text:
The Web Economy/26

Licensing and Business Models
This is a lesson part of the module The Web Economy out of the Open Source ERP/Executive Masters course conducted by a private university.

Topics

 * Licensing is a necessary evil for Open Source. The basic principles of free software is to always be free and never owned by a party that might control it.
 * Open Source began to differentiate from Free Software with semi open licensing.
 * Commercial tendencies began to crop into licensing and business models. Some uses dual licensing models but it has been dificult to capitalise.

Assignment Tasks

 * What are the various licensing models used by Free and Open Source Software and what are their differences with each other?
 * What projects have succesfully made money out of Open Source? How?
 * How can community open source survive when its business model is non existent?

Activities

 * Publish your works in an attractive manner in your user page, stating your own views and findings providing links to your sources. Use the talk page here to score marks.
 * Create or edit sub-pages of course materials within wikiversity.
 * Discuss in the forum (link shall be provided) by offering your ideas and answering or comment on others' postings.

Introduction

 * During the cave age or the beginning of humanity’s first memory, any creation of art or industry was free and not opyrighted nor patented nor subjected to any license agreement. Fire, the wheel and spears are not copyrighted where another caveman need to seek permission or acknowledge the creator before using them. Today the concept of patents and copyrights in the proprietary world is a common reality of life. It is all about making excluding others from making the same money from such an exclusive right. In the FOSS world it is all about freedom. But that doesn’t mean it cannot make money. In fact, money has caught up with it.

Lend Me Your Pen

 * When man began to write, it was for the pursuit of knowledge and sharing the joy of it. Today you make money if you can copyright and sell it. For FOSS advocates, they believe in the moral right of been recognised for it but not keeping it under the condition of payments. It is indeed to be shared for all its worth. So licensing for FOSS is more for protecting that freedom.

All Wrongs Reversed

 * In 1976, a programmer mischievously put the first known copyleft notice in his code (see figure top centre). It was not a proper licensing notice and has no legal significance until in the mid 80s Richard Stallman used a proper licensing model that spells out the specific condition of freedom to share and dsitribute ad infinitum without anyone locking up creative works again.

GNU Public License

 * Stallman introduced the GNU GeneralPublic License or GPL which is used in the first GNU projects, and then naturally adopted by Linux that gave it a serious future in the licensed sense. The GPL is a copyleft license that requires derived works to be available under the same copyleft. Under this philosophy, the GPL grants the recipients of a software the rights of the free software definition and uses copyleft to ensure the freedoms are preserved, even when the work is changed or added to.

Go Forth And Publish

 * The most important clause in GPL is that any derivatives of the sourcecode must be made public again. In this way any creative innovation is always available and can be further improved upon perpetually. This is a basic fundamental by which FOSS lives on.

Standing on Giants

 * By such a clause, small coders or ‘dwarfs’ can now stand on the shoulders of giants before them and see higher than before . This principle is what spearhead Linux development in an exponential manner without fear of the sourcecode been taken off the public domain or worse, its new coders sued and obstructed from continuing their creative pursuits.

Viral Law

 * The GPL license is thus by definition a viral license because any modfied or added code to it has to be GPLed too. This is good news for FOSS advocates as more and more sourcecode enter the public domain. However commercial interests that produces proprietary software become restricted to participate, and cannot take any such GPL code or they will get infected!

The Mozilla Pass

 * When Netscape was freefalling due to the onslaught from the free Internet Explorer agressively taking over its territory, it decided to go open source as a desperate attempt to avoid sudden death. However parts of its code are using proprietary software and so using the GPL license would be incompatible. It thus created the Mozilla license which allow the intermixing of both free and proprietary code. It would allow selective blackboxing. Interestingly the FSF recognises this as a free software license.

The Loophole

 * Thus with commercial fundings and success stories such as SugarCRM and Pentaho, Open Source started to become like a new mini dotcom bubble wave. Even Compiere was not spared. Many of these projects are individual owned and played the dual license game. As owner they can issue out one GPL license version of their code and another commercial version of their code usually called EULA to request from commercial users who wants to pay to obtain better warranty and value on paper.

Non-Warranty Clause

 * All licenses such as Apache and GPL has clauses which says that no warranty is given or is there any liability by the creator over any damage or loss of data from the use of the software. This is not only present in FOSS but most likely present in proprietary ones too. However this clause can be an excuse to engage the software developers as consultants or commercial entities to provide standby or on-site warranty in case of need.

15. Disclaimer of Warranty. THERE IS NO WARRANTY FOR THE PROGRAM, TO THE EXTENT PERMITTED BY APPLICABLE LAW. EXCEPT WHEN OTHERWISE STATED IN WRITING THE COPYRIGHT HOLDERS AND/OR OTHER PARTIES PROVIDE THE PROGRAM “AS IS” WITHOUT WARRANTY OF ANY KIND, EITHER EXPRESSED OR IMPLIED, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE. THE ENTIRE RISK AS TO THE QUALITY AND PERFORMANCE OF THE PROGRAM IS WITH YOU. SHOULD THE PROGRAM PROVE DEFECTIVE, YOU ASSUME THE COST OF ALL NECESSARY SERVICING, REPAIR OR CORRECTION.

Business Purpose

 * Since commercial interest has cropped into Open Source territory and with the above stated tactic, there have been opinion from our community that the commercial ones are often breaking the Open Source stipulations of not putting conditions on the sourcecode. They started imposing restrictions as to what version of (crippled) sourcecode is open to public and what (latest) is only open to paying customers. In that manner we dispute that as not truly open source but as a wolf in sheep clothing! Our argument is that it is ok to conduct paid licensed software business, but not to abuse the term Open Source when it is not.

Lesser GPL

 * Many regard the Richard Stallman been overly strict and for been critcising the restrictiveness of GPL for commercial interests. It is said to be in between the stronger copyleft of GPL and the more permissive ones like BSD and MIT. It is more used for APIs or libraries that link to other software not requiring the viral effect.

The Fallacy of Paid Software

 * With FOSS running around how would one be able to sell any software? The only way possible is when the software is overwhelmingly superior then FOSS which most are not. In fact many of the FOSS ones are highly competitive. They only lack packaging and strong corporate presence. SAP is still leader in the ERP market as the market base is very well established with SAP. But in time to come with ADempiere improving unabatedly the scenario might change.

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title: The Web Economy/26


text:
The Web Economy/27

Community OS vs Commercial OS
This is a lesson part of the module The Web Economy out of the Open Source ERP/Executive Diploma course conducted by a private university.

Topics

 * A self increasing community saves a project alot from investing in marketing, branding, testing and development as well as sustenance.
 * Commercial interests began to control open source projects as much money can be made quickly with minimal investment due to the energy from communities that form around them.
 * Communities are allergic to commercial tendencies and often fork to maintain its openness.
 * Forking is an important and necessary character of Open Source.

Discussion

 * What are the important forks and how are they doing compared with their parent projects?
 * Can commercial open source be as open as community open source?

Activities

 * Publish your works in an attractive manner in your user page, stating your own views and findings providing links to your sources. Use the talk page here to score marks.
 * Create or edit sub-pages of course materials within wikiversity.
 * Discuss in the forum (link shall be provided) by offering your ideas and answering or comment on others' postings.

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

title: The Web Economy/27


text:
The Web Economy/28

Business As Usual
This is a lesson part of the module The Web Economy out of the Open Source ERP/Executive Diploma course conducted by a private university.

Topics

 * There were business pressures onto projects like Linux such as to buy back Linux Inc and control its trademark.
 * Linux has to consider user friendliness for non geek users.
 * Community members have to survive and often argue of their monetary needs not be neglected so that they continue contributing to open source.

Assignment Tasks

 * Can business and hobby mix together in open source?
 * How can contributors make money in a certain manner without compromising the culture and principles of community open source.

Activities

 * Publish your works in an attractive manner in your user page, stating your own views and findings providing links to your sources. Use the talk page here to score marks.
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title: The Web Economy/28


text:
The Web Economy/29

Institutionalising Open Source
This is a lesson part of the module The Web Economy out of the Open Source ERP/Executive Diploma course conducted by a private university.

Topics

 * Open Source been still new and formless need institutions such as Certification, Centre of Excellence, Academic Training Centres and Reference Sites.
 * It has to be social business where dividends are not taken but help advance further its cause.
 * It can have a foundation behind it and the people running it has to be well chosen and accepted by the community.
 * Perhaps the big turning point for Open Source is when commercial giants began to invest into it for their own good. IBM pours in USD1 billion per annum to Linux and Apache via development and further innovation. IBM got more than USD1b revenue per annum from the services it build around it. This can be seen as others trying to assist in enemy territory of Microsoft. All this is good as it is healthy competition and ultmiately give the users a choice.

Discussion

 * What are the forms of institutions that Open Source can adopt and how should their sustenance models be?
 * Study examples in existent today.

Activities

 * Publish your works in an attractive manner in your user page, stating your own views and findings providing links to your sources. Use the talk page here to score marks.
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title: The Web Economy/29


text:
User:Eml4500.f08.team.dwyer.jd/HW4 Oct 17 (Meeting 23)

 HW Redo: Further explanation for the math is added. The third motivation for the PVW is added. Initial derivation of the PVW is added.

For a comparison between this page, and the originally submitted page, click here

Eml4500.f08.team.foskey.ckf 00:35, 3 November 2008 (UTC)

Equilibrium of Node 2:
Now, the sum of the forces in the x and the y directions are put into equation form in the following equations. Then, using the force displacement relationship, this forces in the equation are replaced with the product of the stiffness and the corresponding displacement degree of freedom.

 * Eq (1):

$$f_3^{(1)}+f_1^{(2)}=0 \quad (1)$$

 * Eq (2):

$$f_4^{(1)}+f_2^{(2)}=\underline{P}\quad (2)$$

 * From p 8-3:

$$d_3=d_3^{(1)}=d_1^{(2)}$$ $$d_4=d_4^{(1)}=d_4^{(2)}$$

Assembly Rows 3 & 4 From Eq (1) & (2) into:
Recall that the global stiffness matrix has the following values. Replacing the forces in equation one with the product of the stiffness and displacement degrees of freedom results in the following new equation one. Thus, this new equation one is consequently the third row of the global stiffness matrix on 9.1. Also see 8.4 for the global force displacement relationship. The new equation two is similarly derived.

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

$$\left [ k_{31}^{(1)}*d_1+k_{32}^{(1)}*d_2+k_{33}^{(1)}*d_3+k_{34}^{(1)}*d_4 \right ]+\left [ k_{11}^{(2)}*d_3+k_{12}^{(2)}*d_4+k_{13}^{(2)}*d_5+k_{14}^{(2)}*d_6 \right ]=0\quad (1)$$

$$\left [ k_{41}^{(1)}*d_1+k_{42}^{(1)}*d_2+k_{43}^{(1)}*d_3+k_{44}^{(1)}*d_4 \right ]+\left [ k_{21}^{(2)}*d_3+k_{22}^{(2)}*d_4+k_{23}^{(2)}*d_5+k_{24}^{(2)}*d_6 \right ]=\underline{P}\quad (2)$$

Now, an expression can be written to describe the assembly of a general (an arbitrary number of elements) global stiffness matrix using general elemental stiffness matrices. Assembly of $$\underline{k^{(e)}}$$, e = 1, ....., nel (number of elements), into global stiffness matrix K is as follows. It is necessary to denote a symbol other than $$\sum$$ to allow the below assembly expression to be possible and true. This is because the global stiffness matrix and the elemental stiffness matrices are of different dimensions where the dimensions of the global stiffness matrix can be millions by millions. Thus, simple matrix addition cannot be utilized here and another symbol other than $$\sum$$ must be used. Let "A" be this designated assembly operator.

Principle of Virtual Work (PVW)
There are three motivations for applying the principle of virtual work. 1) (p 10-1): Elimination of rows corresponding boundary conditions to obtain $$\underline{K}$$.

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

2)(p 12-1): Element Force-Displacement Relationship described by axial forces and displacements and the transformation matrix T.

$$ \mathbf{\underset{2 \times 1}q^{(e)}}=\mathbf{\underset{2 \times 4}T^{(e)}}\mathbf{\underset{4 \times 1}d^{(e)}} $$

This leads to the following relationship:

$$	\underbrace{\begin{pmatrix}

q^{(e)}_1 \\

q^{(e)}_2 \end{pmatrix}}_{\underset{2 \times 1}q^{(e)}} = \underbrace{\begin{bmatrix}

l^{(e)} & m^{(e)} & 0 & 0 \\

0 & 0 & l^{(e)} & m^{(e)} \\ \end{bmatrix}}_{\underset{2 \times 4}T^{(e)}}\underbrace{\begin{pmatrix}

d^{(e)}_1 \\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4 \end{pmatrix}}_{\underset{4 \times 1}d^{(e)}}$$

$$\mathbf k^{e} = \mathbf {T^{(e)^T} \hat k^{(e)} T^{(e)}}$$ 3) Deriving the finite element method for partial differential equations (PDEs). Let's begin by stating the force displacement relationship for a bar of a spring. $$\displaystyle Kd=F, Kd-F=0...(3)$$ $$\displaystyle w(Kd-F)=0$$ for all w $$\displaystyle ...(4)$$ Equation four is the "weak form" which is the principle of virtual work. The proof is as follows. a) The fact that equation 3 implies equation 4 is trivial. One can just plug in equation 3 into 4 and get zero. b) However equation 4 implying equation 3is not trivial since equation 4 is valid for all w. You can select any arbitrary value say w=1 then equation 4 becomes 1(Kd-F)=0. Therefore, equation 3 is obtained.


text:
User:Eas4200c.f08.blue.a/Lecture 21

As a continuation, the 3rd Ad Hoc point in the engineering derivation is the inconsistency in the assumption on the size of angle $$\alpha$$ : To get line PP', assume $$\alpha$$ is finite and $$\rho = r\cos \alpha$$, then reintroduce the small angle assumption afterwards.

Returning back to the formal derivation:

$$\bar{\epsilon} = \begin{bmatrix} \epsilon_{xx} & \epsilon_ {xy}&\epsilon_{xz} \\ \epsilon_{yx} & \epsilon_ {yy}&\epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} &\epsilon_{zz} \end{bmatrix}$$

This matrix can also be represented by

$$\bar{\epsilon} = \begin{bmatrix} \epsilon_{11} & \epsilon_ {12}&\epsilon_{13} \\ \epsilon_{21} & \epsilon_ {22}&\epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} &\epsilon_{33} \end{bmatrix} = [\epsilon_{ij}]$$ where i,j = 1,2,3

i is the row index and j is the column index

$$\epsilon _{ij} = \frac{1}{2}\left( \frac{\partial u_{i}}{\partial x_{j}} + \frac{\partial u_{j}}{\partial x_{i}}\right)$$ where $$x\Leftrightarrow x_{1}$$, $$y\Leftrightarrow x_{2}$$, and ,$$z\Leftrightarrow x_{3}$$

For example $$\epsilon _{11} = \frac{1}{2}\left( \frac{\partial u_{1}}{\partial x_{1}} + \frac{\partial u_{1}}{\partial x_{1}}\right) = \frac{\partial u_{1}}{\partial x_{1}} = \frac{\partial u_{x}}{\partial x_{x}}$$

Hence: only 6 independent components of $$\epsilon$$. Similarly for the stress tensor $$\bar{\sigma }= \left[ \sigma_{ij}\right]$$

Also, there are only 6 independent components of $$\bar{\sigma}$$ in 3-D

A question asked in class was whether or not $$\bar{\epsilon}$$ is related to isotropy of the mater. The answer is NO, isotropic elasticity is related to material behavior.

One last keynote to emphasis is the fact that

$$\epsilon _{xx} = \epsilon _{yy} = \epsilon _{zz} = \gamma _{xy} = 0$$

and from the stress-strain relationship

$$\sigma _{xx}=\sigma _{yy}=\sigma _{zz}=\tau _{yz}=0$$

title: User:Eas4200c.f08.blue.a/Lecture 21


text:
The Web Economy/30

Era of Convergence
This is a lesson part of the module The Web Economy out of the Open Source ERP/Executive Diploma course conducted by a private university.

Topics

 * Technology including software tries to converge into the human experience.
 * The TV and Phone are long considered killer appliances, now joining into a single even more encompassing application. The normal PC can now communicate in high definition and fidelity for a similar phone experience, and follows via login ID wherever the user goes.
 * High bandwidth and high storage made cheap defines a more richer user experience that removes the limitations from devices and applications.
 * Present advancement is allowing players to revive the ASP model as the present SaaS (Software as a Service) where users pay per unit usage rather than setup a whole ERP.

Assignment Tasks

 * Name examples of killer apps now present in the PC.
 * Research the future trends in software over the web is delivering funtionality and features right to any device convenient to the user.
 * In which devices or applications will it most likely to create a new wave of impact?
 * Will ERP which is so complex and unique for individual users be effecively served via a configurable remote system?

Activities

 * Publish your works in an attractive manner in your user page, stating your own views and findings providing links to your sources. Use the talk page here to score marks.
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title: The Web Economy/30


text:
The Web Economy/31

Google ERP
This is a lesson part of the module The Web Economy out of the Open Source ERP/Executive Diploma course conducted by a private university.

Topics

 * Google been the most pervasive application covering many appealing applications such as GoogleEarth, Android Phone project, Google Widget toolkit and Chrome browser is fast becoming the Single Window by which all users can rely on for everything else.
 * By having the visual geographical interface of Google Earth representing location in dynamic dimensions can then incorporate location data of ERP Warehousing which will give ERP a very immersed human feel.

Assignment Tasks

 * If Google is to incorporate the idea above, what will happen to conventional ERP or even open source ones?
 * How would such an application appeal to today's users?
 * What other features should it have that can make it even more appealing?

Activities

 * Publish your works in an attractive manner in your user page, stating your own views and findings providing links to your sources. Use the talk page here to score marks.
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title: The Web Economy/31


text:
User:Eas4200c.f08.aero.E/Matlab 1: NACA Airfoil

Matlab: NACA 4 Digit Airfoil Series
This results in a total area of 0.1027 m2 with the centroid at (0.4197, 0.0160).

% filename: airfoil % % PURPOSE: %      To plot an airfoil, calculate it average area and %      to calculate and plot the centroid. % % AUTHOR: %      Stephen Roberts function airfoil % Estabolish chord length and NACA parameters: % c = chord length % m = maximum camber in percentage of chord length c % p = position of maximum camber in tenths of the chord % ta = maximum airfoil thickness in percentage of the chord c = 1; m = 0.02*c; p = 0.4*c; ta = 0.15*c; % ns = number of segments ns = 100; % y = axis along the chord % zc_before = camber mean line before maximum camber % zc_after = camber mean line after maximum camber % ddy_zc_before = derivative of zc_before with respect to y % ddy_zc_after = derivative of zc_after with respect to y % zt_positive = thickness distribution above the camber mean line % zt_negative = thickness distribution below the camber mean line % theta_before = arc tangent of ddy_zc_before % theta_after = arc tangent of ddy_zc_after y = linspace(0,c,ns+1)'; zc_before = (m/(p^2))*(2*p*y - y.^2); zc_after = (m/(1-p)^2)*((1-2*p) + 2*p*y - y.^2); ddy_zc_before = -2*m*(y - p).*1/(p.^2); ddy_zc_after = -2*m*(y-p).*1/((p - 1).^2); zt_positive = (5*ta)*(0.2969*(y.^(1/2)) - 0.1260*(y) - 0.3516*(y.^2) + 0.2843*(y.^3) - 0.1015*(y.^4)); zt_negative = -(5*ta)*(0.2969*(y.^(1/2)) - 0.1260*(y) - 0.3516*(y.^2) + 0.2843*(y.^3) - 0.1015*(y.^4)); theta_before = atan(ddy_zc_before); theta_after = atan(ddy_zc_after); % determine location of maximum camber in terms of i i = 1; while y(i) < (0.4*c) %y(i) %i i = i + 1; end i=i-1; % input the nodal coordinates for j = 1:1:i position(:,j) = [y(j) - zt_positive(j)*sin(theta_before(j)); zc_before(j) + zt_positive(j)*cos(theta_before(j)); 0]; end for k = (i+1):1:(ns+1) position(:,k) = [y(k) - zt_positive(k)*sin(theta_after(k)); zc_after(k) + zt_positive(k)*cos(theta_after(k)); 0]; end for r = (ns+2):1:(ns+2+ns-i-1) position(:,r) = [y(2*ns-r+2) - zt_negative(2*ns-r+2)*sin(theta_after(2*ns-r+2)); zc_after(2*ns-r+2) + zt_negative(2*ns-r+2)*cos(theta_after(2*ns-r+2)); 0]; end for q = (ns+2+ns-i):1:(2*ns) position(:,q) = [y(2*ns-q+2) - zt_negative(2*ns-q+2)*sin(theta_before(2*ns-q+2)); zc_before(2*ns-q+2) + zt_negative(2*ns-q+2)*cos(theta_before(2*ns-q+2)); 0]; end % set up the nodal coordinate arrays for s = 1:1:(2*ns) x(s) = position(1,s); v(s) = position(2,s); z(s) = position(3,s); end % set up the element connectivity array node_connect for t = 1:1:(2*ns-1) node_connect(1,t) = t;    node_connect(2,t) = t+1; end node_connect(1,2*ns) = 2*ns; node_connect(2,2*ns) = 1; % Calculate total area for u = 1:1:(2*ns-1) dL(:,u) = (position(:,u+1) - position(:,u)); dA(:,u) = cross(position(:,u),dL(:,u)); end dL(:,2*ns) = (position(:,1) - position(:,2*ns)); dA(:,2*ns) = cross(position(:,2*ns),dL(:,2*ns)); dummy = sum(dA,2); Total_Area = abs((1/2)*dummy(3)) % Calculate centroid x_centroid(1) = 0; y_centroid(1) = 0; for u = 2:1:(2*ns) x_centroid(u) = (2/3)*(norm(position(:,u)))*position(1,u)/norm(position(:,u)); y_centroid(u) = (2/3)*(norm(position(:,u)))*position(2,u)/norm(position(:,u)); end X = sum(x_centroid.*abs((1/2)*dA(3,:)))/Total_Area Y = sum(y_centroid.*abs((1/2)*dA(3,:)))/Total_Area % plot the whole system for u = 1:1:(2*ns) node_1 = node_connect(1,u); node_2 = node_connect(2,u); yy = [x(node_1),x(node_2)]; zz = [v(node_1),v(node_2)]; xx = [z(node_1),z(node_2)]; plot3(yy,zz,xx,'-',X,Y,0,'+') hold on end % display the figure the way I want to axis([0,1,-.5,.5]) view([0 0 1]) grid title('NACA 2415')
 * }


text: File:Element1 Qwiki.jpg

title: File:Element1 Qwiki.jpg


text: File:Circle Convergence Study.JPG

title: File:Circle Convergence Study.JPG


text: User:Eas4200c.f08.aero.e/Matlab 2: Circle convergence


text:
User:La comadreja/Interested?

I am looking for new collaborators and volunteers, and am active in the Mentors program. I prefer to work with people whose interests are similar to mine, as described on the main page and in the Research Projects section. Feel free to write to me (reseal05@gmail.com).

I try to be responsive to you and to provide valuable feedback on your work. With an eclectic background in Biology, Writing and Computer Science, I am fairly knowledgeable about a broad range of topics and enjoy communicating with other Wikimedians. You will usually get a response from me within a few days, and often within a day—especially if you communicate by email. However, I am occasionally away from the computer for several weeks at a time. If I expect to be offline for more than about two or three weeks, I will normally post the dates.

I have a few expectations for colleagues:
 * 1) First and foremost, I expect you to care about whether you produce valid results, complete your projects in a timely fashion, and maintain a collegial and positive environment for others. My standards are basically the same as those detailed in Research and Civility.
 * 2) Please try to spend at least a few hours a week on your research, or on research-related content development. I put a lot of time and energy into helping other Wikimedians, and you should definitely reciprocate!

title: User:La comadreja/Interested?


text:
User:Eml4500.f08.gravy.bje/Notes4

HW Proof
$$ \underline{\tilde{f}}^{(e)} = \underline{\tilde{T}}^{(e)} \underline{f}^{(e)} $$

$$ \underline{\tilde{T}}^{(e)} = \begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0 \\ m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & -m^{(e)} \\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix}_{4x4}\,$$

$$\ l^{(e)}= \vec \tilde{i} \cdot \vec i= cos\theta^{(e)}\,$$

$$\ m^{(e)}= \vec i \cdot \vec j= cos\left (\frac{\pi}{2}- \theta^{(e)} \right )= sin\left (\theta^{(e)} \right )\,$$

$$\ \tilde{f}_{1}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \end{Bmatrix} $$

$$\ \tilde{f}_{2}^{(e)} = \begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} \begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \end{Bmatrix} $$

$$\begin{Bmatrix}\ \tilde{f}_{1}^{(e)} \\ \tilde{f}_{2}^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \end{Bmatrix} $$

$$\begin{Bmatrix}\ \tilde{f}_{1}^{(e)} \\ \tilde{f}_{2}^{(e)} \\ \tilde{f}_{3}^{(e)} \\ \tilde{f}_{4}^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0 \\ m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & -m^{(e)} \\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix} \begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix} $$

For θ(e) = 0 we have

$$\ l^{(e)} = cos\theta^{(e)} = 1\,$$

$$\ m^{(e)} = sin\theta^{(e)} = 0\,$$

$$\begin{Bmatrix}\ \tilde{f}_{1}^{(e)} \\ \tilde{f}_{2}^{(e)} \\ \tilde{f}_{3}^{(e)} \\ \tilde{f}_{4}^{(e)} \end{Bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix} $$

$$\tilde{f}_{1}^{(e)} = f_{1}^{(e)}$$

$$\tilde{f}_{2}^{(e)} = f_{2}^{(e)}$$

$$\tilde{f}_{3}^{(e)} = f_{3}^{(e)}$$

$$\tilde{f}_{4}^{(e)} = f_{4}^{(e)}$$

Since $$x = \tilde{x}$$ at θ(e) = 0

Eigenvector Homework
$$\ \underline{k} = \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}_{6x6}\,$$

$$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{49}{16} & \frac{3\sqrt{3} - 40}{16} & -\frac{5}{2} & \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3} - 40}{16} & \frac{43}{16} & \frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} \\ 0 & 0 & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$

MATLAB Code

Values of the stiffness matrix are assigned to matrix K >> K= [9/16 (3*sqrt(3))/16 -9/16 -(3*sqrt(3))/16 0 0; (3*sqrt(3))/16 3/16 -(3*sqrt(3))/16 -3/16 0 0; -9/16 -(3*sqrt(3))/16 49/16 ((3*sqrt(3)-40)/16) -5/2 5/2; -(3*sqrt(3))/16 -3/16 ((3*sqrt(3)-40)/16) 43/16 5/2 -5/2; 0 0 -5/2 5/2 5/2 -5/2; 0 0 5/2 -5/2 -5/2 5/2]

Values of K in decimal form K = 0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

Matlab function to calculate eigenvectors and eigenvalues >> [V,D] = eigs(K)

Eigenvectors of the matrix V = -0.0139   0.6174    0.0117    0.0155   -0.7776   -0.1162   -0.0080    0.3565    0.0016    0.1013    0.1481    0.9169    0.5123   -0.5409   -0.0835   -0.2936   -0.4947    0.3273   -0.4904   -0.4329    0.1665    0.6366   -0.3420    0.1486   -0.4984   -0.0765    0.5583   -0.6459   -0.0723    0.1075    0.4984    0.0765    0.8083    0.2843    0.0803   -0.0712

Eigenvalues of the matrix shown in the diagonal D = 10.0295        0         0         0         0         0         0    1.4705         0         0         0         0         0         0    0.0000         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0   -0.0000

title: User:Eml4500.f08.gravy.bje/Notes4


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User:Eml4500.f08.RAMROD.A/MyContributionHW4

A Closer Look at the Force Displacement Relationship
Equation One:   $$\displaystyle F_3^{(1)}+F_1^{(2)}=0 $$

Equation Two:   $$ F_4^{(1)}+F_2^{(2)}=P $$

From the Force Displacement relationship: k(1)*d(1) = F(1), We can break the forces above into their k*d form:

$$ f_3^{(1)} = \begin{bmatrix} k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)}+k_{33}^{(1)}d_3^{(1)}+k_{34}^{(1)}d_4^{(1)} \end{bmatrix} $$

$$ f_1^{(2)} = \begin{bmatrix} k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)}+k_{13}^{(2)}d_3^{(2)}+k_{14}^{(2)}d_4^{(2)} \end{bmatrix} $$

Equation One Becomes:

$$ \begin{bmatrix} k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)}+k_{33}^{(1)}d_3^{(1)}+k_{34}^{(1)}d_4^{(1)} \end{bmatrix} +  \begin{bmatrix} k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)}+k_{13}^{(2)}d_3^{(2)}+k_{14}^{(2)}d_4^{(2)} \end{bmatrix} = 0 $$

And when we convert to Global dof's:

$$ \begin{bmatrix} k_{31}^{(1)}d_1 + k_{32}^{(1)}d_2+k_{33}^{(1)}d_3+k_{34}^{(1)}d_4 \end{bmatrix} +  \begin{bmatrix} k_{11}^{(2)}d_3 + k_{12}^{(2)}d_4 + k_{13}^{(2)}d_5 + k_{14}^{(2)}d_6 \end{bmatrix} = 0 $$

We can take the k coefficients of d3 and d4 and reduce the Matrices

This Proves our Global Stiffness K, from our earlier learning's of the Finite Element Method:

$$ K = \begin{bmatrix} k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} \\ k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} \end{bmatrix} $$

Another way to arrive at Global K
K= Ae * ke, where K is an n x n Matrix.

n is the total # of global dof's before eliminating the boundary conditions.

ned is the # of element dof's

Then:      Kn x n = A(e)undefined * k(e)ned x ned

A is the Assembly Operator

Virtual Work
The General Equation $$ P * \upsilon * \omega  $$

Define q:

q2x1 = T(e) * d(e)

Define k

k(e) = T(e) T * k(e) * T(e)

Deriving Finite Element Method PDE's
Equation (3) is Implied: Equation (3): $$ k * d - F = 0 $$

Equation (4) is derived and is not tirvial:

Equation (4): $$ w(k * d -F) = 0 $$, for all w


text:
User:Eas4200c.f08.aero.E/Matlab 2: Convergence and Multi Cell

{| class="toccolours collapsible collapsed" style="width:100%" !Single Cell Airfoil Convergence

% filename: airfoil_convergence_study %    % PURPOSE: %      To find the value of J such that the difference between the %      current value and the previous value is less than 1% %      (Also plot graph showing J vs ns) %   % AUTHOR: %      Stephen Roberts function airfoil_convergence_study % NS = final value of number of segments % ns = intermediate value of number of segments End_num_Segments = 100; Start_num_Segments = 5; for ns = Start_num_Segments:1:End_num_Segments % Estabolish chord length and NACA parameters: % c = chord length % m = maximum camber in percentage of chord length c       % p = position of maximum camber in tenths of the chord % ta = maximum airfoil thickness in percentage of the chord c = .5; m = 0.02*c; p = 0.4*c; ta = 0.15*c; % y = axis along the chord % zc_before = camber mean line before maximum camber % zc_after = camber mean line after maximum camber % ddy_zc_before = derivative of zc_before with respect to y       % ddy_zc_after = derivative of zc_after with respect to y        % zt_positive = thickness distribution above the camber mean line % zt_negative = thickness distribution below the camber mean line % theta_before = arc tangent of ddy_zc_before % theta_after = arc tangent of ddy_zc_after y_original = linspace(0,c,ns+1)'; y = y_original/c; zc_before = (m/(p^2))*(2*p*y - y.^2); zc_after = (m/(1-p)^2)*((1-2*p) + 2*p*y - y.^2); ddy_zc_before = -2*m*(y - p).*1/(p.^2); ddy_zc_after = -2*m*(y-p).*1/((p - 1).^2); zt_positive = (5*ta)*(0.2969*(y.^(1/2)) - 0.1260*(y) - 0.3516*(y.^2) + 0.2843*(y.^3) - 0.1015*(y.^4)); zt_negative = -(5*ta)*(0.2969*(y.^(1/2)) - 0.1260*(y) - 0.3516*(y.^2) + 0.2843*(y.^3) - 0.1015*(y.^4)); theta_before = atan(ddy_zc_before); theta_after = atan(ddy_zc_after); % determine location of maximum camber in terms of i       i = 1; while y(i) < (0.4*c) %y(i) %i i = i + 1; end i=i-1; % input the nodal coordinates for j = 1:1:i position(:,j) = [y(j) - zt_positive(j)*sin(theta_before(j)); zc_before(j) + zt_positive(j)*cos(theta_before(j)); 0]; end for k = (i+1):1:(ns+1) position(:,k) = [y(k) - zt_positive(k)*sin(theta_after(k)); zc_after(k) + zt_positive(k)*cos(theta_after(k)); 0]; end for r = (ns+2):1:(ns+2+ns-i-1) position(:,r) = [y(2*ns-r+2) - zt_negative(2*ns-r+2)*sin(theta_after(2*ns-r+2)); zc_after(2*ns-r+2) + zt_negative(2*ns-r+2)*cos(theta_after(2*ns-r+2)); 0]; end for q = (ns+2+ns-i):1:(2*ns) position(:,q) = [y(2*ns-q+2) - zt_negative(2*ns-q+2)*sin(theta_before(2*ns-q+2)); zc_before(2*ns-q+2) + zt_negative(2*ns-q+2)*cos(theta_before(2*ns-q+2)); 0]; end for i = 1:1:(ns+1) position(1,i) = y_original(i); end for i = (ns+2):1:(2*ns) position(1,i) = y_original(2*ns+2-i); end % set up the nodal coordinate arrays for s = 1:1:(2*ns) x(s) = position(1,s); v(s) = position(2,s); z(s) = position(3,s); end % set up the element connectivity array node_connect for t = 1:1:(2*ns-1) node_connect(1,t) = t;           node_connect(2,t) = t+1; end node_connect(1,2*ns) = 2*ns; node_connect(2,2*ns) = 1; % Calculate total area for u = 1:1:(2*ns-1) dL(:,u) = (position(:,u+1) - position(:,u)); dA(:,u) = cross(position(:,u),dL(:,u)); end dL(:,2*ns) = (position(:,1) - position(:,2*ns)); dA(:,2*ns) = cross(position(:,2*ns),dL(:,2*ns)); dummy = sum(dA,2); Total_Area = abs((1/2)*dummy(3)); % Calculate total length of exterior for i = 1:1:(2*ns) Length_Vector(i) = norm(dL(:,i)); end Total_Length = sum(Length_Vector); J(ns) = 4*((Total_Area)^2)*0.002/Total_Length; end % Fix J vector to start at Start_num_Segments for i = 1:1:(End_num_Segments-Start_num_Segments) j(i) = J(i+Start_num_Segments-1); end J=j; % Create vector "Number_of_segments" to represent the number of segments for i = 1:1:(End_num_Segments-Start_num_Segments) Number_of_segments(i) = Start_num_Segments+i-1; end % Calculate the percent difference between each J value for i = 1:1:(End_num_Segments-Start_num_Segments-1) Percent_Difference(i) = (((J(i+1))-(J(i)))/(((J(i+1))+(J(i)))/2))*100; end i = 1; while abs(Percent_Difference(i)) > 1 i = i+1; end % Plot J versus the number of segments plot(Number_of_segments,J,Number_of_segments(i+1),J(i+1),'+') xlabel('Number of Segments') ylabel('J') title('Single Cell Airfoil: Convergence Study') grid

 * }


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Talk:Observational astronomy/Planning

well, I'm an interested astronomical observer and would love to kick around ideas

my first one is comets

Dead links
A few dead links have been fixed. I've cleared the edit by User:Mu301Bot from this talk page to verify they work properly. See the page history for details. --mikeu talk 15:46, 15 October 2012 (UTC)

title: Talk:Observational astronomy/Planning


text:
Health mentoring course

The course is currently under development in the German Wikiversity.

Khan Academy

 * Glucose Insulin and Diabetes : The basics of Type I and Type II diabetes
 * Blood Sugar Levels : Normal and elevated blood sugar levels before (preprandia) and after (postprandial) a meal
 * A1C Levels : Why measuring Hemoglobin A1c levels is an indicator of blood sugar levels (and diabetes)
 * Heart Disease and Heart Attacks : Understanding coronary artery disease and myocardial infarctions
 * Thrombo-emboli and Thromboembolisms : Clarifying difference between a thrombus and an embolus
 * Stenosis, ischemia and heart failure : Clarifying a bunch of medical terms around heart disease
 * Strokes : Basics of strokes

Literature

 * Baylor University's Community Mentoring for Adolescent Development (CMAD) Mentor Trainer's Manual, Chapter 9. Health Concerns of Adolescents (PDF)
 * Baylor University's Community Mentoring for Adolescent Development (CMAD) Mentor Trainer's Manual, Chapter 9. Health Concerns of Adolescents (PDF)

title: Health mentoring course


text:
Talk:Health mentoring course

Learning trails
Learning trails can be used to distinguish disease-endemic areas.

title: Talk:Health mentoring course


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User:EML4500.F08.RAMROD.B/Homework 4

=Connectivity Array "conn"=

The following array is the array that should be used in Matlab to represent the global node numbers corresponding to each element of the truss system. The rows of this array represent the element number and the columns represent the local node number.

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

Row one of this array corresponds to element one while row two corresponds to element two. Column one represents local node one and column two represents local node two. So to interpret this array we will do an example. Lets say we want to find the global node number for local node one of element one. Since the columns represent the local node number we go to column one and since the rows represent the element number we go to row one. In spot $$\displaystyle (1,1)$$ of the conn array it can be determined that the global node number of element one, local node one is simply one. In general;

$$\displaystyle conn(e,j) = $$ global node number of local node j of element

=Location Matrix Master Array "Lmm"=

The following array is similar to that of the array above with a minor difference. It is used to determine the global degrees of freedom when the local degrees of freedom and element number are known. The rows represent the element number and instead of representing the local node number, the columns represent the local degrees of freedom. Since the problem being analyzed is a two truss problem, each element has four degrees of freedom so the array has four columns and two rows representing each element.

$$\displaystyle Lmm = \begin{bmatrix} 1 & 2 &3 &4 \\  3& 4 & 5 & 6 \end{bmatrix}$$

To interpret this array we will do an example. Lets say we want to find the global degree of freedom of element two at local degree of freedom three. With this information we go to row two (element two) and column three (local degree of freedom). At this location we find the global degree of freedom corresponding to element two, local degree of freedom three. In general;

$$\displaystyle Lmm(e,j)=$$ the global degree of freedom number for element $$\displaystyle e$$ corresponding to the $$\displaystyle j^{th}$$ local degree of freedom

title: User:EML4500.F08.RAMROD.B/Homework 4


text:
User:Eml4500.f08.RAMROD.A/HW4

 One member needed a little extra time to complete his section: the section titled, "The Transfer Function," so it was uploaded by him and then placed in the right position by the team coordinator. Although this comment box was added today (Eml4500.f08.RAMROD.A 04:37, 29 October 2008 (UTC)) the two versions of HW 4 were completed within a few minutes. One version is on time and the other approximately 13 minutes late. [Here is the Comparison]

Connectivity Array "conn"
The following array is the array that should be used in Matlab to represent the global node numbers corresponding to each element of the truss system. The rows of this array represent the element number and the columns represent the local node number.

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

Row one of this array corresponds to element one while row two corresponds to element two. Column one represents local node one and column two represents local node two. So to interpret this array we will do an example. Lets say we want to find the global node number for local node one of element one. Since the columns represent the local node number we go to column one and since the rows represent the element number we go to row one. In spot $$\displaystyle (1,1)$$ of the conn array it can be determined that the global node number of element one, local node one is simply one. In general;

$$\displaystyle conn(e,j) = $$ global node number of local node j of element

Location Matrix Master Array "Lmm"
The following array is similar to that of the array above with a minor difference. It is used to determine the global degrees of freedom when the local degrees of freedom and element number are known. The rows represent the element number and instead of representing the local node number, the columns represent the local degrees of freedom. Since the problem being analyzed is a two truss problem, each element has four degrees of freedom so the array has four columns and two rows representing each element.

$$\displaystyle Lmm = \begin{bmatrix} 1 & 2 &3 &4 \\ 3& 4 & 5 & 6 \end{bmatrix}$$

To interpret this array we will do an example. Lets say we want to find the global degree of freedom of element two at local degree of freedom three. With this information we go to row two (element two) and column three (local degree of freedom). At this location we find the global degree of freedom corresponding to element two, local degree of freedom three. In general;

$$\displaystyle Lmm(e,j)=$$ the global degree of freedom number for element $$\displaystyle e$$ corresponding to the $$\displaystyle j^{th}$$ local degree of freedom

The Transfer Function
Goal - We want to find T that transfer the set of local elemental dofs (degrees of freedom), d, to another set of local elemental dofs, d, such that T is invertible.

$$\tilde{\underline{d}}^{(e)}_{4x1} = \tilde{\underline{T}}^{(e)}_{4x4}\;\tilde{\underline{d}}^{(e)}_{4x1}$$

Equation (1): $$\tilde{d}_1^{(e)}=\begin{bmatrix} l^{(e)}&m^{(e)} \end{bmatrix} = \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)} \end{Bmatrix}$$

$$\tilde{d}_2^{(e)} = \overrightarrow{d}_1^{(e)} \overrightarrow{\tilde{j}}$$

$$\tilde{d}_2^{(e)} = -\sin \theta^{(e)}d_1^{(e)} + \cos \theta^{(e)}d_2^{(e)}$$

Equation (2): $$\tilde{d}_2^{(e)} = \begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} = \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)} \end{Bmatrix}$$

Put Equations (1) and (2) together in matrix form:

$$\begin{Bmatrix} \tilde{d}_1^{(e)}\\ \tilde{d}_2^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)}& l^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)} \end{Bmatrix}$$

$$\begin{Bmatrix} \tilde{d}_1^{(e)}\\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)}\\ \tilde{d}_4^{(e)} \end{Bmatrix} = \begin{bmatrix} \underline{R}_{2x2}^{(e)} & \underline{O}_{2x2}\\ \underline{O}_{2x2}& \underline{R}_{2x2}^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)} \end{Bmatrix}$$

$$\tilde{\underline{f}}^{(e)} = k^{(e)}\begin{vmatrix} 1 &0 &-1  &0 \\ 0 &0  & 0 &0 \\ -1& 0 & 1 & 0\\ 0 & 0 &0  &0 \end{vmatrix} \tilde{d}^{(e)}$$

$$\tilde{\underline{f}}_{4x1}^{(e)} = \tilde{\underline{k}}_{4x4}^{(e)}\;\tilde{\underline{d}}_{4x1}^{(e)}$$

Matrix Diagonals
Note: Consider case where $$\tilde{d}_{4}^{(e)} \ne 0$$

$$\tilde{d}_{1}^{(e)}=\tilde{d}_{2}^{(e)}=\tilde{d}_{3}^{(e)}= 0$$

$$\tilde{\mathbf{f}}_{4x1}^{(e)}=\tilde{\mathbf{k}}_{4x4}^{(e)}\tilde{\mathbf{d}}_{4x1}^{(e)}=\mathbf{0}_{4x1}$$

The $$\mathbf{0}_{4x1}$$ is the 4th column of $$\mathbf{k}^{(e)}$$

Interpretation of the the transverse degrees of freedom

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

Similarly: $$\tilde{\mathbf{f}}^{(e)}=\tilde{\mathbf{T}}^{(e)}{f}^{(e)}$$

Also:

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

Therefore:

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

If $$\tilde{\mathbf{T}}^{(e)}$$ is invertible, then:

$$ [\tilde{\mathbf{T}}^{(e)-1}{\tilde{\mathbf{k}}^{(e)}}\tilde{\mathbf{T}}^{(e)}]\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

$$\tilde{\mathbf{T}}^{(e)} $$ block diagonal matrix: $$\begin{bmatrix} \mathbf{R}_{2x2}^{(e)} & \mathbf{0}_{2x2} \\ \mathbf{0}_{2x2} & \mathbf{R}_{2x2}^{(e)} \end{bmatrix}$$

Consider a general block diagonal matrix: $$ \mathbf{A}=\begin{bmatrix} \mathbf{D}_{1}& \mathbf{0} \\ \mathbf{0} & \mathbf{D}_{s} \end{bmatrix}$$

Question: What is $$\mathbf{A}^{-1}$$?

Here is a simple example

$$ \mathbf{B}=\begin{bmatrix} \mathbf{d}_{11}& \mathbf{0}& \mathbf{0} \\ \mathbf{0} & \mathbf{d}_{22} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{d}_{nn} \end{bmatrix}$$

$$\mathbf{B}=Diag[d_{11}, d_{22}, ..., d_{nn}]$$ (This is the shortened notation for $$ \mathbf{B} $$)

$$ \mathbf{B}^{-1}=Diag[1/d_{11}, 1/d_{22}, ..., 1/d_{nn}]$$

Assumming $$d_{ii}\ne 0$$ for $$ i=1, ..., n$$

For a block diagonal matrix A:

$$\mathbf{A}=Diag[d_{1} ..., d_{s}]$$

$$\mathbf{A}^{-1}=Diag[d_{1}^{-1} ..., d_{s}^{-1}]$$

$$\tilde{\mathbf{T}}^{(e)-1}=Diag[\mathbf{T}^{(e)-1} ..., R^{(e)-1}]$$

We know:

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

Also

$$\mathbf{R}_{2x2}^{(e)}\mathbf{R}_{2x2}^{(e)T}=\begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix}_{2x2}=\mathbf{I}_{2x2}$$ (This is the Identity matrix)

In conclusion: $$\mathbf{R}^{(e)-1}=\mathbf{R}^{(e)T}$$

So $$\tilde{\mathbf{T}}^{(e)-1}=(Diag[\mathbf{R}^{(e)} ,\mathbf{R}^{(e)}])^{T}$$ where $$Diag[\mathbf{R}^{(e)} ,\mathbf{R}^{(e)}]=\tilde{\mathbf{T}}^{(e)}$$

This shows that: $$\tilde{\mathbf{T}}^{(e)-1}=\tilde{\mathbf{T}}^{(e)T}$$

HW: Eigenvectors for the zero Eigenvalues of K
Using Matlab, the eigenvalues for K were found (see HW 3) and the eigenvectors were plotted for the zero values. The eigenvectors were found to be:

$$E_1 = \begin{Bmatrix} -0.4494 & 0.6754 & 0.1682 & -0.3942 & 0.2812 & -0.2812 \end{Bmatrix}$$

$$E_2 = \begin{Bmatrix} 0.3877&-0.4281&0.3877&-0.4281&0.4163&-0.3995 \end{Bmatrix}$$

$$E_3 = \begin{Bmatrix} 0.5158&0.4825&0.5158&0.4825&-0.0116&-0.0449 \end{Bmatrix}$$

$$E_4 = \begin{Bmatrix} 0.0160&0.0244&0.0160&0.0244&0.7023&0.7107 \end{Bmatrix}$$

The plots are shown below.

These plots show the mode shapes corresponding to each of the zero eigenvalues and are representative of a linear combination of the four pure mode shapes (x-translation, y-translation, rotational, and mechanism).

For the eigenvalue/eigenvector problem:

$$Kv = \lambda v$$

We can let $$\begin{Bmatrix} u_1 & u_2 & u_3 & u_4 \end{Bmatrix}$$ be the pure eigenvetors corresponding to the four zero eigenvalues. Assuming this tells us that:

$$Ku_i = 0u_i$$ for i = 1,2,3,4.

To get the true eigenvector, w, for each zero eigenvalue, we take the sum of the four contributors, each multiplied by a constant $$\alpha _{1}$$, giving us:

$$ w = \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3 + \alpha_4 u_4$$

Because w consists of the combination of eigenvectors for zero eigenvalues, w too represents an eigenvector corresponding to a zero eigenvalue. That is that:

"$\bar{K}\bar{w} = K(\Sigma \alpha _iu_i) = \bar{0} = 0*\bar{w}$"

 Wednesday October 15

Element Stiffness Matrix to Global Stiffness Matrix
Justification of assembly of element stiffness matrices into global stiffness matrix K.

$$K^{(e)} \; \; e=1, ... ,n_{el}\;$$

Consider the 2 bar truss example. Recall Force displacement equation.

$$ K^{(e)}_{4x4} \cdot d^{(e)}_{4x1} = F^{(e)}_{4x1}$$

You then can use the Euler cut method to establish the equilibrium of global node 2

Using the free body diagram of element one and of element 2 you can describe the elements degrees of freedom $$d^{(e)}_2 $$

$$d_2=d_2^{(1)}$$

$$d_3=d_3^{(1)}=d_1^{(2)} $$

$$d_4=d_4^{(1)}=d_2^{(3)}$$

$$d_5=d_3^{(2)}$$

$$d_6=d_4^{(2)}$$

Equilibrium of node 2

$$\sum{F_x}=-f_3^{(1)}-F_1^{(2)}=0$$

$$\sum{F_y}=P-f_4^{(1)}-F_2^{(2)}=0$$

Then you would use element force displacement relation $$K^{(e)}_{4x4} \cdot d^{(e)}_{4x1} = F^{(e)}_{4x1}$$

A Closer Look at the Force Displacement Relationship
Equation One:   $$\displaystyle F_3^{(1)}+F_1^{(2)}=0 $$

Equation Two:   $$ F_4^{(1)}+F_2^{(2)}=P $$

From the Force Displacement relationship: k(1)*d(1) = F(1), We can break the forces above into their k*d form:

$$ f_3^{(1)} = \begin{bmatrix} k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)}+k_{33}^{(1)}d_3^{(1)}+k_{34}^{(1)}d_4^{(1)} \end{bmatrix} $$

$$ f_1^{(2)} = \begin{bmatrix} k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)}+k_{13}^{(2)}d_3^{(2)}+k_{14}^{(2)}d_4^{(2)} \end{bmatrix} $$

Equation One Becomes:

$$ \begin{bmatrix} k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)}+k_{33}^{(1)}d_3^{(1)}+k_{34}^{(1)}d_4^{(1)} \end{bmatrix} +  \begin{bmatrix} k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)}+k_{13}^{(2)}d_3^{(2)}+k_{14}^{(2)}d_4^{(2)} \end{bmatrix} = 0 $$

And when we convert to Global dof's:

$$ \begin{bmatrix} k_{31}^{(1)}d_1 + k_{32}^{(1)}d_2+k_{33}^{(1)}d_3+k_{34}^{(1)}d_4 \end{bmatrix} +  \begin{bmatrix} k_{11}^{(2)}d_3 + k_{12}^{(2)}d_4 + k_{13}^{(2)}d_5 + k_{14}^{(2)}d_6 \end{bmatrix} = 0 $$

We can take the k coefficients of d3 and d4 and reduce the Matrices

This Proves our Global Stiffness K, from our earlier learning's of the Finite Element Method:

$$ K = \begin{bmatrix} k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} \\ k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} \end{bmatrix} $$

Another way to arrive at Global K
K= Ae * ke, where K is an n x n Matrix.

n is the total # of global dof's before eliminating the boundary conditions.

ned is the # of element dof's

Then:      Kn x n = A(e)undefined * k(e)ned x ned

A is the Assembly Operator

Virtual Work
The General Equation $$ P * \upsilon * \omega  $$

Define q:

q2x1 = T(e) * d(e)

Define k

k(e) = T(e) T * k(e) * T(e)

Deriving Finite Element Method PDE's
Equation (3) is Implied: Equation (3): $$ k * d - F = 0 $$

Equation (4) is derived and is not tirvial:

Equation (4): $$ w(k * d -F) = 0 $$, for all w

5 Truss MATLAB code
The 5 bar truss system in example 1.4 in the book can be solved by the below MATLAB code. analysis and explanation of the MATLAB code is shown below. This code and the the functions were downloaded from the books accompanying website:

http://bcs.wiley.com/he-bcs/Books?action=index&itemId=0471648086&bcsId=2256

Functions
The functions PlaneTrussElement.m, NodalSoln.m, PlaneTrussResults.m were all called in the code for the 5 truss system. They are all necessary for the completion of the truss problem.

PlaneTrussElement.m

This function take the Young's Modulus (e), the Area of the element (A) and the coordinates of the element ends (coord) and generates the stiffness matrix for a plane truss element. The function first calculates the lengths of the the elements. Once those are calculated the stiffness matrix is calculated.

NodalSoln.m

The function takes the global coefficient matrix (K), the global right hand side vector (R), the list of degrees of freedom with specified values, and the specified values to determine the displacements and reactions at each node. The dof of the the system is first found by using the  command which finds the longest dimension of R. df then finds the difference between the dofs that have known values (a value of zero) and the dof that were found in the previous line. The displacements and the reactions are then calculated.

PlaneTrussResults.m

This function computes the plane truss element results. It takes in the modulus of elasticity (e), the area of the cross-section (A), the coordinates at the element ends (coord), and the displacements at the elemtent ends (disps) and calculates the axial strain (eps), stress(sigma) and force (force). The results are stored in the variable  and sent back to the main program.

Results from the 5 Truss MATLAB example
The results from the 5 Truss System MATLAB code is shown below. As stated in the PlaneTrussResults.m function the  columns go in the order of axial strain, axial stress, and last axial force.

Code for the plot of the undeformed and deformed five truss system
Below is the code for the plot of the five bar system. Using the results that were obtained from the above MATLAB solution the deformed plot was created. The dotted lines are the undeformed system and the solid lines are the deformed system. The the displacements were multiplied by a factor of 500 so that they would appear on the plot.

Three Bar Truss System
A program was also written to solve the three bar truss system. The code follows the model code that was given to us in HW 2. The functions PlaneTrussElement.m, NodalSoln.m, and PlaneTrussResults.m were once again used in the code to obtain the results. Descriptions of these functions can be found in the section for the Five bar truss system.

Below are the results from the above code.

Plot of undeformed and deformed three bar truss system
Below is the code and the plot of the three bar truss system. The plot was made from the results that were found in the above code. The results demonstrate that only node two had a displacement of (-15.167, 48.231). The undeformed system is plotted with the dotted lines and the deformed system is plotted with solid lines. The deformed displacements have been scaled down by a tenth in order to enhance in the plot.

Contributing Team Members
Christopher Pergola EML4500.F08.RAMROD.B 18:10, 23 October 2008 (UTC)

Eml4500.f08.RAMROD.A 18:46, 23 October 2008 (UTC)

Chris Fontana Eml4500.f08.RAMROD.F 18:52, 24 October 2008 (UTC)

Eml4500.f08.ramrod.D 19:07, 23 October 2008 (UTC)

EML4500.f08.RAMROD.E 22:10, 23 October 2008 (UTC)

Eml4500.f08.ramrod.c 21:03, 24 October 2008 (UTC)

title: User:Eml4500.f08.RAMROD.A/HW4


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User:Eml4500.f08.wiki1.aguilar/TeamHW4

Continued Assembly Instructions for a Three-Bar Truss
To continue, we're going to need to be familiar with a "connectivity array" and a "location matrix master array."

A Connectivity array, "conn" is shown below. (Consider the two-bar truss system from your class notes, p.5-6)

Conn(e,j) = global node number of local node j of element e

We also want to establish a Location Matrix Master array: "lmm".

lmm (i,j) = equation number (global degree of freedom number) for element stiffness coefficient corresponding to the jth local degree of freedom number.

Going back to p.12-1, Method 2 to derive $$\mathbf{k}^{(e)}_{4x4} $$

Finding a $$\tilde {\mathbf{T}}^{(e)}$$ that is Invertible
Goal: Our goal is to find a $$\mathbf{\tilde{T}}^{(e)}_{4x4}$$ that can transform the set of local element degrees of freedom, $$\mathbf{d}^{(e)}_{4x1} $$ to another set of local element degrees of freedom, $$\mathbf{\tilde{d}}^{(e)}_{4x1} $$, such that $$\mathbf{\tilde{T}}^{(e)}_{4x4}$$ is invertible.

$${\tilde{d}^{(e)}}_{1}= \begin{bmatrix} l^{(e)} &m^{(e)} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\end{Bmatrix}(1)$$

We illustrate this below:

$${\tilde{d}^{(e)}}_{1}={\vec{d}^{(e)}}_{[1]}\cdot \vec{\tilde{j}}$$ This is the component of vector $${\vec{d}^{(e)}}_{[1]}$$ along $$\vec{\tilde{j}}$$, i.e. $$\tilde{y}$$ axis. $${\tilde{d}^{(e)}}_{1}=-\sin \theta ^{(e)} d^{(e)}_{1}+\cos\theta ^{(e)} d^{(e)}_{2}$$

This is also done for $${\tilde{d}^{(e)}}_{2}$$. $${\tilde{d}^{(e)}}_{2}= \begin{bmatrix} l^{(e)} &m^{(e)} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\end{Bmatrix}(2)$$

$${\tilde{d}^{(e)}}_{2}={\vec{d}^{(e)}}_{[2]}\cdot \vec{\tilde{j}}$$ This is the component of vector $${\vec{d}^{(e)}}_{[2]}$$ along $$\vec{\tilde{j}}$$, i.e. $$\tilde{y}$$ axis. $${\tilde{d}^{(e)}}_{2}=-\sin \theta ^{(e)} d^{(e)}_{1}+\cos\theta ^{(e)} d^{(e)}_{2}$$

If we put equations (1) and (2) in matrix form:

 $$ \begin{Bmatrix}{\tilde{d}^{(e)}}_{1} \\ {\tilde{d}^{(e)}}_{2} \\ {\tilde{d}^{(e)}}_{3} \\ {\tilde{d}^{(e)}}_{4}\end{Bmatrix}= \begin{bmatrix} \mathbf{R}^{(e)}_{2x2} & 0 \\ 0 & \mathbf{R}^{(e)}_{2x2} \end{bmatrix} \begin{Bmatrix}d^{(e)}_{1} \\ d^{(e)}_{2} \\ d^{(e)}_{3} \\ d^{(e)}_{4}\end{Bmatrix} $$

$${\tilde{\mathbf{d}}^{(e)}}_{4x1}={\tilde{\mathbf{T}}^{(e)}}_{4x4} {\mathbf{d}}^{(e)}_{4x1}$$

$$\mathbf{\tilde f}^{(e)}=k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\mathbf{\tilde d}^{(e)}$$ Note: the zeroes arise because there is no stretching in the transverse direction.

$$\mathbf{\tilde f}^{(e)}_{4x1}=\mathbf{\tilde k}^{(e)}_{4x4} \mathbf{\tilde d}^{(e)}_{4x1}$$

Note: Consider the case: $${\tilde{d}^{(e)}}_{4}\neq 0$$ $${\tilde{d}^{(e)}}_{1}={\tilde{d}^{(e)}}_{2}={\tilde{d}^{(e)}}_{3}=0$$ $$\mathbf{\tilde f}^{(e)}_{4x1}=\mathbf{\tilde k}^{(e)}_{4x4} \mathbf{\tilde d}^{(e)}_{4x1}=\mathbf{0}$$ Here, $$\mathbf{0}_{4x1}$$ corresponds to the 4th column of $$\mathbf{\tilde k}^{(e)}$$ This is the interpolation of transverse degrees of freedom.

From our notes on p.19-3, we recall that:

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

We use the same argument for:

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

Also:

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

Which leads to:

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

If $${\tilde{\mathbf{T}}^{(e)}}$$ is invertible, then:

$$[{\tilde{\mathbf{T}}^{(e)^{-1}}}{\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}}]{\mathbf{d}}^{(e)}={\mathbf{f}}^{(e)}$$

$${\tilde{\mathbf{T}}^{(e)}}$$ is a block diagonal matrix, as you can see by referring back to your notes on p. 19-3.

Consider a general block diagonal matrix:

Q: What is $$\mathbf{A}^{-1}$$?

Simple example:

$$\mathbf{B}=Diag[d_{11},d_{22},...,d_{nn}]$$

So,
$$\mathbf{B}^{-1}=Diag[\frac{1}{d_{11}},\frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$ Assuming $$d_{ii}\neq 0$$ for i=1,...,n.

For a block diagonal matrix A:

$$\mathbf{A}=Diag[\mathbf{D}_{1},\mathbf{D}_{2},...,\mathbf{D}_{s}]$$

So,
$$\mathbf{A}^{-1}=Diag[\mathbf{D}_{1}^{-1},\mathbf{D}_{2}^{-1},...,\mathbf{D}_{s}^{-1}]$$

$$\tilde{\mathbf{T}}^{(e)^{-1}} = Diag[\mathbf{R}^{(e)^{-1}} ,\mathbf{R}^{(e)^{-1}}  ]$$ And from p19-2,
$$\tilde{\mathbf{R}}^{(e)^{T}}=\begin{bmatrix}l^{(e)} &-m^{(e)} \\ m^{(e)} &l^{(e)} \end{bmatrix}$$

And,
$$\tilde{\mathbf{R}}^{(e)^{T}}_{2x2}\tilde{\mathbf{R}}^{(e)}_{2x2}=\begin{bmatrix}l^{(e)} &-m^{(e)} \\ m^{(e)} &l^{(e)} \end{bmatrix}\begin{bmatrix}l^{(e)} &m^{(e)} \\ -m^{(e)} &l^{(e)} \end{bmatrix}$$ Thus:$$\tilde{\mathbf{R}}^{(e)^{T}}_{2x2}\tilde{\mathbf{R}}^{(e)}_{2x2}=\begin{bmatrix}l^{(e)^{2}} m^{(e)^{2}} &0 \\ 0 &l^{(e)^{2}} m^{(e)^{2}} \end{bmatrix}$$

Which simplifies to:
$$\tilde{\mathbf{R}}^{(e)^{T}}_{2x2}\tilde{\mathbf{R}}^{(e)}_{2x2}=\begin{bmatrix}1&0 \\ 0 &1\end{bmatrix}$$

Which gives us $$\mathbf{I}_{2x2}$$, an identity matrix.

We can see that:
$$ \mathbf{R}^{(e)^{-1}} = \tilde{\mathbf{R}}^{(e)^{T}}$$

$$\tilde{\mathbf{T}}^{(e)^{-1}} = Diag[\mathbf{R}^{(e)^{T}} ,\mathbf{R}^{(e)^{T}} ]$$

And then,
$$\tilde{\mathbf{T}}^{(e)^{-1}} = (Diag[\mathbf{R}^{(e)} ,\mathbf{R}^{(e)} ])^{T}$$

And then we get:

$$\tilde{\mathbf{T}}^{(e)^{-1}} = \tilde{\mathbf{T}}^{(e)^{T}}$$

Going back to our notes on p.20-1: $${\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}} {\mathbf{d}}^{(e)}= {\tilde{\mathbf{T}}^{(e)}} {\mathbf{f}}^{(e)}$$ which is equal to :$$[{\tilde{\mathbf{T}}^{(e)^{T}}}{\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}}]{\mathbf{d}}^{(e)}={\mathbf{f}}^{(e)}$$

and comparing it with $${\tilde{\mathbf{k}}^{(e)}} {\tilde{\mathbf{d}}^{(e)}} ={\tilde{\mathbf{f}}^{(e)}}$$,

we finally get:

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

The roles of Eigenvectors and Eigenvalues
Eigenvalue problem: $$\mathbf{kv} = \lambda \mathbf{v}$$

Let $$\left\{\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}, \mathbf{u_4} \right\}$$ be the pure eigenvectors corresponding to the 4 zero eigenvalues:

$$\mathbf{ku_i} = 0\mathbf{u_i} = \mathbf{0}\left(i = 1,2,3,4 \right)$$

Linear combination of $$\left\{u_i, i = 1,2,3,4 \right\} $$

 = \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W}
$$ \sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \equiv \mathbf{W} $$ (equal by definition)

$$ \alpha_i \Rightarrow $$(real numbers)

W is also an eigenvector corresponding to a zero eigenvalue:

$$ \mathbf{kW} = \mathbf{k}\left(\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \right) $$

$$= \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W}$$

Justification of assembly of element stiffness matrix $$\left(\mathbf{k^{(e)}}, e = 1,..., nel \right)\Rightarrow nel =$$ number of element

into global stiffness matrix $$\left(\mathbf{K}\right)$$

Consider example of 2-bar truss, (Mtg. 9):

Recall elemental FD relation: $$\mathbf{k}^{(e)}_{4\times 4}\mathbf{d}^{(e)}_{4\times 1} = \mathbf{f}^{(e)}_{4\times 1}$$

Mtg. 11: Euler cut principle, Method 2 (equilibrium of global node 2)

Mtg. 4: Free Body Diagrams of element 1 and element 2 with element degrees of freedom $$\mathbf{d}^{(e)}_{4\times 1}$$

Mtg. 8: For node 2, identify the relationships between the global degrees of freedom and the element degrees of freedom for elements 1 and 2

Equilibrium of node 2:

 Free Body Diagram Using Euler Cut Principle

$$(1) \sum{F_x} = 0 = -f^{(1)}_3 - f^{(2)}_1 = 0$$

$$(2) \sum{F_y} = 0 = P - f^{(1)}_4 - f^{(2)}_2 = 0$$

Next we use the FD relation: $$\mathbf{k}^{(e)}_{4\times 4}\mathbf{d}^{(e)}_{4\times 1} = \mathbf{f}^{(e)}_{4\times 1}$$

$$\mathbf{Eq. (1) \ }\Rightarrow f^{(1)}_3 + f^{(2)}_1 = 0  \mathbf{\left( 1\right)}$$

$$\mathbf{Eq. (2) \ }\Rightarrow f^{(1)}_4 + f^{(2)}_2 = P \mathbf{\left( 2\right)}$$

$$\mathbf{(1):}\left[k^{(1)}_{31}d^{(1)}_1 + k^{(1)}_{32}d^{(1)}_2 + ... + k^{(1)}_{34}d^{(1)}_4\right] \equiv f^{(1)}_3$$
$$+ \left[k^{(1)}_{11}d^{(2)}_1 + ... + k^{(2)}_{14}d^{(2)}_4\right] \equiv f^{(2)}_1 = 0$$

Mtg. 8: Rewrite local degrees of freedom to global degrees of freedom:

$$\mathbf{\left(1 \right):} \left[k^{(1)}_{31}d_1 + k^{(1)}_{32}d_2 + k^{(1)}_{33}d_3 + k^{(1)}_{34}d_4\right]$$

$$+ \left[k^{(2)}_{11}d_3 + k^{(2)}_{12}d_4 + k^{(2)}_{13}d_5 + k^{(2)}_{14}d_6\right] = 0$$

$$\Rightarrow $$ Thus obtain the 3rd row of $$\mathbf{k}$$ (Mtg. 9) (See also Mtg. 8)

Assembly of $$\mathbf{k^{(e)}}, e = 1,...,nel $$, into global stiffness matrix $$\mathbf{K}$$:

$$\mathbf{K_{n\times n}= }A^{nel}_{e=1}\mathbf{k^{(e)}_{ned\times ned}}$$

A: assembly operator
n: total number of global dof's before eliminating boundary conditions
ned: number of element dof's
(ned<<n)

Principle of Virtual Work (PVW)
Mtg. 10: Elimination of rows corresponding to the boundary conditions to obtain $$\mathbf{\overline{k}_{2\times 2}} $$

Mtg. 12: $$\mathbf{q^{(e)}_{2\times 1}=T^{(e)}_{2\times 4}d^{(e)}_{4\times 1}}$$

Mtg. 14: $$\mathbf{k^{(e)}=T^{(e)^T}\hat{k}^{(e)}T^{(e)}}$$

Deriving FEM for partial differential equations (PDE's):

FD relation for a bar or a spring: $$kd = F$$

$$\Rightarrow kd - F = 0\;\;\;\;\; (3)$$

$$\Leftrightarrow W(kd - F) = 0\;\;\;\left(for \, all\, W\, \right) \;\;\;\;\;(4)$$

W: weighting coefficient
"weak form" $$\equiv $$ PVW

 Proof :

$$A) \; \left(3 \right)\Rightarrow (4): trivial$$

$$B) \; \left(4 \right)\Rightarrow (3): \mathbf{NOT\; TRIVIAL}$$

Since $$\left(4 \right)$$ is valid for all $$W$$, select $$W=1$$ Then $$\left(4 \right)$$ becomes:

$$1\times (kd - F) = 0 \Rightarrow \left(3 \right)$$

Other Resources
ADINA FEA Eigen Frequency Models and Animations Eigenvalues and Eigenvectors on Wikipedia

Contributors
Eml4500.f08.wiki1.schaet 20:09, 24 October 2008 (UTC) Eml4500.f08.wiki1.aguilar 19:03, 23 October 2008 (UTC) Eml4500.f08.wiki1.oatley 20:37, 21 October 2008 (UTC) Eml4500.f08.wiki1.ambrosio 20:40, 23 October 2008 (UTC) Eml4500.f08.wiki1.brannon 01:52, 24 October 2008 (UTC) Eml4500.f08.wiki1.handy 21:27, 23 October 2008 (UTC)

title: User:Eml4500.f08.wiki1.aguilar/TeamHW4


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Eas4200c.f08.ZYX.O'Mahoney 18:30, 23 October 2008 (UTC)

Circle Airfoil

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User:Eml4500.f08.gravy.mmm/ECM

Euler cut principle

Method 1 is the normal statics method to sovle for 2 bar truss. Method 2 is the Euler cut principle method to solve for 2 bar truss.

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User:Eml4500.f08.gravy.jad/Notes4.g

FEA Notes and Homework 4

Solving Truss systems in Matlab
Connectivity Array

Here we are using a connectivity array "conn" (from 2 bar truss system)

$$\ conn = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \,$$

Where the columns represent the local node number, but the numbers inside the matrix state the global node number. The rows tell the element number. (ie. row 1 represents the first element, row 2 represents the second element, ect.)

conn(e,j)= global node # of local node j of element e.

Location Matrix Master Array

Here we are using a location matrix master array "lmm" (from 2 bar truss system)

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

Where the columns represent the local dof's, but the numbers inside the matrix state the global dof's or equation number in matrix k. The rows tell the element number. (ie. row 1 represents the first element, row 2 represents the second element, ect.)

lmm(i,j)= eq # (global dof's #) for the elem. stiffness coeff. corresponding to the jth local dof #.

Solving Truss systems in the $$\tilde{x}$$ and $$\tilde{y}$$ coordinate axis
Back to Derivation of element Force Displacement with respect to global coordinate system: Method 2 to desire k(e)4x4 : transform a system with 4 dofs into a system with also 4 dofs (instead of only 2 dofs) so that the transformation matrix is now 4x4 and inverted. We want a system of equations that follows a coordinate system which correlates to the axial direction of the bar.

Goal : Want to find $$\ T^{(e)}_{4x4} \,$$ that transform the set of local (elem) dofs $$\ d^{(e)}_{4x4} \,$$ to another set of local (elem) dofs $$\ \underline{\tilde{d}}^{(e)}_{4x4} \,$$ such that $$\ \underline{\tilde{T}}^{(e)} \,$$  is invertible.

$$\ \underline{\tilde{d}}^{(e)}_{4x1} = \underline{\tilde{T}}^{(e)}_{4x4} \tilde{d}^{(e)}_{4x1} \,$$

Continuing the example from Force Displacement matrices calculations:

$$\ \tilde{d}^{(e)}_{1} = \left \lfloor \ l^{e} m^{e} \right \rfloor \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$  where $$\ \underline{\tilde{ d }}^{(e)}_{1} \,$$ also equals $$\ q^{(e)}_{1} \,$$ (Equation 1)

$$\ \tilde{d}^{(e)}_{2} = \tilde{d}^{(e)}_{1} = \vec{d}^{(e)}_{1} \cdot \vec{\tilde{j}} \,$$ (component of $$\ \tilde{d}^{(e)}_{1} \,$$ along $$\ \vec{\tilde{j}} \,$$ j, ie. $$\ \tilde{y}-axis \,$$) (of the local nodes)

$$\ = - \sin\theta^{(e)} d^{(e)}_{1} + \cos \theta^{(e)} d^{(e)}_{2} \,$$

$$\ = \left \lfloor \ -m^{e} l^{e} \right \rfloor \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$ (Equation 2)

Deriving the $$\tilde{d}^{(e)}$$ equation
Put (Equation 1) and (Equation 2) in matrix form:

$$\ \begin{Bmatrix} \tilde{d}^{(e)}_{1} \\ \tilde{d}^{(e)}_{2} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$

where $$\ \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \,$$

A complete matrix form of all dofs

$$\ \underbrace{ \begin{Bmatrix} \tilde{d}^{(e)}_{1} \\ \tilde{d}^{(e)}_{2} \\ \tilde{d}^{(e)}_{3} \\ \tilde{d}^{(e)}_{4} \end{Bmatrix} }_{\underline{\tilde{d}}^{(e)}_{4x1}} = \underbrace{ \begin{bmatrix} \underline{R}^{(e)}_{2x2} & \underline{0}_{2x2} \\ \underline{0}_{2x2} & \underline{R}^{(e)}_{2x2} \end{bmatrix} }_{\underline{\tilde{T}}^{(e)}_{4x4}} \underbrace{ \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \\d^{(e)}_{3} \\d^{(e)}_{4} \end{Bmatrix} }_{\underline{d}^{(e)}_{4x1}} \,$$

$$\ \underline {\tilde{f}}^{(e)}_{4x1}= k^{(e)} \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \underline{\tilde{d}}^{(e)} \,$$

In conclusion:

$$\ \underline{\tilde{f}}^{(e)}_{4x1} = \underline{\tilde{k}}^{(e)}_{4x4} \underline{\tilde{d}}^{(e)}_{4x1} \,$$

Special Case

$$\ \tilde{d}^{(e)}_{4} \ne 0 \,$$, $$\ \tilde{d}^{(e)}_{1} =\tilde{d}^{(e)}_{2} =\tilde{d}^{(e)}_{3} = 0 \,$$

Therefore: $$\ \underline{\tilde{f}}^{(e)}_{4x1} = \underline{\tilde{k}}^{(e)}_{4x4} \underline{\tilde{d}}^{(e)}_{4x1} = \underline{0}_{4x1} \,$$ where the zero matrix is the 4th column of $$\ \underline{\tilde{k}}^{(e)} \,$$

Now interpret the transverse of the dofs: $$\ \underline{\tilde{d}}^{(e)} = \underline{\tilde{T}}^{(e)} \tilde{d}^{(e)} \,$$

Similarly: $$\ \underline{\tilde{f}}^{(e)} = \underline{\tilde{T}}^{(e)} \tilde{f}^{(e)} \,$$

Prove of $$ \mathbf{\tilde{f}}^{(e)}$$ equation
Also: $$\ \underline{\tilde{k}}^{(e)} \underline{\tilde{d}}^{(e)} = \tilde{f}^{(e)} \Rightarrow \underline{\tilde{k}}^{(e)} \underline{\tilde{T}}^{(e)} {d}^{(e)} \,$$

Hence, if $$\ \underline{\tilde{T}}^{(e)} \,$$ is invertible then: $$\ \left [ {\underline{\tilde{T}}^{(e)}}^{-1} \underline{\tilde{k}}^{(e)} \underline{\tilde{T}}^{(e)} \right ] \underline{\tilde{d}}^{(e)} = \underline{f}^{(e)} \,$$

$$\ \underline{\tilde{T}}^{(e)} \,$$ is a block diagonal matrix (from earlier in this report).

Now consider the general block-diagonal matrix: $$\ A = \begin{bmatrix} \underline{D}_{1} & & \underline{0} \\ & \ddots &  \\ \underline{0} &  & \underline{D}_{s} \end{bmatrix} \,$$

Question: Then what is $$\ A^{-1} \,$$?

Answer: Lets make a simpler case- $$\ B = \begin{bmatrix} d_{11} & & & \underline{0} \\ & d_{22} & &  \\ & & \ddots & \\ \underline{0} &  & & D_{s} \end{bmatrix} = Diag \left [ d_{11},d_{22}, \ldots, d_{nn} \right ] \,$$

$$\ B^{-1}= Diag \left [ \frac{1}{d_{11}},\frac{1}{d_{22}}, \ldots, \frac{1}{d_{nn}} \right ] \,$$ assuming that $$\ d_{ij} \ne 0 \,$$ for $$\ i=1 \ldots n \,$$.

$$\ B \cdot B^{-1} = I \,$$

Now going back to the A-matrix:

$$\ A= Diag \left [ \underline{D}_{1},\underline{D}_{2}, \ldots, \underline{D}_{s} \right ] \,$$

$$\ A^{-1}= Diag \left [ \frac{1}{\underline{D}_{1}},\frac{1}{\underline{D}_{2}}, \ldots, \frac{1}{\underline{D}_{s}} \right ] = Diag \left [ \underline{D}^{-1}_{1},\underline{D}^{-1}_{2}, \ldots, \underline{D}^{-1}_{s} \right ] \,$$

Therefore: $$\ {\underline{\tilde{T}}^{(e)}}^{-1} = Diag \left [ {\underline{R}^{(e)}}^{-1}, {\underline{R}^{(e)}}^{-1} \right ] \,$$

Recall from earlier in this report $$\ \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \,$$

Now: $$\ {\underline{R}^{(e)}}^{T} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix} \,$$

Therefore: $$\ {\underline{R}^{(e)}}^{T} \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix}_{2x2} \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix}_{2x2} = \begin{bmatrix} {l^{(e)}}^{2} + {m^{(e)}}^{2} & 0 \\ 0 & {l^{(e)}}^{2} + {m^{(e)}}^{2} \end{bmatrix} \,$$

since $$\ {l^{(e)}}^{2} + {m^{(e)}}^{2} = 1 \,$$, then $$\ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_{2x2} \,$$ the Identity matrix.

In conclusion: $$\ \Rightarrow {R^{(e)}}^{-1} = {R^{(e)}}^{T} \,$$ and $$\ \Rightarrow {\underline{\tilde{T}}^{(e)}}^{-1} = {\underline{\tilde{T}}^{(e)}}^{T} \,$$

Prove of $$ \mathbf{k}^{(e)}$$ equation
Verify that the equation: $$ \mathbf k^{(e)} = \tilde \mathbf T^{(e)^{T}}\tilde \mathbf k^{(e)}\tilde \mathbf T^{(e)}$$ is true.

Solving for the eigenvalues of the system
Closing the loop, infinitesimal displacement, eigenvalues problems, zero eigenvalues, rigid body mechanisms.

Solving for the eigenvalues of a 2 bar truss system using Matlab
With the Equation

$$\ \underline{k} = \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}_{6x6}\,$$

$$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{49}{16} & \frac{3\sqrt{3} - 40}{16} & -\frac{5}{2} & \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3} - 40}{16} & \frac{43}{16} & \frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} \\ 0 & 0 & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$

The following Matlab code was created

To evaluate the problem $$\  \mathbf{kv} = \boldsymbol{\lambda}  \mathbf{v}\,$$

Let {u1,u2,u3,u4} be the pure eigenvalues corresponding to four zero eigenvalues.

$$\ \mathbf{k}_{6x6} \mathbf{u}_i =0 \cdot \mathbf{u}_i \,$$, i=1,2,3,4

where $$ 0 \cdot u_i $$ is the zero vector of size 6x1.

Linear combination of {u_i, i=1,2,3,4}

$$ \sum_{i=1}^4 \alpha_i u_i =: \mathbf{W}_{6x1}$$

=: means equal by definition and $$\alpha$$ is a real number so it is 1x1 and u_1 is 6x1.

W is also an eigenvalue vector corresponding to a zero eigenvalue:

$$\ \mathbf{KW}= \mathbf{K}( \sum_{i=1}^4 \boldsymbol{\alpha}_i \mathbf{W}_i ) =  \sum_{i=1}^4 \boldsymbol{\alpha}_i (\mathbf{Ku}_i) \,$$ where $$(\mathbf{Ku}_i)$$ is a zero matriz of size 6x6. $$\ =\mathbf{0}_{6x1} = 0 \cdot \mathbf{W}_{6x1} \,$$

Solving for the eigenvalues of square bar truss system (a) using Matlab

$$\ \mathbf{kv} = \boldsymbol{\lambda}  \,$$

a=b=1, E=2,  A=3

Assembling the global stiffness matrix
Justification of assembly of element stiffness matrix $$k^{(e)} e=1, number of elements$$ into global stiffness matrix K

consider example of 2 bar truss from previous lecture.

Recall element force displacement relation

$$\ k^{(e)}_{4x4}d^{(e)}_{4x1}=f^{(e)}_{4x1} \,$$

from previous lecture: Euler cut principal method 2 (equilibrium of global node 2) Euler cut principal method

from previous lecture: Free body diagram of element 1 and element 2 element degrees of freedom d(e) Trusses, matrix method

from previous lecture: for node 2, identify global degrees of freedom to element degrees of freedom for both element 1 and 2. Global_F-D_Relationship

$$\ \sum F_x= 0-f_3^{(1)}- f_1^{(2)}= 0 \,$$  (Equation 1)

$$ \sum F_y= P-f_4^{(1)}- f_2^{(2)}= 0$$  (Equation 2)

Next use element Force Displacement relation $$\mathbf{k}^{(e)} \mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

Equation 1 becomes $$\ f_3^{(1)} + f_1^{(2)}=0 \,$$

Equation 2 becomes $$\ f_4^{(1)} + f_2^{(2)}=P \,$$

So substituting for the f's in equation 1 gives

$$\ \underbrace{[ k_{31}^{(1)} d_{1}^{(1)} + k_{32}^{(1)} d_{2}^{(1)} + k_{33}^{(1)} d_{3}^{(1)} + k_{34}^{(1)} d_{4}^{(1)} ]}_{f_3^{(1)}} + \underbrace{[k_{11}^{(2)} d_{1}^{(2)} + k_{12}^{(2)} d_{2}^{(2)} + k_{13}^{(2)} d_{3}^{(2)}+ k_{14}^{(2)} d_{4}^{(2)}]}_{f_1^{(2)}}=0 \,$$

Replacing the local d's with the global d's as explained in Global_F-D_Relationship You end up with

$$[ k_{31}^{(1)} d_{1} + k_{32}^{(1)} d_{2} + k_{33}^{(1)} d_{3} + k_{34}^{(1)} d_{4} ]

+ [k_{11}^{(2)} d_{3} + k_{12}^{(2)} d_{4} + k_{13}^{(2)} d_{5}+ k_{14}^{(2)} d_{6}  ]=0 $$

Which is the third row of the $$\ \underline{K} \underline{d} = \underline{F} \,$$ equation.

Combine like d's, and convert to resulting equation into a multiplication of matrices to form:

$$\ \begin{bmatrix} K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)}   \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix} = 0 \,$$

Derivation of equation 2 and the assembly of row 4 of the global stiffness matrix
The assembly of K(e) e=1 to the number of elements, into the global stiffness matrix K can be represented as

$$\ K_{nxn}=Ak^{(e)}_{mxm}\,$$

Where A is the Assembly operator, n is the total number of global degrees of freedom, m is the number of element degrees of freedom and m is much less than n. Which is the principal of virtual work insert link here

Elimination of rows corresponding to the boundary conditions lets you obtain $$\mathbf{\bar{q}}_{2x2}$$

Remember that:

$$\ \mathbf{q}^{(e)}_{2x1} = \mathbf{T}^{(e)}_{2x4} d^{(e)} _{4x1} \,$$

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

Deriving FEM of Partial differential Equations

Force Displacement relation of a bar or spring: Kd=F implies that Kd-F=0  (Equation 3)

which is equivalent to w(Kd-F)=0 for all w  (Equation 4)

Equation 4 is a "weak form" which is Principal of virtual work

Prove:

A) Equation 3 implies equation 4 which is a trivial conclusion

B) Equation 4 implies equation 3 which is not a trivial conclusion

Since equation 4 is valid for all w, first select w=1 and equation 4 becomes 1(Kd-F)=0 which is equation 3

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 15:43, 24 October 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 23:19, 23 October 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 02:32, 24 October 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 02:37, 24 October 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 03:47, 24 October 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 04:32, 24 October 2008 (UTC)

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User:Eas4200c.f08.WIKI.E/hw4

 Friday, October 17

Normal strains

$$\varepsilon _{xx}=\frac{\sigma _{xx}}{E}-\frac{\nu \sigma _{yy}}{E}-\frac{\nu\sigma _{zz} }{E} $$

$$\varepsilon _{yy}=-\frac{\nu\sigma _{xx}}{E}+\frac{ \sigma _{yy}}{E}-\frac{\nu\sigma _{zz} }{E} $$

$$\varepsilon _{zz}=-\frac{\nu\sigma _{xx}}{E}-\frac{\nu \sigma _{yy}}{E}+\frac{\sigma _{zz} }{E} $$

Shear Strains

$$\gamma _{xy} =2 \zeta _{xy}= \frac{\tau_{xy}}{G}=\frac{\sigma_{xy}}{G}$$

$$\gamma _{yz} =2 \zeta _{yz}= \frac{\tau_{yz}}{G}=\frac{\sigma_{yz}}{G}$$

$$\gamma _{zx} =2 \zeta _{zx}= \frac{\tau_{zx}}{G}=\frac{\sigma_{zx}}{G}$$

Unroll this matrix along the diagonal and then up and around counterclockwise.

$$\epsilon = \begin{vmatrix} \varepsilon_ {11} & \varepsilon_{12} & \varepsilon _{13}\\ \varepsilon _{21} &\varepsilon _{22} &\varepsilon _{23} \\ \varepsilon _{31} &\varepsilon _{32} &\varepsilon _{33} \end{vmatrix}$$

Due to the symmetry, you may rewrite this unrolled matrix into a $$6_x 1$$ with a Voigt Notation:

$$ \left\{\varepsilon _{ij} \right\}_{6x1}= \begin{Bmatrix} \varepsilon _{11}\\ \varepsilon _{22}\\ \varepsilon _{33}\\ \varepsilon _{23}\\ \varepsilon _{13}\\ \varepsilon _{12} \end{Bmatrix}$$

$$\left\{\sigma _{ij} \right\}_{6x1}= \begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{13}\\ \sigma _{12} \end{Bmatrix}$$

Hooke's Law for isotropic elasticity:

$$\begin{Bmatrix} \varepsilon _{11}\\ \varepsilon _{22}\\ \varepsilon _{33}\\ \gamma _{23}\\ \gamma _{13}\\ \gamma _{12} \end{Bmatrix}=

\begin{vmatrix} \frac{1}{E}& \frac{-\nu }{E} & \frac{-\nu }{E} & 0 &0 &0 \\ \frac{-\nu }{E}& \frac{1}{E} &\frac{-\nu }{E}  & 0 & 0 & 0\\ \frac{-\nu }{E} & \frac{-\nu }{E} & \frac{1}{E} & 0 &0  &0 \\ 0& 0 & 0 & \frac{1}{G} & 0 & 0\\ 0& 0 & 0 & 0 & \frac{1}{G} &0 \\ 0& 0 & 0 & 0 & 0 & \frac{1}{G} \end{vmatrix} \begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{13}\\ \sigma _{12} \end{Bmatrix}$$

It can also be written as

$$\begin{Bmatrix} \varepsilon _{11}\\ \varepsilon _{22}\\ \varepsilon _{33}\\ \varepsilon _{23}\\ \varepsilon _{13}\\ \varepsilon _{12} \end{Bmatrix}=

\begin{vmatrix} \frac{1}{E}& \frac{-\nu }{E} & \frac{-\nu }{E} & 0 &0 &0 \\ \frac{-\nu }{E}& \frac{1}{E} &\frac{-\nu }{E}  & 0 & 0 & 0\\ \frac{-\nu }{E} & \frac{-\nu }{E} & \frac{1}{E} & 0 &0  &0 \\ 0& 0 & 0 & \frac{1}{2G} & 0 & 0\\ 0& 0 & 0 & 0 & \frac{1}{2G} &0 \\ 0& 0 & 0 & 0 & 0 & \frac{1}{2G} \end{vmatrix} \begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{13}\\ \sigma _{12} \end{Bmatrix}$$

Poisson's ratio values for different materials
Poisson's Ratio table []

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User:Eml4500.f08.Lulz.js/HW4

Array Definitions
Consider the connectivity array "conn" for the two-bar truss system shown in Figure ##:

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

The connectivity array describes the local nodes of a truss element in terms of the global node numbering convention. The rows correspond to individual truss elements, and the columns to local nodes 1 and 2 of those elements. In other words, conn(e,j) refers to the global node number of local node j for element e.

A related concept is the location master matrix lmm:

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

The location master matrix describes the local degrees of freedom for a truss element in terms of the global degrees of freedom. Like with the connectivity array, the rows correspond to individual truss elements, and the columns refer to local degrees of freedom 1-4 for those elements. In other words, lmm(i,j) refers to the global degree of freedom of local degree of freedom j of element i.

Transformation Matrix
A previous discussion centered on the inability to invert the transformation matrix $$\mathbf{T^{(e)}}$$ used in the axial displacement relationship $$\mathbf{q^{(e)}} = \mathbf{T^{(e)}}\cdot\mathbf{d^{(e)}}$$. Our goal is to convert the 2x4 transformation matrix into a 4x4 matrix $$\mathbf{\tilde{T}^{(e)}}$$ that can be easily inverted.

Figure ## shows a general diagram of axial/transvere displacements $$\tilde{d}^{(e)}_{j}$$ on a truss element e. The relationship describing this case is $$\mathbf{\tilde{d}^{(e)}} = \mathbf{\tilde{T}^{(e)}}\cdot\mathbf{d^{(e)}}$$.

Through the same geometric process used to determine $$\mathbf{q^{(e)}}$$ in a previous homework, it can be shown that the two expressions below are true:

$$\tilde{d}^{(e)}_{1} = \begin{bmatrix} l^{(e)}&m^{(e)} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2} \end{Bmatrix}$$

$$\tilde{d}^{(e)}_{2} = \begin{bmatrix} -m^{(e)}&l^{(e)} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2} \end{Bmatrix}$$

Combining these two expressions yields the following relationship for the axial/transverse displacements at one local node:

$$\begin{Bmatrix} \tilde{d}^{(e)}_{1}\\\tilde{d}^{(e)}_{2}\end{Bmatrix} = \begin{bmatrix} l^{(e)}&m^{(e)}\\-m^{(e)}&l^{(e)}\end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2} \end{Bmatrix}$$

This describes a new displacement relationship given by $$\mathbf{\tilde{d}^{(e)}} = \mathbf{R^{(e)}}\cdot\mathbf{d^{(e)}}$$. Extending this to include both local nodes of a truss element yields:

$$\begin{Bmatrix} \tilde{d}^{(e)}_{1}\\\tilde{d}^{(e)}_{2}\\\tilde{d}^{(e)}_{3}\\\tilde{d}^{(e)}_{4}\end{Bmatrix} = \begin{bmatrix} l^{(e)}&m^{(e)}&0&0\\-m^{(e)}&l^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\\0&0&-m^{(e)}&l^{(e)}\end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix} = \begin{bmatrix} \mathbf{R^{(e)}}&\mathbf{0}\\\mathbf{0}&\mathbf{R^{(e)}}\end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix}$$

Figure ## shows the axial/transverse forces for a truss element. The axial/transverse force-displacement relationship is given by $$\mathbf{\tilde{f}^{(e)}} = \mathbf{\tilde{k}^{(e)}}\cdot\mathbf{\tilde{d^{(e)}}}$$. The modified element stiffness matrix $$\mathbf{\tilde{k}^{(e)}}$$is defined below:

$$\mathbf{\tilde{k}^{(e)}} = k^{(e)}\begin{bmatrix} 1&0&-1&0\\0&0&0&0\\-1&0&1&0\\0&0&0&0\end{bmatrix}$$

Consider the case where $$\tilde{d}^{(e)}_{1} = \tilde{d}^{(e)}_{2} = \tilde{d}^{(e)}_{3}= 0$$, but $$\tilde{d}^{(e)}_{4}\neq 0$$. Because the 4th column of matrix $$\mathbf{\tilde{k}^{(e)}}$$ (corresponding to transverse displacement $$\tilde{d}^{(e)}_{4}$$) is a vector of zeros, the following is true:

$$\mathbf{\tilde{f}^{(e)}} = 0$$.

This is the interpretation of transverse degrees of freedom.

Finally, we have determined that $$\mathbf{\tilde{d}^{(e)}} = \mathbf{\tilde{T}^{(e)}}\cdot\mathbf{d^{(e)}}$$ is valid. A corresponding relationship can be defined for the axial/transverse forces:

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

We also know that $$\mathbf{\tilde{f}^{(e)}} = \mathbf{\tilde{k}^{(e)}}\cdot\mathbf{\tilde{d^{(e)}}}$$ is valid, leading to the following logic:

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

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

Assuming that $$\mathbf{\tilde{T}^{(e)}}$$ is an invertible matrix, the following is true:

$$[\mathbf{\tilde{T}^{(e)^{-1}}}\cdot\mathbf{\tilde{k}^{(e)}}\cdot\mathbf{\tilde{T}^{(e)}}]\mathbf{d^{(e)}} = \mathbf{f^{(e)}}$$

Inversion of Diagonal Matrices
Consider a general $$n\;x\;n$$ block diagonal matrix $$\mathbf{A}$$:

In order to explore the inverse, first consider an example matrix $$\mathbf{B}$$:

This matrix can alternately be described using the following notation for diagonal matrices:

$$\mathbf{B} = Diag[d_{11},d_{22},...,d_{nn}]$$

If $$d_{ii} \neq 0$$ for $$i\,=\,1,...,n$$, the inverse of this matrix can be expressed as $$\mathbf{B^{-1}} = Diag[\frac{1}{d_{11}},\frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$. Applying this procedure to $$\mathbf{A}$$ yields the following:

$$\mathbf{A} = Diag[\mathbf{D_{1}},\mathbf{D_{2}},...,\mathbf{D_{n}}]$$

$$\mathbf{A^{-1}} = Diag[\mathbf{D_{1}^{-1}},\mathbf{D_{2}^{-1}},...,\mathbf{D_{n}^{-1}}]$$

Transformation Matrix Continued
Returning to the modified transformation matrix $$\mathbf{\tilde{T}^{(e)}}$$ and applying the inversion methods discussed above, the matrix $$\mathbf{\tilde{T}^{(e)^{-1}}}$$ can be written as follows:

$$\mathbf{\tilde{T}^{(e)^{-1}}} = Diag[\mathbf{R^{(e)^{-1}}},\mathbf{R^{(e)^{-1}}}]$$, where $$\mathbf{R^{(e)}} = \begin{bmatrix} l^{(e)}&-m^{(e)}\\m^{(e)}&l^{(e)}\end{bmatrix}$$

Additionally, the transpose of matrix $$\mathbf{R^{(e)}}$$ is $$\begin{bmatrix} l^{(e)}&m^{(e)}\\-m^{(e)}&l^{(e)}\end{bmatrix}$$, leading to the following relation:

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

Recalling that $$l^{(e)} = \cos{\mathbf{\theta}^{(e)}}$$ and $$m^{(e)} = \sin{\mathbf{\theta}^{(e)}}$$, and the trigonometric identity $$\cos{^{2}\mathbf{\theta}} + \sin{^{2}\mathbf{\theta}} = 1$$, the following is true:

$$\mathbf{R^{(e)^{T}}}\cdot\mathbf{R^{(e)}} = \begin{bmatrix} 1&0\\0&1\end{bmatrix} = \mathbf{I}$$, where $$\mathbf{I}$$ is the 2 x 2 identity matrix.

Since it is a known property that the product of a matrix and it's inverse (e.g. $$\mathbf{M}\cdot\mathbf{M^{-1}} = \mathbf{I}$$), it is apparent that $$\mathbf{R^{(e)^{T}}} = \mathbf{R^{(e)^{-1}}}$$. Thus,

$$\mathbf{\tilde{T}^{(e)^{-1}}} = Diag[\mathbf{R^{(e)^{-1}}},\mathbf{R^{(e)^{-1}}}] = Diag[\mathbf{R^{(e)^{T}}},\mathbf{R^{(e)^{T}}}] = (Diag[\mathbf{R^{(e)}},\mathbf{R^{(e)}}])^{T} = \mathbf{\tilde{T}^{(e)^{T}}}$$

The expression for the modified force-displacement relationship derived earlier can now be rewritten as:

$$[\mathbf{\tilde{T}^{(e)^{T}}}\cdot\mathbf{\tilde{k}^{(e)}}\cdot\mathbf{\tilde{T}^{(e)}}]\mathbf{d^{(e)}} = \mathbf{f^{(e)}}$$

Not only is the expression inside the brackets easily determined, it also resembles the form of the general force-displacement relationship, and the following can be shown:

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

Discussion of Eigenvalues and Eigenvectors
Consider the eigenvalue problem $$\mathbf{kv} = \mathbf{\lambda v}$$. Let $$[\mathbf{u_{1}},\mathbf{u_{2}},\mathbf{u_{3}},\mathbf{u_{4}}]$$ be the eigenvectors for the 4 zero eigenvalues:

$$\mathbf{ku_{i}} = 0\cdot\mathbf{u_{i}} = \mathbf{0}$$, for i = [1,2,3,4].

A linear combination of {$$\mathbf{u_{i}}$$} can be defined as $$\sum_{i=1}^{4} \alpha_{i}\mathbf{u_{i}} = \mathbf{W}$$, where $$\alpha_{i}\;$$ refers to real number constants.

The vector $$\mathbf{W}$$ is also an eigenvector corresponding a zero eigenvalue, as shown below:

$$\mathbf{kW} = \mathbf{k}(\sum_{i=1}^{4} \alpha_{i}\mathbf{u_{i}}) = \sum_{i=1}^{4} \alpha_{i}\mathbf{ku_{i}} = \mathbf{0}$$

Justification of Assembly of Element Stiffness Matrices into Global Stiffness Matrix
ERIC YOUR STUFF GOES HERE

Figure ## shows the generic two-bar truss system, with global node 2 isolated for analysis. Two equations are immediately evident from the diagram:

$$(1)\;\Sigma F_{x} = -f_{3}^{(1)} - f_{1}^{(2)} = 0$$

$$(2)\;\Sigma F_{y} = \mathbf{P} - f_{4}^{(1)} - f_{2}^{(2)} = 0$$

Equation 2 can also be written as $$\mathbf{P} = f_{4}^{(1)} + f_{2}^{(2)}$$ Using the general force-displacement relationship $$\mathbf{k^{(e)}}\cdot\mathbf{d^{(e)}}= \mathbf{f^{(e)}}$$, Equation 1 can be rewritten as follows:

$$0 = [k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{32}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)}] + [k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)}]$$

Replacing the local displacement degrees of freedom with the global degrees of freedom produces the following:

$$0 = [k_{31}^{(1)}d_{1} + k_{32}^{(1)}d_{2} + k_{32}^{(1)}d_{3} + k_{34}^{(1)}d_{4}] + [k_{11}^{(2)}d_{3} + k_{12}^{(2)}d_{4} + k_{13}^{(2)}d_{5} + k_{14}^{(2)}d_{6}]$$

Comparison of the above with the third row of $$\mathbf{K}$$ as described in Meeting 8 shows that the two are identical.

An expression for the assembly of $$ned\;x\;ned$$ element stiffness matrices $$\mathbf{k^{(e)}}$$ for $$e\;=\; 1,...,nel$$, where $$nel$$ refers to the number of elements, into the $$n\;x\;n$$ global stiffness matrix $$\mathbf{K}$$ can be expressed as follows:

$$ \mathbf{K} = \begin{matrix} nel\\A\\e=1\end{matrix}\mathbf{k^{(e)}}$$

$$n\;$$ is the number of global degrees of freedom before the application of boundary conditions. $$ned\;$$ is the number of element degrees of freedom. $$ned\;$$ must be less than $$n\;$$. $$A\;$$ is a symbol representing the operation of assembly.

Principle of Virtual Work (PVW)
PVW was previously used as the validation for elimination of rows in the global stiffness matrix that corresponded to zero-valued boundary conditions to produce $$\mathbf{\bar{k}}$$. It was also used to derive $$\mathbf{q^{(e)}} = \mathbf{T^{(e)}} \cdot \mathbf{d^{(e)}}$$, the axial displacement relationship, and also $$\mathbf{k^{(e)}} = \mathbf{T^{(e)^{T}}} \cdot \mathbf{\hat{k}^{(e)}} \cdot \mathbf{T^{(e)}}$$, the alternate expression for the element stiffness matrix based on the transformation matrix.

Deriving FEM for PDEs
The force-displacement relationship for a bar of spring is given by $$\mathbf{kd}=\mathbf{F}$$. This relationship implies the following:

$$(3)\;\mathbf{kd} - \mathbf{F} = 0$$

An equivalent expression to Equation 3 is also implied:

$$(4)\;\mathbf{W}(\mathbf{kd} - \mathbf{F}) = 0$$ for all $$\mathbf{W}$$.

Proof:

$$(3)\leftrightarrow(4)$$ -  trivial $$(4)\leftrightarrow(3)$$ -  NOT TRIVIAL

Since $$(4)\;$$ is valid for all $$\mathbf{W}$$ selecting $$\mathbf{W}\,=\,1$$ is an acceptable operation.

$$(4)\;$$ then becomes $$\mathbf{kd} - \mathbf{F} = 0$$, which is the same as $$(3)\;$$.

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User:Eml4500.f08.bottle.brockmiller/hw4

Plotting eigenvectors corresponding to zero eigenvectors for a 3 bar truss system
Assume the following case of a 3 bar truss system:

Where a=b=1, and the three bar properties consist of; E=2 and A=3. This means for all elements, $$\ k=\frac{(EA)}{L}=\frac{(2*3)}{1}=6 $$

The system can be described as $$\ \mathbf{K} * \mathbf{v}= \mathbf{\lambda\ }* \mathbf{v} $$

Where $$\ \mathbf{K} $$ is the 6x6 stiffness matrix for the contrained system.

title: User:Eml4500.f08.bottle.brockmiller/hw4


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title: File:Hw4 multicell section.jpg


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User:Filiberto/monobook.css

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User:Eml4500.f08.RAMROD.F/HW4

 Wednesday October 15

Justification of assembly of element stiffness matrices into global stiffness matrix K. $$K^{(e)} \; \; e=1, ... ,n_{el}\;$$

Consider the 2 bar truss example. Recall Force displacement equation.

$$ K^{(e)}_{4x4} \cdot d^{(e)}_{4x1} = F^{(e)}_{4x1}$$

You then can use the Euler cut method to establish the equilibrium of global node 2

Using the free body diagram of element one and of element 2 you can describe the elements degrees of freedom $$d^{(e)}_2 $$

$$d_2=d_2^{(1)}$$

$$d_3=d_3^{(1)}=d_1^{(2)} $$

$$d_4=d_4^{(1)}=d_2^{(3)}$$

$$d_5=d_3^{(2)}$$

$$d_6=d_4^{(2)}$$

Equilibrium of node 2

$$\sum{F_x}=-f_3^{(1)}-F_1^{(2)}=0$$

$$\sum{F_y}=P-f_4^{(1)}-F_2^{(2)}=0$$

Then you would use element force displacement relation $$K^{(e)}_{4x4} \cdot d^{(e)}_{4x1} = F^{(e)}_{4x1}$$

title: User:Eml4500.f08.RAMROD.F/HW4