text:
Engineering Experience 4: Design a Small Solar Vehicle/Nl/Team PM3

teamleden
Dieter Denruyter: big boss en sankey-deskundige Jeroen Bertels: research en development Jens De Klerck: research en development Hendrik Geysen: hoofd elektronica Hannah Daled: directeur afdeling transmissie Matthias Kellens: chief sankey Alexander Croes: mechanicien

De website
http://www.everyoneweb.com/soulcar/ Sinds kort heeft ons team ook een website! Er wordt nog aan gesleuteld, maar als je meer wilt weten over ons project, zijn we ook daar te volgen en zelfs te contacteren als je vragen hebt.

De Ganttchart
http://upload.wikimedia.org/wikipedia/commons/2/2c/Gantt_Chart_EE4_invulbestand.pdf Op deze link kan je volgen wie wat gedaan heeft, en wat gaat doen. Wekelijkse update!

Plan Van Aanpak
http://upload.wikimedia.org/wikipedia/commons/c/c8/Plan_Van_Aanpak_EE4.pdf

Zo gaan we dat aanpakken!

WBS
http://upload.wikimedia.org/wikipedia/commons/f/f0/WBS_1.0.jpg

De structuur afgebroken

Samenwerkingscontract
http://upload.wikimedia.org/wikipedia/commons/d/dc/Samenwerkingscontract_EE4.pdf

We zijn officieel een team!

week 1
De eerste keer dat we elkaar zien. Spannend! Eerst een naamspelletje, wat praten en eerste ideeën worden op tafel gesmeten. We hebben elkaars emailadressen. Tijd om erin te vliegen!

week 2
O jee! we zijn al een teamlid kwijt... Off to "elektronica"! Wij laten ons niet doen en gaan gestaag verder. Het lichtpunt van de week: we kregen ons zonnepaneel en motor! Dit project krijgt vorm... Tijd om ideeën in praktijk te brengen (nog even wachten!!!)

1/03
Iedereen gaat met veel goesting van start: Alexander focust zich op de wiki. Hendrik, Jeroen en Jens denken na over de strategie die we gaan kiezen. Hannah, Dieter en Matthias zijn de zware denkers en buigen zich over de berekeningen.

2/03
We denken dan toch voor een sturing te kiezen. Nog wat opzoeken en overwegen maar het idee zit vastgebakken in onze geesten. Ook qua vormgeving denken we niet puur voor efficientie te gaan. WIJ WILLEN NE MOOIE! Ook het materiaal is zo goed al gekozen. Als basis gaan we voor hout.

week 4
Dinsdag is EE dag! we hadden om half 4 een afspraak met onze begeleider. de gantt chart is definitief goedgekeurd. Ook de berekeningen gemaakt door Hannah en Dieter voor de overbrengingsverhouding was heel goed! Enkel werd geen rekening gehouden met het effect van de lamp: de zon is 3 X sterker! dus stroom ook 3 X. De Besturing is ook grotendeels uitgewerkt. Donderdag beslissen we definitief over de vorm van onze wagen.

week 5
Twee van onze teamleden konden door onfortuinlijke omstandigheden de vergadering van vandaag niet bijwonen. Vandaag werd echter beslist om het chassis van de wagen vast te leggen. Het team heeft besloten dat het onderstel van ons zonnewagentje gaat lijken op een formule 1 wagen uit de jaren 1950. Het zonnepaneel zal dan voor de cockpit bevestigt worden op het voertuig onder de vorm van een vleugel.,Binnenkort komen er de eerste foto's! Jens en jeroen zijn hard aan het werken aan simulink. Stap voor stap komen ze tot de beste benadering voor onze zonnewagen. Hannah werkt gestaag verder aan de overbrengingsverhouding en steekt een volledig verslag ineen die haar berekeningen uitleggen. Ook Dieter en Matthias zijn goed bezig met het Sanky diagram. Alle verliezen worden nauwkeurig berekend en in rekening gebracht. hier is dan de eerste foto! De romp is zo goed als klaar en ook de wielen zijn in ontwikkeling. http://upload.wikimedia.org/wikipedia/commons/5/55/DSC_0028_%281%29.JPG

week 6
Deze week moest de deadline gehaald worden. Dus een laatste sprint was aangewezen. Dinsdag werd er stevig verder gerekend. Jeroen en Jens kregen de eerste resultaten te zien in de simulaties. Ook werd er gezocht naar tandwielen. Op donderdag kwam de voorlopige zonnewagen mee naar groepT. Zo bleek bij zwaarder uit te vallen als verwacht... Hij weegt nu zonder motor en sturing 1,2 kg. Dit moest natuurlijk aangepast worden in de berekeningen! Zo kregen we een overbrengingsverhouding van 1:16. Nu kunnen we definitief naar de juiste tandwielen gaan zoeken. http://upload.wikimedia.org/wikipedia/commons/9/96/ZonnewagenPM3.JPG Zo zag de zonnewagen er dinsdag uit. Ondertussen heeft hij al een bevestigingsmethode voor het zonnepaneel.

week 7
Vandaag punten gehad! SSV 1 kreeg een 18/20 en simulink een 16/20!! BRAVO souljah's!

week 8
Na de grote overwinning van vorige week, is het tijd om de draad terug op te pikken. We hebben kennis gemaakt met fablab en ons logo laten laseren. Sankey werd verder geanalyseerd en sturing/overbrenging zijn zo goed als uitgedacht. Nu nog de uitvoering! Maar we zijn er zeker van dat we na de paasvakantie zullen kunnen testen.

week 9
Er moest niet veel meer gedaan worden, maar het werk is essentieel. de tandwielen en sturing moeten er deze week inzitten en perfect werken. De steunen voor de assen van de tandwielen zorgen voor wat problemen. Ook de spelingen en bevestiging van de andere onderdelen bezitten nog wat foutjes. Tijd om ze op te lossen!

week 10
Dinsdag zonnewagenrace! Maandag nog hard gewerkt om hem af te krijgen... Maar het heeft zijn nut gehad. Snel waren we niet, maar hij reed vlotjes tot boven en stuurde om alle obstakels heen (of beter gezegd bleef mooi bovenop de rails). Ook zag hij er heel prachtig uit! We zetten onze hoop op de schoonheidsprijs! Hier volgt alvast een link naar een online album. geeft een klein sfeerbeeld van de race en de competitie: http://www.flickr.com/photos/62598368@N02/sets/72157626672679786/


text:
User talk:Petré

 Hello Petré, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 16:38, 22 February 2011 (UTC)

Files Missing License Information
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MaintenanceBot (discuss • contribs) 03:51, 11 November 2013 (UTC)

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

title: User talk:Petré


text:
User talk:Colapeninsula

 Hello Colapeninsula, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 16:40, 22 February 2011 (UTC)

title: User talk:Colapeninsula


text:
User talk:Ottava Rima

Welcome to my user talk page. Material from before June 20, 2010 can be found here, after here.

School of Mathematics
Hi, you deleted School of Mathematics but failed to update all the references to it. Now on the main page there's a link to non-existent article: http://en.wikibooks.org/wiki/Wikiversity

Custodianship
Can you make me a custodian? ---Draubb
 * Sorry, but I am not even a custodian myself since the project was hijacked a few years ago. Ottava Rima (talk) 03:42, 18 May 2013 (UTC)

Abd
How is Abd, is he blocked or something, I heard you blocked Abd for multiple accounts. I am schocked because he warned me not to create multiple accounts. What's wrong with Abd? Please give me an answer on my discussion page. See ya around! Draubb (discuss • contribs)
 * Abd is a proponent of the fringe theory of Cold Fusion, and will create multiple accounts and do whatever he can to wage war against 99.9999% of the world who realizes that his ideas are nonsense. Ottava Rima (talk) 23:27, 18 May 2013 (UTC)

Why do you hate him so much? Please answer on my talk page Draubb (discuss • contribs) 00:36, 19 May 2013 (UTC)

Matter concerning you at []. - Sidelight12 Talk 01:23, 10 September 2014 (UTC)

Contributions
Hello Ottava Rima, Can you please send me a message to my talk page or Candidates for Custodianship/Draubb. Can you please check on my contributions section and please leave a message on one of the two pages and say if my contributions are helping or not and explaining what I can do? Thank you very much Ottava Rima! --Draubb (discuss • contribs) 22:49, 27 May 2013 (UTC)

Protected status of FAQ/Editing
Please consider unprotecting FAQ/Editing, which was protected since 2010.

Alternatively, please correct FAQ/Editing: in "?action=purge" the last character is now not bold, unlike the rest of the text.

Nyq (discuss • contribs) 05:22, 14 August 2013 (UTC)

Requests for comment/Global ban for Ottava Rima
Per the Global bans global policy, you are informed of the discussion above. Please comment there and feel free to appropriately distribute more widely in prominent community venues in order to «Inform the community on all wikis where the user has edited». Nemo 10:10, 24 August 2013 (UTC)

Question about blocked user
Hello, sorry to bother you about this. Abd asked about a banned user markoril, who was blocked about two years ago. I don't know why this still matters. Can you clarify what happened then, with that blocked user? You can see the inquiry on my talkpage. Thanks. - Sidelight12 Talk 21:34, 4 October 2013 (UTC)
 * . The guy was a long term sock puppet associated with the site that he spammed, and he did it under multiple names across the wikimedia sites. Clear vandal, and something that Abd has encouraged in the past. Ottava Rima (talk) 22:14, 4 October 2013 (UTC)

 * Alright then. Thanks. I actually saw that page. So there's clear evidence that, that user used other names? Ok then, I'll take your word for it. The Wikipedia custodian also made that statement in the block explanation. - Sidelight12 Talk 22:30, 4 October 2013 (UTC)

Welcome back!
Good to see you back Ottava, you want to do a little work around here? --Goldenburg111 (talk|contribs) 20:16, 22 December 2013 (UTC)
 * Which topic? And I never really left. I always watch the recent contribs. Ottava Rima (talk) 20:17, 22 December 2013 (UTC)
 * Ottava, you may do any topic you desire to. --Goldenburg111 (talk|contribs) 20:23, 22 December 2013 (UTC)
 * Oh, I know that. I just thought you might have had one in particular you needed help with. After the desysopping, I stopped producing content and limited my role. Too many games were played and I have a real life to focus on. Ottava Rima (talk) 20:29, 22 December 2013 (UTC)
 * I figured you were watching this site without necessarily logging in. Greetings. - Sidelight12 Talk 20:24, 22 December 2013 (UTC)
 * Hola. Ottava Rima (talk) 20:29, 22 December 2013 (UTC)
 * Ottava, you said Abd's ideas are nonsense. Don't you find that insulting to Abd??? --Goldenburg111 (talk|contribs) 20:31, 22 December 2013 (UTC)
 * He was banned from Wikipedia for promoting what was revealed to be absolutely false claims about cold fusion. Dealing with Abd's obsession with fringe topics, with people who appropriately prohibited him based on his actions, and from the trouble he caused here is what I feel justifies my comments. One of the reasons why I don't care to bother too much with producing content like I did is his activity. My material is part of my academic research, is mainstream, and connected to my reputation. Ottava Rima (talk) 20:35, 22 December 2013 (UTC)

Enydmion Society
Please use a real account rather than anonymous IP, and tag all of whatever you want deleted. Day-by-day requests are just hiding the scope of your intent and giving the impression that we should reinitiate RFD discussion Archive 9. -- Dave Braunschweig (discuss • contribs) 00:30, 27 July 2019 (UTC)
 * The Endymion Society page was originally created as part of a step by step process on how to build a book with the pages. The book was already built and so the original instruction classroom page became redundant. I've been going slowly through them to make sure everything was copied over and included. Ottava Rima (talk) 00:37, 27 July 2019 (UTC)
 * All of the files here and the general page  can be deleted due to no longer being active and no longer pertinent. There is only one author, and the fragmentary nature of it makes it uneducational now that the project is long done. Ottava Rima (talk)  00:46, 27 July 2019 (UTC)
 * As a side note - this level of vandalism is partly why leaving up the unneeded material is a problem. No one is monitoring it. Ottava Rima (talk)  01:06, 27 July 2019 (UTC)

title: User talk:Ottava Rima


text:
Template talk:Engineering-list

Electro-Mechanical Engineering? This is a technologist field of engineering, but I do not see it listed here.

title: Template talk:Engineering-list


text:
User talk:Farmdawg

 Hello Farmdawg, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 19:19, 22 February 2011 (UTC)

title: User talk:Farmdawg


text:
User talk:Rainesilva

 Hello Rainesilva, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 19:47, 22 February 2011 (UTC)

Are you taking the CPACE?

title: User talk:Rainesilva


title: File:Nm1.s11.mtg22.djvu


title: File:Nm1.s11.mtg23.djvu


text:
User:Fem.tm7/HW4

Problem Description
Re do Problem 2.9, fixing $$\mathbf{\phi=\pi/2}$$ $$\Rightarrow \mathbf{F} = \left \{ 1, sinx, sin2x, sin3x, \cdot \cdot\cdot\cdot \right \}$$

For n=2
PART 1 From G1D1M.1.0/D2 $$P(u)=0 = \frac{\partial }{\partial x}\left[ 2\frac{\partial u}{\partial x}\right] + 3$$ therefore the exact solution is $$u(x)=\frac{-3}{4}x^2-4x+\frac{19}{4} $$ $$ d=\left \{ d_0, d_1, d_2 \right \}$$ $$ b=\left \{ 1,sinx,sin2x \right \}$$ $$ u^h(x) = \sum_{j=0}^2{d_jb_j(x)}=d_0 + d_{2}sin(2x) + d_{1}sin(x)$$ $$ (u^h)' =\frac{d{u}^{h}}{dx}(x) = \sum_{j=0}^2{d_jb_j'(x)}=d_{1}cos(x) + 2d_{2}cos(2x)$$ PART 2 The boundary conditions are enforced to generate two equations as follows: $$\displaystyle u^h(x=1) =0=d_{0} + 0.8414d_{1} + 0.9092d_{2}$$ $$\displaystyle (u^h)'(x=0) =-4=d_{1} + 2d_{2}$$  PART 3  An additional equation is generated by projected the residue onto a basis function as follows: $$\int_{0}^{1}b_{k}(x)P(u^h)dx=-0.9193d_1 - 5.664d_2=-3$$ PART 4  The equations are assembled into stiffness coefficient and force matrices in the form $$\mathbf{Kd}=\mathbf{f}$$:

$$ \begin{bmatrix} 1.0000 &   0.8415 &   0.9093\\         0   & 1.0000  &  2.0000\\         0  & -0.9194  & -5.6646   \end{bmatrix} \begin{bmatrix} d_0\\ d_1\\ d_2 \end{bmatrix} = \begin{bmatrix} 0\\ -4\\ -3 \end{bmatrix} $$ Note that the stiffness matrix, $$\mathbf{K}$$, is not symmetric PART 5  The coefficients are generated by solving the matrices above yielding:

$$ d=\begin{pmatrix} 4.7162\\  -7.4908\\    1.7454 \end{pmatrix} $$

For n=4
$$ d=\begin{pmatrix} 4.7491\\ -13.1144\\    8.2087\\   -3.6116\\    0.8829 \end{pmatrix} $$

For n=6
$$ d=\begin{pmatrix} 4.7500\\ -257.2135\\ 341.5393\\ -261.6484\\  128.7013\\  -38.5824\\    5.5312 \end{pmatrix} $$

Author
Eml5526.s11.team7.Srilalithkumar 05:05, 2 March 2011 (UTC)

Problem Description
“A First Course in Finite Elements” by Fish & Belytschko P73 Problem 3.5

Obtain the weak form of the equations of heat conduction with the following boundary conditions: The essential boundary condition defined at $$x=0$$ $$T(0) = 100$$ and the natural boundary condition defined at $$x=10$$ $$\bar q(10) = hT$$. The natural boundary condition is a convection condition.

Solution
The strong form of the heat problem is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac{d}(AK\frac) + s = 0

$$     (4.2.1)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

T(0) = 100

$$     (4.2.2)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\bar q(10) = hT

$$     (4.2.3) So the weak form is obtained as follows:
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int_0^{10} {w(\frac{d}(AK\frac) + s)} dx = 0

$$     (4.2.4)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\Rightarrow wAK\frac|_0^{10} - \int_0^{10} {\fracAK\fracdx + \int_0^{10} {wsdx} } = 0

$$     (4.2.5)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\Rightarrow w(10)AK\frac - w(0)AK\frac - \int_0^{10} {\fracAK\fracdx + \int_0^{10} {wsdx} } = 0

$$     (4.2.6)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\Rightarrow w(10)AhT - 0 - \int_0^{10} {\fracAK\fracdx + \int_0^{10} {wsdx} } = 0\;\;at\;\;w(0) = 0

$$     (4.2.7)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\Rightarrow w(10)AhT + \int_0^{10} {wsdx} = \int_0^{10} {\fracAK\fracdx}\;\;\;\; \forall w\;|\;w(0) = 0

$$     (4.2.8 )
 *
 * }

To sum up, the weak form of the heat problem is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

w(10)AhT + \int_0^{10} {wsdx} = \int_0^{10} {\fracAK\fracdx}\;\;\;\; \forall w\;|\;w(0) = 0

$$     (4.2.9)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

T(0) = 100

$$     (4.2.10)
 *

 * }

Author
Zongyi Yang

Brandon Hua

Problem Description

 * [From Fish & Belytschko, Chapter 3, Problem 3.6]

Given the strong form for the heat conduction problem in a circular plate:

 * $$k \frac{d}{dr}\left( r \frac{dT}{dr}\right)+rs=0 \qquad \text{on} \qquad 0 < r \leq R.$$

 * natural boundary condition: $$\displaystyle \frac{dT}{dr}(r=0) = 0,$$

 * essential boundary condition: $$\displaystyle T(r=R) = 0,$$

 * where R is the total radius of the plate, s is the heat source per unit length along the plate radius, T is the temperature and k is the conductivity. Assume that k, s, and R are given.

Find

 * 1) Construct the weak form for the above strong form.
 * 2) Use quadratic trial (candidate) solutions of the form $$\displaystyle T = \alpha_0 + \alpha_1 r + \alpha_2 r^2$$ and weight functions of the same form to obtain a solution of the weak form.
 * 3) Solve the differential equation with the boundary conditions and show that the temperature distribution along the radius is given by
 * $$ \displaystyle T = \frac{s}{4k} (R^2 - r^2)$$.

Part 1
First, the governing equation and natural boundary condition are multiplied by the weight function, $$\displaystyle w(r) $$, and integrated over the domain ($$ 0 < r \leq R$$). The restriction on the weight function is that $$\displaystyle w(R)=0$$.

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

$$ \int_{0}^{R}w(r) \left [ k \frac{d}{dr} \left ( r \frac{dT}{dr} \right ) + rs \right ] dr = 0 \qquad \text{and} \qquad w(r) \frac{dT}{dr} = 0.$$ (4.3.1)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The first equation can be expanded as

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

$$ \int_{0}^{R}w(r) k \frac{d}{dr} \left ( r \frac{dT}{dr} \right ) dr + \int_{0}^{R}w(r) r \ s \ dr = 0 \qquad \forall w.$$ (4.3.2)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The following integration by parts is simplified if the substitution $$\displaystyle f = r \frac{dT}{dr}$$ is made. Also note that the weight function will subsequently be referred to as $$\displaystyle w$$ with the dependence on the spatial domain, $$\displaystyle r $$, being implied. Equation 4.3.2 can therefore be written as,

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

$$ \int_{0}^{R}w k \frac{df}{dr} dr + \int_{0}^{R}wrs \ dr = 0.$$ (4.3.3)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Integration by parts yields,

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

$$ \int_{0}^{R} \frac{d}{dr} (wkf)dr - \int_{0}^{R} fk \frac{dw}{dr} dr + \int_{0}^{R}wrs \ dr = 0$$ (4.3.4)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

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

$$ wkf|_{0}^{R} - \int_{0}^{R} kf \frac{dw}{dr} dr + \int_{0}^{R}wrs \ dr = 0.$$ (4.3.5)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The substitution in equation 4.3.3, $$\displaystyle f = r \frac{dT}{dr}$$, can now be reintroduced, and equation 4.3.5 becomes

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

$$ \left( wkr \frac{dT}{dr} \right)_{r=R} - \left( wkr \frac{dT}{dr} \right)_{r=0} - \int_{0}^{R} kr \frac {dT}{dr} \frac{dw}{dr} dr + \int_{0}^{R}wrs \ dr = 0.$$ (4.3.6)
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The first term reduces to zero due to the prescribed weight function at $$\displaystyle r=R $$. In other words, $$\displaystyle w(R)=0 $$. The second term also reduces to zero since the natural boundary condition defines the heat flux at $$\displaystyle r=0 $$ to equal zero. Therefore, the weak form of the heat conduction problem becomes

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$$ \int_{0}^{R} \frac{dw}{dr} kr \frac {dT}{dr} dr = \int_{0}^{R}wrs \ dr \qquad \forall w \quad \text{with} \quad w(R)=0.$$ (4.3.7)
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Part 2
Consider the candidate and weight solutions of the following forms:

 * Candidate: $$\displaystyle T(r) = \alpha_0 + \alpha_1 r + \alpha_2 r^2$$,
 * Weight: $$\displaystyle w(r) = \beta_0 +\beta_1 r + \beta_2 r^2 $$.

For the candidate solution to be admissible, it must satisfy the essential boundary condition T(R) = 0. Likewise, the weight function must vanish at r = R. Implementing these conditions yields expressions with respect to the current and total radii of the plate.

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$$ \displaystyle T(R) = \alpha_0 + \alpha_1 R + \alpha_2 R^2 = 0 \qquad \Rightarrow \qquad \alpha_0 = - \alpha_1 R - \alpha_2 R^2 $$     (4.3.8)
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$$ \displaystyle w(R) = \beta_0 +\beta_1 R + \beta_2 R^2 = 0 \qquad \Rightarrow \qquad \beta_0 = - \beta_1 R - \beta_2 R^2 $$     (4.3.9)
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 * }

Substituting these expressions for $$\displaystyle \alpha_0$$ and $$\displaystyle \beta_0$$ back into the candidate solution and weight function,

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$$ \displaystyle T(r) = \alpha_1 \left( r-R \right) + \alpha_2 \left( r^2 - R^2 \right) $$     (4.3.10)
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$$ \displaystyle w(r) = \beta_1 \left( r-R \right) + \beta_2 \left( r^2 - R^2 \right) $$     (4.3.11)
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The derivatives of equations 4.3.10 and 4.3.11

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$$ \frac{dT}{dr} = \alpha_1 + 2 \alpha_2 r,$$ (4.3.12)
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$$ \frac{dw}{dr} = \beta_1 + 2 \beta_2 r. $$ (4.3.13)
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Equations 4.3.10 through 4.3.13 are substituted into the weak form, equation 4.3.7:

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$$ \int_{0}^{R} \left( \beta_1 + 2 \beta_2 r \right) kr \left( \alpha_1 + 2 \alpha_2 r \right) dr = \int_{0}^{R} \left[ \beta_1 (r-R) + \beta_2 (r^2 - R^2) \right] rs \ dr $$ (4.3.14)
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Integrating the both sides of the equation over the domain yields

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$$ \text{LHS:} \quad \int_{0}^{R} \left( \beta_1 + 2 \beta_2 r \right) kr \left( \alpha_1 + 2 \alpha_2 r \right) dr \quad = \quad \frac{2}{3} \alpha_1 \beta_2 k R^3 \ + \ \frac{1}{2} \alpha_1 \beta_1 k R^2 \ + \ \frac{2}{3} \alpha_2 \beta_1 k R^3 \ + \ \alpha_2 \beta_2 k R^4 $$     (4.3.15)
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$$ \text{RHS:} \quad \int_{0}^{R} \left[ \beta_1 (r-R) + \beta_2 (r^2 - R^2) \right] rs \ dr \quad = \quad - \frac{1}{4} \beta_2 s R^4 \ - \ \frac{1}{6} \beta_1 s R^3 $$     (4.3.16)
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Equations 4.3.15 and 4.3.16 are substituted into 4.3.14, and the entire equation is rewritten to collect the terms $$\displaystyle \beta_1$$ and $$\displaystyle \beta_2$$.

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$$ \beta_1 \left( \frac{1}{2} \alpha_1 k R^2 + \frac{2}{3} \alpha_2 k R^3 + \frac{1}{6} s R^3 \right ) + \beta_2 \left( \frac{2}{3} \alpha_1 k R^3 + \alpha_2 k R^4 + \frac{1}{4} s R^4 \right) =0 $$     (4.3.17)
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The coefficients $$\displaystyle \alpha_1$$ and $$\displaystyle \alpha_2$$ can be determined in terms of the known components s, k, and R.

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$$ \displaystyle \begin{bmatrix} {\frac{1}{2}R} & { \frac{2}{3}R^3} \\ \\ {\frac{2}{3}R^3} & {R^4} \end{bmatrix} \cdot \begin{bmatrix} \alpha_1\\ \\ \alpha_2 \end{bmatrix} = \begin{bmatrix} {- \frac{1}{6}sR^3} \\ \\ {- \frac{1}{4}sR^4} \end{bmatrix}
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$$     (4.3.18)
 *

 * }

 * Therefore,

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$$ \alpha_1 = - \frac{ks}{12} \left( 2R^7 + R^5 \right) \qquad \text{and} \qquad \alpha_2 = - \frac{ks}{36} \left( 9R^8 + 4R^6 \right) $$     (4.3.19)
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Substituting the values of $$\displaystyle \alpha_1$$ and $$\displaystyle \alpha_2$$ into equation 4.3.10 yields the candidate solution in terms of known values. There is some concern regarding the validity of this equation, since the system units do not correspond to an appropriate unit measurement of temperature.

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

$$ T(r) = - \frac{ks}{12} \left( 2R^7 + R^5 \right)(r-R) - \frac{ks}{36} \left( 9R^8 + 4R^6 \right)(r^2 - R^2) $$     (4.3.20)
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The candidate (trial) solution above is substituted into the governing equation to confirm equilibrium.

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$$ k \frac{d}{dr}\left( r \frac{dT}{dr}\right)+rs=0 \qquad \Rightarrow \qquad -k \left[ \frac{ks(2R^7 + R^5)}{12} + \frac{rks(9R^8 + 4R^6)}{9} \right] +rs = 0 $$     (4.3.21)
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 * Solving for $$\displaystyle r$$ and simplifying, we are given an expression for when $$\displaystyle r$$ is valid for given values of k and R. Note that the validity of $$\displaystyle r$$ is independent of the heat source per unit length, s.

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$$ r = - \frac{6R^7 k^2 + 3R^5k^2}{36R^8k^2 + 16R^6k^2 - 36} $$     (4.3.22)
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Part 3
The differential - or governing - equation below will now be solved for $$\displaystyle T(r)$$.

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$$ k \frac{d}{dr}\left( r \frac{dT}{dr}\right)+rs=0 \qquad \text{on} \qquad 0 < r \leq R $$ (4.3.23)
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The equation is rewritten with $$\displaystyle rs$$ and $$\displaystyle k$$ moved to the right hand side. Then both sides can be integrated over $$\displaystyle r$$.

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

$$ \int \frac{d}{dr} \left( r \frac{dT}{dr} \right) dr = - \int \frac{s}{k} r dr $$ (4.3.24)
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 * }

This reduces to

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$$ r \frac{dT}{dr} = - \frac{s}{2k} r^2 + constant $$     (4.3.25)
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Recalling the natural boundary condition and applying it to equation 4.3.25,

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$$ \frac{dT}{dr}(0) = 0 \quad \Rightarrow \quad r \frac{dT}{dr} = - \frac{s}{2k} r^2 + constant $$     (4.3.26)
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$$ 0 = 0 + constant \qquad \text{therefore} \qquad r \frac{dT}{dr}  = - \frac{s}{2k} r^2. $$     (4.3.27)
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 * }

Rewrite as

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

$$ dT = - \frac{s}{2k} r dr \qquad \Rightarrow \qquad \int dT  = - \int \frac{s}{2k} r dr. $$ (4.3.28)
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 *

 * }

Integrate recalling the essential boundary condition, $$\displaystyle T(R) = 0$$.

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$$ T = - \frac{s}{4k} r^2 + constant \quad \Rightarrow \quad T(R) = - \frac{s}{4k} R^2 + constant = 0 $$     (4.3.29)
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 * }

 * $$\quad \Rightarrow constant = \frac{s}{4k} R^2$$

This confirms that the temperature distribution along the radius is given by

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$$ T(r) = \frac{s}{4k} \left( R^2 - r^2 \right). $$     (4.3.30)
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Authors
Philip Flater Braden Snook

Problem Description
HW 4.4 p.21.1 Problem 3.7 from "A First Course in Finite Elements" by Fish and Belytschko

Given the strong form for the circular bar in torsion:
 * {| style="width:100%" border="0"

$$ \displaystyle \frac{d}{dx}\left ( JG\frac{d\phi }{dx} \right )+m=0, 0\leq x\leq l $$ (4.4.1)
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Essential Boundary Condition:
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$$ \displaystyle \phi \left ( x=0 \right )=\overline{\phi } $$       (4.4.2)
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Natural Boundary Condition:
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$$ \displaystyle M\left ( x=l \right )=\left ( JG\frac{d\phi }{dx} \right )_{l}=\overline{M} $$       (4.4.3)
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Where:

$$ \displaystyle m(x) $$ is a distributed moment per unit length $$ \displaystyle M $$ is the torsion moment $$ \displaystyle \phi $$ is the angle of rotation $$ \displaystyle G $$ is the shear modulus $$ \displaystyle C $$ is the radius of the circular shaft $$ \displaystyle J $$ is the polar moment of inertia given by the equation

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$$ \displaystyle J=\frac{\pi C^{4}}{2}$$ (4.4.4)
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a) Contruct the weak form for the circular bar in torsion. b) Assume that (m=0) and integrate the differential equation given above (4.4.1). Find the integration constants using the boundary conditions (4.4.2) and (4.4.3).

Hint: "I suggest to use this opportunity to solve the G1DM1.0 / D2 (at least a slightly modified version of it) as stated on p.9-2 of Mtg 9 (e); you just need to swap the boundary locations around to conform to the statement for HW 4.4, i.e., let x=0 be the location for the ess. b.c., and x=1 the location for the nat. b.c. So use the same data as stated in G1DM1.0 / D2, with the boundary locations swapped around; we can call this problem G1DM1.0 / D2b (data set 2b). This way, you can plot the exact solution, which can then be used to verify the results for future problems." -Prof Vu Quoc

Part a
Mulitply (4.4.1) by an arbitrary function $$ \displaystyle w(x) $$ and integrate over the domain $$ \displaystyle \Omega = ]0,l[ $$

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$$ \displaystyle \int_{0}^{l}w\left [ \frac{d}{dx}JG\frac{d\phi}{dx} \right ]dx+\int_{0}^{l}wMdx=0, \forall w $$ (4.4.5)
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$$ \displaystyle \frac{d}{dx}\left ( wf \right )=\frac{d(wf)}{dx}-f\frac{dw}{dx} $$       (4.4.6)
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$$ \displaystyle \int_{0}^{l}w\frac{dw}{dx}=\int_{0}^{l}\frac{d}{dx}(wf)dx-\int_{0}^{l}f\frac{dw}{dx}dx $$       (4.4.7)
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$$ \displaystyle \int_{0}^{l}w\frac{dw}{dx}=(wf)_{x=l}-(wf)_{x=0}-\int_{0}^{l}f\frac{dw}{dx}dx $$       (4.4.8)
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Let
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$$ \displaystyle f=JG\frac{d\phi}{dx} $$       (4.4.9)
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(4.4.5) becomes
 * {| style="width:100%" border="0"

$$ \displaystyle \left (wJG\frac{d\phi}{dx} \right )_{x=l}-\left (wJG\frac{d\phi}{dx}  \right )_{x=0}-\int_{0}^{l}\frac{dw}{dx}JG\frac{d\phi}{dx}dx-\int_{0}^{l}wmdx=0,\forall w, w(l)=0 $$       (4.4.10)
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$$ \displaystyle G\frac{d\phi}{dx}=\tau $$       (4.4.11)
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(4.4.10) becomes
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$$ \displaystyle \left (wJ \tau  \right )_{x=l}-\left (wJ \tau   \right )_{x=0}-\int_{0}^{l}\frac{dw}{dx}JG\frac{d\phi}{dx}dx-\int_{0}^{l}wmdx=0,\forall w $$ (4.4.10)
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Recall:

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$$ \displaystyle w(0)=0 $$       (4.4.12)
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Hint: "The problem stated on p.9-2 of Mtg 9 (e) corresponds to the G1DM1.0 / D1 (data set 1, in which the coefficient a_2 is a linear function of x), and NOT the G1DM1.0 / D2 (data set 2, in which the coefficient a_2 is a constant). To be precise, the problem that I referred to on p.9-2 of Mtg 9 (e) is stated in Eq.(3) on that page.  Hence, let's define G1DM1.0 / D1b (data set 1b, NOT 2b) as the problem when you swap the boundary conditions around, i.e., let x=0 be the location for the ess. b.c., and x=1 the location for the nat. b.c." -Prof Vu-Quoc From (3) 9-2

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$$ \displaystyle \frac{d}{dx}\left [ \left ( 2+3x \right ) \frac{du}{dx} \right ]+5x=0,\forall x\varepsilon ]0,1[ $$ $$ \displaystyle \Rightarrow JG=2+3x $$ $$ \displaystyle \Rightarrow m(x)=5x $$       (4.4.13)
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$$ \displaystyle u(0)=4 $$ $$ \displaystyle \Rightarrow \phi(0)=4 $$       (4.4.14)
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$$ \displaystyle \frac{du}{dx}(x=1)=-6  $$ $$ \displaystyle \Rightarrow \frac{d\phi}{dx}(x=1)=-6 $$       (4.4.15)
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Let
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$$ \displaystyle w(0)=0 $$ (4.4.10) becomes:
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$$ \displaystyle \left (w(2+3x)\frac{d\phi}{dx} \right )_{x=l}-\int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx-\int_{0}^{1}w(5x)dx=0,\forall w, w(0)=0 $$     (4.4.16)
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$$ \displaystyle w(2+3*1)(-6) -\int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx-\int_{0}^{1}w(5x)dx=0 $$     (4.4.17)
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$$ \displaystyle 30w -\int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx-\int_{0}^{1}w(5x)dx=0 $$     (4.4.18)
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$$ \displaystyle \int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx=30w -\int_{0}^{1}w(5x)dx $$     (4.4.19)
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$$ \displaystyle \int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx=30w(x) -\int_{0}^{1}w(x)(5x)dx $$       (4.4.20)
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Part b
From (4.4.23)
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$$ \displaystyle \frac{d}{dx}\left ( (2+3x)\frac{d\phi}{dx} \right )+5x=0 $$     (4.4.21)
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$$ \displaystyle \int d(2+3x)\frac{d\phi}{dx}+\int5xdx=0 $$
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$$ \displaystyle (2+3x)\frac{d\phi}{dx}+ \frac{5x^{2}}{2}+C_{1}=0 $$     (4.4.22)
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$$ \displaystyle \int (2+3x)d\phi= \frac{- \frac{5x^{2}}{2}+C_{1}}{(2+3x)} $$     (4.4.23)
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$$ \displaystyle \phi= \int - \frac{5x^{2}}{2(2+3x)}+\int \frac{C_{1}}{(2+3x)} $$     (4.4.24)
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From 2.4.4
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$$ \displaystyle \int \frac{x^2}{2+3x}dx=\frac{9x^{2}-12x+8log(2+3x)}{54} $$     (4.4.25)  (4.4.24) becomes:
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$$ \displaystyle \phi= -\frac{5}{2}\left (\frac{9x^{2}-12x+8log(2+3x)}{54} \right ) + \frac{C_{1}}{3} {log(2+3x)}+C_{2} $$       (4.4.26) Apply (4.4.15) to (4.4.22)
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$$ \displaystyle (2+3*1)*-6+ \frac{5*1^{2}}{2}+C_{1}=0$$ $$ \displaystyle C_{1}=-27.5 $$     (4.4.27)
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Apply (4.4.27) to (4.4.26)
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$$ \displaystyle \phi= -\frac{5}{2}\left (\frac{9x^{2}-12x+8log(2+3x)}{54} \right ) + \frac{-27.5}{3} {log(2+3x)}+C_{2} $$     (4.4.28)  Apply (4.4.14) to (4.4.28)
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$$ \displaystyle \phi(x=0)=4= -\frac{5}{2}\left (\frac{9*0^{2}-12*0+8log(2+3*0)}{54} \right ) + \frac{-27.5}{3} {log(2+3*0)}+C_{2} $$ $$ \displaystyle C_{2}=6.8709342 $$     (4.4.29)  The final solution is:
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$$ \displaystyle \phi= -\frac{5}{2}\left (\frac{9x^{2}-12x+8log(2+3x)}{54} \right ) + \frac{-27.5}{3} {log(2+3x)}+ 6.8709342 $$       (4.4.26)
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Author
Brandonhua 04:06, 2 March 2011 (UTC)

Problem Description
Possible basis functions are given as:

Polynomial: $${{F}_{P}}=\left\{ {{x}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Cosine: $${{F}_{C}}=\left\{ \cos jx,\text{ }j=0,1,2,\ldots \right\}$$

Sine: $${{F}_{S}}=\left\{ 1,\text{ }\sin jx,\text{ }j=1,2,\ldots \right\}$$

Fourier: $${{F}_{F}}=\left\{ \cos jx,\text{ }j=0,1,2,\ldots ;\text{ }\sin kx,\text{ }k=1,2,\ldots \right\}$$

Exponential: $${{F}_{E}}=\left\{ {{e}^{jx}},\text{ }j=0,1,2,\ldots \right\}$$

For $${{\Gamma }_{g}}=\left\{ \beta \right\}$$, and each of the above families of basis functions, i.e., $${{\Gamma }_{I}}=\left\{ {{b}_{j}}(x),\text{ }j=0,1,\ldots \right\},\text{ }I\in \left\{ P,C,S,F,E \right\}$$

Find
Find corresponding family $${{\bar{F}}_{I}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ satisfying constraint breaking solution (CBS), i.e., $${{b}_{0}}(\beta )\ne 0$$ and $${{b}_{i}}(\beta )=0,\text{ }i=1,2,\ldots ,n$$. Let $$\Omega =\left] \alpha ,\beta \right[=\left] -2,4 \right[$$.

1) $${{\bar{b}}_{0}}(x)=\text{const}$$, plot $${{b}_{j}}$$ and $${{\bar{b}}_{j}}$$ for $$j=1,2,3$$.

2) Show $$\left\{ {{e}^{jx}},\text{ }j=0,1,2 \right\}$$ linear independent on $$\Omega $$.

Part 1
The corresponding family $${{\bar{F}}_{I}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ satisfying CBS:

Polynomial:

$${{F}_{P}}=\left\{ {{\left( x-4 \right)}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Cosine:

$${{F}_{C}}=\left\{ 1,\text{ }\cos \left( j\left( x-4 \right)+\frac{\pi }{2} \right),\text{ }j=1,2,\ldots \right\}$$

Sine:

$${{F}_{S}}=\left\{ 1,\text{ }\sin \left( j\left( x-4 \right) \right),\text{ }j=1,2,\ldots \right\}$$

Fourier:

$${{F}_{F}}=\left\{ 1,\text{ }\cos \left( j\left( x-4 \right)+\frac{\pi }{2} \right),\text{ }j=1,2,\ldots ;\text{ }\sin \left( k\left( x-4 \right) \right),\text{ }k=1,2,\ldots \right\}$$

Exponential:

$${{F}_{E}}=\left\{ 1,\text{ }{{e}^{j\left( x-4 \right)}}-1,\text{ }j=1,2,\ldots \right\}$$

The following figures show the polynomial basis functions (j=1,2,3):

The following figures show the cosine basis functions (j=1,2,3):

The following figures show the sine basis functions (j=1,2,3):

The following figures show the fourier basis functions (j=1,2,3):

The following figures show the exponential basis functions (j=1,2,3):

Part 2
The exponential basis function is $${{b}_{j}}=\left\{ {{e}^{jx}},\text{ }j=0,1,2 \right\}$$,

Thus the Gram matrix is:

$$\Gamma =\left[ \begin{matrix} 6 & {{e}^{4}}-{{e}^{-2}} & \frac{1}{2}\left( {{e}^{8}}-{{e}^{-4}} \right) \\ {{e}^{4}}-{{e}^{-2}} & \frac{1}{2}\left( {{e}^{8}}-{{e}^{-4}} \right) & \frac{1}{3}\left( {{e}^{12}}-{{e}^{-6}} \right) \\ \frac{1}{2}\left( {{e}^{8}}-{{e}^{-4}} \right) & \frac{1}{3}\left( {{e}^{12}}-{{e}^{-6}} \right) & \frac{1}{4}\left( {{e}^{16}}-{{e}^{-8}} \right) \\ \end{matrix} \right]$$

And the determinant of Gram matrix is:

$$\det \left( \Gamma \right)=1.1145\times {{10}^{9}}\ne 0$$

Thus, the exponential basis function is linear independent on $$\Omega $$.

Author
Jiang Jin

Problem Description
An elastic bar is subjected along its length to a variable distributed spring p(x) as seen in the figure below. An axial force proportional to the displacement is imposed by the spring on the bar. Adapted from "A First Course in Finite Elements" by Fish and Belytschko p. 74 pb.3.9

a. Construct the strong form.

b. Construct the weak form.

Given
This bar is of length l, cross-sectional area A(x), Young's modulus E(x) with body force b(x) and with boundary conditions:

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$$\displaystyle u(0)=0$$
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$$\displaystyle E(l)A(l)\frac{du(l)}{dx}=t$$
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As seen in the figure above.

Solution
Assumptions:
 * The bar must be in equilibrium
 * It satisfies Hooke's law
 * Compatible Displacement Field
 * The Strain-Displacement Equation must be Satisfied

a. Construct the Strong Form
Following a similar solution method as Team 2 used here with Eq. 1.1-1.10. Using the following diagram from Team 7 homework 1 to start creating the strong form:

Performing a force balance on the rod using the free body diagram above, the net force is equal to the internal forces plus the external, or applied, forces. Note, in this specific problem we will designate our body force as b(x+dx/2) (instead of f(x+dx/2 as seen in the diagram from Team 7) and have the additional body force in the opposite direction caused by the springs (as seen by the figure in the problem description) designated as p(x+dx/2)

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$$ \displaystyle \Sigma \text{ Internal Force } + \text{ Applied Force } = \text{ Net Force}$$ (4.6.1)
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$$ \displaystyle \Sigma \text{ Internal Force }=N(x) + N(x+\Delta x)$$ (4.6.2)
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$$ \displaystyle \text{ Applied Force }=b(x+\frac{dx}{2})dx-p(x+\frac{dx}{2})dx$$ (4.6.3)
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Since we are looking at the static case:

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$$ \displaystyle \text{ Net Force}= m*a =0$$ (4.6.4)
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Substituting equations 4.6.2-4.6.4 into the force balance equation, 4.6.1:

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$$ \displaystyle N(x) + N(x+dx) + b(x+\frac{dx}{2})dx-p(x+\frac{dx}{2})dx =0$$ (4.6.5)
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Plugging in for the direction of the surface normal, the balance of forces at position x can be expressed as:

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$$ \displaystyle N(x)=n(x)\sigma (x)A(x)=-\sigma (x)A(x)$$ (4.6.6)
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Also plugging in for the surface normal, the corresponding force at x + dx:

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$$ \displaystyle N(x+dx)=n(x+dx)\sigma (x+dx)A(x+dx)=\sigma (x+dx)A(x+dx)$$ (4.6.7)
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Using a Taylor Series:

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$$ \displaystyle \sigma (x+dx)=\sigma (x)+\frac{d\sigma (x)}{dx}dx+...$$ (4.6.8)
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$$ \displaystyle A(x+dx)=A(x)+\frac{dA(x)}{dx}dx+...$$ (4.6.9)
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Additionally, in Eq. 4.6.3 the body force $$b(x+\frac{dx}{2})$$ has been averaged for the differential volume we are considering to formulate the governing differential equation. The same has been done for the force caused by the variable distributed spring, given as $$p(x+\frac{dx}{2})$$. Taylor expanding for these as well:

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$$\displaystyle b(x+\frac{dx}{2})=b(x)+\frac{db(x)}{dx}\frac{dx}{2}....$$. (4.6.10)
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$$\displaystyle p(x+\frac{dx}{2})=p(x)+\frac{dp(x)}{dx}\frac{dx}{2}....$$. (4.6.11)
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Plugging in Eq.'s 4.6.8 and 4.6.9 into Eq. 4.6.7, and then plugging this and Eq.'s 4.6.6, 4.6.10 and 4.6.11 into Eq. 4.6.5 gives:
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$$\displaystyle \sigma (x)A(x) + \sigma (x)\frac{dA(x)}{dx}dx+A(x)\frac{d\sigma (x)}{dx}-\sigma (x)A(x)+b(x)dx-p(x)dx$$. (4.6.12)
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Using the product rule, canceling out terms, and dividing through by dx gives:
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$$\displaystyle \frac{d}{dx}\sigma (x)A(x)+b(x)-p(x)=0$$. (4.6.13)
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Noting that the strain-displacement equation is
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$$ \displaystyle
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\varepsilon \left( x \right) = \frac

$$     (4.6.14) And by Hooke ‘ s Law
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$$ \displaystyle
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\sigma \left( x \right) = E\left( x \right)\varepsilon \left( x \right) = E\left( x \right)\frac

$$     (4.6.15)
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Gives the following strong form:
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$$\displaystyle \frac{d}{dx}E(x)A(x)\frac{du}{dx}+b(x)-p(x)=0$$. (4.6.16)
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b. Construct the Weak Form
Following the same solutions as Team 7's previous work here.

In order to reduce the requirements or constraints on what trial solutions could be substituted in for u, we will generate the weak form which requires the basis solutions only be once differentiable. This occurs from the natural boundary condition being absorbed in the weak form, as will be seen shortly. Also, this creates a symmetric stiffness matrix.

As with Team 7's solution, the strong form and natural boundary condition are used in an inner product with an arbitrary weighting function giving the following equations:

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$$\displaystyle \int_{0}^{l}w(x)\big [\frac{d}{dx}E(x)A(x)\frac{du}{dx}+b(x)-p(x)\big ]dx=0$$. (4.6.17)
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$$\displaystyle w(l)E(l)A(l)\frac{du(l)}{dx}-w(l)t=0$$. (4.6.18)
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Distributing the weighting function w(x) and the integral of Eq. 4.6.17 and looking at the first term:

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$$\displaystyle \int_{0}^{l}w\frac{d}{dx}E(x)A(x)\frac{du}{dx}dx=w(x)E(x)A(x)|_{0}^{l}-int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx$$. (4.6.19)
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In Eq. 4.6.19 integration by parts (derived and example of use here) was used. Now to remove the unknown flux at the essential boundary location (x=0), we will make the arbitrary weighting function be zero at x equal to zero, or

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$$\displaystyle w(0)=0$$. (4.6.20) Using Eq. 4.6.7 and 4.6.9, Eq. 4.6.8 becomes:
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$$\displaystyle w(x)E(x)A(x)|_{0}^{l}-\int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx=w(l)t-\int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx$$. (4.6.21)
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Combining Equations 4.6.18 and 4.6.21 gives the following weak form:

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$$\displaystyle w(l)t-\int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx+\int_{0}^{l}w(x)p(x)dx+\int_{0}^{l}w(x)b(x)dx$$. (4.6.22)
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Note in equation 4.6.22 the weighting function must be zero at the essential boundary condition (w(0)=0). Also, the weighting function and the displacement are both first order derivatives and now a symmetric stiffness matrix can be formed.

Authors
Braden Snook, Zongyi Yang

Problem Description
A calculix manual for novice/dummies

Author
Eml5526.s11.team7.Srilalithkumar

Problem Description
Consider a trial solution of the form $$u(x) = {\alpha _0} + {\alpha _1}(x - 3) + {\alpha _2}{(x - 3)^2} $$and a weight function of the same form. Obtain a solution to the weak form in Problem 3.1, "A First Course in Finite Elements", by Fish & Belytschko P72. Find $$\tilde M$$, assuming $$A = 1,E = 2,\bar m = 3$$, Do for $$n = 3$$. For dynamics, with $${u^h}(\beta ,t) = g(t) = \sin 2t$$, find $$F(t)$$ The weak form of Problem 3.1 is
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$$ \displaystyle
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\int\limits_1^3 {\fracAE\frac} dx = - 0.1{(wA)_{x = 1}} + \int\limits_1^3 {2xwdx\forall w(3) = 0}

$$     (4.8.1)
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$$ \displaystyle
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u(3) = 0.001

$$
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Solution
Part 1: The basis functions $${b_i} = \{ 1,(x - 3),{(x - 3)^2},{(x - 3)^3}\} i = 0,1,2,3$$
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$$ \displaystyle
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\tilde M = \left[ {\begin{array}{ccccccccccccccc} & \\  & \end{array}} \right]

$$     (4.8.2)
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$$ \displaystyle
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{M_{EE}} = {\tilde M_{00}} = \tilde m({b_0},{b_0}) = \int\limits_1^3 {{b_0}\bar m{b_0}dx} = \int\limits_1^3 {3dx}  = 6

$$     (4.8.3)
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$$ \displaystyle
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{M_{EF}} = \left[ {{M_{0j}};j = 1,2,3} \right] = \tilde m({b_0},{b_j})

$$     (4.8.4)
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$$ \displaystyle
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\tilde m({b_0},{b_1}) = \int\limits_1^3 {{b_0}\bar m{b_1}dx} = \int\limits_1^3 {3(x - 3)dx}  =  - 6

$$     (4.8.5)
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$$ \displaystyle
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\tilde m({b_0},{b_2}) = \int\limits_1^3 {{b_0}\bar m{b_2}dx} = \int\limits_1^3 {3{{(x - 3)}^2}dx}  = 8

$$     (4.8.6)
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$$ \displaystyle
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\tilde m({b_0},{b_3}) = \int\limits_1^3 {{b_0}\bar m{b_3}dx} = \int\limits_1^3 {3{{(x - 3)}^3}dx}  =  - 12

$$     (4.8.7)
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$$ \displaystyle
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{M_{EF}} = \left[ {\begin{array}{ccccccccccccccc} { - 6}&8&{ - 12} \end{array}} \right]

$$     (4.8.8)
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$$ \displaystyle
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{M_{FE}} = {M_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} { - 6} \\  8 \\   { - 12} \end{array}} \right]

$$     (4.8.9)
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$$ \displaystyle
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{M_{FF}} = \left[ {{M_{ij}};i,j = 1,2,3} \right] = \tilde m({b_i},{b_j})

$$     (4.8.10)
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$$ \displaystyle
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\tilde m({b_1},{b_1}) = \int\limits_1^3 {{b_1}\bar m{b_1}dx} = 3\int\limits_1^3 {(x - 3)(x - 3)dx}  = 8

$$     (4.8.11)
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$$ \displaystyle
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\tilde m({b_1},{b_2}) = \int\limits_1^3 {{b_1}\bar m{b_2}dx} = 3\int\limits_1^3 {(x - 3){{(x - 3)}^2}dx}  =  - 12

$$     (4.8.12)
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$$ \displaystyle
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\tilde m({b_1},{b_3}) = \int\limits_1^3 {{b_1}\bar m{b_3}dx} = 3\int\limits_1^3 {(x - 3){{(x - 3)}^3}dx}  = \frac{5}

$$     (4.8.13)
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$$ \displaystyle
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\tilde m({b_2},{b_1}) = \int\limits_1^3 {{b_2}\bar m{b_1}dx} = 3\int\limits_1^3 {{{(x - 3)}^2}(x - 3)dx}  =  - 12

$$     (4.8.14)
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$$ \displaystyle
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\tilde m({b_2},{b_2}) = \int\limits_1^3 {{b_2}\bar m{b_2}dx} = 3\int\limits_1^3 {{{(x - 3)}^2}{{(x - 3)}^2}dx}  = \frac{5}

$$     (4.8.15)
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$$ \displaystyle
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\tilde m({b_2},{b_3}) = \int\limits_1^3 {{b_2}\bar m{b_3}dx} = 3\int\limits_1^3 {{{(x - 3)}^2}{{(x - 3)}^3}dx}  =  - 32

$$     (4.8.16)
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$$ \displaystyle
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\tilde m({b_3},{b_1}) = \int\limits_1^3 {{b_3}\bar m{b_1}dx} = 3\int\limits_1^3 {{{(x - 3)}^3}(x - 3)dx}  = \frac{5}

$$     (4.8.17)
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$$ \displaystyle
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\tilde m({b_3},{b_2}) = \int\limits_1^3 {{b_3}\bar m{b_2}dx} = 3\int\limits_1^3 {{{(x - 3)}^3}{{(x - 3)}^2}dx}  =  - 32

$$     (4.8.18)
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$$ \displaystyle
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\tilde m({b_3},{b_3}) = \int\limits_1^3 {{b_3}\bar m{b_3}dx} = 3\int\limits_1^3 {{{(x - 3)}^3}{{(x - 3)}^3}dx}  = \frac{7}

$$     (4.8.19)
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$$ \displaystyle
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\therefore {M_{FF}} = \left[ {\begin{array}{ccccccccccccccc} 8&{ - 12}&{\frac{5}} \\ { - 12}&{\frac{5}}&{ - 32} \\ {\frac{5}}&{ - 32}&{\frac{7}} \end{array}} \right]

$$     (4.8.20)
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$$ \displaystyle
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\therefore {M_{FF}} = \left[ {\begin{array}{ccccccccccccccc} 6&{ - 6}&8&{ - 12} \\  { - 6}&8&{ - 12}&{\frac{5}} \\ 8&{ - 12}&{\frac{5}}&{ - 32} \\ { - 12}&{\frac{5}}&{ - 32}&{\frac{7}} \end{array}} \right]

$$     (4.8.21) Part 2:
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$$ \displaystyle
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F = {F_F} - \left( {{M_{FE}}{g^{(2)}} + {K_{FE}}g} \right)

$$     (4.8.22)
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$$ \displaystyle
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{F_F} = \left\{ {{F_i},i = 1,...,n} \right\} = \left\{ {\tilde f\left( \right),i = 1,...,n} \right\}

$$     (4.8.23)
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$$ \displaystyle
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\because \tilde f({b_i}) = {b_i}(\alpha )h + \int\limits_\alpha ^\beta {{b_i}fdx}

$$     (4.8.24)
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$$ \displaystyle
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\because f = 0

$$     (4.8.25)
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$$ \displaystyle
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\therefore \tilde f({b_i}) = {b_i}(1)h

$$     (4.8.26)
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$$ \displaystyle
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h = {a_2}(\alpha )\frac = AE\frac = 0.1

$$     (4.8.27)
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$$ \displaystyle
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\tilde f({b_1}) = {b_1}(1)h = - 2 \times 0.1 =  - 0.2

$$     (4.8.28)
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$$ \displaystyle
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\tilde f({b_2}) = {b_2}(1)h = {\left( { - 2} \right)^2} \times 0.1 = 0.4

$$     (4.8.29)
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$$ \displaystyle
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\tilde f({b_3}) = {b_3}(1)h = {\left( { - 2} \right)^3} \times 0.1 = - 0.8

$$     (4.8.30)
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$$ \displaystyle
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\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} { - 0.2} \\  {0.4} \\   { - 0.8} \end{array}} \right]

$$     (4.8.31)
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$$ \displaystyle
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{g^{(2)}} = - 4\sin (2t)

$$     (4.8.32)
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$$ \displaystyle
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{K_{FE}} = \left\{ {{K_{0j}};j = 1,...,n} \right\} = \left\{ {\tilde k\left( {{b_0},{b_j}} \right),j = 1,...,n} \right\}

$$     (4.8.33)
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$$ \displaystyle
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\because \tilde k({b_i},{b_j}) = \int\limits_1^3 {b{'_i}{a_2}b{'_j}dx} = \int\limits_1^3 {b{'_i}AEb{'_j}dx}

$$     (4.8.34) So
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$$ \displaystyle
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\tilde k({b_0},{b_1}) = \int\limits_1^3 {b{'_0}{a_2}b{'_1}dx} = \int\limits_1^3 {b{'_0}AEb{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$     (4.8.35)
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$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \end{array}} \right]

$$     (4.8.36) And：
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\because {M_{FE}} = {M_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} { - 6} \\  8 \\   { - 12} \end{array}} \right]

$$     (4.8.37)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

F = {F_F} - \left( {{M_{FE}}{g^{(2)}} + {K_{FE}}g} \right) = \left[ {\begin{array}{ccccccccccccccc} { - 0.2 - 24\sin (2t)} \\ {0.4 + 32\sin (2t)} \\ { - 0.8 - 48\sin (2t)} \end{array}} \right]

$$     (4.8.38)
 *

 * }

Authors
Zongyi Yang

Jiang Jin

title: User:Fem.tm7/HW4


text:
User:Volunteer Marek

We'll see how this works out.

title: User:Volunteer Marek


text:
User talk:Volunteer Marek

 Hello Volunteer Marek, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 22:28, 22 February 2011 (UTC)

title: User talk:Volunteer Marek


text:
User talk:Robynjay~enwikiversity

 Hello Robynjay, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 23:34, 22 February 2011 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:52, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:20, 19 April 2015 (UTC)

title: User talk:Robynjay~enwikiversity