all age levels. Some ideas towards such change are here.

"experimenting with various wikis attempting to create a collaborative project" <-- Wikiversity can build on the success and traffic of Wikipedia. As soon as we have some learning resources at Wikiversity for a particular topic, we can link to those Wikiversity pages from related pages at Wikipedia. This is a good source of traffic for Wikiversity and provides a reasonable basis for the future growth of Wikiversity.

"create what is thought of as 'required learning' for teaching" <-- Such content would certainly be welcome at Wikiversity.

"my text, it is not yet ready for Wiki" <-- The great thing about wiki is that it provides tools for starting with a rough draft that can then be improved upon. You can even explicitly label wiki pages as "under construction". Most current Wikiversity pages are very rough, so you should not think that you need to have polished content before you put it on a Wikiversity page. Much of Wikiversity is currently just content development projects, pages where participants make plans for learning resources. You could certainly start a Wikiversity content development project for future collaboration aimed at making educational resources oriented towards teaching middle school science.

"I am very interested in project-based learning" <-- We want Wikiversity to emphasize active, "learn by doing" approaches. A basic type of "learning resource" that is welcome here at Wikiversity is the "learning project", which should provide a framework for activities that learners can become engaged in. Since wiki is all about making collaboration possible, Wikiversity learning projects can naturally build upon the idea of groups of learns working together to learn something.

"the Wikipedia 'Project-Based Learning' article" <-- Wikiversity can go into such topics in more detail than Wikipedia. Wikibooks could also have a whole book on the topic.--JWSchmidt 17:13, 1 August 2007 (UTC)

Gizmo
"Gizmo and its underlying SIP protocol" <-- this is the first I've ever heard of "Gizmo". Is it this? Some folks at Wikiversity have been experimenting with VOIP as described at Wiki Campus Radio.

"Middle school science.....I should just go ahead and create this page?" <-- Yes, I see no reason not to just jump in and start a page at Wikiversity. --JWSchmidt 17:20, 2 August 2007 (UTC)

collaborators
I think Wikiversity is about 5 years behind Wikipedia. At first, almost nobody knew that Wikipedia existed and few people knew about the idea of using wiki technology to create an encyclopedia. It took about 5 years for the world to "catch on" and "discover" Wikipedia. Wikiversity is still "pre-discovery", people do not know that it exists and people do not understand how to use wiki technology to support online learning. At this point, making Wikiversity grow and making Wikiversity a center for online education-oriented collaborations requires some "evangelism" and social networking. I'll take a look at Middle School Science. What do you think about starting a portal page that could link to Middle School Science? I was previously thinking about portals for primary and secondary education. I'm not sure how middle school would fit into that kind of plan. --JWSchmidt 18:14, 17 August 2007 (UTC)

 * Hi John van v and welcome. Your Middle School Science project looks both interesting and useful.  If I understand it properly it is intended to help teachers self educate in their professional qualifications.   I wonder if you (and/or other collaborators) would be interested in evaluating and providing feedback regarding a couple of learning projects and how they could improved and made more attractive to middle schoolers and/or their teachers.   and .   Both projects were intended to encourage participation but both have failed to attract much interest.   Much of this is due to low site traffic but any suggestions on how to make the projects more attractive for immediate participation would be greatly appreciated.  Thanks!  Mirwin 08:14, 16 September 2007 (UTC)

Contacting you?
How do people contact you? Robert Elliott 06:51, 23 September 2007 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

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

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

title: User talk:John van v~enwikiversity


text:
User talk:Damërung

Hello Damërung, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 03:27, 1 August 2007 (UTC)

title: User talk:Damërung


text:
C Programming/Functions

= Objective =
 * Familiarise yourself with functions
 * How to define them
 * How to invoke them
 * Prototyping

= Introduction to functions in C = In today's world we believe that Time is Money. The less time you spend in writing a program, the more efficient use of resources. Functions helps a programmer avoid rewritting code over and over, even more so than looping, and hence save time, cost, and effort.

Let's say we would like to write a program that calculates the square of a number. The application that we are asked to design requires that you calculate the square of several numbers again and again. So, you will write the following code to calculate the square of the number wherever needed.

If you have to calculate the square of the number at 3 different places in your program, you will have to write these lines of code over and over again.

Luckily, we do not need to do this, because C, like most programming languages, allows us to declare functions. A function is named a block of code that, when invoked, is executed. This is what a function definition looks like: returntype name(parameters ...) {    statements ... }

A C function may specify any primitive type or a struct (a special C type that will be explored further in a later lesson) as its return type, or void if the function does not return anything. Functions with  as a return type is also termed procedures, which is old-fashioned nowadays (the term procedures is still used in Pascal). If a function defines a return type, it must  a variable, constant, or literal of that type before the function-closing   is encountered, or a compiler error will occur. A function definition must also specify its parameters, a list of comma-separated variable declarations, or  if it does not have any parameters. If a function does not specify any parameters, then its parameter list is implicitly, and does not need to be declared as such. Here is a C function that calculates the square of an integer :

Once a function has been defined, we can now use this function wherever we would like to use it. For example:

Thus, instead of writing the entire code over and over again, we can simply call the function here.

It's also important to note that whitespace is ignored - so the following snippets are all the same: int func_name(void) { } int func_name (void) { }

Each of these snippets incorporating elements of possible styles for writing C code. The main thing to note about whitespace is it only matters when it would be between two different things which could then make a single "symbol" (a symbol is a name representing a certain value - be it a function/function location or a variable). With proper syntax in all other ways, this only can cause problems when declaring variables/functions, or when casting. Just remember to place some whitespace between each specifier, qualifier, modifier, the type and the symbol name (specifiers, qualifiers and modifiers explained in that page). Programmers prefer to use a space, but these have the same meaning: int x=1 int x	=

1

= Function prototype = A function prototype or function interface is a declaration of a function that specifies the function’s name and type signature (data types of parameters, and return type), but omits the function body.

Functions are the backbone of any C algorithm. All C data types can be passed to a function, even with an array (if passed they will be converted to pointers, which will be discussed in later lesson) or function (as callback function), and all may be returned.

Functions can also call other functions:

Function arguments as void: A function with no parameter can accept a.

= main function = main, the center of every C program, can have many forms: int main{} int main(void){} int main(int argc, char **argv){} int main(int argc, char **argv, char **envp){} //Note that this syntax is not specified in POSIX, but is used by all three main OSes. GCC compiler expects main to return an int. If the code finishes with no problems, main should return 0. Otherwise, it should return some other value - normally 1. The syntax for a main function that simply returns 0 is: int main(void) { return 0; }

= Command line arguments with main =

It is possible to pass some values from the command line to your C programs when they are executed. These values are called command line arguments. Example Input: ./a.out Hello
 * argc: number of arguments passed
 * argv[]: pointer array which points to each argument passed to the program

Output

To enter a string in, use  :

Input:

Output

= Assignments =

Converts the given Fahrenheit temperature to Celsius: fahrenheit_to_celsius.c (Source code from the Programming Fundamentals course)

 * Previous Lesson: Operators
 * Next Lesson: Flow Control
 * Course Home Page

title: C Programming/Functions


text:
User talk:Cmptrgy412~enwikiversity

Hello Cmptrgy412, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 04:49, 1 August 2007 (UTC)

Your account will be renamed
Hello,

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

Unfortunately, your account clashes with another account also called Cmptrgy412. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Cmptrgy412~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:32, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:05, 19 April 2015 (UTC)

title: User talk:Cmptrgy412~enwikiversity


text:
BCP/Albizia julibrissin/Logs

2007 Logs

 * --SB_Johnny | talk 11:10, 4 July 2007 (UTC)
 * --SB_Johnny | PA! 09:29, 1 August 2007 (UTC)

2008 Logs

 * (no archived logs yet)

Recent Logs

 * --SB_Johnny | talk 07:29, 29 June 2008 (UTC)
 * --SB_Johnny | talk 09:05, 14 July 2008 (UTC)
 * --SB_Johnny | talk 13:00, 28 July 2008 (UTC)
 * --SB_Johnny talk 10:29, 17 August 2008 (UTC)
 * --SB_Johnny talk 09:23, 5 September 2008 (UTC)

title: BCP/Albizia julibrissin/Logs


title: Category:BCP/NHAMP/7


text: Category:BCP/NHAMP/IP

title: Category:BCP/NHAMP/IP


text: Category:Bloom Clock/New Hampshire Categories


text: Category:BCP/NHAMP/NP

title: Category:BCP/NHAMP/NP


text: Category:Policy discussions

title: Category:Policy discussions


text:
Wikiversity:Colloquium/archives/August 2006/Blocking policy

Blocking policy
I've set up the shell of a blocking policy at WV:BLOCK, and I'd appreciate input and expansion! --digital_metalk 22:31, 18 August 2006 (UTC)

 * I will say that as a personal policy, I block anybody that adds profanity, particularly if it is in the name of the user. That is immediate and usually infinite, especially if that is about the only edit the user has made.  As for anon users, I tend to follow a policy of starting out with modest blocks (about a year) and move up over time.  There is also a general policy of blocking open proxies with infinite blocks that has been suggested on Meta.
 * It would be nice to have some people with Check User rights here, but that is very unlikely at the moment for reasons I won't get into here (it is a part of my political beef with the WMF right now and I will be spewing some very POV stuff here). I think that we can work around that, however.  --Robert Horning 05:30, 19 August 2006 (UTC)

I don't think we'll need to use CheckUser at this stage, unless we come under fire from some serious, concentrated vandalism, in which case, we can ask a steward for assistance. robchurch | talk 22:15, 20 August 2006 (UTC)


text:
Wikiversity:Colloquium/archives/September 2006/Deletion of userpages

Deletion of userpages
I disagree with this policy; I think it would be good to have templates for these etc. What does everyone think?? --Hotrocks 16:34, 8 September 2006 (UTC)
 * We do not need shrines for vandals. Nor do we need pages so they can have something to link to as proof of their "great deeds". And we certainly do not need custom templates for each vandal so they can be even more excited about vandalizing here.
 * Vandalism is a nuisance and will be addressed as such but it's not a big issue, nor should it be made into one. -- sebmol ? 17:07, 8 September 2006 (UTC)
 * I completely agree here. I have agressively deleted these vandalism user templates when found, and certainly don't think we should be glorifying individuals who are counter-productive to the mission of Wikiversity and Wikimedia projects in general.  --Robert Horning 18:34, 8 September 2006 (UTC)
 * Just throwing in my lot with sebmol and Robert Horning, I agree that we shouldn't be glorifying vandalism, and I have deleted several of these templates.-- digital  _  me   22:22, 8 September 2006 (UTC)
 * I've taken a moment to write up an essay describing the position on vandals. It's suitably located at WV:SHRINE. -- sebmol ? 21:43, 9 September 2006 (UTC)


text:
Wikiversity:Colloquium/archives/September 2006/Email

Email
We have a few people who provided their email on the wiki pages, like School of Mathematics:Introductory Real Analysis. What should we do with them, email them as ask them to help or delete it so that the spammer don't get to them?--Rayc | (Talk) 05:07, 9 September 2006 (UTC)

 * Good question. Maybe change the addresses so spammer bots can't read them as easily and put a warning above it not to post e-mail addresses. -- sebmol ? 08:37, 9 September 2006 (UTC)

 * What if it's an email address set up for using this Wiki only?? --82.42.145.158 09:56, 9 September 2006 (UTC)

 * They might still get spammed. There are constantly programs running through web sites collecting e-mail addresses to spam. This page is definitely on the sites to visit for those programs because of its association with Wikipedia. -- sebmol ? 11:04, 9 September 2006 (UTC)

 * This wiki has a emailing function. We should permote it. -- Matteo 05:19, 15 September 2006 (UTC)


text: Category:Colloquium archives by thread

title: Category:Colloquium archives by thread


text: Category:Policy discussions/Vandalism

title: Category:Policy discussions/Vandalism


text:
Filmmaking Basics/3D Storyboard/Corrections for Mok

Rough Draft by Sir Mok

Corrections


text:
Portal:Liberal arts

Welcome
Welcome to the Liberal Arts Portal! This page connects browsers to the learning materials that have been developed by the various Wikiversity Liberal Arts content development projects. This portal features exciting examples of Liberal Arts learning resources.

The content development projects are the schools, departments and divisions linked to from this page. Wikiversity participants who are interested in the Liberal Arts are invited to create and develop learning projects and learning resources and help organize access to them by developing this portal.

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title: Portal:Liberal arts


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Python Concepts/Functions

Objective

 * Discuss the purpose of functions.
 * Learn how to create a simple function.
 * Understand the use of parameters.
 * Work with "default parameter values".
 * Work with positional arguments.
 * Work with keyword arguments.
 * Use the return statement within a function.
 * Understand the location and abstraction of a function.

Functions
A function is a way of reusing a block of code so that it can be used more than once. Functions give the programmer the ability to reduce code redundancy and eliminate hours of wasted work. Each function allows the code to be centralized, where one change to the function will affect everything that depends on it. Functions can be used to divide and conquer tasks that would seem otherwise impossible to accomplish!

Creating A Function
To create a function, use the  keyword, followed by the function's name, with an empty pair of parentheses, ending with a colon. Functions make up a code block, so you'll need to indent any code that's within the function. Don't forget to indent, otherwise you'll get the pesky "IndentationError".

The code above assigns and creates the function called. It acts much like the the command ; both the given examples create a new variable,   and   respectively.

Now that you've created the function nothing should visibly happen. To actually invoke or call the function, you'll need to use the function's name followed by the parentheses as shown in the script below.

The output should be:

Hello, world!

Adding the command  again will result in the words   being displayed twice. This demonstrates a small fraction of the function's uses and abilities.

Function Parameters
At this point, you might be wondering why we need to have those pointless parentheses when we call the function. Wouldn't  be a lot easier than. Believe it or not, but we can pass arguments to the function when we include them within the parentheses. This allows us to create functions that do things based on their given arguments. First, the arguments need to be declared when the function is first defined and then we need to call it with the correct number of arguments.

The following output should be: 4 9 16

now acts like a variable within the function. You should note that x will only exist within the function. Trying something like the example below will definitely cause a.

A function can take multiple arguments with any two adjacent arguments separated by a comma if more than one exists as demonstrated in the next example.

The following output should be: 7 11

Default Parameters
Having parameters is great, but it can be repetitive to continue to use them over and over, again. We can define a default value for a parameter by setting its value when you define the parameter. Take the script below for example.

The output should be: 1 0 23

Did you see the second time we called ? We didn't give an argument, yet it still printed out. If no argument is given, a default is automatically assigned.

Positional Arguments
A positional argument is an argument that is not a keyword argument. In an argument list it occupies a position that corresponds to a parameter in the function definition. In all four function calls above numbers  are positional arguments. Number  occupies the position that corresponds to arg1 in the function definition; number   occupies the position that corresponds to arg2 in the function definition; number   occupies the position that corresponds to arg3 in the function definition. All four of the above function calls produce the same output: The following syntax accepts a variable number of arguments: Python's syntax accepts required and optional arguments and a variable number of arguments: Parameters are part of the function definition. When the function is invoked, arguments are passed to the function and each argument passed must correspond to a parameter in the definition of the function.

Keyword Arguments
Keyword arguments are arguments that are supplied to a dictionary within the function. They may be supplied as the sole list of arguments or they may follow other arguments: or We saw keyword arguments used in Dictionaries:Other operations on dictionaries This lesson shows by examples how they may be passed and received. The result is:

The following example shows how (optional) positional arguments and (optional) keyword arguments may be passed to a given function upon invocation.

Returning A Value
If we wanted to use the value that the square function gives us, we would rewrite it to use the  command. Which can then be used to do such fun things as: or or even in statements such as: and If the function is to return a value, the  statement is required. Without it the function returns

Assignments

 * Previous Lesson:Directories
 * This Lesson:Functions
 * Next Lesson:Variable Scoping
 * Course Home Page

title: Python Concepts/Functions


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Acceptable Use Policies (AUP)

The following is an example.

Acceptable Use Policy (AUP)

The use of the Internet must be in support of education and consistent with the educational objectives of the Board of Education. Users must also comply with the rules and regulations of the Educational Telecomputing policies.

As a computer user I agree to follow the rules in all of my work with computers while attending this school: I recognize that all computer users have the same rights to use the equipment; therefore,
 * I will not play games or use the computer resources for non-academic activities when others require the system for academic purposes;
 * I will not waste or take supplies such as paper, printer ribbons, and diskettes that are provided by the school;
 * I will avoid excessive use of network hard disk space by removing any files that has not been used within one month.

I recognize that software is protected by copyright laws; therefore,
 * I will not make unauthorized copies of software found on school computer by copying them onto my own diskettes or onto other computers through electronic mail or bulletin boards;
 * I will not give, lend, or sell copies of software to others unless I have the written permission of the copyright owner or the original software is clearly identified as shareware or in the public domain.

I recognize that the work of all users is valuable; therefore,
 * I will protect the privacy of others by not trying to learn their passwords;
 * I will not copy, change, read, or use files in another user’s area without that user’s permission;
 * I will not attempt to gain unauthorized access to system programs or computer equipment;
 * I will not create or introduce any virus which may destroy files or disrupt service to other users;
 * I will not download information onto the hard drives of any computer for permanent storage.

User, Parent, and Teacher must sign agreement form in order for student to have access to computers in the school system.

Acceptable Use Policy

USER – Please sign and return

I understand and will abide by the above Acceptable Use Policy when using the school’s computer network. I further understand that any violation of the regulations above is unethical and may constitute a criminal offense. Should I commit any violation, my access privileges may be revoked, school disciplinary action may be taken, and/or appropriate legal action may be pursued.

User’s name (Please print):________________________________________________

Signature:_____________________________________________________________

Date: __________________________________ Grade:________________________

Parent – MUST be signed for Internet Access

I, the parent/guardian of _________________________________________, have read and discussed the above Acceptable Use Policy (AUP) for computer networks with my child. I understand that this access is designed for educational purposes. The school has taken precautions to eliminate controversial material. However, I also recognize it is impossible to restrict access to all controversial materials, and I will not hold them responsible for materials acquired on the network. Further, I accept full responsibility for supervision when my child’s use is not in a school setting. I hereby give permission to issue an account for my child and certify that the information contained on this form is correct.

Parent/Guardian’s Name (Please print):______________________________________

Signature:_____________________________________________________________

Date: _________________________________________________________________

Teacher – MUST be signed for Internet Access

I agree to promote the AUP with this user. Because the user may use the network for individual work or in the context of another class, I cannot be held responsible for the user’s use of the network. As the sponsoring teacher, I do agree to instruct the user on the acceptable use of the network and proper network etiquette.

Sponsoring Teacher’s Name (Please print):___________________________________

Signature: _____________________________________________________________

Date: ________________________________________________________________

title: Acceptable Use Policies (AUP)


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Talk:C++/Simple Math

Just an FYI the last example has a typo:

"pow(myValue,myPowner);   // myValue is the base and myPower is the exponent" ^ the n should be removed or            ^ the n should be added

This is a great article, and I am enjoying reading it.

Thanks, Russell

Hi!

Could it be that the the variable  in the third line of the code should be renamed in ?

AND I think the  2.00 should  be replaced by an   2. At least my Dev-C++ (version 4.9.9.2) can't handle a  as the second parameter of the   function.

BTW, nice tutorial so far.

GreetZ

title: Talk:C++/Simple Math


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User talk:RobboA

Hello RobboA, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --HappyCamper 16:12, 1 August 2007 (UTC)

Image:AnotherRobert.jpg
Please provide a source and license template for this image. Thanks. – Mike.lifeguard | @en.wb 22:48, 13 February 2008 (UTC)

title: User talk:RobboA


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Wikiversity:Colloquium/archives/August 2006/Logo and slogan

Logo and slogan
OK, we are the "The limitless learning center" according to wikipedia. After about a week (to get partisipents) we should start up a slogan and logo contest.--Rayc 00:21, 18 August 2006 (UTC)

 * "The limitless learning center" might be a good slogan, but it's a poor description. (It tells readers nothing about the project's nature.)  I changed the text to "Free learning materials."  —David Levy 07:24, 18 August 2006 (UTC)

 * I've used "Free Learning tools" on the other Wikimedia websites. I am still not sure what should go here, but it is important that people are made aware of this project.  --Robert Horning 13:18, 18 August 2006 (UTC)

 * How about "Open learning centre"? Or is that too similar to "Open University" (a household name in the UK)? And on the logo contest, if someone wants to organise these, please be bold and do so. It should even help generate further momentum in the project. Cormaggio 13:24, 18 August 2006 (UTC)

Is there enough room to make it "Free learning materials and activities"? --JWSchmidt 13:42, 18 August 2006 (UTC)

 * JW, I like the one you mention. This catches the dual nature of textual study and engaged learning. To tweak: How about "free learning resources and activities." Materials sounds more limiting than resources. Reswik 01:21, 19 August 2006 (UTC)

 * How about the "Free Learning Community"? Is inclusive and stresses the concept of learning. Awolf002 01:51, 19 August 2006 (UTC)
 * Maybe we could have a list of good slogans and switch between them on a regular basis. --JWSchmidt 02:02, 19 August 2006 (UTC)

On second thought and taking a step back... A consensus mission statement could be very helpful. Perhaps a consensus mission statement needs to be developed first. Then, a slogan could reflect that. But can the mission be settled while the status and extent of research in WV is up in the air? If extensive research sponsorship is eventually a part of WV, then a slogan might read like: "creating and sharing free knowledge." And this is really the nature of a university or any academy of scholars: creating and sharing knowledge. The free culture movement adds freedom and co-creation of content to that mission. Sharing can imply co-creation, so one can add freedom to "creating and sharing knowledge." Sharing also implies service another aspect of the life of universities. There may be actually too much emphasis on learning in some WV discussions at the cost of service and knowledge creation -- which are also primary motivations of scholars. Reswik 02:49, 19 August 2006 (UTC)
 * That's a fine idea, Doug. Please, everyone, help out at Wikiversity mission. -- Cormaggio 07:47, 19 August 2006 (UTC)
 * Thanks. I made extensive edits to the mission page to outline a Delphi method planning process for co-creating a mission statement. What do ya think? Reswik 13:25, 19 August 2006 (UTC)
 * We need to lose the beta on the logo. This is not a software project.  The project is fundamentally about building online learning communities.  Would you spend your precious social/learning time joining a club of card players, fraternity, or writing circle with a loud "Beta, planned shutdown next week." painted on the front door? Mirwin 07:22, 23 August 2006 (UTC)

 * Marking Wikiversity as "beta" is just a warning to the outside world that nobody should expect to find a finished "product" on any Wikiversity page. Speaking bluntly, I suspect it (the beta period) is also a way to "cover one's ass". I know that during the next six months Jimbo will run into people who say, "I saw Wikiversity and, you know, there is nothing useful there." But Jimbo will be able to say, "Ya, well, it is in a beta phase, just getting on its feet. It is not really open for serious use yet." Jimbo should have that kind of "out". --JWSchmidt 20:04, 23 August 2006 (UTC)


text: Category:Policy discussions/Logo, slogan, and motto


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User:MosheA

About Me
I'm very active on the English Wikipedia, on which I have +1000 edits. Check out my page there.

title: User:MosheA


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User:Wikicollege creator DS07

Hi, i am the creator of Wikicollege, the exclusive but payed college online you can go on once you graduate from Wikiversity. I go by my stage names, Metamorphosis or DS07 or just call me Wikiman, for I have users on Wikipedia, and other sister projects of Wikiversity.

title: User:Wikicollege creator DS07


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User talk:Wikicollege creator DS07

I am a Wikicolleague!
Assignment 1 Lesson:Formatting the script From Wikiversity Jump to: navigation, search

This school is: Wikiversity Film School - Narrative film production This course is: The basics of narrative filmmaking This lesson is: Formatting the script

Pages of this lesson:

Page 1 - Lesson summary & the story

Page 2 - Analyze the story

Page 3 - Learn Final Draft

Page 4 - Type the script

The completed assignments (formatted scripts)

Analyze the story and format the movie script using a script formatting program

For the first lesson of this course, you are given a short story and you are told to format the story into the proper format for a motion picture script.

A golden opportunity - Final Draft FREE demo version Now is an excellent time to learn script formatting. This is a task that people do every day in Hollywood so this lesson give you excellent practice.

A wonderful tool You can download a free yet working demo version of Final Draft which is a script typing and script formatting program. This is the ideal program for learning script formatting. There are versions for both the Macintosh and Windows.

An interesting alternative I prefer that you use Final Draft FREE demo version. But if that is not possible, you can download a free program called CeltX which is still under construction (version 0.99) and the manual is incomplete. However, it looks very interesting, it is free, and it works on Linux as well as Macintosh and Windows.

Why such a simple story?

This is an extremely simple story. It seems trivial. You say, "I want something more challenging!"

I have heard this a thousand times from beginning filmmakers. Filmmaking looks so easy.

Wrong! This movie will be anything but easy. By the time we turn this simple script into a motion picture, this tiny movie will require a huge amount of effort. Filmmaking is never as simple as it looks. With this very simple scene, you will have the opportunity to do it all.

The steps of this lesson

1. Look at the story To the right, you see the story for "Seduced by the Dark Side!". Even though this is a very short story, it still needs proper formatting because this movie will be worked on by hundreds of people. Everyone in Hollywood expects to see your script typed in the proper format.

2. Identify the elements On page 2, you see how the story is broken down into the different elements such as CHARACTER and DIALOG. Before you turn to this page, think about what elements of the story are needed for the script.

3. Download and learn to use Final Draft demo version On page 3, you get a brief introduction to Final Draft. Final Draft is both a very professional program with lots of features and a simple typing program. First, you need to learn all the elements of a formatted script. Your second step is to learn how to navigate inside of Final Draft.

Important: Even though Final Draft has a tutorial, you will learn navigation mostly by trial and error using the TAB and the RETURN keys.

4. Typing the script - Line by Line On page 4, you as shown all the steps of typing this script using Final Draft. Try doing this without looking. If you have studied the script and learned how to format using Final Draft, you should not need to read this page… but I include it just in case. HYPERLINKS Course #1: 	Basic Filmmaking Course #2: 	Film Editing Introduction Course #3: 	Film Scoring for filmmakers Course #4: 	Film Scoring for musicians Exercise: 	Script Writing Exercise Project: 	The Storyboard Artwork Project Future: 	Digital Puppet Animatio

title: User talk:Wikicollege creator DS07


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General stress functions in polar coordinates

Related Content
Introduction to Elasticity


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Stresses in wedges

Wedge with Boundary Tractions
Suppose Then
 * The tractions on the boundary vary as $$r^n$$.
 * No body forces.
 * $$\text{(1)} \qquad

\varphi = r^{n+2} f(\theta) $$ To find $$f(\theta)$$ plug into $$\nabla^4{\varphi} = 0$$.
 * $$\text{(2)} \qquad

\left(\frac{d^2}{d\theta^2} + (n+2)^2\right) \left(\frac{d^2}{d\theta^2} + n^2\right) f(\theta) = 0 $$ If $$n \ne 0$$ and $$n \ne -2$$,
 * $$\text{(3)} \qquad

\varphi = r^{n+2} \left[ a_1 \cos\{(n+2)\theta\} + a_2 \cos(n\theta) + a_3 \sin\{(n+2)\theta\} + a_4 \sin(n\theta)\right] $$ The corresponding stresses and displacements can be found from Michell's solution. We have to take special care for the case where $$n = 0$$, i.e., the traction on the surface is constant.

Williams' Asymptotic Solution
''Ref: M.L. Williams, ASME J. Appl. Mech., v. 19 (1952), 526-528.''
 * Stress concentration at the notch.
 * Singularity at the sharp corner, i.e, $$\sigma_{ij}\rightarrow\infty$$.
 * William's solution involves defining the origin at the corner and expanding the stress field as an asymptotic series in powers of r.
 * If the stresses (and strains) vary with $$r^{\alpha}$$ as we approach the point $$r = 0$$, the strain energy is given by
 * $$\text{(4)} \qquad

U = \frac{1}{2}\int_0^{2\pi}\int_0^r\sigma_{ij}\varepsilon_{ij}~r~dr~d\theta = C\int_0^r r^{2a+1} dr  $$ This integral is bounded only if $$a > -1$$. Hence, singular stress fields are acceptable only if the exponent on the stress components  exceeds $$-1$$.

Stresses near the notch corner

 * Use a separated-variable series as in equation (3).
 * Each of the terms satisfies the traction-free BCs on the  surface of the notch.
 * Relax the requirement that $$n$$ in equation (3) is an integer. Let $$ n = \lambda-1$$.
 * $$\begin{align}

\text{(5)} \qquad \varphi = r^{\lambda+1} \left[\right. & a_1 \cos\{(\lambda+1)\theta\} + a_2 \cos{(\lambda-1)\theta} +\\ & a_3 \sin\{(\lambda+1)\theta\} + a_4 \sin{(\lambda-1)\theta}\left.\right] \end{align}$$ The stresses are
 * $$\begin{align}

\sigma_{rr} = r^{\lambda-1} \left[\right. & -a_1\lambda(\lambda+1)\cos\{(\lambda+1)\theta\} -a_2\lambda(\lambda-3)\cos\{(\lambda-1)\theta\} \\ & -a_3\lambda(\lambda+1)\sin\{(\lambda+1)\theta\} -a_4\lambda(\lambda-3)\sin\{(\lambda-1)\theta\}\left.\right] \text{(6)} \qquad \\ \sigma_{r\theta}= r^{\lambda-1} \left[\right. & +a_1\lambda(\lambda+1)\sin\{(\lambda+1)\theta\} +a_2\lambda(\lambda-1)\sin\{(\lambda-1)\theta\} \\ & -a_3\lambda(\lambda+1)\cos\{(\lambda+1)\theta\} -a_4\lambda(\lambda-1)\cos\{(\lambda-1)\theta\}\left.\right] \text{(7)} \qquad \\ \sigma_{\theta\theta}= r^{\lambda-1} \left[\right. & +a_1\lambda(\lambda+1)\cos\{(\lambda+1)\theta\} +a_2\lambda(\lambda+1)\cos\{(\lambda-1)\theta\} \\ & +a_3\lambda(\lambda+1)\sin\{(\lambda+1)\theta\} +a_4\lambda(\lambda+1)\sin\{(\lambda-1)\theta\}\left.\right]\text{(8)} \qquad \end{align}$$ The BCs are $$\sigma_{r\theta} = \sigma_{\theta\theta} = 0$$ at $$\theta = \alpha$$.Hence,
 * $$\begin{align}

0= r^{\lambda-1} \lambda \left[\right. & +a_1(\lambda+1)\sin\{(\lambda+1)\alpha\} +a_2(\lambda-1)\sin\{(\lambda-1)\alpha\} \\ & -a_3(\lambda+1)\cos\{(\lambda+1)\alpha\} -a_4(\lambda-1)\cos\{(\lambda-1)\alpha\}\left.\right] \text{(9)} \qquad \\ 0= r^{\lambda-1} \lambda \left[\right. & -a_1(\lambda+1)\sin\{(\lambda+1)\alpha\} -a_2(\lambda-1)\sin\{(\lambda-1)\alpha\} \\ & -a_3(\lambda+1)\cos\{(\lambda+1)\alpha\} -a_4(\lambda-1)\cos\{(\lambda-1)\alpha\}\left.\right] \text{(10)} \qquad \end{align}$$ The BCs are $$\sigma_{r\theta} = \sigma_{\theta\theta} = 0$$ at $$\theta = -\alpha$$.Hence,
 * $$\begin{align}

0= r^{\lambda-1} \lambda \left[\right. & +a_1(\lambda+1)\cos\{(\lambda+1)\alpha\} +a_2(\lambda+1)\cos\{(\lambda-1)\alpha\} \\ & +a_3(\lambda+1)\sin\{(\lambda+1)\alpha\} +a_4(\lambda+1)\sin\{(\lambda-1)\alpha\}\left.\right] \text{(11)} \qquad \\ 0= r^{\lambda-1} \lambda \left[\right. & +a_1(\lambda+1)\cos\{(\lambda+1)\alpha\} +a_2(\lambda+1)\cos\{(\lambda-1)\alpha\} \\ & -a_3(\lambda+1)\sin\{(\lambda+1)\alpha\} -a_4(\lambda+1)\sin\{(\lambda-1)\alpha\}\left.\right]\text{(12)} \qquad \end{align}$$ The above equations will have non-trivial solutions only for certain eigenvalues of $$\lambda$$, one of which is $$\lambda = 0$$. Using the symmetries of the equations, we can partition the coefficient matrix.

Eigenvalues of λ
Adding equations (9) and (10),
 * $$\text{(13)} \qquad

a_3(\lambda+1)\cos\{(\lambda+1)\alpha\} + a_4(\lambda-1)\cos\{(\lambda-1)\alpha\} = 0 $$ Subtracting equation (10) from (9),
 * $$\text{(14)} \qquad

a_1(\lambda+1)\sin\{(\lambda+1)\alpha\} + a_2(\lambda-1)\sin\{(\lambda-1)\alpha\} = 0 $$ Adding equations (11) and (12),
 * $$\text{(15)} \qquad

a_1(\lambda+1)\cos\{(\lambda+1)\alpha\} + a_2(\lambda+1)\cos\{(\lambda-1)\alpha\} = 0 $$ Subtracting equation (12) from (11),
 * $$\text{(16)} \qquad

a_3(\lambda+1)\sin\{(\lambda+1)\alpha\} + a_4(\lambda+1)\sin\{(\lambda-1)\alpha\} = 0 $$ Therefore, the two independent sets of equations are
 * $$\text{(17)} \qquad

\begin{bmatrix} (\lambda+1)\sin\{(\lambda+1)\alpha\} & (\lambda-1)\sin\{(\lambda-1)\alpha\} \\ (\lambda+1)\cos\{(\lambda+1)\alpha\} & (\lambda+1)\cos\{(\lambda-1)\alpha\} \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ and
 * $$\text{(18)} \qquad

\begin{bmatrix} (\lambda+1)\cos\{(\lambda+1)\alpha\} & (\lambda-1)\cos\{(\lambda-1)\alpha\} \\ (\lambda+1)\sin\{(\lambda+1)\alpha\} & (\lambda+1)\sin\{(\lambda-1)\alpha\} \end{bmatrix} \begin{bmatrix} a_3 \\ a_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Equations (17) have a non-trivial solution only if
 * $$\text{(19)} \qquad

\lambda\sin(2\alpha) + \sin(2\lambda\alpha) = 0 $$ Equations (18) have a non-trivial solution only if
 * $$\text{(20)} \qquad

\lambda\sin(2\alpha) - \sin(2\lambda\alpha) = 0 $$
 * From equation (4), acceptable singular stress fields must have $$\lambda > 0$$.Hence, $$\lambda = 0$$ is not acceptable.
 * The term with the smallest eigenvalue of $$\lambda$$ dominates the  solution.  Hence, this eigenvalue is what we seek.
 * $$\lambda = 1$$ leads to $$\varphi = a_4 \sin(0)$$. Unacceptable.
 * We can find the eiegnvalues for general wedge angles using  graphical methods.

Special case : α = π = 180°
In this case, the wedge becomes a crack.In this case,
 * $$\text{(21)} \qquad

\lambda = \frac{1}{2}, 1, \frac{3}{2}, $$ The lowest eigenvalue is $$1/2$$. If we use, this value in equation (17), then the two equations will not be linearly independent and we can express them as one equation with the substitutions
 * $$\text{(22)} \qquad

a_1 = \frac{A}{2}\sin\left(\frac{\alpha}{2}\right) ~; a_2 = -\frac{3A}{2}\sin\left(\frac{3\alpha}{2}\right) $$ where $$A$$ is a constant. The singular stress field at the crack tip is then
 * $$\begin{align}

\sigma_{rr} & = \frac{K_I}{\sqrt{2\pi r}} \left[\frac{5}{4}\cos\left(\frac{\theta}{2}\right) - \frac{1}{4}\cos\left(\frac{3\theta}{2}\right)\right] \text{(23)} \qquad \\ \sigma_{\theta\theta} & = \frac{K_I}{\sqrt{2\pi r}} \left[\frac{3}{4}\cos\left(\frac{\theta}{2}\right) + \frac{1}{4}\cos\left(\frac{3\theta}{2}\right)\right] \text{(24)} \qquad \\ \sigma_{r\theta} & = \frac{K_I}{\sqrt{2\pi r}} \left[\frac{1}{4}\sin\left(\frac{\theta}{2}\right) + \frac{1}{4}\sin\left(\frac{3\theta}{2}\right)\right] \text{(25)} \qquad \end{align}$$ where, $$K_I$$ is the { Mode I Stress Intensity Factor.}
 * $$\text{(26)} \qquad

K_I = 3A\sqrt{\frac{\pi}{2}} $$ If we use equations (18) we can get the stresses due to a mode II loading.
 * $$\begin{align}

\sigma_{rr} & = \frac{K_{II}}{\sqrt{2\pi r}} \left[-\frac{5}{4}\sin\left(\frac{\theta}{2}\right) + \frac{3}{4}\sin\left(\frac{3\theta}{2}\right)\right] \text{(27)} \qquad \\ \sigma_{\theta\theta} & = \frac{K_{II}}{\sqrt{2\pi r}} \left[-\frac{3}{4}\sin\left(\frac{\theta}{2}\right) - \frac{3}{4}\sin\left(\frac{3\theta}{2}\right)\right] \text{(28)} \qquad \\ \sigma_{r\theta} & = \frac{K_{II}}{\sqrt{2\pi r}} \left[\frac{1}{4}\cos\left(\frac{\theta}{2}\right) + \frac{3}{4}\cos\left(\frac{3\theta}{2}\right)\right] \text{(29)} \qquad \end{align}$$

Axially Loaded Wedge
The BCs at $$\theta = \pm \beta$$ are
 * $$\text{(30)} \qquad

t_r = t_{\theta} = 0 ~; \widehat{\mathbf{n}}{} = \pm \widehat{\mathbf{e}}{\theta} \Rightarrow \sigma_{r\theta} = \sigma_{\theta\theta} = 0 $$

What about the concentrated force BC?

 * What is $$\widehat{\mathbf{n}}{}$$ at the vertex ?
 * The traction is infinite since the force is applied on zero area. Consider equilibrium of a portion of the wedge.

At $$r = a$$, the BCs are
 * $$\text{(31)} \qquad

\widehat{\mathbf{n}}{} = \widehat{\mathbf{e}}{r} \Rightarrow \sigma_{r\theta} = t_r ~; \sigma_{\theta\theta} = t_{\theta} $$ For equilibrium, $$\sum F_1 = \sum F_2 = \sum M_3 = 0$$. Therefore,
 * $$\begin{align}

P_1 + \int_{-\beta}^{\beta} \left[ \sigma_{rr}(a,\theta)\cos\theta - \sigma_{r\theta}(a,\theta)\sin\theta\right] a~d\theta = 0 \text{(32)} \qquad \\ \int_{-\beta}^{\beta} \left[ \sigma_{rr}(a,\theta)\sin\theta + \sigma_{r\theta}(a,\theta)\cos\theta\right] a~d\theta = 0 \text{(33)} \qquad \\ \int_{-\beta}^{\beta} \left[ a \sigma_{r\theta}(a,\theta) \right] a~d\theta = 0 \text{(34)} \qquad \end{align}$$ These constraint conditions are equivalent to the concentrated force BC.

Solution Procedure
Assume that $$\sigma_{r\theta}(r,\theta) = 0$$. This satisfies the traction BCs on $$\theta = \pm\beta$$ and equation (34). Therefore,
 * $$\text{(35)} \qquad

\sigma_{r\theta} = \frac{\partial }{\partial} {}{r}\left(\frac{1}{r}\frac{\partial }{\partial} {\varphi}{\theta}\right) = 0 \Rightarrow \varphi = r\eta(\theta) + \zeta(r) $$ Hence,
 * $$\text{(36)} \qquad

\sigma_{\theta\theta} = \frac{\partial^2 }{\partial \varphi \partial r} = \zeta^{''}(r) $$ That means $$\sigma_{\theta\theta}$$ is independent of $$\theta$$. Therefore, in order to satisfy the BCs, $$\sigma_{\theta\theta} = 0$$, i.e.,
 * $$\text{(37)} \qquad

\zeta(r) = C_1 r + C_2 \Rightarrow \varphi = r\eta(\theta) + C_1 r = r[\eta(\theta)+C_1] = r\xi(\theta) $$ Checking for compatibility, $$\nabla^4{\varphi} = 0$$, we get
 * $$\text{(38)} \qquad

\xi^{(IV)}(\theta) + 2\xi^{''}(\theta) + \xi(\theta) = 0 $$ The general solution is
 * $$\text{(39)} \qquad

\xi(\theta) = A\sin\theta + B\cos\theta + C\theta\sin\theta + D\theta\cos\theta $$ Therefore,
 * $$\text{(40)} \qquad

{ \varphi = r\left[A\sin\theta + B\cos\theta + C\theta\sin\theta + D\theta\cos\theta\right] } $$ The only non-zero stress is $$\sigma_{rr}$$.
 * $$\text{(41)} \qquad

\sigma_{rr} = \frac{1}{r}\left[2C\cos\theta - 2D\sin\theta\right] $$ Plugging into equation (33), we get
 * $$\text{(42)} \qquad

-D\left[2\beta - \sin(2\beta)\right] = 0 \Rightarrow D = 0 $$ Hence,
 * $$\text{(43)} \qquad

\sigma_{rr} = \frac{2C}{r}\cos\theta $$ Plugging into equation (32), we get
 * $$\text{(44)} \qquad

-P = C\left[2\beta + \sin(2\beta)\right] \Rightarrow C = \frac{-P}{2\beta+\sin(2\beta)} $$ Therefore,
 * $$\text{(45)} \qquad

{ \varphi = Cr\theta\sin\theta = \frac{-P r\theta\sin\theta}{2\beta+\sin(2\beta)}} $$ The stress state is
 * $$\text{(46)} \qquad

{ \sigma_{rr} = -\frac{2P\cos\theta}{r[2\beta+\sin(2\beta)]} ~; \sigma_{r\theta} = 0 ~; \sigma_{\theta\theta} = 0 } $$

Special Case : β = π/2
A concentrated point load acting on a half plane.

 * $$\text{(47)} \qquad

{ \sigma_{rr} = -\frac{2P\cos\theta}{\pi r} ~; \sigma_{r\theta} = 0 ~; \sigma_{\theta\theta} = 0 } $$

Displacements

 * $$\begin{align}

2\mu u_r & = -\frac{\partial }{\partial} {\varphi}{r} + \alpha r \frac{\partial }{\partial} {\psi}{\theta} \\ 2\mu u_{\theta} & = -\frac{1}{r}\frac{\partial }{\partial} {\varphi}{\theta} + \alpha r^2 \frac{\partial }{\partial} {\psi}{r} \end{align}$$ where
 * $$\begin{align}

\nabla^2{\psi} = 0 \\ \frac{\partial }{\partial} {}{r}\left(r\frac{\partial }{\partial} {\psi}{\theta}\right) = \nabla^2{\varphi} \end{align}$$ Plug in $$\varphi = Cr\theta\sin\theta$$,
 * $$\begin{align}

& \frac{\partial }{\partial} {}{r}\left(r\frac{\partial }{\partial} {\psi}{\theta}\right)= \nabla^2{\varphi} \\ \Rightarrow & \frac{\partial }{\partial} {}{r}\left(r\frac{\partial }{\partial} {\psi}{\theta}\right)= \frac{2C}{r}\cos\theta \\ \Rightarrow & r\frac{\partial }{\partial} {\psi}{\theta}= 2C \ln r\cos\theta + A(\theta) \\ \Rightarrow & \frac{\partial }{\partial} {\psi}{\theta}= 2C \frac{\ln r}{r} \cos\theta + \frac{A(\theta)}{r} \\ \Rightarrow & \psi= 2C \frac{\ln r}{r} \sin\theta + \frac{\eta(\theta)}{r} + \xi{r} \end{align}$$ Plug $$\psi$$ into $$\nabla^2{\psi} = 0$$,
 * $$\begin{align}

& \frac{1}{r^3}\eta^{''}(\theta) + \frac{1}{r^3}\eta(\theta) + \xi^{''}(r) + \frac{1}{r}\xi^{'}(r) -\frac{4C}{r^3}\sin\theta= 0 \\ \Rightarrow & \eta^{}(\theta) + \eta(\theta) + r^3\xi^{}(r) + r^2\xi^{'}(r) - 4C\sin\theta= 0 \end{align}$$ Hence,
 * $$\begin{align}

\eta^{''}(\theta) + \eta(\theta) - 4C\sin\theta & = b\\ r^3\xi^{''}(r) + r^2\xi^{'}(r) & = -\frac{b}{r^3} \end{align}$$ Solving,
 * $$\begin{align}

\eta(\theta) & = -2C\theta\cos\theta + d\cos\theta + e\sin\theta + b\\ \xi^{'}(r) & = \frac{f}{r} + \frac{b}{r^2} \end{align}$$ Therefore,
 * $$\begin{align}

2\mu u_r & = 2\alpha C \ln r \cos\theta + (2\alpha-1)C\theta\sin\theta + \alpha(e - 2C)\cos\theta - \alpha d\sin\theta \\ 2\mu u_{\theta} & = -2\alpha C \ln r \sin\theta + (2\alpha-1)C\sin\theta + (2\alpha - 1)C\theta\cos\theta - \alpha d\cos\theta -\alpha e \sin\theta + \alpha f r \end{align}$$ To fix the rigid body motion, we set $$u_{\theta} = 0$$ when $$\theta = 0$$, and set $$u_r = 0$$ when $$\theta = 0$$ and $$r = L$$.Then,
 * $$\begin{align}

u_r & = \frac{\alpha C}{\mu}\ln\left(\frac{r}{L}\right)\cos\theta + \frac{(2\alpha-1)C}{2\mu}\theta\sin\theta \\ u_{\theta} & = -\frac{\alpha C}{\mu}\ln\left(\frac{r}{L}\right)\sin\theta + \frac{(2\alpha-1)C}{2\mu}\theta\cos\theta - \frac{C}{2\mu}\sin\theta \end{align}$$ The displacements are singular at $$r = 0$$ and $$r = \infty$$. At $$\theta = 0$$,
 * $$\begin{align}

u_r & = \frac{\alpha C}{\mu}\ln\left(\frac{r}{L}\right)\\ u_{\theta} & = 0 \end{align}$$ Is the small strain assumption satisfied ?

The Flamant Solution
From Michell's solution, pick terms containing $$1/r$$ in the stresses. Then,
 * This problem is also self-similar (no inherent length scale).
 * All quantities can be expressed in the separated-variable form  $$\sigma = f(r)g(\theta)$$.
 * The stresses vary as $$(1/r)$$ (the area of action of the force  decreases with increasing $$r$$).  How about a conical wedge ?

\varphi = C_1 r \theta\sin\theta + C_2 r\ln r \cos\theta + C_3 r \theta\cos\theta + C_4 r\ln r \sin\theta $$ Therefore, from Tables,
 * $$\begin{align}

\sigma_{rr} & = C_1\left(\frac{2\cos\theta}{r}\right) + C_2\left(\frac{\cos\theta}{r}\right) + C_3\left(\frac{2\sin\theta}{r}\right) + C_4\left(\frac{\sin\theta}{r}\right) \\ \sigma_{r\theta} & = C_2\left(\frac{\sin\theta}{r}\right) + C_4\left(\frac{-\cos\theta}{r}\right) \\ \sigma_{\theta\theta} & = C_2\left(\frac{\cos\theta}{r}\right) + C_4\left(\frac{\sin\theta}{r}\right) \end{align}$$ From traction BCs, $$C_2 = C_4 = 0$$. From equilibrium,
 * $$\begin{align}

F_1 + 2\int_{\alpha}^{\beta} \left(\frac{C_1\cos\theta - C_3\sin\theta}{a}\right)a\cos\theta d\theta & = 0 \\ F_2 + 2\int_{\alpha}^{\beta} \left(\frac{C_1\cos\theta - C_3\sin\theta}{a}\right)a\sin\theta d\theta & = 0 \end{align}$$ After algebra,

\sigma_{rr} = \frac{2C_1\cos\theta}{r} + \frac{2C_3\sin\theta}{r} ~; \sigma_{r\theta} = 0 ~; \sigma{\theta\theta} = 0 $$

Special Case : α = -π, β = 0

C_1 = - \frac{F_1}{\pi} ~; C_2 = \frac{F_2}{\pi} $$ The displacements are
 * $$\begin{align}

u_1 & = - \frac{F_1(\kappa+1)\ln|x_1|}{4\pi\mu} + \frac{F_2(\kappa+1)\text{sign}(x_1)}{8\mu} \\ u_2 & = - \frac{F_2(\kappa+1)\ln|x_1|}{4\pi\mu} - \frac{F_1(\kappa+1)\text{sign}(x_1)}{8\mu} \end{align}$$ where
 * $$\begin{align}

\kappa = 3 - 4\nu & & \text{plane strain} \\ \kappa = \frac{3 - \nu}{1+\nu} & & \text{plane stress} \\ \end{align}$$ and

\text{sign}(x) = \begin{cases} +1 & x > 0 \\ -1 & x < 0 \end{cases} $$

Related Content
Introduction to Elasticity

title: Stresses in wedges


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Torsion

About the problem:

 * Circular Cylinder.
 * Centroidal axis thru the center of each c.s.
 * Length $$L$$, Outer radius $$c$$.
 * Applied torque $$T$$.
 * Angle of twist $$\phi$$.

Assumptions:

 * Each c.s. remains plane and undistorted.
 * Each c.s. rotates through the same angle.
 * No warping or change in shape.
 * Amount of displacement of each c.s. is proportional to  distance from end.

Find:

 * Shear strains in the cylinder ($$\gamma$$).
 * Shear stress in the cylinder ($$\tau$$).
 * Relation between torque ($$T$$) and angle of twist ($$\phi$$).
 * Relation between torque ($$T$$) and shear stress ($$\tau$$).

Solution:
If $$\gamma$$ is small, then
 * $$\text{(1)} \qquad

L\gamma = r\phi \Rightarrow {\gamma = \frac{r\phi}{L}} $$ Therefore,
 * $$\text{(2)} \qquad

\gamma_{\text{max}} = \frac{c\phi}{L} \Rightarrow \gamma = \frac{r} \gamma_{\text{max}} $$ If the deformation is elastic,
 * $$\text{(3)} \qquad

\tau = G\gamma \Rightarrow {\tau = \frac{r\phi G}{L}} $$ Therefore,
 * $$\text{(4)} \qquad

\tau_{\text{max}} = \frac{r\phi G}{L} \Rightarrow \tau = \frac{r} \tau_{\text{max}} $$ The torque on each c.s. is given by
 * $$\text{(5)} \qquad

T = \int_A \tau r dA = \frac{\phi G}{L}\int_A r^2 dA = \frac{G\phi J}{L} $$ where $$J$$ is the polar moment of inertia of the c.s.
 * $$\text{(6)} \qquad

J = \begin{cases} \frac{1}{2} \pi c^4 & \text{solid circular c.s.} \\ \frac{1}{2} \pi (c_2^{~4}-c_1^{~4}) & \text{ annular circular c.s.} \end{cases} $$ Therefore,
 * $$\text{(7)} \qquad

{\tau = \frac{Tr}{J}} \Rightarrow \tau_{\text{max}} = \frac{Tc}{J} $$ and
 * $$\text{(8)} \qquad

{\phi = \frac{TL}{JG}} $$

About the problem

 * Solution first found by St. Venant.
 * Tractions at the ends are statically equivalent to equal and opposite torques $$\pm \mathbf{T} = \pm T \widehat{\mathbf{e}}{3}$$.
 * Lateral surfaces are traction-free.

Assumptions:

 * An axis passes through the center of twist ($$x_3$$ axis).
 * Each c.s. projection on to the $$x_1-x_2$$ plane rotates,but remains undistorted.
 * The rotation of each c.s. ($$\phi$$) is proportional to $$x_3$$. :$$\text{(9)} \qquad    \phi = \alpha x_3  $$  where $$\alpha$$ is the  twist per unit length.
 * The out-of-plane distortion (warping) is the same for each c.s. and is proportional to $$\alpha$$.

Find:

 * Torsional rigidity ($$T/\alpha$$).
 * Maximum shear stress.

Displacements

 * $$\begin{align}

u_1 & = r\cos(\phi+\theta) - r\cos\theta = x_1(\cos\phi-1)-x_2\sin\phi \\ u_2 & = r\sin(\phi+\theta) - r\sin\theta = x_1\sin\phi+x_2(\cos\phi-1)\\ u_3 & = \alpha\psi(x_1,x_2) \end{align}$$ where $$\psi(x_1,x_2)$$ is the { warping function}.\\ If $$\phi = \alpha x_3 << 1$$ (small strain),
 * $$\text{(10)} \qquad

{ u_1 \approx -\alpha x_2 x_3 ~; u_2 \approx \alpha x_1 x_3 ~; u_3 = \alpha\psi(x_1,x_2) } $$

Strains

\varepsilon_{ij} = \frac{1}{2}\left(u_{i,j} + u_{j,i}\right) $$ Therefore,
 * $$\begin{align}

\varepsilon_{11} & = \frac{1}{2}\left(0 + 0\right) = 0 \\ \varepsilon_{22} & = \frac{1}{2}\left(0 + 0\right) = 0 \\ \varepsilon_{33} & = \frac{1}{2}\left(0 + 0\right) = 0 \\ \varepsilon_{kk} & = \varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33} = 0 \\ \varepsilon_{12} & = \frac{1}{2}\left(-\alpha x_3 + \alpha x_3 \right) = 0 \\ \varepsilon_{23} & = \frac{1}{2}\left(\alpha\psi_{,2} + \alpha x_1\right) \text{(11)} \qquad \\ \varepsilon_{31} & = \frac{1}{2}\left(\alpha\psi_{,1} - \alpha x_2\right) \text{(12)} \qquad \end{align}$$

Stresses

\sigma_{ij} = 2\mu\varepsilon_{ij} + \lambda\varepsilon_{kk}\delta_{ij} $$ Therefore,
 * $$\begin{align}

\sigma_{11} & = 0 \\ \sigma_{22} & = 0 \\ \sigma_{33} & = 0 \\ \sigma_{kk} & = 0 \\ \sigma_{12} & = 0 \\ \sigma_{23} & = \mu\alpha(\psi_{,2} + x_1) \text{(13)} \qquad \\ \sigma_{31} & = \mu\alpha(\psi_{,1} - x_1) \text{(14)} \qquad \end{align}$$

Equilibrium

\sigma_{ji,j} = 0 ~ \text{no body forces.} $$ Therefore,
 * $$\begin{align}

\sigma_{11,1} + \sigma_{21,2} + \sigma_{31,3} = 0 & \Rightarrow 0 = 0 \\ \sigma_{12,1} + \sigma_{22,2} + \sigma_{32,3} = 0 & \Rightarrow 0 = 0 \\ \sigma_{13,1} + \sigma_{23,2} + \sigma_{33,3} = 0 & \Rightarrow \mu\alpha(\psi_{,11}+\psi_{,22}) = \mu\alpha\nabla^2{\psi} = 0 \text{(15)} \qquad \end{align}$$

Internal Tractions

 * Normal to cross sections is $$\widehat{\mathbf{n}}{} = \widehat{\mathbf{e}}{3}$$.
 * Normal traction $$t^n = \mathbf{t}\bullet\widehat{\mathbf{n}}{} = 0$$.
 * Projected shear traction is  $$t^s = \sqrt{\sigma_{13}^2 + \sigma_{23}^2}$$.
 * Traction vector at a point in the cross section is { tangent} to the cross section.

Boundary Conditions on Lateral Surfaces
We parameterize the boundary curve $$\partial S$$ using
 * Lateral surface traction-free.
 * Unit normal to lateral surface appears as an in-plane unit  normal to the boundary $$\partial S$$.

\mathbf{x} = \tilde{\mathbf{x}}(s) ~, 0 \le s \le l; \tilde{\mathbf{x}}(0) = \tilde{\mathbf{x}}(l) $$ The tangent vector to $$s$$ is

\widehat{\boldsymbol{\nu}} = \frac{d\mathbf{x}}{ds} ~\text{and} \widehat{\mathbf{n}}{} = \widehat{\boldsymbol{\nu}}\times\widehat{\mathbf{e}}_{3} \Rightarrow \widehat{\mathbf{n}}{} = \frac{dx_2}{ds} \widehat{\mathbf{e}}{1} - \frac{dx_1}{ds} \widehat{\mathbf{e}}_{2} $$ The tractions $$t_1$$ and $$t_2$$ on the lateral surface are identically zero. However, to satisfy the BC $$t_3 = 0$$, we need

t_3 = n_1 \sigma_{13} + n_2 \sigma_{23} = 0 \Rightarrow \left(\psi_{,1} - x_2\right) n_1 + \left(\psi_{,2} + x_1\right) n_2= 0 $$ or,
 * $$ \text{(16)} \qquad

\left(\psi_{,1} - x_2\right) \frac{dx_2}{ds} + \left(\psi_{,2} + x_1\right) \frac{dx_1}{ds}= 0 $$

Boundary Conditions on End Surfaces
The traction distribution is statically equivalent to the torque $$\mathbf{T}$$. At $$x_3 = L$$,

t_1 = \sigma_{13}~; t_2 = \sigma_{23}~; t_3 = \sigma_{33} = 0 $$ Therefore,

F_1 = \int_S \sigma_{13}~dS = \mu\alpha\int_S(\psi_{,1}-x_2)~dS $$ From equilibrium,
 * $$\begin{align}

\nabla^2{\psi} = 0 \Rightarrow \psi_{,1}-x_2 & = (\psi_{,1}-x_2) + x_1(\psi_{,11} + \psi_{,22}) \\ & = \psi_{,1} + x_1\psi_{,11} - x_2 + x_1\psi_{,22} \\ & = (x_1\psi_{,1} - x_1x_2)_{,1} + (x_1\psi_{,2} + x_1x_1)_{,2} \\ & = \left[x_1(\psi_{,1} - x_2)\right]_{,1} + \left[x_1(\psi_{,2} + x_1)\right]_{,2} \end{align}$$ Hence,
 * $$\text{(17)} \qquad

F_1 = \mu\alpha\int_S\left[x_1(\psi_{,1} - x_2)\right]_{,1} + \left[x_1(\psi_{,2} + x_1)\right]_{,2} dS $$

The Green-Riemann Theorem
If $$P = f(x_1,x_2)$$ and $$Q = q(x_1,x_2)$$ then

\int_S (Q_{,1} - P_{,2}) dS = \oint_{\partial S} (P dx_1 + Q dx_2) $$ with the integration direction such that $$S$$ is to the left.

Applying the Green-Riemann theorem to equation (17), and using equation (16)
 * $$\text{(18)} \qquad

F_1 = \mu\alpha\oint_{\partial S} -x_1(\psi_{,2} + x_1)dx_1 + x_1(\psi_{,1} - x_2)dx_2 = 0 $$ Similarly, we can show that $$F_2 = 0$$. $$F_3 = 0$$ since $$t_3 = 0$$.

The moments about the $$x_1$$ and $$x_2$$ axes are also zero.

The moment about the $$x_3$$ axis is

M_3 = \int_S (x_1\sigma_{23} - x_2\sigma_{13}) dS = \mu\alpha\int_S(x_1\psi_{,2} + x_1^2 - x_2\psi_1 + x_2^2) dS = \mu\alpha\tilde{J} $$ where $$J$$ is the torsion constant. Since $$M_3 = T$$, we have

\alpha = \frac{T}{\mu\tilde{J}} $$ If $$\psi = 0$$, then $$\tilde{J} = J$$, the polar moment of inertia.

The Torsion Problem Summarized

 * Find a warping function $$\psi(x_1,x_2)\,$$ that is harmonic. and satisfies the traction BCs.
 * Compatibility is not an issue since we start with displacements.
 * The problem is independent of applied torque and the material properties of the cylinder.
 * So it is just a geometrical problem. Once $$\psi$$ is known, we can calculate
 * The displacement field.
 * The stress field.
 * The twist per unit length.

Related Content
Introduction to Elasticity

title: Torsion


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text:
Warping functions

Warping Function and Torsion of Non-Circular Cylinders
The displacements are given by

u_1 = -\alpha x_2 x_3 ~; u_2 = \alpha x_1 x_3 ~; u_3 = \alpha\psi(x_1,x_2) $$ where $$\alpha$$ is the twist per unit length, and $$\psi$$ is the warping function.

The stresses are given by

\sigma_{13} = \mu\alpha(\psi_{,1} - x_2) ~; \sigma_{23} = \mu\alpha(\psi_{,2} + x_1) $$ where $$\mu$$ is the shear modulus.

The projected shear traction is

\tau = \sqrt{(\sigma_{13}^2 + \sigma_{23}^2)} $$

Equilibrium is satisfied if

\nabla^2{\psi} = 0 ~ \forall (x_1, x_2) \in \text{S} $$

Traction-free lateral BCs are satisfied if

(\psi_{,1} - x_2) \frac{dx_2}{ds} - (\psi_{,2} + x_1) \frac{dx_1}{ds} = 0 ~ \forall (x_1, x_2) \in \partial\text{S} $$ or,

(\psi_{,1} - x_2) \hat{n}_{1} + (\psi_{,2} + x_1) \hat{n}_{2} = 0 ~ \forall (x_1, x_2) \in \partial\text{S} $$

The twist per unit length is given by

\alpha = \frac{T}{\mu \tilde{J}} $$ where the torsion constant

\tilde{J} = \int_S (x_1^2 + x_2^2 + x_1\psi_{,2} - x_2\psi_{,1}) dA $$

Example 1: Circular Cylinder
Choose warping function

\psi(x_1,x_2) = 0 $$

Equilibrium ($$\nabla^2{\psi} = 0$$) is trivially satisfied.

The traction free BC is

(0 - x_2) \frac{dx_2}{ds} - (0 + x_1) \frac{dx_1}{ds} = 0 ~ \forall (x_1, x_2) \in \partial\text{S} $$ Integrating,

x_2^2 + x_1^2 = c^2 ~ \forall (x_1, x_2) \in \partial\text{S} $$ where $$c$$ is a constant.

Hence, a circle satisfies traction-free BCs.

There is no warping of cross sections for circular cylinders

The torsion constant is

\tilde{J} = \int_S (x_1^2 + x_2^2) dA = \int_S r^2 dA = J $$

The twist per unit length is

\alpha = \frac{T}{\mu J} $$

The non-zero stresses are

\sigma_{13} = -\mu\alpha x_2 ~; \sigma_{23} = \mu\alpha x_1 $$

The projected shear traction is

\tau = \mu\alpha\sqrt{(x_1^2 + x_2^2)} = \mu\alpha r $$

Compare results from Mechanics of Materials solution

\phi = \frac{TL}{GJ} \Rightarrow \alpha = \frac{T}{GJ} $$ and

\tau = \frac{Tr}{J} \Rightarrow \tau = G\alpha r $$

Example 2: Elliptical Cylinder
Choose warping function

\psi(x_1,x_2) = k x_1 x_2 \, $$ where $$k$$ is a constant.

Equilibrium ($$\nabla^2{\psi} = 0$$) is satisfied.

The traction free BC is

(kx_2 - x_2) \frac{dx_2}{ds} - (kx_1 + x_1) \frac{dx_1}{ds} = 0 ~ \forall (x_1, x_2) \in \partial\text{S} $$ Integrating,

x_1^2 + \frac{1-k}{1+k}x_2^2 = a^2 ~ \forall (x_1, x_2) \in \partial\text{S} $$ where $$a$$ is a constant.

This is the equation for an ellipse with major and minor axes $$a$$ and $$b$$, where

b^2 = \left(\frac{1+k}{1-k}\right) a^2 $$

The warping function is

\psi = -\left(\frac{a^2-b^2}{a^2 + b^2}\right) x_1 x_2 $$

The torsion constant is

\tilde{J} = \frac{2b^2}{a^2+b^2} I_2 + \frac{2a^2}{a^2+b^2} I_1 = \frac{\pi a^3 b^3}{a^2 + b^2} $$ where

I_1 = \int_S x_1^2 dA = \frac{\pi a b^3}{4} ; I_2 = \int_S x_2^2 dA = \frac{\pi a^3 b}{4} $$

'''If you compare $$\tilde{J}$$ and $$J$$ for the ellipse, you will find that $$\tilde{J} < J$$. This implies that the torsional rigidity is less than that predicted with the assumption that plane sections remain plane.'''

The twist per unit length is

\alpha = \frac{(a^2+b^2)T}{\mu\pi a^3 b^3} $$

The non-zero stresses are

\sigma_{13} = -\frac{2\mu\alpha a^2 x_2}{a^2 + b^2} ; \sigma_{23} = -\frac{2\mu\alpha b^2 x_1}{a^2 + b^2} $$

The projected shear traction is

\tau = \frac{2\mu\alpha}{a^2+b^2}\sqrt{b^4 x_1^2 + a^4 x_2^2} \Rightarrow \tau_{\text{max}} = \frac{2\mu\alpha a^2 b}{a^2+b^2} (b < a) $$

For any torsion problem where $$\partial$$S is convex, the maximum projected shear traction occurs at the point on $$\partial$$S that is nearest the centroid of S

The displacement $$u_3$$ is

u_3 = -\frac{(a^2-b^2)Tx_1x_2}{\mu\pi a^3b^3} $$

Example 3: Rectangular Cylinder
In this case, the form of $$\psi$$ is not obvious and has to be derived from the traction-free BCs

(\psi_{,1} - x_2) \hat{n}_{1} + (\psi_{,2} + x_1) \hat{n}_{2} = 0 ~ \forall (x_1, x_2) \in \partial\text{S} $$

Suppose that $$2a$$ and $$2b$$ are the two sides of the rectangle, and $$a > b$$. Also $$a$$ is the side parallel to $$x_1$$ and $$b$$ is the side parallel to $$x_2$$. Then, the traction-free BCs are

\psi_{,1} = x_2 \text{on} x_1 = \pm a ,\text{and} \psi_{,2} = -x_1 \text{on} x_2 = \pm b $$

A suitable $$\psi$$ must satisfy these BCs and $$\nabla^2{\psi} = 0$$.

We can simplify the problem by a change of variable

\bar{\psi} = x_1x_2 - \psi $$

Then the equilibrium condition becomes

\nabla^2{\bar{\psi}} = 0 $$

The traction-free BCs become

\bar{\psi}_{,1} = 0 \text{on} x_1 = \pm a ,\text{and} \bar{\psi}_{,2} = 2x_1 \text{on} x_2 = \pm b $$

Let us assume that

\bar{\psi}(x_1,x_2) = f(x_1) g(x_2) $$

Then,

\nabla^2{\bar{\psi}} = \bar{\psi}_{,11} + \bar{\psi}_{,22} = f^{}(x_1) g(x_2) + g^{}(x_2) f(x_1) = 0 $$ or,

\frac{f^{}(x_1)}{f(x_1)} = - \frac{g^{}(x_2)}{g(x_2)} = \eta $$

Case 1: η > 0 or η = 0
In both these cases, we get trivial values of $$C_1 = C_2 = 0$$.

Case 2: η < 0
Let

\eta = -k^2 ; k > 0 $$

Then,
 * $$\begin{align}

f^{''}(x_1) + k^2 f(x_1) = 0 \Rightarrow & f(x_1) = C_1 \cos(kx_1) + C_2 \sin(kx_1) \\ g^{''}(x_2) - k^2 g(x_2) = 0 \Rightarrow & g(x_2) = C_3 \cosh(kx_2) + C_4 \sinh(kx_2) \end{align}$$

Therefore,

\bar{\psi}(x_1,x_2) = \left[ C_1 \cos(kx_1) + C_2 \sin(kx_1) \right] \left[ C_3 \cosh(kx_2) + C_4 \sinh(kx_2) \right] $$

Apply the BCs at $$x_2 = \pm b$$ ($$\bar{\psi}_{,2} = 2x_1$$), to get
 * $$\begin{align}

\left[ C_1 \cos(kx_1) + C_2 \sin(kx_1) \right] \left[ C_3 \sinh(kb) + C_4 \cosh(kb) \right] & = 2x_1\\ \left[ C_1 \cos(kx_1) + C_2 \sin(kx_1) \right] \left[-C_3 \sinh(kb) + C_4 \cosh(kb) \right] & = 2x_1 \end{align}$$ or,

F(x_1) G^{'}(b) = 2 x_1 ; F(x_1) G^{'}(-b) = 2 x_1 $$ The RHS of both equations are odd. Therefore, $$F(x_1)$$ is odd. Since, $$\cos(kx_1)$$ is an even function, we must have $$C_1 = 0$$.

Also,

F(x_1) \left[ G^{'}(b) - G^{'}(-b)\right] = 0 $$ Hence, $$G'(b)$$ is even. Since $$\sinh(kb)$$ is an odd function, we must have $$C_3 = 0$$.

Therefore,

\bar{\psi}(x_1,x_2) = C_2 C_4 \sin(kx_1) \sinh(kx_2) = A \sin(kx_1) \sinh(kx_2) $$

Apply BCs at $$x_1 = \pm a$$ ($$\bar{\psi}_{,1} = 0$$), to get

A k \cos(ka) \sinh(kx_2) = 0 $$ The only nontrivial solution is obtained when $$\cos(ka) = 0$$, which means that

k_n = \frac{(2n+1)\pi}{2a}, n = 0,1,2,... $$

The BCs at $$x_1 = \pm a$$ are satisfied by every terms of the series

\bar{\psi}(x_1,x_2) = \sum_{n=0}^{\infty} A_n \sin(k_n x_1) \sinh(k_n x_2) $$

Applying the BCs at $$x_1 = \pm b$$ again, we get

\sum_{n=0}^{\infty} A_n k_n \sin(k_n x_1) \cosh(k_n b) = 2 x_1 \Rightarrow \sum_{n=0}^{\infty} B_n \sin(k_n x_1) = 2 x_1 $$

Using the orthogonality of terms of the sine series,

\int_{-a}^a \sin(k_n x_1) \sin(k_m x_1) dx_1 = \begin{cases} 0 & {\rm if}~ m \ne n \\ a & {\rm if}~ m = n \end{cases} $$ we have

\int_{-a}^a \left[\sum_{n=0}^{\infty} B_n \sin(k_n x_1)\right] \sin(k_m x_1) dx_1 = \int_{-a}^a \left[2 x_1\right] \sin(k_m x_1) dx_1 $$ or,

B_m a = \frac{4}{k_m^2} \sin(k_m a) $$ Now,

\sin(k_m a) = \sin\left(\frac{(2m+1)\pi}{2}\right) = (-1)^m $$ Therefore,

A_m = \frac{B_m}{k_m\cosh(k_m b)} = \frac{(-1)^m 32a^2}{(2m+1)^3\pi^3\cosh(k_m b)} $$

The warping function is

\psi = x_1 x_2 - \frac{32a^2}{\pi^3} \sum_{n=0}^{\infty} \frac{(-1)^n \sin(k_n x_1) \sinh(k_n x_2)}{(2n+1)^3\cosh(k_n b)} $$ The torsion constant and the stresses can be calculated from $$\psi$$.

Prandtl Stress Function (φ)
The traction free BC is obviously difficult to satisfy if the cross-section is not a circle or an ellipse.

To simplify matters, we define the Prandtl stress function $$\phi(x_1,x_2)$$ using

{ \sigma_{13} = \phi_{,2} ; \sigma_{23} = -\phi_{,1} } $$ You can easily check that this definition satisfies equilibrium.

It can easily be shown that the traction-free BCs are satisfied if

{ \frac{d\phi}{ds} = 0 \forall~(x_1,x_2) \in \partial\text{S} } $$ where $$s$$ is a coordinate system that is tangent to the boundary.

If the cross section is simply connected, then the BCs are even simpler:

{ \phi = 0 \forall~(x_1,x_2) \in \partial\text{S} } $$

From the compatibility condition, we get a restriction on $$\phi$$

{ \nabla^2{\phi} = C \forall~(x_1,x_2) \in \text{S} } $$ where $$C$$ is a constant.

Using relations for stress in terms of the warping function $$\psi$$, we get

{ \nabla^2{\phi} = -2\mu\alpha \forall~(x_1,x_2) \in \text{S} } $$

Therefore, the twist per unit length is

{ \alpha = -\frac{1}{2\mu} \nabla^2{\phi} } $$

The applied torque is given by

{ T = -\int_{S} (x_1 \phi_{,1} + x_2 \phi_{,2}) dA } $$ For a simply connected cylinder,

{ T =2 \int_{S} \phi dA } $$

The projected shear traction is given by

{\tau = \sqrt{(\phi_{,1})^2+ (\phi_{,2})^2}} $$

The projected shear traction at any point on the cross-section is tangent to the contour of constant $$\phi$$ at that point.

The relation between the warping function $$\psi$$ and the Prandtl stress function $$\phi$$ is

{ \psi_{,1} = \frac{1}{\mu\alpha} \phi_{,2} + x2 ~; \psi_{,2} = -\frac{1}{\mu\alpha} \phi_{,1} - x1 } $$

Membrane Analogy
The equations

\nabla^2{\phi} = -2\mu\alpha \forall~(x_1,x_2) \in \text{S}; \phi = 0 \forall~(x_1,x_2) \in \partial\text{S} $$ are similar to the equations that govern the displacement of a membrane that is stretched between the boundaries of the cross-sectional curve and loaded by an uniform normal pressure.

This analogy can be useful in estimating the location of the maximum shear stress and the torsional rigidity of a bar.

 * The stress function is proportional to the displacement of the membrane from the plane of the cross-section.
 * The stiffest cross-sections are those that allow the maximum volume to be developed between the deformed membrane and the plane of the   cross-section for a given pressure.
 * The shear stress is proportional to the slope of the membrane.

Solution Strategy
The equation $$\nabla^2{\phi} = -2\mu\alpha$$ is a Poisson equation. Since the equation is inhomogeneous, the solution can be written as

\phi = \phi_p + \phi_h $$ where $$\phi_p$$ is a particular solution and $$\phi_h$$ is the solution of the homogeneous equation.

Examples of particular solutions are, in rectangular coordinates,

\phi_p = -\mu\alpha x_1^2 ; \phi_p = -\mu\alpha x_2^2 $$ and, in cylindrical co-ordinates,

\phi_p = -\frac{\mu\alpha r^2}{2} $$

The homogeneous equation is the Laplace equation $$\nabla^2{\phi}=0$$, which is satisfied by both the real and the imaginary parts of any { analytic} function ($$f(z)$$) of the complex variable

z = x_1 + i x_2 = r e^{i\theta} $$ Thus,

\phi_h = \text{Re}(f(z)) \text{or} \phi_h = \text{Im}(f(z)) $$

Suppose $$f(z) = z^n$$.

Then, examples of $$\phi_h$$ are

\phi_h = C_1 r^n\cos(n\theta) ; \phi_h = C_2 r^n\sin(n\theta) ; \phi_h = C_3 r^{-n}\cos(n\theta) ; \phi_h = C_4 r^{-n}\sin(n\theta) $$ where $$C_1$$, $$C_2$$, $$C_3$$, $$C_4$$ are constants.

Each of the above can be expressed as polynomial expansions in the $$x_1$$ and $$x_2$$ coordinates.

Approximate solutions of the torsion problem for a particular cross-section can be obtained by combining the particular and homogeneous solutions and adjusting the constants so as to match the required shape.

Only a few shapes allow closed-form solutions. Examples are There are a few other papers which propose closed-form or semi-closed-form solutions to the torsion problem for cross-sections with irregular shapes.
 * Circular cross-section.
 * Elliptical cross-section.
 * Circle with semicircular groove.
 * Equilateral triangle.

Example: Equilateral Triangle
The equations of the three sides are
 * $$\begin{align}

\text{side}~\partial S^{(1)} ~: & f_1(x_1,x_2) = x_1 - \sqrt{3} x_2 + 2a = 0 \\ \text{side}~\partial S^{(2)} ~: & f_2(x_1,x_2) = x_1 + \sqrt{3} x_2 + 2a = 0\\ \text{side}~\partial S^{(3)} ~: & f_3(x_1,x_2) = x_1 - a = 0 \end{align}$$

Let the Prandtl stress function be

\phi = C f_1 f_2 f_3 $$

Clearly, $$\phi = 0$$ at the boundary of the cross-section (which is what we need for solid cross sections).

Since, the traction-free boundary conditions are satisfied by $$\phi$$, all we have to do is satisfy the compatibility condition to get the value of $$C$$. If we can get a closed for solution for $$C$$, then the stresses derived from $$\phi$$ will satisfy equilibrium.

Expanding $$\phi$$ out,

\phi = C (x_1 - \sqrt{3} x_2 + 2a)(x_1 + \sqrt{3} x_2 + 2a)(x_1 - a) $$

Plugging into the compatibility condition

\nabla^2{\phi} = 12 C a = -2\mu\alpha $$ Therefore,

C = -\frac{\mu\alpha}{6a} $$ and the Prandtl stress function can be written as

\phi = -\frac{\mu\alpha}{6a} (x_1^3+3ax_1^2+3ax_2^2-3x_1x_2^2-4a^3) $$

The torque is given by

T = 2\int_S \phi dA = 2\int_{-2a}^{a} \int_{-(x_1+2a)/\sqrt{3}}^{(x_1+2a)/\sqrt{3}} \phi dx_2 dx_1 = \frac{27}{5\sqrt{3}} \mu\alpha a^4 $$

Therefore, the torsion constant is

\tilde{J} = \frac{27 a^4}{5\sqrt{3}} $$

The non-zero components of stress are
 * $$\begin{align}

\sigma_{13} = \phi_{,2} & = \frac{\mu\alpha}{a}(x_1-a)x_2 \\ \sigma_{23} = -\phi_{,1} & = \frac{\mu\alpha}{2a}(x_1^2+2ax_1-x_2^2) \end{align}$$

The projected shear stress

\tau = \sqrt{\sigma_{13}^2+ \sigma_{23}^2} $$ is plotted below The maximum value occurs at the middle of the sides. For example, at $$(a,0)$$,

\tau_{\text{max}} = \frac{3\mu\alpha a}{2} $$

The out-of-plane displacements can be obtained by solving for the warping function $$\psi$$. For the equilateral triangle, after some algebra, we get

u_3 = \frac{\alpha x_2}{6a} (3x_1^2 - x_2^2) $$ The displacement field is plotted below

Thin-walled Open Sections
Examples are I-beams, channel sections and turbine blades.

We assume that the length $$b$$ is much larger than the thickness $$t$$, and that $$t$$ does not vary rapidly with change along the length axis $$\xi$$.

Using the membrane analogy, we can neglect the curvature of the membrane in the $$\xi$$ direction, and the Poisson equation reduces to

\frac{d^\phi}{d\eta^2} = -2\mu\alpha $$ which has the solution

\phi = \mu\alpha\left(\frac{t^2}{4}-\eta^2\right) $$ where $$\eta$$ is the coordinate along the thickness direction.

The stress field is

\sigma_{3\xi} = \frac{\partial }{\partial} {\phi}{\eta} = -e\mu\beta\eta ; \sigma_{3\eta} = 0 $$ Thus, the maximum shear stress is

\tau_{\text{max}} = \mu\beta t_{\text{max}} $$

Thin-walled Closed Sections
The Prandtl stress function $$\phi$$ can be approximated as a linear function between $$\phi_1$$ and $$0$$ on the two adjacent boundaries.

The local shear stress is, therefore,

\sigma_{3s} = \frac{\phi_1}{t} $$ where $$s$$ is the parameterizing coordinate of the boundary curve of the cross-section and $$t$$ is the local wall thickness.

The value of $$\phi_1$$ can determined using

\phi_1 = \frac{2\mu\alpha A}{\oint_S \frac{dS}{t}} $$ where $$A$$ is the area enclosed by the mean line between the inner and outer boundary.

The torque is approximately

T = 2A\phi_1 $$

Related content
Introduction to Elasticity

title: Warping functions


text:
Wikiversity:Activity creation

 * A suggestion from Wikimania discussions: a curriculum about contributing activities to Wikiversity

Who this is for: Any student in middle school or high school who has ever said "I wish there was a class like this..."

Wikiversity stub creation activity
How to capture a great activity that you've enjoyed or seen, and describe it for others so that they can try it for themselves or with their friends and classmates.

0) Think about a great activity that you have done or a great way that you have learned, and write a brief description. Have you ever wished school was more like... well, something else? Do you have something you and your friends enjoy doing that you'd love to be able to do in your classrooms? This lesson will help you create a Wikiversity module for running your own class; you can share with your teacher and classmates at school.

1) Make an account, learn how to use the wiki

Links: to wikieducator tutorials*, to other welcome pages

2) Starting a new activity : visit browse and search the site for related materials. Select a school that reflects your subject area, and go to the relevant topic page.

3) Pick a title for your page, add a link from the topic page to the new title.

4) Start a stub at this new page, and add category:activity-stub.

5) Write a brief description of the activity, try to link to existing pages on related topics [you may have to search for each link, or create redirects from what you expect to what you want]

 * If you need help with the description, see these brainstorming pages / guideline templates (some templates exist already by type of activity). If it helps, write a vignette about the specific experience you had on the discussion page, describing what it was like without analyzing it.

5.5) Form or subst:template for a description.

6) identify ways to engage in what you just described -- variations on a theme, tie-ins to other activities; different hooks for learning about a number of other topics [though it may be categorized under a primary topic]

 * Capture the moments of insight that you had during the activity -- what made it interesting? Was there a point where everyone involved suddenly understood something new?  Did you see the activity differently after it was over than when you had first started?

6.5) Form or template for related variations and activities.

7) Choose images to capture the process you have in mind. [You might want to search for photos from Commons, take short videos of key actions or preparation.]  Is there anything left that is hard to describe with words alone?  Ask for help from an artist if necessary. [link]

7.5) Link to easy image-upload form [suggest commonsist as well?]*

8) Categorization : find relevant categories to add to your pages, and help others find them more easily.

9) Find portals and other pages where you might add links to the new activity.

10) Look into peer review : How to associate this with a stream or group interested in making it better?  If this is your first time...

11) Link to help on the creation of outputs of this material, whether PDF, XML, DOC, MediaWiki, HTML

12) followup : what happened after the activity was over? did you find ways to extend your activity into other things you were doing?  suggest some related activities that might be used in combination or in series.

Extras
(Help) : WV tutorials, like wikieducator but in context

(Code) : Make a wikiversity bot that parses the form and uploads directly to commons with info on the wv:user and context making the original upload

(Articles) : How to clone existing activity templates and styles; look through related activities for templates and processes that match what you have in mind. If it's not clear how to clone something you like, find the authors and projects involved and ask for help.
 * likewise for components of an activity -- a plan, list of materials, table, todo list, exercise set, &c.

title: Wikiversity:Activity creation


text:
Wikiversity talk:Activity creation

for cormac: add link to the new activity or object from welcome

title: Wikiversity talk:Activity creation