text:
Talk:Northern Arizona University/Environmental Ethics/A Land Ethic

Copyright issues
Please could the person who posted the material on this page be clearer about its copyright status? I think this may be a breach of copyright, but it would be great if you could clarify this and show the materials can be published under the GFDL. McCormack 04:36, 30 August 2007 (UTC)
 * Just some more info. No decision about the status. Will write the course teacher too.
 * Aldo Leopold (January 11, 1887 - April 21, 1948)
 * text from A Sand County Almanac, year 1948.

 * regarding Wikipedia's Copyright article (Copyright. (2007, September 7). In Wikipedia, The Free Encyclopedia. Retrieved 17:14, September 7, 2007, from http://en.wikipedia.org/w/index.php?title=Copyright&oldid=156254978):

 * "In most of the world the default length of copyright for many works is generally the life of the author plus either 50 or 70 years. Copyright in general always expires at the end of the year concerned, rather than on the exact date of the death of the author.
 * In the United States, all books and other works published before 1923 have expired copyrights and are in the public domain.In addition, works published before 1964 that did not have their copyrights renewed 28 years after first publication year also are in the public domain, except that books originally published outside the US by non-Americans are exempt from this requirement, if they are still under copyright in their home country ...
 * But if the intended exploitation of the work includes publication (or distribution of derivative work, such as a film based on a book protected by copyright) outside the U.S., the terms of copyright around the world must be considered. If the author has been dead more than 70 years, the work is in the public domain in most, but not all, countries. In Italy and France, there are wartime extensions that could increase the term by approximately 6 years in Italy and up to about 14 in France. Some works are covered by copyright in Spain for 80 years after the author's death."
 * But if the intended exploitation of the work includes publication (or distribution of derivative work, such as a film based on a book protected by copyright) outside the U.S., the terms of copyright around the world must be considered. If the author has been dead more than 70 years, the work is in the public domain in most, but not all, countries. In Italy and France, there are wartime extensions that could increase the term by approximately 6 years in Italy and up to about 14 in France. Some works are covered by copyright in Spain for 80 years after the author's death."
 * But if the intended exploitation of the work includes publication (or distribution of derivative work, such as a film based on a book protected by copyright) outside the U.S., the terms of copyright around the world must be considered. If the author has been dead more than 70 years, the work is in the public domain in most, but not all, countries. In Italy and France, there are wartime extensions that could increase the term by approximately 6 years in Italy and up to about 14 in France. Some works are covered by copyright in Spain for 80 years after the author's death."

 * One could come to the conclusion then, that since:
 * the book is from 1948 >1923
 * author died in 1948 => 1948+70=2018 >2007
 * regarding "works published before 1964 that did not have their copyrights renewed 28 years after first publication year also are in the public domain": this is unknown

 * it would be better/safer for Wikiversity that the course gives a link to the external website ?

 * Just to be sure I will also ask at Wikipedia:Copyright problems and tell the course teacher here in Wikiversity. Erkan Yilmaz ( evaluate me!, discussion ) 17:51, 7 September 2007 (UTC)
 * I found this discussion through en:Wikipedia:Copyright problems. Without evidence that the material is in the public domain or released under the GFDL it should be deleted. Garion96 20:48, 13 October 2007 (UTC)

 * I got from a member of the Aldo Leopold Foundation Inc. following information:
 * The copyright to “A Sand County Almanac” by Aldo Leopold is held by Oxford University Press. The copyright is in effect until 2024.  Please contact Oxford University Press for additional information.
 * Therefore removing the content and letting only the link to the external page, Erkan Yilmaz ( my talk page, wiki blog ) 15:45, 15 October 2007 (UTC)

Other talk
Jacob Slade PHI 331 Aug. 29th, 2007 Critical Questions #1 Aldo Leopold makes the claim that one of the main reasons why most people have not developed a land ethic, is because the importance of such an ideal is not inherently within us. By claiming that in one of the highest moral contexts, religion, the land holds little to no importance, he is stating that religious beliefs don’t speak of protecting the land. “The proof that conservation has not yet touched these foundations of conduct, lies in the fact that philosophy and religion have not heard of it.” In this passage he is referring to loyalties, affections, and convictions when he says “these foundations”. Leopold is saying that when it comes to a person’s belief system, the most powerful determinants are that person’s religious and philosophical ideals. The problem with this statement is that many different religions speak of protecting the land as a way of striving to become a morally good person. Tikkun Olam is a phrase that is spoken in the Jewish religion quite often. It’s most basic translation is, “to protect the earth”. One thing that tends to occur when people begin to follow a religion though, is that person’s interpretation of the religious text can become skewed. It is a common trait in humans to essentially hear what they want to hear when listening to anything. That is my contention. Instead of Leopold blaming religion and philosophy for not addressing land ethics, it should be stated that people have the power to choose to listen or ignore certain aspects of their religion. Many people have chosen not to place as much importance on the land as they do on personal aspects of their morality. That is the power the religion wields. It is only the blueprint that guides people towards their own decisions regarding everything, and in this case many people have decided that protecting the land is not on the top of their list of priorities.

Ben Britton August 29, 2007 PHI 331 Critical Question #1

First paragraph of the Ethical Sequence… “This extension of ethics, so far studied, so far studied only by philosophers, is actually a process in ecological evolution. … An ethic ecologically is a limitation on freedom of action in the struggle for existence. An ethic, philosophically, is a differentiation between social from anti social conduct. … (Ethics) has its origin in the tendency of interdependent individuals or groups to evolve modes of cooperation. Ecologist calls it symbiosis. Politics and economics are advanced symbioses in which the original free-for-all competition has been replaced in part, by co-operative mechanisms with an ethical content.”

I can understand how the ethics of land have changed and evolved much like the natural has. Leopold tells us that the ecological ethic is a limitation of freedom, but it is a conscious one. We know that some of us do now follow these innate ethics. For example, we ethically are bound to some stewardship of the land just so it will be fertile enough to grow food for us to eat. No matter how much we overlook the Earth’s abundance, we will never be able to overlook our dependence. This is one of the reasons we can never let a nuclear war happen, all of the land would be ruined. I have just read in my ENV 230 class that this justification of ethic is called utilitarian, in which the land has value because it benefits individuals economically or is a direct necessity for human survival. This is just the start of land ethic. An optimist see the potential, in not just sustaining the land, but letting enough go untouched so that it may grow to be something greater one day.

Evolution I have noticed to be driven by hardship. Man probably first learned how to walk because he ran out of food and had to move to another place. This eventually lead him to a less tree populated area in which he spent more time on the ground, crawling. Often evolution finds itself between a rock and a hard place. In the Book The Klamath Knot, the author explains how lichen came about. It was on the surface rocks cracked over time by water expanding and contracting. Neither algae of fungi could dominate the space, so somehow the co mingled and now lichen exists. This symbiosis of organism created the blue green algae which are found on almost every environment on Earth. Leopold states the Political and Economical world are robot like machines with artificial intelligence know as ethics content (...replaced in part, by co-operative mechanisms with an ethical content.”) Since when does the market place or the government have any ethics? True, democracy consist of many interdependent branches that function as a whole, but as long as we have one man/woman as head of the country there shall always be a knot is the symbiotic relationship of the political world, making it harder for any evolution to occur.

Kimberly Cook August 30, 2007, Environmental Ethics Critical Question #1 In response to Jacob Slade:

In response to critical question # 1 I would like to refer to the point where Leopold states “The answer, if there is any, seems to be in a land ethic, or some other force which assigns more obligation to the private landowner.” This text describes how Leopold believes the one solution to immature or underdeveloped land ethics is through the individual. The questions referred to by “The answer,” are asking if the government of a society where conservation requires monetary incentive will find trouble inevitably. He is saying the responsibility of conservation belongs to the individual, “private landowner” and not the government or religion. In this portion of The Land Ethic text he indicates there are economic undertones to American conservation. He also identifies a problem in a democracy which allows its government to be responsible for the conservation of land. Leopold goes on to say, “An ethical obligation on the part of the private owner is the only visible remedy for these situations.” This reaffirms his belief that it is the individuals’ responsibility to conserve the land beneath his feet.

These quotes lead me to discuss how Leopold was not blaming religion or philosophy for lack of land ethics. He was simply stating that the land is not given ample credit; a land ethic has not yet been adequately addressed by anyone, even religious figures and philosophers. Leopold emphasized that a single land owner should take on conservation regulations, but also assume the main responsibility for conservation. As far as protecting the earth goes it is not common to see religious rights campaigns integrating conservational ideas. Referring to the religious reference there are many things spoken or written that are not applied i.e. the references in Exodus of the Old Testament to slavery. Exodus 20:17"Thou shalt not covet thy neighbor's house, thou shalt not covet thy neighbor's wife, nor his manservant, nor his maidservant, nor his ox, nor his ass, nor any thing that is thy neighbor's." Though this has been said and read, it was certainly not practiced by all those who believe in the Old Testament. Today a majority of the people having faith in the Old Testament would still not have slaves. Point being although someone has faith and believes in a religious text doesn’t mean they practice all things in the text. Therefore they are not credited for practicing conservation just by talking about it as they are not punished for having slaves though their religious text may says it’s ok.


text:
User:Elmedio~enwikiversity

Hi my name is Germán Scully, I am a student at the Wikiversity. My main interests are in the domains of New Media Art, Math for computer programming, Music, Media Theory, Cultural Studies,Graphic Design, Literature and Creative Writing.

At first sight, what is "New Media Arts" to you?
... New Media Arts is the extension of manifestations in 20th century Art, Design, Music, Film and Computer Science brought bypost-World War II technological advancements, namely in the field of digital technology and the progressive development and evolution of the personal computer and other means of individual technological expression. it is a cross-disciplinary research encompassing Computer Science, Fine Arts and Filmmaking/Storytelling as a spectrum. The term could be considered a misnomer since many of the basic principles of computer aided multimedia art and design have been in practice for almost half a century now. The fact that the details of the practice and the technologies in question are ever-changing and thus 'new' is also a matter of discussion since the same is not absent to the other media deemed traditional, the impact of the development of acrylic paint to painting in the 20th century for example. Other suggested names by which the otherwise new media field has gone by are: Computer Art/Design, Design and Technology, Digital Media... these keep changing almost resisting definition suggesting more of a concern for the technological novelties than a concrete paradigm. It is very good that the department was founded by the determination that New Media Art is not a discipline but an area of research and this is what interested me about the school :)

Elmedio 15:07, 29 August 2007 (UTC)

What do you understand in these words? Do we need capital letters?
...

Capital letters are a formality of Modern English titling practices, the title could be

What is the technical specificities of new media arts, what differs from "old medias"?
...

 * List the arts forms, medias and existing bodies of works that can be qualified so and explain.

...

 * Is there a rupture?

...

I think this is the most important question. It could be more of a question of principles and convictions, one mindset separates the new artist from the old while another could just consider a the artist as an unity throughout time and history, I'm more in the later group than with the former. The fact that with a single computer the artist can be a musician, illustrator, photographer, filmmaker... without having to be a Leonardo Da Vinci is an interesting, very current phenomenon. Another question is for how long will this be New Media? That's a problem of permanence that we should consider If we want to built something that creates a precedence and is not otherwise ephemeral.Elmedio 15:23, 29 August 2007 (UTC)

What is new media arts functions?
...

For yourself, for the society?
...

Is new medias a vision of our everyday future?
...

What does the student of new media arts needs to learn?
...

What are the fundamental softwares, cultural references he/she needs to acquire?
...

Why?
...

How should new media arts be taught?
...

For now i guess its ok to go through an quick overview, eventually the best thing to do its have a more specific and coherent division of departments :

Animation - Digital Video - Computer Art - Web Design etc.Elmedio 15:42, 29 August 2007 (UTC)

How can Wikiversity bring an interesting approach?
...

Wikiversity could make for the perfect setting for the development of a truly new media department, ever growing changing keeping itself alive, wiki technology is a very current technological and human forefront of teaching, due to its dynamic and modular format.Elmedio 15:42, 29 August 2007 (UTC)

How should the teaching be structured, scheduled, granted?
...

The ideal I think would be to combine a selected history of the development of the fields in question with practical (hands on) exercises that take the student through the work of the pioneers/forefathers of these fields by experimenting with the same techniques into pieces of original work. Separating the theory from the practice is a mistake that I feel create a disjointed and incomplete learning process.Elmedio 15:42, 29 August 2007 (UTC)

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

title: User:Elmedio~enwikiversity


text:
Statistics/OpenStat/Research projects

 * /Copyright and the legal status of statistical material/
 * /Documentation and evaluation standards of open statistical material/
 * /Formatting rules and function for importing into R/

title: Statistics/OpenStat/Research projects


text:
Strength of materials/Lesson 3

Lesson 3: Stress transformation and Mohr's circle
Last time we talked about Hooke's law and plane stress. We also discussed how the normal and shear components of stress change depending on the orientation of the plane that they act on. In this lecture we will talk about stress transformations for plane stress.

For the rest of this lesson we assume that we are dealing only with plane stress, i.e., there are only three nonzero stress components $$\sigma_x$$, $$\sigma_y$$, $$\tau_{xy}$$. We also assume that these three components are known.

We want to find the planes on which the stresses are most severe and the magnitudes of these stresses.

Stress transformation rules
Let us consider an arbitrary plane inside an infinitesimal element. Let this plane be inclined at an angle $$\theta$$ to the vertical face of the element. A free body diagram of the region to the left of this plane is shown in the figure below.

A balance of forces on the free body in the $$n$$-direction gives us

\sum f_n = \sigma_n~dA - (\sigma_y~dA~\sin\theta)~\sin\theta - (\tau_{xy}~dA~\sin\theta)~\cos\theta - (\sigma_x~dA~\cos\theta)~\cos\theta - (\tau_{xy}~dA~\cos\theta)~\sin\theta = 0 $$ or,

\sigma_n = \sigma_x~\cos^2\theta + \sigma_y~\sin^2\theta + 2~\tau_{xy}~\sin\theta\cos\theta $$ Using the trigonometric identities

\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta ~; \sin 2\theta = 2\sin\theta\cos\theta $$ we get

\sigma_n = \sigma_x~\left(\cfrac{1 + \cos 2\theta}{2}\right) + \sigma_y~\left(\cfrac{1 - \cos 2\theta}{2}\right) + \tau_{xy}~\sin 2\theta $$ or,
 * $$ \text{(1)} \qquad

\sigma_n = \frac{1}{2}~(\sigma_x + \sigma_y) + \frac{1}{2}~(\sigma_x - \sigma_y)~\cos 2\theta + \tau_{xy}~\sin 2\theta $$

Similarly, a balance of forces in the $$t$$-direction leads to

\sum f_t = \tau_{nt}~dA - (\sigma_y~dA~\sin\theta)~\cos\theta - (\tau_{xy}~dA~\sin\theta)~\sin\theta - (\sigma_x~dA~\cos\theta)~\sin\theta - (\tau_{xy}~dA~\cos\theta)~\cos\theta = 0 $$ or

\tau_{nt} = -(\sigma_x + \sigma_y)~\sin\theta\cos\theta + \tau_{xy}~(\cos^2\theta - \sin^2\theta) $$ or,
 * $$ \text{(2)} \qquad

\tau_{nt} = -\frac{1}{2}~(\sigma_x + \sigma_y)~\sin 2\theta + \tau_{xy}~\cos 2\theta $$

Now let us look at a section that is perpendicular to the one we have looked at. This situation is shown in the figure below.

In this case, a balance of forces on the free body in the $$n$$-direction gives us

\sum f_n = \sigma_n~dA - (\sigma_y~dA~\cos\theta)~\cos\theta + (\tau_{xy}~dA~\cos\theta)~\sin\theta - (\sigma_x~dA~\sin\theta)~\sin\theta + (\tau_{xy}~dA~\sin\theta)~\cos\theta = 0 $$ or,

\sigma_n = \sigma_x~\sin^2\theta + \sigma_y~\cos^2\theta - 2~\tau_{xy}~\sin\theta\cos\theta $$ or,

\sigma_n = \sigma_x~\left(\cfrac{1 - \cos 2\theta}{2}\right) + \sigma_y~\left(\cfrac{1 + \cos 2\theta}{2}\right) - \tau_{xy}~\sin 2\theta $$ or,
 * $$ \text{(3)} \qquad

\sigma_n = \frac{1}{2}~(\sigma_x + \sigma_y) - \frac{1}{2}~(\sigma_x - \sigma_y)~\cos 2\theta - \tau_{xy}~\sin 2\theta $$

A balance of forces in the $$t$$-direction gives

\sum f_t = \tau_{nt}~dA - (\sigma_y~dA~\cos\theta)~\sin\theta - (\tau_{xy}~dA~\cos\theta)~\cos\theta + (\sigma_x~dA~\sin\theta)~\cos\theta + (\tau_{xy}~dA~\sin\theta)~\sin\theta = 0 $$ or,

\tau_{nt} = -(\sigma_x - \sigma_y)~\sin\theta\cos\theta + \tau_xy~(\cos^2\theta - \sin^2\theta) $$ or,
 * $$ \text{(4)} \qquad

\tau_{nt} = -\frac{1}{2}~(\sigma_x + \sigma_y)~\sin 2\theta + \tau_{xy}~\cos 2\theta $$ From equations (2) and (4) we see that the shear stresses are equal. However the normal stresses on the two planes are different as you can see from equations (1) and (3).

You can think of the two cuts as just the faces of a new infinitesimal element which is at an angle $$\theta$$ to the original element as can be seen form the following figure.

If we label the new normal stresses as $$\sigma_x^{'}$$ and $$\sigma_y^{'}$$ and the shear stresses as $$\tau_{xy}^{'}$$, then we can write

\begin{align} \sigma_x^{'} & = \frac{1}{2}~(\sigma_x + \sigma_y) + \frac{1}{2}~(\sigma_x - \sigma_y)~\cos 2\theta + \tau_{xy}~\sin 2\theta \\ \sigma_y^{'} & = \frac{1}{2}~(\sigma_x + \sigma_y) - \frac{1}{2}~(\sigma_x - \sigma_y)~\cos 2\theta - \tau_{xy}~\sin 2\theta \\ \tau_{xy}^{'} & = -\frac{1}{2}~(\sigma_x - \sigma_y)~\sin 2\theta + \tau_{xy}~\cos 2\theta \end{align} $$

Maximum normal stresses
What is the orientation of the infinitesimal element that produces the largest normal stress and the largest shear stress? This information can be useful in predicting where failure will occur.

To find angle at which we get the maximum/minimum normal stress we can take the derivatives of $$\sigma_x^{'}$$ and $$\sigma_y^{'}$$ with respect to $$\theta$$ and set them to zero. So we have

\begin{align} \cfrac{d\sigma_x^{'}}{d\theta} = 0 = -(\sigma_x - \sigma_y)~\sin 2\theta + 2~\tau_{xy}~\cos 2\theta \\ \cfrac{d\sigma_y^{'}}{d\theta} = 0 = (\sigma_x - \sigma_y)~\sin 2\theta - 2~\tau_{xy}~\cos 2\theta \end{align} $$ or,

2~\theta = \tan^{-1} \left(\cfrac{2~\tau_{xy}}{\sigma_x - \sigma_y}\right) $$ The angle at which $$\theta$$ is a maximum or a minimum is called a  principal angle or $$\theta_p$$.

Now, from the identities (or we can think in terms of a right angled triangle with a rise of $$\tau_{xy}$$ and a run of $$1/2(\sigma_x - \sigma_y)$$)

\cos\left(\tan^{-1}(x)\right) = \cfrac{1}{\sqrt{1+x^2}} ~; \sin\left(\tan^{-1}(x)\right) = \cfrac{x}{\sqrt{1+x^2}} $$ we have

\begin{align} \cos 2\theta_p & = \cfrac{\cfrac{\sigma_x - \sigma_y}{2}} {\sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}} \\ \sin 2\theta_p & = \cfrac{\tau_{xy}} {\sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}} \end{align} $$ Taking another derivative with respect to $$\theta$$ we have

\begin{align} \cfrac{d^2\sigma_x^{'}}{d\theta^2} = -2~(\sigma_x - \sigma_y)~\cos 2\theta_p - 4~\tau_{xy}~\sin 2\theta_p \\ \cfrac{d^2\sigma_y^{'}}{d\theta^2} = 2~(\sigma_x - \sigma_y)~\cos 2\theta_p + 4~\tau_{xy}~\sin 2\theta_p \end{align} $$ Plugging in the expressions for $$\cos 2\theta_p$$ and $$\sin 2\theta_p$$ we get

\begin{align} \cfrac{d^2\sigma_x^{'}}{d\theta^2} = -4~ \sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \le 0 \\ \cfrac{d^2\sigma_y^{'}}{d\theta^2} = 4 \sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \ge 0 \end{align} $$ Clearly $$\sigma_x^{'}$$ is a maximum while $$\sigma_y^{'}$$ is a minimum value.

Principal stresses
The normal stresses corresponding to the principal angle $$\theta_p$$ are called the  principal stresses.

We have

\begin{align} \sigma_x^{'} & = \frac{1}{2}~(\sigma_x + \sigma_y) + \frac{1}{2}~(\sigma_x - \sigma_y)~\cos 2\theta_p + \tau_{xy}~\sin 2\theta_p \\ \sigma_y^{'} & = \frac{1}{2}~(\sigma_x + \sigma_y) - \frac{1}{2}~(\sigma_x - \sigma_y)~\cos 2\theta_p - \tau_{xy}~\sin 2\theta_p \end{align} $$ Plugging in the expressions for $$\cos 2\theta_p$$ and $$\sin 2\theta_p$$ we get

\begin{align} \sigma_x^p & = \frac{1}{2}~(\sigma_x + \sigma_y) + \sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \\ \sigma_y^p & = \frac{1}{2}~(\sigma_x + \sigma_y) - \sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \end{align} $$ These principal stresses are often written as $$\sigma_{p1}$$ and $$\sigma_{p2}$$ or $$\sigma_1$$ and $$\sigma_2$$ where $$\sigma_1 > \sigma_2$$.

The value of the shear stress $$\tau_{xy}^{'}$$ for an angle of $$\theta_p$$ is

\tau_{xy}^{'} = -\frac{1}{2}~(\sigma_x - \sigma_y)~\sin 2\theta_p + \tau_{xy}~\cos 2\theta_p $$ Plugging in the expressions for $$\cos 2\theta_p$$ and $$\sin 2\theta_p$$ we get

\tau_{xy}^p = 0 $$ Hence there are no shear stresses in the orientations where the stresses are maximum or minimum.

Maximum shear stresses
Similarly, we can find the value of $$\theta$$ which makes the shear stress a maximum or minimum. Thus

\cfrac{d\tau_{xy}^{'}}{d\theta} = 0 = -(\sigma_x - \sigma_y)~\cos 2\theta - 2~\tau_{xy}~\sin 2\theta $$ or

2~\theta = \tan^{-1} \left(- \cfrac{2~\tau_{xy}}{\sigma_x - \sigma_y}\right) $$ In that case

\begin{align} \cos 2\theta_s & = \cfrac{\cfrac{\sigma_x - \sigma_y}{2}} {\sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}} \\ \sin 2\theta_s & = -\cfrac{\tau_{xy}} {\sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}} \end{align} $$ The value of the shear stress $$\tau_{xy}^{'}$$ for an angle of $$\theta_s$$ is

\tau_{xy}^{'} = -\frac{1}{2}~(\sigma_x - \sigma_y)~\sin 2\theta_s + \tau_{xy}~\cos 2\theta_s $$ Plugging in the expressions for $$\cos 2\theta_s$$ and $$\sin 2\theta_s$$ we get

\tau_{xy}^{\text{max}} = \sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} $$ We can show that this is the maximum value of $$\tau_{xy}$$.

Note that, at the value of $$\theta$$ where $$\tau_{xy}$$ is maximum, the normal stresses are not zero.

Mohr's circle
Mohr's idea was to express these algebraic relations in geometric form so that a physical interpretation of the idea became easier. The idea was based on the recognition that for an orientation equal to the principal angle, the stresses could be represented as the sides of a right-angled triangle.

Recall that

\begin{align} \cos 2\theta_p & = \cfrac{\cfrac{\sigma_x - \sigma_y}{2}} {\sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}} \\ \sin 2\theta_p & = \cfrac{\tau_{xy}} {\sqrt{\left(\cfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}} \end{align} $$ We can represent this in graphical form as shown in the figure below. In general, the locus of all points representing stresses at various orientations lie on a circle which is called  Mohr's circle.

Notice that we can directly find the largest normal stress and the small normal stress as well as the maximum shear stress directly from the circle. In three-dimensions there are two more Mohr's circles.

Negative shear stress
Also note that there is a region where the shear stress $$\tau$$ is negative. The convention that we follow is that if the shear stress rotates the element clockwise then it is a positive shear stress. If the element is rotated counter-clockwise then the shear stress is negative.

In the next lecture we will get into some more detail about actually plotting Mohr's circles.

title: Strength of materials/Lesson 3


text: File:MohrsCircle.png

title: File:MohrsCircle.png


title: File:StressTransform2D.png


text: File:StressTransform2D 1.png

title: File:StressTransform2D 1.png


text:
Statistics/OpenStat/External links

R - the statistical language

title: Statistics/OpenStat/External links


text: File:StressTransform2D 2.png

title: File:StressTransform2D 2.png


text:
Menu Planning/Menu contents/Breakfast

Breakfast is the most important meal of the day for many people, because it gives the carbohydrates needed for energy throughout the day; thus, it should ideally be the largest. Included in it should be a grain, fruit, meat, dairy product, and a starch. Simple breakfasts (as with donuts and coffee or a slice of toast, etc.) are not suitable for long-term fueling.

A traditional Western breakfast might include:
 * Pancakes or waffles
 * Grits, oatmeal, or other cereals
 * Biscuits and gravy
 * Cinnamon rolls
 * Burritos (egg, sausage, cheese, optional chopped onions and green peppers, etc.)
 * A variety of fruit juices or milk

Back home–Lunch

title: Menu Planning/Menu contents/Breakfast


text:
Menu Planning/Menu contents/Lunch

Included in a lunch should be a protein, grains, a fruit and a vegetable. Lunch offers a midday break during work hours, and it can thus be leisurely or very fast. What works best for your establishment should be determined on how long workers have to eat during their lunch breaks. Regardless of the speed required, however, quality should not come down.

Lunch ideas include:
 * Sandwiches; plain, toasted, deep-fried or grilled
 * Chicken, beef or seafood (grilled, roasted, broiled, baked, etc.)
 * Rice
 * Salads (Cesar, Greek, chef, coleslaw, chicken, fruit, antipasto, others)
 * Various kinds of fruits, especially ones that can be knocked about if the meal is going out of the restaurant
 * Soups
 * Tacos

Breakfast–Dinner

title: Menu Planning/Menu contents/Lunch


text:
Applied linear operators and spectral methods/Lecture 4

More on spectral decompositions
In the course of the previous lecture we essentially proved the following theorem:

Theorem:
1) If a $$n\times n$$ matrix $$\mathbf{A}$$ has $$n$$ linearly independent real or complex eigenvectors, the $$\mathbf{A}$$ can be diagonalized. 2) If $$\mathbf{T}$$ is a matrix whose columns are eigenvectors then $$\mathbf{T}\mathbf{A}\mathbf{T}^{-1}= \boldsymbol{\Lambda}$$ is the diagonal matrix of eigenvalues.

The factorization $$\mathbf{A} = \mathbf{T}^{-1}\boldsymbol{\Lambda}\mathbf{T}$$ is called the  spectral representation of $$\mathbf{A}$$.

Application
We can use the spectral representation to solve a system of linear homogeneous ordinary differential equations.

For example, we could wish to solve the system

\cfrac{d\mathbf{u}}{dt} = \mathbf{A}\mathbf{u} = \begin{bmatrix}-2 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} $$ (More generally $$\mathbf{A}$$ could be a $$n \times n$$ matrix.)

Comment:
Higher order ordinary differential equations can be reduced to this form. For example,

\cfrac{d^2u_1}{dt^2} + a~\cfrac{du_1}{dt} = b~u_1 $$ Introduce

u_2 = \cfrac{du_1}{dt} $$ Then the system of equations is

\begin{align} \cfrac{du_1}{dt} & = u_2 \\ \cfrac{du_2}{dt} & = b~u_1 - a~u_2 \end{align} $$ or,

\cfrac{d\mathbf{u}}{dt} = \begin{bmatrix} 0 & 1 \\ b & -a \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \mathbf{A} \mathbf{u} $$

Returning to the original problem, let us find the eigenvalues and eigenvectors of $$\mathbf{A}$$. The characteristic equation is

det(\mathbf{A} - \lambda~\mathbf{I}) = 0 $$ o we can calculate the eigenvalues as

(2+\lambda)(2+\lambda) - 1 = 0 \quad \implies \quad \lambda^2 + 4\lambda + 3 = 0 \qquad \implies \qquad \lambda_1 = -1, \qquad \lambda_2= -3 $$ The eigenvectors are given by

(\mathbf{A} - \lambda_1~\mathbf{I})\mathbf{n}_1 = \mathbf{0} ~; (\mathbf{A} - \lambda_2~\mathbf{I})\mathbf{n}_2 = \mathbf{0} $$ or,

-n_1^1 + n_2^1 = 0 ~; n_1^1 - n_2^1 = 0 ~; n_1^2 + n_2^2 = 0 ~; n_1^2 + n_2^2 = 0 $$ Possible choices of $$\mathbf{n}_1$$ and $$\mathbf{n}_2$$ are

\mathbf{n}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} ~; \mathbf{n}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$ The matrix $$\mathbf{T}$$ is one whose columd are the eigenvectors of $$\mathbf{A}$$, i.e.,

\mathbf{T} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ and

\boldsymbol{\Lambda} = \mathbf{T}^{-1} \mathbf{A} \mathbf{T} = \begin{bmatrix} -1 & 0 \\ 0 & -3 \end{bmatrix} $$ If $$\mathbf{u} = \mathbf{T} \mathbf{u}^{'}$$ the system of equations becomes

\cfrac{d\mathbf{u}^{'}}{dt} = \mathbf{T}^{-1} \mathbf{A} \mathbf{T} \mathbf{u}^{'} = \boldsymbol{\Lambda}~\mathbf{u}^{'} $$ Expanded out

\cfrac{du_1^{'}}{dt} = - u_1^{'} ~; \cfrac{du_2^{'}}{dt} = - 3~u_2^{'} $$ The solutions of these equations are

u_1^{'} = C_1~e^{-t} ~; u_2^{'} = C_2~e^{-3t} $$ Therefore,

\mathbf{u} = \mathbf{T}~\mathbf{u}^{'} = \begin{bmatrix} C_1~e^{-t} + C_2~e^{-3t} \\ C_1~e^{-t} - C_2~e^{-3t} \end{bmatrix} $$ This is the solution of the system of ODEs that we seek.

Most "generic" matrices have linearly independent eigenvectors. Generally a matrix will have $$n$$ distinct eigenvalues unless there are symmetries that lead to repeated values.

Theorem
If $$\mathbf{A}$$ has $$k$$ distinct eigenvalues then it has $$k$$ linearly independent eigenvectors.

 Proof:

We prove this by induction.

Let $$\mathbf{n}_j$$ be the eigenvector corresponding to the eigenvalue $$\lambda_j$$. Suppose $$\mathbf{n}_1, \mathbf{n}_2, \dots, \mathbf{n}_{k-1}$$ are linearly independent (note that this is true for $$k$$ = 2). The question then becomes: Do there exist $$\alpha_1, \alpha_2, \dots, \alpha_k$$ not all zero such that the linear combination

\alpha_1~\mathbf{n}_1 + \alpha_2~\mathbf{n}_2 + \dots + \alpha_k~\mathbf{n}_k = 0 $$ Let us multiply the above by $$(\mathbf{A} - \lambda_k~\mathbf{I})$$. Then, since $$\mathbf{A}~\mathbf{n}_i = \lambda_i~\mathbf{n}_i$$, we have

\alpha_1~(\lambda_1 - \lambda_k)~\mathbf{n}_1 + \alpha_2~(\lambda_2 - \lambda_k)~\mathbf{n}_2 + \dots + \alpha_{k-1}~(\lambda_{k-1} - \lambda_k)~\mathbf{n}_{k-1} + \alpha_k~(\lambda_k - \lambda_k)~\mathbf{n}_k = \mathbf{0} $$ Since $$\lambda_k$$ is arbitrary, the above is true only when

\alpha_1 = \alpha_2 = \dots = \alpha_{k-1} = 0 $$ In thast case we must have

\alpha_k~\mathbf{n}_k = \mathbf{0} \quad \implies \quad \alpha_k = 0 $$ This leads to a contradiction.

Therefore $$\mathbf{n}_1, \mathbf{n}_2, \dots, \mathbf{n}_k$$ are linearly independent. $$\qquad \square$$

Another important class of matrices which are diagonalizable are those which are self-adjoint.

Theorem
If $$\boldsymbol{A}$$ is self-adjoint the following statements are true

 * 1) $$\langle \boldsymbol{A}\mathbf{x}, \mathbf{x} \rangle$$ is real for all $$\mathbf{x}$$.
 * 2) All eigenvalues are real.
 * 3) Eigenvectors of distinct eigenvalues are orthogonal.
 * 4) There is an orthonormal basis formed by the eigenvectors.
 * 5) The matrix $$\boldsymbol{A}$$ can be diagonalized (this is a consequence of the previous statement.)

 Proof

1) Because the matrix is self-adjoint we have

\langle \boldsymbol{A}\mathbf{x}, \mathbf{x} \rangle = \langle \mathbf{x}, \boldsymbol{A}\mathbf{x} \rangle $$ From the property of the inner product we have

\langle \mathbf{x}, \boldsymbol{A}\mathbf{x} \rangle = \overline{\langle \boldsymbol{A}\mathbf{x}, \mathbf{x} \rangle} $$ Therefore,

\langle \boldsymbol{A}\mathbf{x}, \mathbf{x} \rangle = \overline{\langle \boldsymbol{A}\mathbf{x}, \mathbf{x} \rangle} $$ which implies that $$\langle \boldsymbol{A}\mathbf{x}, \mathbf{x} \rangle$$ is real.

2) Since $$\langle \boldsymbol{A}\mathbf{x}, \mathbf{x} \rangle$$ is real, $$\langle \boldsymbol{I}\mathbf{x}, \mathbf{x} \rangle = \langle \mathbf{x},\mathbf{x} \rangle$$ is real. Also, from the eiegnevalue problem, we have

\langle \boldsymbol{A}\mathbf{x}, \mathbf{x} \rangle = \lambda~\langle \mathbf{x}, \mathbf{x} \rangle $$ Therefore, $$\lambda$$ is real.

3) If $$(\lambda,\mathbf{x})$$ and $$(\mu, \mathbf{y})$$ are two eigenpairs then

\lambda~\langle \mathbf{x}, \mathbf{y} \rangle = \langle \boldsymbol{A}\mathbf{x}, \mathbf{y} \rangle $$ Since the matrix is self-adjoint, we have

\lambda~\langle \mathbf{x}, \mathbf{y} \rangle = \langle \mathbf{x}, \boldsymbol{A}\mathbf{y} \rangle = \mu~\langle \mathbf{x}, \mathbf{y} \rangle $$ Therefore, if $$\lambda \ne \mu \ne 0$$, we must have

\langle \mathbf{x}, \mathbf{y} \rangle = 0 $$ Hence the eigenvectors are orthogonal.

4) This part is a bit more involved. We need to define a manifold first.

Linear manifold
A linear manifold (or vector subspace) $$\mathcal{M} \in \mathcal{S}$$ is a subset of $$\mathcal{S}$$ which is closed under scalar multiplication and vector addition.

Examples are a line through the origin of $$n$$-dimensional space, a plane through the origin, the whole space, the zero vector, etc.

Invariant manifold
An invariant manifold $$\mathcal{M}$$ for the matrix $$\boldsymbol{A}$$ is the linear manifold for which $$\mathbf{x} \in \mathcal{M}$$ implies $$\boldsymbol{A}\mathbf{x} \in \mathcal{M}$$.

Examples are the null space and range of a matrix $$\boldsymbol{A}$$. For the case of a rotation about an axis through the origin in $$n$$-space, invaraiant manifolds are the origin, the plane perpendicular to the axis, the whole space, and the axis itself.

Therefore, if $$\mathbf{x}_1, \mathbf{x}_2, \dots,\mathbf{x}_m$$ are a basis for $$\mathcal{M}$$ and $$\mathbf{x}_{m+1}, \dots, \mathbf{x}_n$$ are a basis for $$\mathcal{M}_\perp$$ (the perpendicular component of $$\mathcal{M}$$) then in this basis $$\boldsymbol{A}$$ has the representation

\boldsymbol{A} = \begin{bmatrix} x & x & | & x & x \\ x & x & | & x & x \\ - & - & - & - & - \\                       0 & 0 & | & x & x \\ 0 & 0 & | & x & x \end{bmatrix} $$ We need a matrix of this form for it to be in an invariant manifold for $$\boldsymbol{A}$$.

Note that if $$\mathcal{M}$$ is an invariant manifold of $$\boldsymbol{A}$$ it does not follow that $$\mathcal{M}_\perp$$ is also an invariant manifold.

Now, if $$\boldsymbol{A}$$ is self adjoint then the entries in the off-diagonal spots must be zero too. In that case, $$\boldsymbol{A}$$ is block diagonal in this basis.

Getting back to part (4), we know that there exists at least one eigenpair ($$\lambda_1, \mathbf{x}_1$$) (this is true for any matrix). We now use induction. Suppose that we have found ($$n-1$$) mutually orthogonal eigenvectors $$\mathbf{x}_i$$ with $$\boldsymbol{A}\mathbf{x}_i = \lambda_i\mathbf{x}_i$$ and $$\lambda_i$$ are real, $$i = 1, \dots, k-1$$. Note that the $$\mathbf{x}_i$$s are invariant manifolds of $$\boldsymbol{A}$$ as is the space spanned by the $$\mathbf{x}_i$$s and so is the manifold perpendicular to these vectors).

We form the linear manifold

\mathcal{M}_k = \{\mathbf{x} | \langle \mathbf{x}, \mathbf{x}_j \rangle = 0 j = 1, 2, \dots, k-1\} $$ This is the orthogonal component of the $$k-1$$ eigenvectors $$\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_{k-1}$$ If $$\mathbf{x} \in \mathcal{M}_k$$ then

\langle \mathbf{x}, \mathbf{x}_j \rangle = 0 \quad \text{and} \quad \langle \boldsymbol{A}\mathbf{x}, \mathbf{x}_j \rangle = \langle \mathbf{x}, \boldsymbol{A}\mathbf{x}_j \rangle = \lambda_j\langle \mathbf{x}, \mathbf{x}_j \rangle = 0 $$ Therefore $$\boldsymbol{A}\mathbf{x} \in \mathcal{M}_k$$ which means that $$\mathcal{M}_k$$ is invariant.

Hence $$\mathcal{M}_k$$ contains at least one eigenvector $$\mathbf{x}_k$$ with real eigenvalue $$\lambda_k$$. We can repeat the procedure to get a diagonal matrix in the lower block of the block diagonal representation of $$\boldsymbol{A}$$. We then get $$n$$ distinct eigenvectors and so $$\boldsymbol{A}$$ can be diagonalized. This implies that the eigenvectors form an orthonormal basis.

5) This follows from the previous result because each eigenvector can be normalized so that $$\langle \mathbf{x}_i, \mathbf{x}_j \rangle = \delta_{ij}$$.

We will explore some more of these ideas in the next lecture.


text:
Menu Planning/Menu contents/Dinner

Dinner is often the time when families come together at restaurants to be together for some time, and have a leisurely meal after the day's work has ended. Therefore, be prepared to have intriguing menu options for your customers. However, care should be taken not to exclude those who do not wish for a large, elaborate meal at the end of their day (who might have already had it at lunch—be cognizant of shift changes and how that affects which meal of the day they will want to eat the most at).

Common dinner options include:
 * Steak
 * Potatoes
 * Cheesecake
 * Cake

Lunch–Snack

title: Menu Planning/Menu contents/Dinner


text:
Menu Planning/Menu contents/Snack

A snack is to be something light. with sufficient caloric/nutritional value to bolster the nutrition benefits if the preceding meal. Additionally, water is an important element to be considered when preparing snacks Dinner–Back home

title: Menu Planning/Menu contents/Snack


text:
File:KinnetiK 3D Storyboards Frame 1.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 2.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 3.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 4.jpg

These pictures are created in the Lesson:3D Storyboard


text:
Statistics/OpenStat/MAMEU

=Data MAMEU=

Description
This is supposed to become a dataset containing various variables concerning migration into the European Union.

Feel free to expand it. For the purpose of machine readability don`t forget to separate every field (year and number,source,verifications) by the character |. If you can verify the data by comparing to the specified source, you can add a verified mark (4 times a ~ at the end of the row).

Variables
Y/m/d/h/m | data | source | 1st verify | 2nd verify

Asylumseekers Germany
1975 | 9627 | baafbrd2002 |

1980 | 107818 | baafbrd2002 |

1985 | 73832 | baafbrd2002 |

1990 | 193063 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 1991 | 256112 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 1992 | 438191 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 1993 | 322599 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 1994 | 127210 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 1995 | 127937 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 1996 | 116367 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 1997 | 104353 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 1998 | 98644 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 1999 | 95113 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 2000 | 78564 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 2001 | 88287 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC) 2002 | 71127 | baafbrd2002 | euro 23:38, 29 August 2007 (UTC)

New citizens Germany
1981|35 878 1982|39 280 1983|39 485 1984|38 046 1985|34 913 1986|36 646 1987|37 810 1988|46 783 1989|68 526 1990|101 377 1991|141 630 1992|179 904 1993|199 443 1994|259 170 1995|313 606 1996|302 830 1997|278 662 1998|291 331 1999|248 206 2000|186 688 2001|178 098 2002|154 547 2003|140 731 2004|127 153 2005|117 241 2006|124 832

Asylums Granted Swizerland
1994 | 2954 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 1995 | 2641 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 1996 | 2278 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 1997 | 2623 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 1998 | 2029 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 1999 | 2063 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 2000 | 2078 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 2001 | 2252 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 2002 | 1738 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 2003 | 1636 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 2004 | 1555 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 2005 | 1497 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 2006 | 1857 | bamejp2007 | euro 12:28, 4 February 2008 (UTC) 2007 | 1561 | bamejp2007 | euro 12:28, 4 February 2008 (UTC)

Deportations Switzerland
1994 | 2109                                                                                1995 | 1405                                                                                1996 | 1350                                                                                1997 | 1731                                                                                1998 | 4348                                                                                1999 | 3149                                                                                2000 | 5349                                                                                2001 | 2258                                                                                2002 | 2751                                                                                2003 | 2858                                                                                2004 | 2330                                                                                2005 | 1281                                                                                2006 | 1193                                                                                2007 | 1092

Asylums granted per Asylumseekers Germany
1975 | 22.2 | baafbrd2002

1980 | 12.0 | baafbrd2002

1985 | 29.2 | baafbrd2002

1990 | 4.4 | baafbrd2002 1991 | 6.9 | baafbrd2002 1992 | 4.3 | baafbrd2002 1993 | 3.2 | baafbrd2002 1994 | 7.3 | baafbrd2002 1995 | 9.0 | baafbrd2002 1996 | 7.4 | baafbrd2002 1997 | 4.9 | baafbrd2002 1998 | 4.0 | baafbrd2002 1999 | 3.0 | baafbrd2002 2000 | 3.0 | baafbrd2002 2001 | 5.3 | baafbrd2002 2002 |1.8 | baafbrd2002

title: Statistics/OpenStat/MAMEU


text:
File:KinnetiK 3D Storyboards Frame 5.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 7.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 6.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 8.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 9.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 10.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 11.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 12.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 13.jpg

These pictures are created in the Lesson:3D Storyboard


text:
File:KinnetiK 3D Storyboards Frame 14.jpg

These pictures are created in the Lesson:3D Storyboard


text:
User:WerewolfOfVulcan

Geek, part time college faculty, non-profit enthusiast, Star Trek fan... }:-)

title: User:WerewolfOfVulcan


text:
User talk:WerewolfOfVulcan

Hello WerewolfOfVulcan, 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.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 00:16, 30 August 2007 (UTC)

title: User talk:WerewolfOfVulcan


text:
Elasticity/Antiplane shear

-- Return to Introduction to Elasticity --

Antiplane shear (or antiplane strain) is the state of strain that is obtained when the displacement field is of the form

u_1 = u_2 = 0, u_3 = u_3(x_1, x_2) $$ There is only an out of plane displacement.

Since the strains are given by

\varepsilon_{ij} = \cfrac{1}{2}(u_{i,j} + u_{j,i}) = \cfrac{1}{2}\left(\cfrac{\partial u_i}{\partial x_j} + \cfrac{\partial u_j}{\partial x_i}\right) $$ we have

\begin{align} \varepsilon_{11} & = \cfrac{\partial u_1}{\partial x_1} = 0 \\ \varepsilon_{22} & = \cfrac{\partial u_2}{\partial x_2} = 0 \\ \varepsilon_{33} & = \cfrac{\partial u_3}{\partial x_3} = 0 \\ \varepsilon_{23} & = \cfrac{1}{2}\left(\cfrac{\partial u_2}{\partial x_3} + \cfrac{\partial u_3}{\partial x_2}\right) = \cfrac{1}{2}~u_{3,2} \\ \varepsilon_{31} & = \cfrac{1}{2}\left(\cfrac{\partial u_3}{\partial x_1} + \cfrac{\partial u_1}{\partial x_3}\right) = \cfrac{1}{2}~u_{3,1} \\ \varepsilon_{12} & = \cfrac{1}{2}\left(\cfrac{\partial u_1}{\partial x_2} + \cfrac{\partial u_2}{\partial x_1}\right) = 0 \end{align} $$ Therefore, for antiplane shear, the only nonzero strains are the out-of-plane shear strains

\varepsilon_{23} = \cfrac{1}{2}\cfrac{\partial u_3}{\partial x_2} ~; \varepsilon_{13} = \cfrac{1}{2}\cfrac{\partial u_3}{\partial x_1} $$

title: Elasticity/Antiplane shear


text:
Wikiversity:Donate

If you are interested in donating specifically to Wikiversity it is greatly appreciated! This page will be where those who wish to donate specifically to Wikiversity can do so. All funds will be allocated to the general fund. However, if you wish to have funds earmarked for a specific project or area of Wikiversity, that can be arranged as well.

You are highly encouraged to give the Wikimedia Foundation priority in your generosity.

title: Wikiversity:Donate


text:
Elasticity/Constitutive relations

Constitutive relations
Any problem in elasticity is usually set up with the following components:
 * 1) A strain-displacement relation.
 * 2) A traction-stress relation.
 * 3) Balance laws for linear and angular momentum in terms of the stress.

To close the system of equations, we need a relation between the stresses and strains. Such a relation is called a constitutive equation.

Isotropic elasticity
The most popular form of the constitutive relation for linear elasticity (see, for example, Strength of materials) is the following relation that holds for isotropic materials:

\begin{bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ \varepsilon_{zz} \\ \gamma_{yz} \\ \gamma_{zx} \\ \gamma_{xy} \end{bmatrix} = \begin{bmatrix} \cfrac{1}{E} & -\cfrac{\nu}{E} & -\cfrac{\nu}{E} & 0 & 0 & 0 \\ -\cfrac{\nu}{E} & \cfrac{1}{E} & -\cfrac{\nu}{E} & 0 & 0 & 0 \\ -\cfrac{\nu}{E} & -\cfrac{\nu}{E} & \cfrac{1}{E} & 0 & 0 & 0 \\ 0 & 0 & 0 & \cfrac{1}{G} & 0 & 0 \\ 0 & 0 & 0 & 0 & \cfrac{1}{G} & 0 \\ 0 & 0 & 0 & 0 & 0 & \cfrac{1}{G} \end{bmatrix} \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{zz} \\ \sigma_{yz} \\ \sigma_{zx} \\ \sigma_{xy} \end{bmatrix} $$ where $$E$$ is the Young's modulus, $$\nu$$ is the Poisson's ratio and $$G$$ is the shear modulus. The shear modulus can be expressed in terms of the Young's modulus and Poisson's ratio as

G = \cfrac{E}{2(1+\nu)} $$ The shear modulus is also often represented by the symbol $$\mu$$. We will use $$ G$$ and $$\mu$$ interchangeably in this discussion.

Engineering shear strain and tensorial shear strain
In the above matrix equation we have used $$\gamma_{ij}$$ to represent the shear strains rather than $$\varepsilon_{ij}$$. You should keep in mind that

\gamma_{ij} = 2~\varepsilon_{ij} $$

An alternative form of the stress-strain relation
An alternative form of the isotropic linear elastic stress-strain relation that is easier to work with is:

\boldsymbol{\sigma} = \lambda~\text{tr}(\boldsymbol{\varepsilon})~\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\varepsilon} $$ or, in index notation,

\sigma_{ij} = \lambda~\varepsilon_{kk}~\delta_{ij} + 2~\mu~\varepsilon_{ij} $$ Here $$\lambda$$ is the Lamé modulus and $$\mu$$ is the shear modulus. In terms of $$E$$ and $$\nu$$:

\lambda = \cfrac{E~\nu}{(1+\nu)(1-2\nu)} ~; \mu = \cfrac{E}{2(1+\nu)} $$

The inverse relationship is

\boldsymbol{\varepsilon} = -\cfrac{\nu}{E}~\text{tr}(\boldsymbol{\sigma})~\boldsymbol{\mathit{1}} + \cfrac{1+\nu}{E}~\boldsymbol{\sigma} $$ or, in index notation,

\varepsilon_{ij} = -\cfrac{\nu}{E}~\sigma_{kk}~\delta_{ij} + \cfrac{1+\nu}{E}~\sigma_{ij} $$

The stiffness and compliance tensors
For hyperelastic materials, the stress and strain of a linear elastic material are such that one can be derived from a stored energy potential function of the other (also called a strain energy density function). Therefore, we can define an elastic material to be one which satisfies

\boldsymbol{\sigma} = \cfrac{\partial w(\boldsymbol{\varepsilon})}{\partial \boldsymbol{\varepsilon}} \qquad \text{or} \qquad \sigma_{ij} = \cfrac{\partial w}{\partial \varepsilon_{ij}} $$ where $$w$$ is the strain energy density function.

If the material, in addition to being elastic, also has a linear stress-strain relation then we can write

\boldsymbol{\sigma} = \boldsymbol{\mathsf{C}}:\boldsymbol{\varepsilon} \qquad \text{or} \qquad \sigma_{ij} = C_{ijkl}~\varepsilon_{kl} $$ The quantity $$\boldsymbol{\mathsf{C}}$$ is called the  stiffness tensor or the elasticity tensor.

Therefore, the strain energy density function has the form (this form is called a quadratic form)

w(\boldsymbol{\varepsilon}) = \cfrac{1}{2}~\boldsymbol{\varepsilon}:\boldsymbol{\mathsf{C}}:\boldsymbol{\varepsilon} = \cfrac{1}{2}~C_{ijkl}~\varepsilon_{ij}~ \varepsilon_{kl} $$

Clearly, the elasticity tensor has 81 components (think of a $$9 \times 9$$ matrix because the stresses and strains have nine components each). However, the symmetries of the stress tensor implies that

C_{ijkl} = C_{jikl}\, $$ This reduces the number independent components of $$C_{ijkl}\,$$ to 54 (6 components for the $$ij$$ term and 3 each for the $$k, l$$ terms.

Similarly, using the symmetry of the strain tensor we can show that

C_{ijkl} = C_{ijlk}\, $$ These are called the minor symmetries of the elasticity tensor and we are then left with only 36 components that are independent.

Since the strain energy function should not change when we interchange $$ij$$ and $$kl$$ in the quadratic form, we must have

C_{ijkl} = C_{klij}\, $$ This reduces the number of independent constants to 21 (think of a symmetric $$6 \times 6$$ matrix). These are called the major symmetries of the stiffness tensor.

The inverse relation between the strain and the stress can be determined by taking the inverse of stress-strain relation to get

\boldsymbol{\varepsilon} = \boldsymbol{\mathsf{S}}:\boldsymbol{\sigma} \qquad \text{or} \qquad \varepsilon_{ij} = S_{ijkl}~\sigma_{kl} $$ where $$\boldsymbol{\mathsf{S}}$$ is the compliance tensor. The compliance tensor also has 21 components and the same symmetries as the stiffness tensor.

Voigt notation
To express the general stress-strain relation for a linear elastic material in terms of matrices (as we did for the isotropic elastic material) we use what is called the Voigt notation.

In this notation, the stress and strain are expressed as $$ 6 \times 1$$ column vectors and the elasticity tensor is expressed as a symmetric $$ 6 \times 6 $$ matrix as shown below.

\begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix} =  \begin{bmatrix} C_{1111} & C_{1122} & C_{1133} & C_{1123} & C_{1131} & C_{1112} \\ C_{2211} & C_{2222} & C_{2233} & C_{2223} & C_{2231} & C_{2212} \\ C_{3311} & C_{3322} & C_{3333} & C_{3323} & C_{3331} & C_{3312} \\ C_{2311} & C_{2322} & C_{2333} & C_{2323} & C_{2331} & C_{2312} \\ C_{3111} & C_{3122} & C_{3133} & C_{3123} & C_{3131} & C_{3112} \\ C_{1211} & C_{1222} & C_{1233} & C_{1223} & C_{1231} & C_{1212} \end{bmatrix} \begin{bmatrix} \varepsilon_{11}\\ \varepsilon_{22}\\ \varepsilon_{33}\\ 2~\varepsilon_{23}\\ 2~\varepsilon_{31}\\ 2~\varepsilon_{12} \end{bmatrix} $$ or

\mathbf{\sigma} = \mathbf{C}~\mathbf{\varepsilon} $$

The inverse relation is

\begin{bmatrix} \varepsilon_{11}\\ \varepsilon_{22}\\ \varepsilon_{33}\\ 2~\varepsilon_{23}\\ 2~\varepsilon_{31}\\ 2~\varepsilon_{12} \end{bmatrix} =  \begin{bmatrix} S_{1111} & S_{1122} & S_{1133} & 2~S_{1123} & 2~S_{1131} & 2~S_{1112} \\ S_{2211} & S_{2222} & S_{2233} & 2~S_{2223} & 2~S_{2231} & 2~S_{2212} \\ S_{3311} & S_{3322} & S_{3333} & 2~S_{3323} & 2~S_{3331} & 2~S_{3312} \\ 2~S_{2311} & 2~S_{2322} & 2~S_{2333} & 4~S_{2323} & 4~S_{2331} & 4~S_{2312} \\ 2~S_{3111} & 2~S_{3122} & 2~S_{3133} & 4~S_{3123} & 4~S_{3131} & 4~S_{3112} \\ 2~S_{1211} & 2~S_{1222} & 2~S_{1233} & 4~S_{1223} & 4~S_{1231} & 4~S_{1212} \end{bmatrix} \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix} $$ or

\mathbf{\varepsilon} = \mathbf{S}~\mathbf{\sigma} $$ We can show that

{\mathbf{S}} = {\mathbf{C}}^{-1} $$

Isotropic materials
We have already seen the matrix form of the stress-strain equation for isotropic linear elastic materials. In this case the stiffness tensor has only two independent components because every plane is a plane of elastic symmetry. In direct tensor notation

\boldsymbol{\mathsf{C}} = \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf{I}} $$ where $$\lambda$$ and $$\mu$$ are the elastic constants that we defined before, $$\boldsymbol{\mathit{1}}$$ is the second-order identity tensor, and $$\boldsymbol{\mathsf{I}}$$ is the symmetric fourth-order identity tensor. In index notation

C_{ijkl} = \lambda~\delta_{ij}~\delta_{kl} + 2~\mu~\cfrac{1}{2}(\delta_{ik}~\delta_{jl} + \delta_{il}~\delta_{jk}) $$ You could alternatively express this equation in terms of the Young's modulus ($$E$$) and the Poisson's ratio ($$\nu$$) or in terms of the bulk modulus ($$K$$) and the shear modulus ($$\mu$$) or any other combination of two independent elastic parameters.

In Voigt notation the expression for the stress-strain law for isotropic materials can be written as

\begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix} =  \begin{bmatrix} C_{11} & C_{12} & C_{12} & 0 & 0 & 0 \\ C_{12} & C_{11} & C_{12} & 0 & 0 & 0 \\ C_{12} & C_{12} & C_{11} & 0 & 0 & 0 \\ 0 & 0 & 0 & (C_{11}-C_{12})/2 & 0 & 0 \\ 0 & 0 & 0 & 0 & (C_{11}-C_{12})/2 & 0 \\ 0 & 0 & 0 & 0 & 0 & (C_{11}-C_{12})/2 \end{bmatrix} \begin{bmatrix} \varepsilon_{11}\\ \varepsilon_{22}\\ \varepsilon_{33}\\ 2~\varepsilon_{23}\\ 2~\varepsilon_{31}\\ 2~\varepsilon_{12} \end{bmatrix} $$ where

C_{11} = \cfrac{E~(1-\nu)}{(1+\nu)(1-2\nu)} ~; C_{12} = \cfrac{E~\nu}{(1+\nu)(1-2\nu)} ~; (C_{11}-C_{12})/2 = \cfrac{E}{2(1+\nu)} = \mu ~. $$

The Voigt form of the strain-stress relation can be written as

\begin{bmatrix} \varepsilon_{11}\\ \varepsilon_{22}\\ \varepsilon_{33}\\ 2~\varepsilon_{23}\\ 2~\varepsilon_{31}\\ 2~\varepsilon_{12} \end{bmatrix} =  \begin{bmatrix} S_{11} & S_{12} & S_{12} & 0 & 0 & 0 \\ S_{12} & S_{11} & S_{12} & 0 & 0 & 0 \\ S_{12} & S_{12} & S_{11} & 0 & 0 & 0 \\ 0 & 0 & 0 & 2~(S_{11}-S_{12}) & 0 & 0 \\ 0 & 0 & 0 & 0 & 2~(S_{11}-S_{12}) & 0 \\ 0 & 0 & 0 & 0 & 0 & 2~(S_{11}-S_{12}) \end{bmatrix} \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix} $$ where

S_{11} = \cfrac{1}{E} ~; S_{12} = -\cfrac{\nu}{E} ~; 2~(S_{11}-S_{12}) = \cfrac{2(1+\nu)}{E} = \cfrac{1}{\mu} ~. $$

The relations between various moduli are shown in the table below:

title: Elasticity/Constitutive relations


text:
User talk:Vladimir39

Welcome
Hello Vladimir39, 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.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! Erkan Yilmaz (evaluate me!, discussion ) 05:52, 30 August 2007 (UTC)

P.S. Alicia gonzalez ceren foundation contains now also an expand template. If there is more content this can be removed.

corrections (also on website)
Hello, I have read a little here and I think these should be corrected on the website:

format: wrong - recommended correction
 * beutiful - beautiful
 * Los Angesle - Los Angeles
 * strugle - struggle

About the first part of Alicia gonzalez ceren foundation:

The first two sentences seem to be a copy of the above mentioned website. Perhaps they could be rephrased a little ? Otherwise I am not sure, if this violates the copyright status, because you paste it here. Erkan Yilmaz (evaluate me!, discussion ) 06:03, 30 August 2007 (UTC)

title: User talk:Vladimir39


text:
Talk:Telemedicine in French-speaking Africa

What percentage of people in Africa have phones, and what percentage of people in Africa have fairly easy access to phones? --Remi 06:56, 30 August 2007 (UTC)


text:
Feynman path integrals for mathematicians

Feynman path integral, or simply path integral, is a gadget whereby physicists calculate probability amplitudes for quantum transitions by "summing over all possible classical histories (or paths)". Naively, it makes no sense, but it is powerful and aesthetically pleasing.

The basic form of the path integral is as follows: (text taken and modified from path integral formulation)

 * $$Z = \int Dx\, e^{i\mathcal{S}[x]/\hbar}$$

where $$\mathcal{S}[x]=\int_0^T \mathrm{d}t L[x(t)]$$

is the classical action, x(t) is a path, i.e. a map from the interval [0,T] to space-time,$$\hbar$$ is the Planck constant divided by $$2\pi$$, and here is the rub: Dx is the measure for integration over the infinite-dimensional space of "all" paths.


text:
User talk:Rad~enwikiversity

Hello Rad, 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.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --Remi 07:07, 30 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 Rad. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Rad~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 23:48, 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) 06:18, 19 April 2015 (UTC)

title: User talk:Rad~enwikiversity