text:
Who Am I?

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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Grade Level: 9th grade:
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Subject: Advisory course
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Sub-Subject: focus on self, culture, community and college
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Length/Duration: 3-4 days
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Technologies Used: Imovie, Pikiwiki, Memory Miner
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Introduction/Specifications
In what ways can people learn about one another? Conversations, letters, art work, and stories are all examples of how we can learn about one another. With the advancement of technology there are many other ways one can share stories about their lives and who they are. Digital storytelling has allowed people to share their stories through the practice of combining narrative with digital content including images, sound, and video, typically with a strong emotional component.

As a member and participant of this Advisory course we will be discussing several personal and controversial topics. It is important that we maintain an environment that is respectful, welcoming, and open-minded.

This digital storytelling project will allow us to share a personal story about ourselves to the class as a way to get aquainted with one another. You may share anything you are comfortable with sharing that will allow us to have a better understanding of who you are, where you come from, and where you are going.

For your digital story you will need to include the following components:

- a narrative piece on one topic about who are that you would like to share with the class.

- images that pertain to your story

- visual effects

- a musical component

 * We will discuss these components further in class including the length of the video and other details you may have questions about. We will also have Mr. Shaw visiting us to assist us in the creation of your digital story

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Ideas/Examples
Some ideas for your digital storyboard can include but are not limited to the following:

- A story about what makes you who you are today

- What your asipirations are for the future

- A significant event that changed your life

Listed below is a link to some digital stories that have been created by students of the Streetside Stories/Tech Tales Organization located in the Bay Area:

http://www.streetside.org/stories/digital-stories.htm (main page)

http://www.streetside.org/stories/tech-tales/2006-07/donnaL.htm

http://www.streetside.org/stories/tech-tales/2006-07/joeyI.htm

http://www.streetside.org/stories/tech-tales/2006-07/arlandoG.htm

http://www.streetside.org/stories/tech-tales/2006-07/angelicaE.htm

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Software
The software that we will be using as a class for the production of your digital story is Imovie. For an introduction to Imovie please visit the link below:

http://www.apple.com/ilife/imovie/

Additional software includes the following:

http://www.pikiwiki.com/ews/demoVideo.jsp (demo video of PikiWiki)

http://memoryminer.com/help/video_gallery.html (gallery of Memory Miner)

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Articles that Support this Lesson

 * Dewey, J. (1902). The child and the cirriculum (pp. 21-30).  Chicago, IL: University of      Chicago Press.

The ideas presented in this article speak about the concept of guidance. We must provide our students with the necessary guidance in order to help them grow but also we have to be sure not to stifle their growth by imposing too many boundaries. Additionally, the article also mentioned the importance of monitoring the growth of the child in their work, not just what they produce at the end. This article supports this lesson/project because the students will be supported throughout the entirety of their process and monitored on the advances that they make as they create their story. Also, the project gives the student a choice about what they want to share about themselves and does not limit them to one possibility of where they can go with their story.

 * Steele, C.M., & Aronson, J. (1995). Stereotype threat and the intellectual test performance of African Americans.  Journal of Personality and Social Psychology, 69 (5), 797-811.

This article supports this lesson by way of overcoming stereotyping. We all make judgements about others on a regular basis. My hope with this lesson is that students genuinely learn about their peers and the anxiety that knowing that everyone is going to hear their story will be alleviated since everyone will be in the same predicament. The design of this course is to embrace differences of others and be open to different perspectives. The article also mentions the importance of addressing stereotypes. In my own experience in the classroom, when stereotypes/sensitive topics are not addressed, they remain ever present. We cannot ignore differences and sweep them under the rug, we must bring them to light. The concept of audience approval arises in projects like these and the article also addresses this issue. When students are put on the spot they are in fear of being wrong and hope to not embarrass themselves. One may feel that they will be wrong and everyone will think they are not smart. I hope that by having everyone in the class share in that feeling that this tension may be decreased.

 * Nasir, N.i.S., McLaughlin, M.W., & Jones, A. (2009). What does it mean to be African American?  Constructions of race and academic identity in an urban public high school.  American Educational Research Journal, 46 (1), 73-114.

This article mentions our nature to stereotype our own stereotype. This project aims for students to share stories about their own lives. These stories may change the minds of other students who originally classified their peers under a certain stereotype they picked up in our society. Additionally, the Nasir, et al. article comments on the sociocutltural perspectives and how there can be a shift in perspective when activities are structured in such a way around values, norms, and expectations. One of the intentions of this activity is to shift the perspective from stereotypes to genuinely getting to know about someone.

 * Tettegah, S.Y., Whang, E.W., Taylor, K.R., & Cash, T.J. (2008). Narratives, virtual environments and identity semiotics: An exploration of pre-service teachers' cognitions.  Journal E-learning, 5 (1), 103-127.

The article specifically talks about how pre-service teachers, through text-based and animated narrative vignettes can provide a way of learning about personal and social identity. The article goes further to speak about how we all have stories to tell about our lives. This connects directly to my own lesson since the students are creating digital narratives about their own lives and experiences they have endured. Technological advances have given us the opportunities to share our stories and the article supports that idea.

 * Jenkins, H., Purushotma, R., Clinton, K., Wiegel, M., & Robison, A. Confronting the challenges of participatory culture: Media education for the 21st century.  Retrieved from http://www.newmedialiteracies.org/files/working/NMLWhitePaper.pdf

This article focuses on the importance of integrating technology into the lives of students and what skills they can acquire from learning these technologies. Not only does this lesson ask kids to share about themselves, it also requires them to learn about various technological skills that will allow them to put their digital story together.

 * Hull, G., & Katz, M. Crafting an agentive self: Case studies on digital storytelling. Retrieved from http://www.udel.edu/paulhyde/ds/readings/hull_agentive_self.pdf

This article summarizes a number of case studies on digital storytelling. The article illustrates how technological innovations can result in powerful forms of representation for various ages. The articles also supports the idea of digital storytelling as a means to convey powerful stories. This essentially supports my reason for incorporating for digital storytelling. It is just not about using technology for the sake of technology, but for a way for students to learn at a deeper level.

LINKS:

http://en.wikiversity.org/wiki/Educational_technologies

http://en.wikiversity.org/wiki/Educational_Technology_Lesson_Plans

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title: Who Am I?


text:
User:Egm6321.f09.Team06.sada/HW2

Problem 8
'''$$ \ F = 8x^{5}p \qquad \qquad G = 2x^{2}p + 20x^{4}(y^{'})^{2} + 4xy $$ '''

Show that equations (4) & (5) on 10-2 for exactness are satisfied for the above equations?

Solution:

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. (4) $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. (5) $$

The partial derivatives are identified as,

$$ F_{x}=40x^{4}p  \qquad $$

$$ F_{xx}=160x^{3}p \qquad $$

$$ F_{xy}=0 \qquad $$

$$ F_{xp}=40x^{4}  \qquad $$

$$ F_{y}=0 \qquad $$

$$ F_{yy}=0 \qquad $$

$$ F_{yp}=0 \qquad $$

$$ G_{x}= 4xp + 80x^{3}p^{2} + 4y \qquad $$

$$ G_{xp}= 4x + 160x^{3}p \qquad $$

$$ G_{y}= 4x \qquad $$

$$ G_{yp}= 0 \qquad $$

$$ G_{p}= 2x^{2} + 40x^{4}p \qquad $$

$$ G_{pp}= 40x^{4} \qquad $$

Substituting in Equ.(4), we get

$$ \ 160x^{3}p + 2p (0) + p^{2} (0) = 4x + 160x^{3}p + p (0) - 4x $$

$$ \ 160x^{3}p = 160x^{3}p $$

Eq. (4) is satisfied.

Applying the results in Eq. (5), we get

$$ \ 40x^{4} + p(0) + 2(0) = 40x^{4} $$

$$ \ 40x^{4} = 40x^{4} $$

Eq. (5) is also satisfied.

Since Equation (4) and Equation (5) are both satisfied, then the 2nd exactness condition is satisfied. It is then concluded that the NL.ODE is exact.

Problem 9
$$ \sqrt[]{x}y^{''} + 2xy^{'} + 3y = 0 $$

Verify exactness of above ODE.

Solution

Exactness Condition 1:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

$$ \ F(x,y,p) = \sqrt[]{x} \qquad \qquad G(x,y,p) = 2xp + 3y $$

The 2nd exactness condition for a second order ODE is,

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. (4) $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. (5) $$

The following partial derivatives are found:

$$ F_{x}= -\frac{1}{2}x^{\frac{-1}{2}}  \qquad $$

$$ F_{xx}= -\frac{1}{4}x^{\frac{-3}{2}}  \qquad $$

$$ F_{xy}= 0 \qquad $$

$$ F_{xp}= 0  \qquad $$

$$ F_{y}= 0 \qquad $$

$$ F_{yy}=0 \qquad $$

$$ F_{yp}=0 \qquad $$

$$  G_{x}= 2p  \qquad $$

$$ G_{xp}= 2 \qquad $$

$$ G_{y}= 3 \qquad $$

$$ G_{yp}= 0 \qquad $$

$$ G_{p}= 2x \qquad $$

$$ G_{pp}= 0 \qquad $$

Substituting these values in Eq. (4):

$$ \frac{-1}{4}x^{\frac{-3}{2}} + 2p(0) + p^{2}(0) = 2 + p(0) -3 $$

$$ \frac{-1}{4}x^{\frac{-3}{2}} = -1 $$

Equation (4) is not satisfied therefore the ODE is not exact.

But it does satisfy Eq.(5), which is shown below

$$ \ 0 = 0 + p(0) + 2(0) $$

$$ \ 0 = 0 $$

But in order for the ODE to be exact it must satisfy both Eq.(4) and Eq.(5), since it doesn't satisfy equ.(4) it is concluded that the ODE is not exact.

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

title: User:Egm6321.f09.Team06.sada/HW2


text:
User:Egm6321.f09.Team6/HW2

Problem 1
Given the Equation hxN-hyM+h(Nx-My)=0

Consider the case where hxN=0

Since N can not be 0, then h(y)

So you get -hyM+h(Nx-My)=0

Rearranging you are left with hy/h=(1/M)(Nx-My)=-g(y)  So what is the final form of $$ h(y)$$ ?--Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 2
Given the Equation y'+(1/x)y=x2

Solve for h(x), where h(x)=

$$\int$$a0dx

and a0=(1/x)

So h(x)=

$$\int$$1/x=elnx=x

Therefore h(x)=x

 This does not verify exactness. You were asked to find the solution $$y(x)$$. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 3
Given the equation hy'+h'y=hb where

h=x and b=x2

The resulting equation is xy'+y=x3

which can also be written as d/dx(xy)=x3

One can now find y(x)=

(1/x)$$\int(x)$$3

=1/x(x4/4+C)

So y(x)=x3/4+C/x

 This is still problem 2 ..... --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

 What happend to the actual problem #3? --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 4:
Statement:

Show the following is an exact N1-ODE:

$$(\frac{1}{3})x^3y^4y'+(5x^3+2)(\frac{1}{5})y^5=0 $$

Solution:

First, we need to see if the ODE has the form

$$M+Ny'=0$$

It does, so the first condition for exactness is satisfied.

Next, we need to see if

$$M_y=N_x$$

We know that

$$M_y=a(x)c(y)$$ $$N_x=b(x)c(y)$$

From the problem statement, we can determine the following: $$a(x)=5x^3+2$$ $$bi(x)=(\frac{1}{3})x^3$$ $$b(x)=x^2$$ $$c(y)=y^4$$ $$ci(y)=(\frac{1}{5})y^5$$

Therefore:

$$M_y=5x^3y^4+2y^4$$ $$N_x=x^2y^4$$

These are not equal! Thus, we must use the Euler integrating factor method. This class of exact N1-ODE is related to case 1. That means $$h_y$$ is equal to 0. Thus: $$(\frac{1}{N})(N_x-M_y)=(\frac{1}{bi(x)c(y)})(b(x)c(y)-a(x)c(y))$$ $$(\frac{1}{N})(N_x-M_y)=(\frac{1}{bi(x)})(b(x)-a(x))$$ $$(\frac{1}{N})(N_x-M_y)=-f(x)$$

We now know that h(x,y) is only a function of x. Therefore: $$h(x)=exp\int{f(x)dx}$$

where: $$f(x)=(\frac{1}{bi(x)})(b(x)-a(x))$$

So: $$h(x)=exp\int{\frac{x^2-5x^3-2}{\frac{1}{3}x^3}}dx$$

$$h(x)=exp\frac{3(x^2ln(x)-5x^3+1)}{x^2}$$

$$h(x)=abs(x^3)exp(\frac{3}{x^2}-15x)$$

The initial equation is now exactly integratable by the Euler integrating factor method: $$hM+hNy'=0$$

 This is not the correct $$h(x)$$. You have an absolute value in the expression? Additionally you have not verified exactness. You must show $$\overline{N}_x=\overline{M}_y$$--Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 5:
Statement:

Show that the second condition is satisfied for the following N2-ODE:

$$(xy)y''+x(y')^2+yy'=0$$

Solution:

We know that in order for the second condition of exactness to be satisfied, the following two equations must hold true:

$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ $$f_{xp}+pf_{yp}+2f_y=g_{pp}$$

Solving for the partial differential equations gives: $$f_x=y$$ → $$f_{xx}=0$$ $$f_x=y$$ → $$f_{xy}=1$$ $$f_y=x$$ → $$f_{yy}=0$$ $$g_x=p^2$$ → $$g_{xp}=2p$$ $$g_y=p$$ → $$g_{yp}=1$$ $$g_y=p$$

When entered into the first equation, we get: $$0+2p(1)+p^2(0)=2p+p(1)-p$$ $$2p=2p$$

The first equation holds true! Now let's check the second equation. First, solving for the partial differential equations gives: $$f_x=y$$ → $$f_{xp}=0$$ $$f_y=x$$ → $$f_{yp}=0$$ $$f_y=x$$ $$g_p=2xp+y$$ → $$g_{pp}=2x$$

When entered into the second equation, we get: $$0+p(0)+2x=2x$$ $$2x=2x$$

The second equation also holds true!

Thus, $$(xy)y''+x(y')^2+yy'=0$$ satisfies the second condition of exactness for N2-ODEs.

 good work. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 6
Problem Statement:

Show that the Second Condition of Exactness is satisfied for the following ordinary differential equation:

$$(xy)y''+x(y')^2+yy'=0$$ (6.1)

Second Condition of Exactness:

$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ (6.2)

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$ (6.3)

Solution:

First, we must solve for all of the partial differentials of Equation 6.1, which are given below.

$$f_x=y \rightarrow f_{xx}=0, f_{xy}=1, f_{xp}=0$$

$$f_y=x \rightarrow f_{yp}=0, f_{yy}=0$$

$$g_x=p^2 \rightarrow g_{xp}=2p$$

$$g_y=p \rightarrow g_{yp}=1$$

$$g_p=2xp+y \rightarrow g_{pp}=2x$$

By substituting all of the partial differentials of Equation 6.1 into Equations 6.2 and 6.3, we find that since the left and right halves of the equations are equal, the Second Condition of Exactness is satisfied for Equation 6.1. This substitution is shown below.

$$0+2p(1)+p^2(0)=2p+p(1)-p \rightarrow 2p=2p$$

$$0+p(0)+2(x)=2x \rightarrow 2x=2x$$

 This appears to be a second solution for problem #5. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

 Where is the real problem #7 ? --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 7
Problem Statement:

Derive Equation 7.1 by differentiating Equation 7.2. Where 7.1 and 7.2 are given as:

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$ (7.1)

$$g(x,y,p):=\phi_x(x,y,p)+\phi_y(x,y,p)p$$ (7.2)

Solution:

First, we take the first derivative with respect to $$p$$ of Equation 7.2, leaving out the arguments for simplicity's sake.

$$\frac{d}{dp}[g=\phi_x+\phi_yp] \rightarrow g_p = \phi_{xp}+\phi_y(1)+\phi_{yp}$$

This results in the following relationship.

$$\rightarrow g_p = f_x +\phi_y +f_yp$$

Now, we take the second derivative of Equation 7.2 with respect to $$p$$.

$$\frac{d^2}{dp^2}[g_p = f_x +\phi_y +f_yp] \rightarrow g_{pp}=f_{xp}+\phi_{yp}+f_y(1)+f_{yp}p$$

$$\rightarrow g_{pp}=f_{xp}+f_y+f_y+f_{yp}p$$

This resulting relationship, Equation 7.3, is equivalent to Equation 7.1 and the derivation is complete.

$$\rightarrow g_{pp}=f_{xp}+f_y+f_{yp}p$$ (7.3)  Good work. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 8
$$ \ F = 8x^{5}p \qquad \qquad G = 2x^{2}p + 20x^{4}(y^{'})^{2} + 4xy $$

Show that equations (4) & (5) on 10-2 for exactness are satisfied for the above equations?

Solution:

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. (4) $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. (5) $$

The partial derivatives are identified as,

$$ F_{x}=40x^{4}p  \qquad $$

$$ F_{xx}=160x^{3}p \qquad $$

$$ F_{xy}=0 \qquad $$

$$ F_{xp}=40x^{4}  \qquad $$

$$ F_{y}=0 \qquad $$

$$ F_{yy}=0 \qquad $$

$$ F_{yp}=0 \qquad $$

$$ G_{x}= 4xp + 80x^{3}p^{2} + 4y \qquad $$

$$ G_{xp}= 4x + 160x^{3}p \qquad $$

$$ G_{y}= 4x \qquad $$

$$ G_{yp}= 0 \qquad $$

$$ G_{p}= 2x^{2} + 40x^{4}p \qquad $$

$$ G_{pp}= 40x^{4} \qquad $$

Substituting in Equ.(4), we get

$$ \ 160x^{3}p + 2p (0) + p^{2} (0) = 4x + 160x^{3}p + p (0) - 4x $$

$$ \ 160x^{3}p = 160x^{3}p $$

Eq. (4) is satisfied.

Applying the results in Eq. (5), we get

$$ \ 40x^{4} + p(0) + 2(0) = 40x^{4} $$

$$ \ 40x^{4} = 40x^{4} $$

Eq. (5) is also satisfied.

Since Equation (4) and Equation (5) are both satisfied, then the 2nd exactness condition is satisfied. It is then concluded that the NL.ODE is exact.

 good work. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 9
$$ \sqrt[]{x}y^{''} + 2xy^{'} + 3y = 0 $$

Verify exactness of above ODE.

Solution

Exactness Condition 1:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

$$ \ F(x,y,p) = \sqrt[]{x} \qquad \qquad G(x,y,p) = 2xp + 3y $$

The 2nd exactness condition for a second order ODE is,

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. (4) $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. (5) $$

The following partial derivatives are found:

$$ F_{x}= -\frac{1}{2}x^{\frac{-1}{2}}  \qquad $$

$$ F_{xx}= -\frac{1}{4}x^{\frac{-3}{2}}  \qquad $$

$$ F_{xy}= 0 \qquad $$

$$ F_{xp}= 0  \qquad $$

$$ F_{y}= 0 \qquad $$

$$ F_{yy}=0 \qquad $$

$$ F_{yp}=0 \qquad $$

$$  G_{x}= 2p  \qquad $$

$$ G_{xp}= 2 \qquad $$

$$ G_{y}= 3 \qquad $$

$$ G_{yp}= 0 \qquad $$

$$ G_{p}= 2x \qquad $$

$$ G_{pp}= 0 \qquad $$

Substituting these values in Eq. (4):

$$ \frac{-1}{4}x^{\frac{-3}{2}} + 2p(0) + p^{2}(0) = 2 + p(0) -3 $$

$$ \frac{-1}{4}x^{\frac{-3}{2}} = -1 $$

Equation (4) is not satisfied therefore the ODE is not exact.

But it does satisfy Eq.(5), which is shown below

$$ \ 0 = 0 + p(0) + 2(0) $$

$$ \ 0 = 0 $$

But in order for the ODE to be exact it must satisfy both Eq.(4) and Eq.(5), since it doesn't satisfy equ.(4) it is concluded that the ODE is not exact.  Correct. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Contributing Team Members
Egm6321.f09.Team6.PaulMoore 22:58, 22 September

Egm6321.f09.Team06.sada 04:59, 23 September 2009 (UTC)

Egm6321.f09.Team6.dianafoster 15:16, 23 September 2009 (UTC)

title: User:Egm6321.f09.Team6/HW2


text:
Talk:Trade Finance/Commercial Risk

Any chance of fixing the typo in the article title and elsewhere here? I feel it kind of lowers the wiki's status as a source of info. Just my tuppenceworth! Berek 22:27, 22 September 2009 (UTC)

Agreed! Just noticed the misspelling in the link at :en:Portal:Wikimedia featured content/Wikiversity and came here to investigate. -- John of Reading (talk) 10:57, 23 July 2012 (UTC)

title: Talk:Trade Finance/Commercial Risk


text:
User:Egm6321.f09.Team6.dianafoster/HW1

Problem 6:
Generating exact nonlinear 1st order ODEs: Let $$ \Phi(x,y)=6x^{4}+2y^{3/2}$$, then $$ M=\Phi_{x}$$ $$ N=\Phi_{y}$$ Complete the details and invent 3 more examples.

a) First, it is known that: $$ M=\Phi_{x}=\frac{\partial \Phi(x,y)}{\partial x}$$ and  $$ N=\Phi_{y}=\frac{\partial \Phi(x,y)}{\partial y}$$

Taking the partial derivative for $$ \Phi_{x}$$ and $$ \Phi_{y}$$ for the following $$ \Phi(x,y)=6x^{4}+2y^{3/2}$$, yields: $$ M=\Phi_{x}=24x^3$$ and $$ N=\Phi_{y}=3y^{1/2}$$

Putting into the form $$ M+Ny'=0$$ so that the nonlinear 1st order ODE is in the desired form yields: $$ 24x^3+3y^{1/2}y'=0$$ b) Let $$ \Phi(x,y)=\frac{1}{3}x^{5}-4y^{3}$$ Taking the partial derivative for $$ \Phi_{x}$$ and $$ \Phi_{y}$$ yields:  $$ M=\Phi_{x}=\frac{5}{3}x^4$$ and  $$ N=\Phi_{y}=-12y^{2}$$ Putting into the form  $$ M+Ny'=0$$ yields:  $$ \frac{5}{3}x^4-12y^{2}y'=0$$ c) Let $$ \Phi(x,y)=x^6y^7+\frac{2}{5}y^{4}$$ Taking the partial derivative for $$ \Phi_{x}$$ and $$ \Phi_{y}$$ yields: $$ M=\Phi_{x}=6x^5y^7$$ and $$ N=\Phi_{y}=7x^6y^6+\frac{8}{5}y^3$$ Putting into the form $$ M+Ny'=0$$ yields: $$ 6x^5y^7+(7x^6y^6+\frac{8}{5}y^3)y'=0$$ d) Let $$ \Phi(x,y)=\frac{1}{2}xy^{3/2}+x^2y^5$$ Taking the partial derivative for $$ \Phi_{x}$$ and $$ \Phi_{y}$$ yields:  $$ M=\Phi_{x}=\frac{1}{2}y^{3/2}+2xy^5$$ and  $$ N=\Phi_{y}=\frac{3}{4}xy^{1/2}+5x^2y^4$$ Putting into the form  $$ M+Ny'=0$$ yields:  $$ (\frac{1}{2}y^{3/2}+2xy^5)+(\frac{3}{4}xy^{1/2}+5x^2y^4)y'=0$$

title: User:Egm6321.f09.Team6.dianafoster/HW1


text:
User:Egm6321.f09.Team6.dianafoster/HW2

Problem 4:
Statement:

Show the following is an exact N1-ODE:

$$(\frac{1}{3})x^3y^4y'+(5x^3+2)(\frac{1}{5})y^5=0 $$

Solution:

First, we need to see if the ODE has the form

$$M+Ny'=0$$

It does, so the first condition for exactness is satisfied.

Next, we need to see if

$$M_y=N_x$$

We know that

$$M_y=a(x)c(y)$$ $$N_x=b(x)c(y)$$

From the problem statement, we can determine the following: $$a(x)=5x^3+2$$ $$bi(x)=(\frac{1}{3})x^3$$ $$b(x)=x^2$$ $$c(y)=y^4$$ $$ci(y)=(\frac{1}{5})y^5$$

Therefore:

$$M_y=5x^3y^4+2y^4$$ $$N_x=x^2y^4$$

These are not equal! Thus, we must use the Euler integrating factor method. This class of exact N1-ODE is related to case 1. That means $$h_y$$ is equal to 0. Thus: $$(\frac{1}{N})(N_x-M_y)=(\frac{1}{bi(x)c(y)})(b(x)c(y)-a(x)c(y))$$ $$(\frac{1}{N})(N_x-M_y)=(\frac{1}{bi(x)})(b(x)-a(x))$$ $$(\frac{1}{N})(N_x-M_y)=-f(x)$$

We now know that h(x,y) is only a function of x. Therefore: $$h(x)=exp\int{f(x)dx}$$

where: $$f(x)=(\frac{1}{bi(x)})(b(x)-a(x))$$

So: $$h(x)=exp\int{\frac{x^2-5x^3-2}{\frac{1}{3}x^3}}dx$$

$$h(x)=exp\frac{3(x^2ln(x)-5x^3+1)}{x^2}$$

$$h(x)=abs(x^3)exp(\frac{3}{x^2}-15x)$$

The initial equation is now exactly integratable by the Euler integrating factor method: $$hM+hNy'=0$$

Problem 5:
Statement:

Show that the second condition is satisfied for the following N2-ODE:

$$(xy)y''+x(y')^2+yy'=0$$

Solution:

We know that in order for the second condition of exactness to be satisfied, the following two equations must hold true:

$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ $$f_{xp}+pf_{yp}+2f_y=g_{pp}$$

Solving for the partial differential equations gives: $$f_x=y$$ → $$f_{xx}=0$$ $$f_x=y$$ → $$f_{xy}=1$$ $$f_y=x$$ → $$f_{yy}=0$$ $$g_x=p^2$$ → $$g_{xp}=2p$$ $$g_y=p$$ → $$g_{yp}=1$$ $$g_y=p$$

When entered into the first equation, we get: $$0+2p(1)+p^2(0)=2p+p(1)-p$$ $$2p=2p$$

The first equation holds true! Now let's check the second equation. First, solving for the partial differential equations gives: $$f_x=y$$ → $$f_{xp}=0$$ $$f_y=x$$ → $$f_{yp}=0$$ $$f_y=x$$ $$g_p=2xp+y$$ → $$g_{pp}=2x$$

When entered into the second equation, we get: $$0+p(0)+2x=2x$$ $$2x=2x$$

The second equation also holds true!

Thus, $$(xy)y''+x(y')^2+yy'=0$$ satisfies the second condition of exactness for N2-ODEs.

title: User:Egm6321.f09.Team6.dianafoster/HW2


text:
User:Lexanjaravata/Description

anf,mansd,mfnm,anfd,masf

title: User:Lexanjaravata/Description


text:
Educational technology/LessonPlanTemplate

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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Technologies Used:
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{Type a question here... - Wrong answer - Wrong answer + CORRECT answer - Wrong answer
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Header 4
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text:
Native American literature reading group/Suggestions

Reading suggestions
To nominate a book, please give a minimum of the title and author and why you think the book would be an appropriate choice for the reading group. It would also be helpful to know the ISBN of a current edition and/or link to Amazon.com or a similar site for those who would like to purchase or get a feel for the book in question. Thanks! Trinity507 05:04, 23 September 2009 (UTC)


text: Category:Native American literature reading group


text:
User:EAS

Name: Eric Ahlqvist Scott

Founder of RE:UP www.reupsf.com

title: User:EAS


text: File:3.A.Magnetic Sensor.pdf

title: File:3.A.Magnetic Sensor.pdf


title: User:David1010


text: Category:Urdu

title: Category:Urdu


text: User:Ali Rana

title: User:Ali Rana


text:
Cornish Language

{| cellpadding="10" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
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Kernewek
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Cornish / ˈkɔːnɪʃ/ (Kernewek) is a Celtic language spoken in Cornwall. It is closely related to Welsh and Breton, and sits - linguistically as well as geographically - in between the two.

Cornish went into decline during the 17th and 18th centuries and ceased to be a community language, but it has now been revived and the number of speakers is growing. There are currently around 3,500 people with knowledge of the language and it is taught in many Cornish schools. Some young people have been brought up to be native Cornish speakers.

The language was protected under the European Charter for Regional and Minority Languages in 2003, and in 2008 a standard written form of the language was agreed, which is used in these lessons.

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Topics

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title: Cornish Language


text:
User:Egm6936.s09/Curling of paper sheet using shell theory

= Small transverse deflections of thin plates =

Consider a load-free plate in which the $$\displaystyle xy$$ plane coincides with plate's midplane and the $$\displaystyle z$$ coordinate is perpendicular to it and is directed downwards. The fundamental assumptions of the linear, elastic, small deflection theory of bending for thin plates may be stated as follows  :

(i) The material of the plate is elastic, homogeneous, and isotropic.

(ii) The plate is initially flat.

(iii) The deflection (the normal component of the displacement vector) of the midplane is small compared with the thickness of the plate. The slope of the deflected surface is therefore very small and the square of the slope is a negligible quantity in comparison with unity.

(iv) The straight lines, initially normal to the middle plane before bending, remain straight and normal to the middle surface during the deformation, and the length of such elements is not altered. This means that the vertical shear strain $$\displaystyle \gamma _{xz}$$ and $$\displaystyle \gamma _{yz}$$ are negligible and the normal strain $$\displaystyle \varepsilon _z$$ may also be omitted. This assumption is referred to as the "hypothesis of straight normals."

(v) The stress normal to the middle plane, $$ \sigma_z$$ is small compared with the other stress components and may be neglected in the stress-strain relations

(vi) Since the displacements of a plate are small, it is assumed that the middle surface remains unstrained after bending.

Strain-curvature relations (Kinematic equations)
Let $$\displaystyle u$$,$$\displaystyle v$$ and $$\displaystyle w$$ be components of the displacement vector of points in the middle surface of the plate occuring in the $$\displaystyle x$$,$$\displaystyle y$$ and $$\displaystyle z$$ directions, respectively. As it follows from the assumption (iv) above

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$$  \displaystyle \begin{array}{l} \varepsilon _z = 0\,\,,\,\,\gamma _{xz}  = 0\,\,,\,\,\gamma _{yz}  = 0 \\ \end{array} $$     (1.1)
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Integrating the expressions (1.1) and taking into account assumptions (vi) we have the following form in the context of Kirchhoff's theory :

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$$   \begin{array}{l} \displaystyle w = w\left( {x,y} \right)\,\,,\,\,u = - z\frac\,\,,\,\,v =  - z\frac \\ \end{array} $$     (1.2)
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 *
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From the definitions, strains are given by

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$$  \displaystyle \begin{array}{l} \displaystyle \varepsilon _x = \frac =  - z\frac = z\kappa _x  \\ \displaystyle \varepsilon _y = \frac =  - z\frac = z\kappa _y  \\ \displaystyle \gamma _{xy} = \frac + \frac =  - 2z\frac = z\kappa _{xy}  \\ \end{array} $$     (1.3) where curvatures are
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$$  \displaystyle \begin{array}{l} \displaystyle \kappa _x =  - \frac \\ \displaystyle \kappa _y =  - \frac \\ \displaystyle \kappa _{xy} =  - 2\frac \\ \end{array} $$     (1.4)
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Stresses, stress resultants, and stress couples
Stress-strain relations are related by the Hooke's law :

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$$  \displaystyle \begin{array}{l} \displaystyle \sigma _x = \frac{E}\left( {\varepsilon _x  + \nu \varepsilon _y } \right) =  - \frac\left( {\frac + \nu \frac} \right) \\ \displaystyle \sigma _y = \frac{E}\left( {\varepsilon _y  + \nu \varepsilon _x } \right) =  - \frac\left( {\frac + \nu \frac} \right) \\ \displaystyle \tau _{xy} = G\gamma _{xy}  =  - \frac\frac \\ \end{array} $$     (1.5) where $$\displaystyle G = \frac{E}$$
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The resultant forces and moments are

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$$   \begin{array}{l} \displaystyle \left\{ {\begin{array}{*{20}c} {Q_x } \\ \displaystyle {Q_y } \\ \end{array}} \right\} = \displaystyle \int\limits_{ - h/2}^{h/2} {\left\{ \begin{array}{l} \displaystyle \tau _{xz} \\ \displaystyle \tau _{yz} \\ \end{array} \right\}dz} \\ \end{array} $$     (1.6)
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$$  \displaystyle \begin{array}{l} \left\{ {\begin{array}{*{20}c} {M_x } \\ \displaystyle {M_y } \\ \displaystyle {M_{xy} } \\ \displaystyle \end{array}} \right\} = \displaystyle \int\limits_{ - h/2}^{h/2} {\left\{ {\begin{array}{*{20}c} \displaystyle {\sigma _x } \\ \displaystyle {\sigma _y } \\ \displaystyle {\tau _{xy} } \\ \displaystyle \end{array}} \right\}zdz} \\ \end{array} $$     (1.7)
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Substituting eqs (1.5) into eqs (1.7) one gets

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$$  \displaystyle \begin{array}{l} \displaystyle M_x =  - D\left( {\frac + \nu \frac} \right) \\ \displaystyle M_y =  - D\left( {\frac + \nu \frac} \right) \\ \displaystyle M_{xy} =  - D\left( {1 - \nu } \right)\frac \\ \end{array} $$     (1.8)
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 *
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where D is the flexural rigidity of the plate as below
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$$  \displaystyle \begin{array}{l} \displaystyle D = \frac \\ \end{array} $$     (1.9)
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 *
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Governing equation for deflection of plates in cartesian coordinates
To get the governing equation for deflection of plates, we imagine to cut out an infinite element and consider its conditions of equilibrium. The force summation in the z axis gives

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$$  \displaystyle \begin{array}{l} \displaystyle \frac + \frac + p\left( {x,y} \right) = 0 \\ \end{array} $$     (1.10)
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 *
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The moment summation about the x and y axises lead to
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$$  \displaystyle \begin{array}{l} \displaystyle Q_x = \frac + \frac \\ \displaystyle Q_y = \frac + \frac \\ \end{array} $$     (1.11)
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Substituting eqs (1.11) into (1.10), one finds the following:
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$$  \displaystyle \begin{array}{l} \displaystyle \frac + 2\frac + \frac = - p\left( {x,y} \right) \\ \end{array} $$     (1.12)
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 *
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Finally, introduction of the expressions from eqs (1.8) into eqs (1.12) yields

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$$  \displaystyle \begin{array}{l} \displaystyle \frac + 2\frac + \frac = \frac{p}{D} \\ \end{array} $$     (1.13)
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 *
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This is the governing differential equation for the deflections for thin plate bending analysis based on Kirchhoff's assumptions. Equation (1.13) can be rewritten as follows :

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$$  \displaystyle \begin{array}{l} \displaystyle \nabla ^4 w = \frac{p}{D} \\ \end{array} $$     (1.14) where
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 *
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$$  \displaystyle \begin{array}{l} \displaystyle \nabla ^4 \equiv \frac + 2\frac + \frac \\ \end{array} $$     (1.15)
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Expressions for the vertical forces easily obtain by substituting eqs (1.8) into eqs (1.11)

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$$  \displaystyle \begin{array}{l} \displaystyle Q_x =  - D\frac{\partial }\left( {\frac + \frac} \right) = - D\left( {\frac + \frac} \right) = - D\frac{\partial }\left( {\nabla ^2 w} \right) \\ \displaystyle Q_y =  - D\frac{\partial }\left( {\frac + \frac} \right) =   - D\left( \frac+ {\frac} \right)= - D\frac{\partial }\left( {\nabla ^2 w} \right) \\ \end{array} $$     (1.16)
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Note that the bending moments $$\displaystyle M_{x},M_{y}$$ and twist $$\displaystyle M_{xy}$$ depend only on $$\displaystyle w$$, not on the in-plane displacements.

Boundary conditions
Considering boundary condition at one edge, for example, edge $$\displaystyle x=a$$. We have three common boundary conditions :

(A) Plates are simple supported at the edge:

The deflection and bending moment are both zero:
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{array}{l} \displaystyle w = M_x = 0|_{x = a}  \\ \end{array} $$     (1.17)
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 *

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(B) Plates are clamped or built in, or fixed at this edge

The deflection and slope are zero:
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{array}{l} \displaystyle w = \frac = 0|_{x = a} \\ \end{array} $$     (1.18)
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 *

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(C) Plates are free at this edge

All forces and moments are zero:

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$$  \displaystyle \begin{array}{l} \displaystyle Q_x = M_x  = M_{xy}  = 0|_{x = a}  \\ \end{array} $$     (1.19)
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 *

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However, three boundary conditions are too many for the governing differential equation. Kirchhoff suggested that the three boundary conditions at the free edge be combined into two by equating to zero the bending moment and the so-called effective shear force.

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$$  \displaystyle \begin{array}{l} \displaystyle M_x = 0|_{x = a} \,\,,\,\,V_x  = Q_x  + \frac = 0|_{x = a}  \\ \end{array} $$     (1.20)
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= Large transverse deflections of thin plates =

Large deformation theory

To calculate deformations of a elastic body, we consider a chance in length of an infinitesimal line PQ before and after deformation. The square of length PQ in the original configuration is given by :

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$$  \displaystyle dS^2 = dX_1^2 + dX_2^2  + dX_3^2  = \delta _{ij} dX_i dX_j  = dX_k dX_k $$     (2.1)
 * style="width:95%" |
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 *

 * }

Due to deformation, the square of length pq in the deformed configuration is:

 * {| style="width:100%" border="0"

\displaystyle ds^2 = dx_1^2  + dx_2^2  + dx_3^2  = \delta _{ij} dx_i dx_j  = dx_k dx_k $$     (2.2)
 * style="width:95%" |$$
 * style="width:95%" |$$
 *

 * }

The differentials $$\displaystyle dx_i$$ can be transformed into coordinate systems $$\displaystyle X_i$$

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\displaystyle dx_k = \fracdX_i $$     (2.3)
 * style="width:95%" |$$
 * style="width:95%" |$$
 *

 * }

Hence

\displaystyle ds^2 = \fracdX_i \fracdX_j  = \frac\fracdX_i dX_j $$

By defitinion, Green's strain tensor $$\varepsilon _{ij}$$ is obtained as following:

 * {| style="width:100%" border="0"

\displaystyle ds^2 - dS^2  = \left( {\frac\frac - \delta _{ij} } \right)dX_i dX_j  = 2\varepsilon _{ij} dX_i dX_j $$     (2.4)
 * style="width:95%" |$$
 * style="width:95%" |$$
 *

 * }

or

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\displaystyle \varepsilon _{ij} = \frac{1}{2}\left( {\frac\frac - \delta _{ij} } \right) $$     (2.5)
 * style="width:95%" |$$
 * style="width:95%" |$$
 *

 * }

From the defition of displacements:

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\displaystyle x_i - X_i  = u_i $$     (2.6)
 * style="width:95%" |$$
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 *

 * }

we have:

\displaystyle \frac = \frac + \frac = \frac + \delta _{ki} $$

Substituting above equation into (2.5):

\displaystyle \varepsilon _{ij} = \frac{1}{2}\left( {\frac\frac - \delta _{ij} } \right) = \frac{1}{2}\left( {\left( {\frac + \delta _{ki} } \right)\left( {\frac + \delta _{kj} } \right) - \delta _{ij} } \right) = \frac{1}{2}\left( {\frac\frac + \frac\delta _{kj}  + \frac\delta _{ki} } \right) $$ or

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$$  \displaystyle \varepsilon _{ij} = \frac{1}{2}\left( {\frac + \frac + \frac\frac} \right) $$     (2.7)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
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 *

 * }

In unabridged notations, in-plane strains, for example, are given by:

\displaystyle \varepsilon _{11} = \frac + \frac{1}{2}\left[ {\left( {\frac} \right)^2  + \left( {\frac} \right)^2  + \left( {\frac} \right)^2 } \right] $$

\displaystyle \varepsilon _{22} = \frac + \frac{1}{2}\left[ {\left( {\frac} \right)^2  + \left( {\frac} \right)^2  + \left( {\frac} \right)^2 } \right] $$

\displaystyle \gamma _{12} = 2\varepsilon _{12}  = \frac + \frac + \left( {\frac\frac + \frac\frac + \frac\frac} \right) $$

Strain-displacement relations
The linear theory of thin plate ignores strains of middle surface of the plate and the corresponding in-plane stresses are neglected. However, if the magnitude of the lateral deflections are large compared to the thickness of plate, these deflections are accompanied by stretching of the middle surface. As a result, the membrane forces are included. The large deflection plate bending theory uses two assumptions:

- The hypothesis of the straight normals: the straight lines, initially normal to the middle plane before deformation remain straight and normal to the middle surface during the deformation therefore $$\displaystyle \gamma _{xz}  = \gamma _{yz}  = \varepsilon _z  = 0$$

- The stress normal to the middle surface, $$\displaystyle \sigma _z $$, is small compared with the other stress components and may be neglected in the stress-strain relations.

The above figure shows a section of the plate normal to the y-axis before and after deformation. The x-y plane is equidistant from the top and bottom surfaces of the plate and is called the midplane of reference plane.

The reference plane displacements $$u_0$$ and $$v_0$$ in the x and y directions and the out of plane displacment w in the z direction are functions of x and y only:

 * {| style="width:100%" border="0"

$$  \displaystyle \begin{array}{l} u_0 = u_0 \left( {x,y} \right) \\ \displaystyle v_0 = v_0 \left( {x,y} \right) \\ w = w\left( {x,y} \right) \\ \end{array} $$     (2.8)
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 *

 * }

The in-plane displacements components of a point of coordinate z are:

 * {| style="width:100%" border="0"

\displaystyle \begin{array}{l} \displaystyle u = u_0 - z\frac \\ \displaystyle v = v_0 - z\frac \\ \end{array} $$     (2.9) where z is the through-thickness coordinate of a general point of the cross section.
 * style="width:95%" |$$
 * style="width:95%" |$$
 *

 * }

Due to large deformations, strain-displacement relations (2.7) must be used. However, we assume that displacements u and v are small, one can neglect $$\left( {\partial u/\partial x} \right)^2 ,\,\left( {\partial u/\partial y} \right)^2 ,\,\left( {\partial v/\partial x} \right)^2 ,\,\left( {\partial v/\partial y} \right)^2 ,\,\left( {\partial u/\partial x} \right)\left( {\partial u/\partial y} \right),\,\left( {\partial v/\partial x} \right)\left( {\partial v/\partial y} \right)$$

Thus strain-displacement relations for large deformation of thin plate are:

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\displaystyle \begin{array}{l} \displaystyle \varepsilon _x = \frac + \frac{1}{2}\left( {\frac} \right)^2  \\ \displaystyle \varepsilon _y = \frac + \frac{1}{2}\left( {\frac} \right)^2  \\ \displaystyle \gamma _{xy} = 2\varepsilon _{xy}  = \frac + \frac + \frac\frac   \\ \end{array} $$     (2.10)
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 *

 * }

Substituting (2.9) into (2.10) we get

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\displaystyle \begin{array}{l} \displaystyle \varepsilon _x = \frac + \frac{1}{2}\left( {\frac} \right)^2  - z\frac = \varepsilon _x^0  + z\kappa _x  \\ \displaystyle \varepsilon _y = \frac + \frac{1}{2}\left( {\frac} \right)^2  - z\frac = \varepsilon _x^0  + z\kappa _y  \\ \displaystyle \gamma _{xy} = \frac + \frac + \frac\frac - 2z\frac = \gamma _{xy}^0  + z\kappa _{xy}  \\ \end{array} $$     (2.11)
 * style="width:95%" |$$
 * style="width:95%" |$$
 *

 * }

where

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\displaystyle \begin{array}{l} \displaystyle \varepsilon _x^0 = \frac + \frac{1}{2}\left( {\frac} \right)^2  \\ \displaystyle \varepsilon _y^0 = \frac + \frac{1}{2}\left( {\frac} \right)^2  \\ \displaystyle \gamma _{xy}^0 = 2\varepsilon _{xy}^0  = \frac + \frac + \frac\frac \\ \displaystyle \kappa _x =  - \frac \\ \displaystyle \kappa _y =  - \frac \\ \displaystyle \kappa _{xy} =  - 2\frac \\ \end{array} $$     (2.12)
 * style="width:95%" |$$
 * style="width:95%" |$$
 *

 * }

or in matrix form

 * {| style="width:100%" border="0"

$$  \displaystyle \begin{array}{l} \displaystyle \left[ {\begin{array}{*{20}c} {\varepsilon _x } \\ \displaystyle {\varepsilon _x } \\ \displaystyle {\gamma _{xy} } \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} \displaystyle {\varepsilon _x^0 } \\ \displaystyle {\varepsilon _y^0 } \\ \displaystyle {\gamma _{xy}^0 } \\ \end{array}} \right] + z\left[ {\begin{array}{*{20}c} \displaystyle {\kappa _x } \\ \displaystyle {\kappa _y } \\ \displaystyle {\kappa _{xy} } \\ \end{array}} \right] \\ \end{array} $$     (2.13)
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 *

 * }

From Hooke's Law, one obtains:

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\displaystyle \left\{ {\begin{array}{*{20}c} {\sigma _x } \\ \displaystyle {\sigma _y } \\ \displaystyle {\sigma _{xy} } \\ \end{array}} \right\} = \frac{E}\left[ {\begin{array}{*{20}c} \displaystyle 1 & \nu & 0  \\ \displaystyle \nu & 1 & 0  \\ \displaystyle 0 & 0 & \displaystyle {\frac{2}} \\ \end{array}} \right]\left\{ {\begin{array}{*{20}c} \displaystyle {\varepsilon _x } \\ \displaystyle {\varepsilon _y } \\ \displaystyle {\tau _{xy} } \\ \end{array}} \right\} = \displaystyle \frac{E}\left[ {\begin{array}{*{20}c} \displaystyle 1 & \nu & 0  \\ \displaystyle \nu & 1 & 0  \\ \displaystyle 0 & 0 & \displaystyle {\frac{2}} \\ \end{array}} \right]\left\{ {\begin{array}{*{20}c} \displaystyle {\varepsilon _x^0 } \\ \displaystyle {\varepsilon _y^0 } \\ \displaystyle {\tau _{xy}^0 } \\ \end{array}} \right\} + \displaystyle \frac\left[ {\begin{array}{*{20}c} \displaystyle 1 & \nu & 0  \\ \displaystyle \nu & 1 & 0  \\ \displaystyle 0 & 0 & \displaystyle {\frac{2}} \\ \end{array}} \right]\left\{ {\begin{array}{*{20}c} \displaystyle {\kappa _x } \\ \displaystyle {\kappa _y } \\ \displaystyle {\kappa _{xy} } \\ \end{array}} \right\} $$
 * style="width:100%" |$$
 * style="width:100%" |$$
 * style= | (2.14)
 * }

Relation (2.14), however, is valid for isotropic materials except for anisotropic materials.

Force resultants and moment resultants

Similar to the small transverse deflection of thin plates, shear forces and moments are the same in (1.6) (1.7):

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$$   \displaystyle \begin{array}{l} \displaystyle \left\{ {\begin{array}{*{20}c} {Q_x } \\ \displaystyle {Q_y } \\ \end{array}} \right\} = \displaystyle \int\limits_{ - h/2}^{h/2} {\left\{ \begin{array}{l} \displaystyle \tau _{xz} \\ \displaystyle \tau _{yz} \\ \end{array} \right\}dz} \\ \end{array} $$     (1.6)
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 *

 * }

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$$  \displaystyle \begin{array}{l} \left\{ {\begin{array}{*{20}c} {M_x } \\ \displaystyle {M_y } \\ \displaystyle {M_{xy} } \\ \displaystyle \end{array}} \right\} = \displaystyle \int\limits_{ - h/2}^{h/2} {\left\{ {\begin{array}{*{20}c} \displaystyle {\sigma _x } \\ \displaystyle {\sigma _y } \\ \displaystyle {\tau _{xy} } \\ \displaystyle \end{array}} \right\}zdz} \\ \end{array} $$     (1.7)
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 *

 * }

The linear theory ignores straining of the middle surface of the plate and the corresponding in-plane stresses are neglected. However, in case of large deflection of thin plates, we have additional in-plane forces as below:

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$$ \displaystyle \left\{ {\begin{array}{*{20}{c}} \\    \\     \\ \end{array}} \right\} = \displaystyle  \int\limits_{ - h/2}^{h/2} {\left\{ {\begin{array}{*{20}{c}} \\    \\     \\ \end{array}} \right\}dz  }$$ (2.15)
 *

 * }

Using (2.14), Eqs (1.7) and (2.15) are rewritten as followings:

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$$   \displaystyle \left\{ {\begin{array}{*{20}{c}} \displaystyle \\ \displaystyle \\ \displaystyle \\ \end{array}} \right\} = D\left[ {\begin{array}{*{20}{c}} \displaystyle  1 & \nu  & 0  \\ \displaystyle  \nu  & 1 & 0  \\ \displaystyle  0 & 0 & \displaystyle{\frac{2}}  \\ \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} \displaystyle    \\ \displaystyle \\ \displaystyle \\ \end{array}} \right\} $$   (2.16)
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 * style="width:95%" |
 *

 * }

Or
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{array}{l} \displaystyle M_x =  - D\left( {\frac + \nu \frac} \right) \\ \displaystyle M_y =  - D\left( {\frac + \nu \frac} \right) \\ \displaystyle M_{xy} =  - D\left( {1 - \nu } \right)\frac \\ \end{array} $$     (1.8)
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 *

 * }

And
 * {| style="width:100%" border="0"

$$   \displaystyle \left\{ {\begin{array}{*{20}{c}} \\    \\     \\ \end{array}} \right\} =  A\left[ {\begin{array}{*{20}{c}} 1 & \nu & 0  \\ \nu & 1 & 0  \\ 0 & 0 & \displaystyle {\frac{2}} \\ \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\varepsilon _x^0} \\ {\varepsilon _y^0} \\ {\gamma _{xy}^0} \\ \end{array}} \right\} $$   (2.17)
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 *

 * }

Or
 * {| style="width:100%" border="0"

$$   \displaystyle \begin{array}{l} {N_x} = A\left( {\varepsilon _x^0 + \nu \varepsilon _y^0} \right) \\ {N_y} = A\left( {\varepsilon _y^0 + \nu \varepsilon _x^0} \right) \\ {N_{xy}} = Gh\gamma _{xy}^0 \\ \end{array}$$ (2.18)
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 *

 * }

where $$\displaystyle A = \displaystyle \frac\,\,,\,\,G = \displaystyle \frac{E}\,\,,\,\,D = \displaystyle \frac $$

The equilibrium of thin plate with large deflections
From six equilibrium equations:

$$\displaystyle \sum {{F_x} = 0\,,\,\,\,} \sum {{F_y} = 0\,,\,\,} \sum {{F_z} = 0\,,\,\,} \sum {{M_x} = 0\,,\,\,} \sum {{M_y} = 0\,,\,\,} \sum {{M_z} = 0\,\,\,}$$

One gets:

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$$ \displaystyle \frac + \frac + p\left( {x,y} \right) = 0$$ (2.19)
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 *

 * }

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{array}{l} \displaystyle {Q_x} = \frac + \frac + {N_x}\frac + {N_{xy}}\frac \\ \displaystyle {Q_y} = \frac + \frac + {N_{xy}}\frac + {N_y}\frac \\ \end{array}$$ (2.20)
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 *

 * }

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$$ \displaystyle \begin{array}{l} \displaystyle \frac + \frac = 0 \\ \displaystyle \frac + \frac = 0 \\ \end{array}$$ (2.21)
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 *

 * }

Substituting (2.20) into (2.19) to obtain the following:

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$$ \displaystyle \frac + 2\frac + \frac = - \left[ {p\left( {x,y} \right) + {N_x}\frac + 2{N_{xy}}\frac + {N_y}\frac} \right]$$ (2.22)
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 *

 * }

Or

 * {| style="width:100%" border="0"

$$ \displaystyle \frac + 2\frac + \frac = \frac{1}{D}\left[ {p\left( {x,y} \right) + {N_x}\frac + 2{N_{xy}}\frac + {N_y}\frac} \right]$$ (2.23)
 * style="width:92%;padding:10px;border:2px solid #8888aa"|
 * style="width:92%;padding:10px;border:2px solid #8888aa"|
 *

 * }

Thus we have 4 unknown functions $$\displaystyle {N_x},{N_y},{N_{xy}}$$ and $$\displaystyle w$$ with three equations (2.21) and (2.23). The rest equation obtained by eliminating the tangential components of displacements $$\displaystyle u_{0}$$ and $$\displaystyle v_{0}$$ from Eqs (2.11) to get the following equation of the compatibility of deformations in the middle surface of the plate:

 * {| style="width:100%" border="0"

$$ \displaystyle \frac + \frac - \frac = {\left( {\frac} \right)^2} - \frac\frac$$ (2.24)
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 *

 * }

The strain-membrane force equations, according to Hooke's law (2.18) are of the form:

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$$ \displaystyle \begin{array}{l} \displaystyle {{\varepsilon }_{x}}=\frac{1}{Eh}\left( {{N}_{x}}-\nu {{N}_{y}} \right) \\ \displaystyle {{\varepsilon }_{y}}=\frac{1}{Eh}\left( {{N}_{y}}-\nu {{N}_{x}} \right) \\ \displaystyle {{\gamma }_{xy}}=\frac{Gh} \\ \end{array}$$ (2.25)
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 * }

Upon substitution of Eqs (2.25) into Eqs (2.24), we derive the following:

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$$ \displaystyle \frac{1}{Eh}\left[ \frac{\partial {{y}^{2}}}\left( {{N}_{x}}-\nu {{N}_{y}} \right)+\frac{\partial {{x}^{2}}}\left( {{N}_{y}}-\nu {{N}_{x}} \right)-2\left( 1+\nu \right)\frac{\partial x\partial y} \right]={{\left( \frac{{{\partial }^{2}}w}{\partial x\partial y} \right)}^{2}}-\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}$$ (2.26)
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 *

 * }

Introducing a stress function $$\displaystyle\phi$$ for in-plane stress components, as follows:

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$$ \displaystyle {{N}_{x}}=\frac{{{\partial }^{2}}\Phi }{\partial {{y}^{2}}},\,\,{{N}_{y}}=\frac{{{\partial }^{2}}\Phi }{\partial {{x}^{2}}},\,\,\,{{N}_{xy}}=-\frac{{{\partial }^{2}}\Phi }{\partial x\partial y}$$ (2.27)
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 *

 * }

Where $$\displaystyle\Phi=\phi h$$

It is easy to find that by doing so, Eqs (2.21) are identically satisfied. Substituting for $$\displaystyle \Phi$$ from (2.27) into Eqs (2.23) we obtain:

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$$ \displaystyle \begin{array}{l} \displaystyle {{\nabla }^{4}}\Phi \equiv \frac{{{\partial }^{4}}\Phi }{\partial {{x}^{4}}}+2\frac{{{\partial }^{4}}\Phi }{\partial {{x}^{2}}\partial {{y}^{2}}}+\frac{{{\partial }^{4}}\Phi }{\partial {{y}^{4}}}=Eh\left[ {{\left( \frac{{{\partial }^{2}}w}{\partial x\partial y} \right)}^{2}}-\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right] \\ \displaystyle {{\nabla }^{4}}w\equiv \frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}+2\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}+\frac{{{\partial }^{4}}w}{\partial {{y}^{4}}}=\frac{1}{D}\left( p\left( x,y \right)+\frac{{{\partial }^{2}}\Phi }{\partial {{x}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}-2\frac{{{\partial }^{2}}\Phi }{\partial x\partial y}\frac{{{\partial }^{2}}w}{\partial x\partial y}+\frac{{{\partial }^{2}}\Phi }{\partial {{y}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right) \\ \end{array}$$ (2.28)
 * style="width:92%;padding:10px;border:2px solid #8888aa"|
 * style="width:92%;padding:10px;border:2px solid #8888aa"|
 *

 * }

These are the governing differential equations for large deflections of thin plates. Equations (2.28) are coupled, nonlinear, partial differential equations, each of fourth order. The first equation can be interpreted as a compatibility equation in terms of the stress function, while the second one is an equilibrium equation. They are often called VON KARMAN the system of equations.

For small deflection, the above P.D.Es become uncoupled and we can use displacement- based method to find solutions as following. And strain-displacement relations of middle surface obtained from (2.12) when ignoring $$\displaystyle \left( \frac{\partial w}{\partial x} \right)^{2},\left( \frac{\partial w}{\partial y} \right)^{2},\frac{\partial w}{\partial x}\frac{\partial w}{\partial y}$$:
 * {| style="width:100%" border="0"

$$ \displaystyle \begin{array}{l} \displaystyle \varepsilon _{x}^{0}=\frac{\partial u_{0}}{\partial x} \\ \displaystyle \varepsilon _{y}^{0}=\frac{\partial v_{0}}{\partial y} \\ \displaystyle \gamma _{xy}^{0}=2\varepsilon _{xy}^{0}=\frac{\partial u_{0}}{\partial x}+\frac{\partial v_{0}}{\partial y} \\ \end{array}$$ (2.29)
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 *

 * }

Substituting Eqs (2.29) into (2.18) yields:

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$$ \displaystyle \begin{array}{l} \displaystyle N_{x}=A\left[ \frac{\partial u_{0}}{\partial x}+\nu \frac{\partial v_{0}}{\partial y} \right] \\ \displaystyle N_{y}=A\left[ \frac{\partial v_{0}}{\partial y}+\nu \frac{\partial u_{0}}{\partial x} \right] \\ \displaystyle N_{xy}=N_{yx}=Gt\left[ \frac{\partial u_{0}}{\partial y}+\frac{\partial v_{0}}{\partial x} \right] \\ \end{array}$$ (2.30)
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 *

 * }

Substituting Eqs (2.30) into Eqs (2.21):

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{array}{l} \displaystyle A\frac{\partial ^{2}u_{0}}{\partial x^{2}}+\nu A\frac{\partial ^{2}v_{0}}{\partial x\partial y}+Gt\frac{\partial ^{2}u_{0}}{\partial y^{2}}+Gt\frac{\partial ^{2}v_{0}}{\partial x\partial y}=0 \\ \displaystyle A\frac{\partial ^{2}v_{0}}{\partial y^{2}}+\nu A\frac{\partial ^{2}u_{0}}{\partial x\partial y}+Gt\frac{\partial ^{2}u_{0}}{\partial x\partial y}+Gt\frac{\partial ^{2}v_{0}}{\partial x^{2}}=0 \\ \end{array}$$ (2.31)
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 *

 * }

The above Eqs can be reduced to:

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{array}{l} \displaystyle \nabla ^{4}u_{0}=0 \\ \displaystyle \nabla ^{4}v_{0}=0 \\ \end{array}$$ (2.32)
 * style="width:92%;padding:10px;border:2px solid #8888aa"|
 * style="width:92%;padding:10px;border:2px solid #8888aa"|
 *

 * }

The boundary conditions
According to

=Classical laminate theory=

The strain-displacement (geometry) relations remain those of equations (2.9), (2.13), the membrane forces, bending moments and twists and vertical shear forces remain those of equations (1.6), (1.7), (2.15). The P.D.E's equations (2.19), (2.21) involving the vertical and membrane forces, respectively, remain valid. The relationship between transverse shear forces and the moments or also accounting for the effect of membrane forces as expressed by equation (2.20) are valid irrespective of whether the response is isotropic or anisotropic. For general orthotropy : $$\displaystyle \left\{ \sigma \right\}=\left[ Q \right]\left\{ \varepsilon  \right\}$$

Thus, the $$\displaystyle \sigma -\varepsilon $$ in the $$\displaystyle k$$ orthotropic ply or lamina of a stacked laminate is (assuming $$\displaystyle \sigma_{z}=\gamma_{xz}=\gamma_{yz}=0$$)
 * {| style="width:100%" border="0"

$$ \displaystyle {{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}_{k}}={{\left[ \begin{matrix} {{Q}_{xx}} & {{Q}_{xy}} & {{Q}_{xs}} \\ {{Q}_{yx}} & {{Q}_{yy}} & {{Q}_{ys}} \\ {{Q}_{sx}} & {{Q}_{sy}} & {{Q}_{ss}} \\ \end{matrix} \right]}_{k}}{{\left\{ \begin{align} & {{\varepsilon }_{x}} \\ & {{\varepsilon }_{y}} \\ & {{\gamma }_{xy}} \\ \end{align} \right\}}_{k}}$$ (3.1)
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 *

 * }

Where $$\displaystyle x, y$$ are general orthotropic coordinates, $$\displaystyle {{\left[ {{Q}_{ij}} \right]}_{k}}$$ are the transformed reduced stiffness of the $$\displaystyle k$$ ply which can be related to $$\displaystyle E_{11}^{k},E_{22}^{k},\nu _{12}^{k}$$ and $$\displaystyle G_{12}^{k}$$ of this $$\displaystyle k$$ ply and the angle between the 1-direction (fiber direction) of this lamina and the $$\displaystyle x$$-axis Since the strains in the $$\displaystyle k$$ ply are given by equation (2.13):
 * {| style="width:100%" border="0"

$$ \displaystyle {{\left[ \begin{matrix} {{\varepsilon }_{x}} \\ {{\varepsilon }_{x}} \\ {{\gamma }_{xy}} \\ \end{matrix} \right]}_{k}}=\left[ \begin{matrix} \varepsilon _{x}^{0} \\ \varepsilon _{y}^{0} \\ \gamma _{xy}^{0} \\ \end{matrix} \right]+z\left[ \begin{matrix} {{\kappa }_{x}} \\ {{\kappa }_{y}} \\ {{\kappa }_{xy}} \\ \end{matrix} \right]$$ (3.2)
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 *

 * }

 * {| style="width:100%" border="0"

$$ \displaystyle {{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}_{k}}={{\left[ \begin{matrix} {{Q}_{xx}} & {{Q}_{xy}} & {{Q}_{xs}} \\ {{Q}_{yx}} & {{Q}_{yy}} & {{Q}_{ys}} \\ {{Q}_{sx}} & {{Q}_{sy}} & {{Q}_{ss}} \\ \end{matrix} \right]}_{k}}\left( \left[ \begin{matrix}  \varepsilon _{x}^{0}  \\   \varepsilon _{y}^{0}  \\   \gamma _{xy}^{0}  \\ \end{matrix} \right]+z\left[ \begin{matrix}   {{\kappa }_{x}}  \\   {{\kappa }_{y}}  \\   {{\kappa }_{xy}}  \\ \end{matrix} \right] \right)$$ (3.3)
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 *

 * }

 * {| style="width:100%" border="0"

$$ \displaystyle \left\{ \begin{align} & {{N}_{x}} \\ & {{N}_{y}} \\ & {{N}_{xy}} \\ \end{align} \right\}=\int\limits_{\begin{smallmatrix} -h/2 \\ \left( z=-{{z}_{0}} \right) \end{smallmatrix}}^{\begin{smallmatrix} \left( z={{z}_{N}} \right) \\ h/2 \end{smallmatrix}}{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}dz=\sum\limits_{k=1}^{N}{{{\int\limits_^{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}}_{k}}dz}=\sum\limits_{k=1}^{N}\left( \int\limits_^{\left[ \begin{matrix}  \varepsilon _{x}^{0}  \\   \varepsilon _{y}^{0}  \\   \gamma _{xy}^{0}  \\ \end{matrix} \right]}dz+\int\limits_^{\left[ \begin{matrix}   {{\kappa }_{x}}  \\   {{\kappa }_{y}}  \\   {{\kappa }_{xy}}  \\ \end{matrix} \right]}zdz \right)$$ (3.4)
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 *

 * }

 * {| style="width:100%" border="0"

$$ \displaystyle \left\{ \begin{align} & {{N}_{x}} \\ & {{N}_{y}} \\ & {{N}_{xy}} \\ \end{align} \right\}=\left[ \begin{matrix} {{A}_{xx}} & {{A}_{xy}} & {{A}_{xs}} \\ {{A}_{yx}} & {{A}_{yy}} & {{A}_{ys}} \\ {{A}_{sx}} & {{A}_{sy}} & {{A}_{ss}} \\ \end{matrix} \right]\left\{ \begin{align} & \varepsilon _{x}^{0} \\ & \varepsilon _{y}^{0} \\ & \gamma _{xy}^{0} \\ \end{align} \right\}+\left[ \begin{matrix} {{B}_{xx}} & {{B}_{xy}} & {{B}_{xs}} \\ {{B}_{yx}} & {{B}_{yy}} & {{B}_{ys}} \\ {{B}_{sx}} & {{B}_{sy}} & {{B}_{ss}} \\ \end{matrix} \right]\left\{ \begin{align} & {{\kappa }_{x}} \\ & {{\kappa }_{y}} \\ & {{\kappa }_{xy}} \\ \end{align} \right\}$$ (3.5)
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 *

 * }

Where
 * {| style="width:100%" border="0"

$$ \displaystyle {{A}_{ij}}=\sum\limits_{k=1}^{N}\int\limits_^{dz}=\sum\limits_{k=1}^{N}\underbrace{\left( {{z}_{k}}-{{z}_{k-1}} \displaystyle \right)}_=\sum\limits_{k=1}^{N}{{t}_{k}}$$ (3.6)
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 *

 * }

Also
 * {| style="width:100%" border="0"

$$ \displaystyle \left\{ \begin{align} & {{M}_{x}} \\ & {{M}_{y}} \\ & {{M}_{xy}} \\ \end{align} \right\}=\int\limits_{\begin{smallmatrix} -h/2 \\ \left( z=-{{z}_{0}} \right) \end{smallmatrix}}^{\begin{smallmatrix} \left( z={{z}_{N}} \right) \\ h/2 \end{smallmatrix}}{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}z}dz=\sum\limits_{k=1}^{N}{{{\int\limits_^{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}}_{k}}zdz}=\sum\limits_{k=1}^{N}\left( \int\limits_^{\left[ \begin{matrix}  \varepsilon _{x}^{0}  \\   \varepsilon _{y}^{0}  \\   \gamma _{xy}^{0}  \\ \end{matrix} \right]z}dz+\int\limits_^{\left[ \begin{matrix}   {{\kappa }_{x}}  \\   {{\kappa }_{y}}  \\   {{\kappa }_{xy}}  \\ \end{matrix} \right]}{{z}^{2}}dz \right)$$ (3.7)
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 * }

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$$ \displaystyle \left\{ \begin{align} & {{M}_{x}} \\ & {{M}_{y}} \\ & {{M}_{xy}} \\ \end{align} \right\}=\left[ \begin{matrix} {{B}_{xx}} & {{B}_{xy}} & {{B}_{xs}} \\ {{B}_{yx}} & {{B}_{yy}} & {{B}_{ys}} \\ {{B}_{sx}} & {{B}_{sy}} & {{B}_{ss}} \\ \end{matrix} \right]\left\{ \begin{align} & \varepsilon _{x}^{0} \\ & \varepsilon _{y}^{0} \\ & \gamma _{xy}^{0} \\ \end{align} \right\}+\left[ \begin{matrix} {{D}_{xx}} & {{D}_{xy}} & {{D}_{xs}} \\ {{D}_{yx}} & {{D}_{yy}} & {{D}_{ys}} \\ {{D}_{sx}} & {{D}_{sy}} & {{D}_{ss}} \\ \end{matrix} \right]\left\{ \begin{align} & {{\kappa }_{x}} \\ & {{\kappa }_{y}} \\ & {{\kappa }_{xy}} \\ \end{align} \right\}$$ (3.8)
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 * }

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & {{B}_{ij}}=\sum\limits_{k=1}^{N}\int\limits_^{zdz}=\frac{1}{2}\sum\limits_{k=1}^{N}{{{\left[ {{Q}_{ij}} \right]}_{k}}\left( z_{k}^{2}-z_{k-1}^{2} \right)} \\ & {{D}_{ij}}=\sum\limits_{k=1}^{N}\int\limits_^{{{z}^{2}}dz}=\frac{1}{3}\sum\limits_{k=1}^{N}{{{\left[ {{Q}_{ij}} \right]}_{k}}\left( z_{k}^{3}-z_{k-1}^{3} \right)} \\ \end{align}$$ (3.9)
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Note that unlike the isotropic case, both the in-plane forces and moments depend on membrane displacements $$\displaystyle u_{0},v_{0}$$ and transverse flexure $$\displaystyle w=w_{0}$$, i.e. there is coupling between in-plane and out-of-plane activity. The coupling vanishes if and only if $$\displaystyle B_{ij}=0$$

By determining $$\displaystyle \tau_{xz} $$ and $$\displaystyle \tau_{yz}$$ from the equilibrium equations:

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$$ \displaystyle \begin{align} & \frac{\partial {{\sigma }_{x}}}{\partial x}+\frac{\partial {{\tau }_{yx}}}{\partial y}+\frac{\partial {{\tau }_{xz}}}{\partial z}=0 \\ & \frac{\partial {{\tau }_{xy}}}{\partial x}+\frac{\partial {{\sigma }_{y}}}{\partial y}+\frac{\partial {{\tau }_{yz}}}{\partial z}=0 \\ \end{align}$$ (3.10)
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and boundary conditions $$\displaystyle \tau_{xz} = \tau_{yz} = 0 $$ at $$\displaystyle z=\pm {}^{t}\!\!\diagup\!\!{}_{2}\;$$ or from (for small deflections)
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$$ \displaystyle \begin{align} & {{Q}_{x}}=\frac{\partial {{M}_{x}}}{\partial x}+\frac{\partial {{M}_{xy}}}{\partial y} \\ & {{Q}_{y}}=\frac{\partial {{M}_{xy}}}{\partial x}+\frac{\partial {{M}_{y}}}{\partial y} \\ \end{align}$$ (3.11)
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One obtains:
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$$ \displaystyle \begin{align} & {{Q}_{x}}=\int\limits_{-h/2}^{h/2}{{{\tau }_{xz}}dz=\left[ {{B}_{11}}\frac{\partial {{x}^{2}}}+2{{B}_{16}}\frac{\partial x\partial y}+{{B}_{66}}\frac{\partial {{y}^{2}}}+{{B}_{16}}\frac{\partial {{x}^{2}}}+\left( {{B}_{12}}+{{B}_{66}} \right)\frac{\partial x\partial y}+{{B}_{26}}\frac{\partial {{x}^{2}}} \right]} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\left[ {{D}_{11}}\frac{{{\partial }^{3}}w}{\partial {{x}^{3}}}+3{{D}_{16}}\frac{{{\partial }^{3}}w}{{\partial }{x}^{2}\partial y}+\left( {{D}_{12}}+2{{D}_{66}} \right)\frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}+{{D}_{26}}\frac{{{\partial }^{3}}w}{\partial {{y}^{3}}} \right] \\ & {{Q}_{y}}=\int\limits_{-h/2}^{h/2}{{{\tau }_{yz}}dz=\left[ {{B}_{16}}\frac{\partial {{x}^{2}}}+\left( {{B}_{12}}+{{B}_{66}} \right)\frac{\partial x\partial y}+{{B}_{26}}\frac{\partial {{y}^{2}}}+{{B}_{66}}\frac{\partial {{x}^{2}}}+2{{B}_{26}}\frac{\partial x\partial y}+{{B}_{22}}\frac{\partial {{x}^{2}}} \right]} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\left[ {{D}_{16}}\frac{{{\partial }^{3}}w}{\partial {{x}^{3}}}+\left( {{D}_{12}}+2{{D}_{66}} \right)\frac{{{\partial }^{3}}w}{{\partial }{x}^{2}\partial y}+3{{D}_{26}}\frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}+{{D}_{22}}\frac{{{\partial }^{3}}w}{\partial {{y}^{3}}} \right] \\ \end{align}$$ (3.12)
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Expressions for membrane forces and moments can be written in a matrix form as follows:

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$$ \displaystyle \left[ \begin{align} & N \\ & M \\ \end{align} \right]=\left[ \begin{matrix} A & B \\ B & D \\ \end{matrix} \right]\left[ \begin{align} & {{\varepsilon }^{0}} \\ & \kappa \\ \end{align} \right]$$ (3.13)
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Governing equation of the classic laminate theory and general solution procedures Solving the problem of the classical laminate theory, one is aiming at finding $$\displaystyle \left\{ N \right\}$$ or $$\displaystyle \left\{ {\varepsilon }^{0} \right\}$$ and $$\displaystyle \left\{ M \right\}$$ or $$\displaystyle \left\{ \kappa \right\}$$ by solving equation (2.19) and (2.21). Using the generalized stress-strain relationship (3.13), all the generalized stresses and strains can be determined. The strains in a lamina at any a given $$\displaystyle z$$ coordinate can be found using Eqs (3.2). The strains found are in the laminate loading coordinate system. The stresses in the loading coordinate system can obtained using (3.1) and can be both transformed to the material principal coordinate system of the lamina, using the coordinate transformation matrix.

There are two typical ways of obtaining the governing equations.

Displacement-based method
If the generalized strains on the right hand side of the generalized stress-strain relationship (3.13) are replaced by displacement derivatives according to their definition in (2.12), all the generalised stresses have been expressed in terms of displacements $$\displaystyle u^{0}, v^{0}$$ and $$\displaystyle w$$. Substituting them into equilibrium equations (2.19) and (2.21), one obtains three partial differentical equations as the governing equations for three displacements $$\displaystyle u^{0}, v^{0}$$ and $$\displaystyle w$$.
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$$ \displaystyle \begin{align} & {{A}_{11}}\frac{\partial {{x}^{2}}}+2{{A}_{16}}\frac{\partial x\partial y}+{{A}_{66}}\frac{\partial {{y}^{2}}}+{{A}_{16}}\frac{\partial {{x}^{2}}}+\left( {{A}_{12}}+{{A}_{66}} \right)\frac{\partial x\partial y}+{{A}_{26}}\frac{\partial {{x}^{2}}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{B}_{11}}\frac{{{\partial }^{3}}w}{\partial {{x}^{3}}}-{{B}_{16}}\frac{{{\partial }^{3}}w}{\partial {{x}^{2}}\partial y}-\left( {{B}_{12}}+2{{B}_{66}} \right)\frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}-{{B}_{26}}\frac{{{\partial }^{3}}w}{\partial {{y}^{3}}}=0 \\ & {{A}_{16}}\frac{\partial {{x}^{2}}}+\left( {{A}_{12}}+{{A}_{66}} \right)\frac{\partial x\partial y}+{{A}_{26}}\frac{\partial {{y}^{2}}}+{{A}_{66}}\frac{\partial {{x}^{2}}}+2{{A}_{26}}\frac{\partial x\partial y}+{{A}_{22}}\frac{\partial {{y}^{2}}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{B}_{16}}\frac{{{\partial }^{3}}w}{\partial {{x}^{3}}}-\left( {{B}_{12}}+2{{B}_{66}} \right)\frac{{{\partial }^{3}}w}{\partial {{x}^{2}}\partial y}-{{B}_{26}}\frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}-{{B}_{22}}\frac{{{\partial }^{3}}w}{\partial {{y}^{3}}}=0 \\ & {{D}_{11}}\frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}+4{{D}_{16}}\frac{{{\partial }^{4}}w}{\partial {{x}^{3}}\partial y}+2\left( {{D}_{12}}+2{{D}_{66}} \right)\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}+4{{D}_{26}}\frac{{{\partial }^{4}}w}{\partial x\partial {{y}^{3}}}+{{D}_{22}}\frac{{{\partial }^{4}}w}{\partial {{y}^{4}}} \\ & -{{B}_{11}}\frac{\partial {{x}^{3}}}-3{{B}_{16}}\frac{\partial {{x}^{2}}\partial y}-\left( {{B}_{12}}+2{{B}_{66}} \right)\frac{\partial x\partial {{y}^{2}}}-{{B}_{26}}\frac{\partial {{y}^{3}}} \\ & -{{B}_{22}}\frac{\partial {{y}^{3}}}-{{B}_{26}}\frac{\partial x\partial {{y}^{2}}}-\left( {{B}_{12}}+2{{B}_{66}} \right)\frac{\partial {{x}^{2}}\partial y}-{{B}_{16}}\frac{\partial {{x}^{3}}}=p\left( x,y \right) \\ \end{align}$$ (3.14) Solving these governing equations, one obtains displacements $$\displaystyle u^{0}, v^{0}$$ and $$\displaystyle w$$. Generalised strains can then be obtained using (2.29). The procedure as described in the previous paragraph then prevails. Finite element method is a special case following this approach. It uses the energy minimization to satisfy equilibrium instead of solving the partial differential governing equations directly.
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Stress function-based method
This approach is mostly used for analytical study. The generalized stress-strain relationship can be reorganized into the following form:
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$$ \displaystyle \left\{ \begin{align} & {{\varepsilon }^{0}} \\ & M \\ \end{align} \right\}=\left[ \begin{matrix} a & -{{b}^{T}} \\ b & d \\ \end{matrix} \right]\left\{ \begin{align} & N \\ & \kappa \\ \end{align} \right\}$$ (3.15) Where
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$$ \displaystyle \left[ a \right]={{\left[ A \right]}^{-1}},\,\,\left[ b \right]=\left[ B \right]{{\left[ A \right]}^{-1}},\,\,\left[ d \right]=\left[ D \right]-\left[ B \right]{{\left[ A \right]}^{-1}}\left[ B \right]$$ (3.16)
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Introducing the Airy stress function as (2.27) one finds the in-plane equilibrium equations (2.21) are satisfied identically. From the membrane strains as defined in (2.29) (for small deflections), a compatibility equation for these strains can be obtained as (a compatibility equation (2.24) is for large deformation):
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$$ \displaystyle \frac{{{\partial }^{2}}\varepsilon _{x}^{0}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}\varepsilon _{y}^{0}}{\partial {{x}^{2}}}=\frac{{{\partial }^{2}}\gamma _{xy}^{0}}{\partial x\partial y}$$ (3.17)
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This should be satisfied by the membrane strains corresponding to the membrane forces derived from the Airy stress function. Replacing $$\displaystyle \left\{ N \right\}$$ and $$\displaystyle \left\{ \kappa \right\}$$ by the expression of $$\displaystyle \phi $$ and $$\displaystyle w$$ in the right hand side of (3.15), $$\displaystyle \left\{ {\varepsilon }^{0} \right\}$$  and $$\displaystyle \left\{ M \right\}$$  have been expressed in terms of $$\displaystyle \phi $$ and $$\displaystyle w$$. Then substituting such $$\displaystyle \left\{ {\varepsilon }^{0} \right\}$$ and $$\displaystyle \left\{ M \right\}$$ into equation (3.17) and (1.12), two fourth order partial differential equations are obtained as the governing equations for $$\displaystyle \phi $$ and $$\displaystyle w$$:
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$$ \displaystyle \begin{align} & {{L}_{a}}\phi +{{L}_{b}}w=0 \\ & {{L}_{b}}\phi +{{L}_{d}}w=p \\ \end{align}$$ (3.18) Where
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$$ \displaystyle \begin{align} & {{L}_{a}}={{a}_{22}}\frac{\partial {{x}^{4}}}-2{{a}_{26}}\frac{\partial {{x}^{3}}\partial y}+\left( 2{{a}_{12}}+{{a}_{66}} \right)\frac{\partial {{x}^{2}}\partial {{y}^{2}}}-2{{a}_{16}}\frac{\partial x\partial {{y}^{3}}}+{{a}_{11}}\frac{\partial {{y}^{4}}} \\ & {{L}_{b}}=-{{b}_{12}}\frac{\partial {{x}^{4}}}-\left( 2{{b}_{62}}-{{b}_{16}} \right)\frac{\partial {{x}^{3}}\partial y}-\left( {{b}_{11}}+{{b}_{22}}-2{{b}_{66}} \right)\frac{\partial {{x}^{2}}\partial {{y}^{2}}}-\left( 2{{b}_{61}}-{{b}_{26}} \right)\frac{\partial x\partial {{y}^{3}}}+{{b}_{21}}\frac{\partial {{y}^{4}}} \\ & {{L}_{d}}={{d}_{11}}\frac{\partial {{x}^{4}}}+4{{d}_{16}}\frac{\partial {{x}^{3}}\partial y}+2\left( {{d}_{12}}+2{{d}_{66}} \right)\frac{\partial {{x}^{2}}\partial {{y}^{2}}}+4{{d}_{26}}\frac{\partial x\partial {{y}^{3}}}+{{d}_{22}}\frac{\partial {{y}^{4}}} \\ \end{align}$$ (3.19)
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=Large deformation of paper under thermal loading=

Remember that if a plate is bent by lateral load only and deflections are so small, the plate middle surface was assumed to be unstrained, hence it is considered as a neutral surface. Occasionally, however, direct forces (acting in the plane of the middle surface of the plate) are applied directly at the boundaries or they arise due to temperature (or moisture) changes. The latter will be considered below in detail.

It was assumed in the foregoing discussions that the temperature of an elastic plate remains constant and has the same value at all points of the plate; whence, temperature effects were not taken into consideration. If the temperature of the plate is raised or lowered it expands or contracts, respectively. Within a certain change, such expansion or contraction, for most structural materials, is directly proportional to the change in temperature. When a free plate made of homogenous isotropic material is heated uniformly, there appear normal strains but no thermal stresses. The thermal stresses will occur in the following cases: first, if the plate experiences a nonuniform temperature field; secondly, if the displacements are prevented from occurring freely because of the restrictions placed on the boundary even with a uniform temperature; and thirdly, if the material displays anisotropy even with uniform heating-for example, if a heated plate consists of several layers of different materials.

Curl is defined as the departure form an ideal flat form of a paper caused by ambient humidity or temperature changes. The deformation of an unconstrained sheet may be expressed by three basic curl curvatures $$\displaystyle{{\kappa }_{x}},{{\kappa }_{y}},{{\kappa }_{xy}}$$ where $$\displaystyle{{\kappa }_{x}},{{\kappa }_{y}}$$ represent bending curvatures while $$\displaystyle{{\kappa }_{xy}}$$ is the twist curvature. Classical laminate theory based on linear kinematics has been used to analyze curl of papers. Such analysis generally predicts a double curvature shape of the sheets. Observations of curled papers and asymmetric composite laminates, however, indicate a single curvature, i.e a cylindrical shape. The reason for this phenomenon was examined by Hyer [??] who showed that laminate theory modeling of curvatures of unsymmetric carbon/epoxy laminates requires incorporation of nonlinear kinematics. The analysis of Hyer showed that, at a certain critical temperature, the cylindrical shape becomes the most energetically favorable.

For simulation of curl in paper incorporating effects of nonlinear kinematics, the finite element code ANSYS was used. A two-layer papers of various stacking sequence,i.e, [0/15], [0/30], [0/45] and [0/90] consisting of orthotropic linear elastic plies were considered in the numerical analyses. The paper has properties:

- Dimension: $$ \displaystyle width \times height=100mm \times 89mm $$ and thickness per layer is $$\displaystyle t=0.31mm$$

- Elastic modulus: $$ \displaystyle {{E}_{11}}=3.8GPa,\,\,{{E}_{22}}=1.7GPa,\,\,{{\nu }_{12}}=0.15,\,\,{{G}_{12}}=1GPa $$

- Thermal coefficients (unit is in/in per temperature change): $$ \displaystyle {{\alpha}_{11}}=4e-6 ,\,\,{{\alpha}_{22}}=1e-5 $$

- Moisture coefficients (unit is in/in per %RH change): $$ \displaystyle {{\beta}_{11}}=4e-5,\,\,{{\beta}_{22}}=1e-4 $$

- Density: $$\displaystyle {\rho}=603kg/m^3 $$

A gird 16x16 was used to discretize a plate.

Two different cases are considering here: the first is to increase the moisture and the second is to increase the temperature.

Curling of paper fixed at the top line due to moisture changes
The paper was moisturized from 50% RH to 90% RH after 12 hours. So the increasing of moisture is 40%. These are the results between experiments (done by J. T. Decker, an undergraduate student from University of Wisconsin,Madison) and ANSYS (done by Tuan Hoang, UW Madison)

Curling of paper fixed at the center due to temperature changes
The reference temperature is $$\displaystyle{{T}_{0}}=0$$ and then paper was heated up to $$\displaystyle{{T}_{end}}=300$$ At the critical temperature, the stiffness matrix is zero and the Ansys program crashes, hence we can obtain the value of $$\displaystyle{{T}_{critical}}=150$$.To find the possible stable cylindrical solution, a three-step solution procedure was performed:

- Firstly, four concentrated loads F=0.01N, but no temperature changes, were applied at four corners of the paper to achieve a slight curvature

- Then the temperature was increased incrementally. When the temperature $$\displaystyle{{T}\ge {T}_{critical}}$$, the concentrated loads were released.

- Finally, the temperature was increased until $$\displaystyle{{T}={T}_{end}}$$ without concentrated loads, the shape of the panel will then remain approximately cylindrical.

Curling of [0-90] (cross-ply) laminate
- Step 1: apply four small loads

- Step 2: increase the temperature from T=0 to T=170 while keeping forces

- Step 3: continue increasing the temperature but remove the forces

- Step 4: keep on increasing temperature until T=300

The final shape for layer [0-90] after heating up to 300 Celsius is shown as below:

Curling of [0/45] laminate
- Step 1: apply four small loads

- Step 2: increase the temperature from T=0 to T=170 while keeping forces

- Step 3: continue increasing the temperature but remove the forces

- Step 4: keep on increasing temperature until T=300

The final shape for layer [0-45] after heating up to 300 Celsius is shown as below:

=Nonlinear buckling analysis in ANSYS=

 * Eccentric loading or geometry/imperfection is required
 * Is more accurate then linear buckling solutions
 * Requires gradual loading to find the point of instability
 * can model "post-buckling" response of a structure

ANSYS nonlinear buckling procedure

 * 1) (OPTIONAL) CLEAR the Options of previous analyses on the same model, see Solution > Reset Options
 * 2) Solve the linear, static analysis of a structure
 * 3) Include large deformation effects NLGEOM,ON from Solution > Analysis Options
 * 4) Use Output Controls to insure data other than the Last Step is available for postprocessing (use ALL SUBSTEPS, if possible)
 * 5) Use gradual (ramped) loading and automatic time stepping from Solution > Time/Frequenc > Time and Substeps
 * 6) Control convergence behavior for number of equilibrium iteration, convergence criteria, etc. under Solution > Nonlinear
 * 7) Difficult problems (or analysis of post-buckling) may require use of the "Arc Length" convergence enhancement tool for a successful run

What to do when it will not converge

 * check the solution output and error log file for messages
 * apply the load in more, smaller substeps
 * allow more equilibrium iterations
 * try alternative material models (bilinear vs. multilinear, different hardening rules, etc.)
 * inspect the mesh in high stress regions for poor element shapes
 * re-evaluate you choice of loads (get rid of point loads, overconstrained edges)
 * look at the last good solution for any irregularities
 * re-check element options (and material properties, and real constants, and load values, and dimensions)
 * experiment with convergence enhancement tools in the code
 * deactivate one or more of the nonlinear conditions to find the problem
 * (rarely) loosen the convergence criteria - to find the problem - then tighten it up again

Commands to calculate curl of papers
For sets of command in Ansys refer to = References =


text:
Portal:Urdu

اُردوُ

Welcome to the Urdu Department at Wikiversity, part of the Center for Foreign Language Learning and the School of Language and Literature.

Urdu is a dialect of the Hindustani language group native to Pakistan and northern India. The standard dialect is the official language of the Pakistani government along with English. Urdu was formed by the injection of Persian into local Indian dialects, especially Sanskrit. It is written in the Nastaliq Script, a derivative of Arabic, and uses Persian, Hindi and some Arabic words to form its vocabulary.

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Wikipedia

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title: Portal:Urdu


text:
User:Ezhuttukari

SEE MALYALAM WIKI

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Cornish Language/Pronunciation

{|cellspacing="5" cellpadding="10" style="margin-left:auto;margin-right:auto;width:99%;background-color:inherit" In this lesson, you will learn:
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 * How to pronounce Cornish

Stress
Stress in Cornish is always on the penultimate syllable (the second to last). This even happens when syllables are added:

Example:
 *  Ker now 'Cornwall, a Cornishman'
 * Ker new ek 'the Cornish language'
 * kerne weg er, 'a Cornish speaker'
 * kerneweg or yon 'Cornish speakers'

There are some exceptions, these will be pointed out as they appear.

Vowel length
Before you can accurately pronounce Cornish, you need to know how to ascertain the length of the vowel. Cornish has rules for this so it is quite regular, but there are exceptions, which will need to be learnt separately.

 * All unstressed vowels are short.

 * In stressed words of one syllable, vowels are:
 * Long at the end of words
 * Long before the graphs <b>, , , , , , , , , <s>, , <th> and <v>
 * Short before the graphs , , , >p<, , ,
 * Short before any other two consonants (except )

 * Stressed vowels in words of more than one syllable are short (there are a few exceptions which will be pointed out as they appear).

Pronunciation

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title: Cornish Language/Pronunciation


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User:Salixa

Web Page Graphic Design

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