text:
Extraction of metals/Iron

Reduction with carbon

 * Iron is extracted from its ore, haematite, by reduction with carbon.
 * This takes place in a blast furnace (named after blasts of air [20% oxygen], heated to 1000 °C, which are blasted into the bottom of the furnace):
 * A mixture of crushed iron ore, coke (an almost pure form of carbon) and limestone is fed into the top of the blast furnace.
 * The coke is oxidised to carbon dioxide:
 * C + O2 → CO2
 * The reaction is exothermic.

 * The carbon dioxide reacts with coke to form carbon monoxide.
 * CO2 + C → 2CO
 * The reaction is endothermic.

 * Iron(III) oxide in the ore is reduced to iron by the carbon monoxide.
 * Fe2O3 + 3CO → 2Fe + 3CO2
 * The reaction is exothermic.

 * The iron falls to the bottom of the blast furnace where it melts

SiO2 and Al2O3 impurities

 * Limestone, or calcium carbonate (CaCO3), is used to removed impurities
 * CaCO3 decomposes in the blast furnace to calcium oxide:
 * CaCO3(s) → CaO(s) + CO2(g)

 * The CaO then reacts with SiO2 and Al2O3 impurities:
 * CaO(s) + SiO2 → CaSiO3(s)
 * CaO(s) + Al2O3 → CaAl2O4(s)

 * The CaSiO3 and CaAl2O4 is less dense than the iron so it forms a slag which floats on top of the iron
 * The slag can then be removed.

Sulphur impurities

 * Magnesium powder is added to the molten pig iron to remove sulphur impurities
 * Magnesium reacts with sulphur to form magnesium sulphides which are removed as slag

title: Extraction of metals/Iron


text:
User talk:Eml4500.f08.echo.mott/Echo.HW1

Yo, what else needs to be done?--Charles 17:39, 12 September 2008 (UTC)

Chris, I added more to the "Sparse Matrix" section.--Charles 17:54, 12 September 2008 (UTC)

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


text: File:Diagram.jpg

title: File:Diagram.jpg


text:
Extraction of metals

This will look into the extraction of 3 metals (iron, aluminium and titanium) from there ores. As well as the chemistry involved, we will look at the economical and environmental factors that need to be considered.

 * /Iron/
 * /Aluminium/
 * /Titanium/

Discussion questions and essay ideas

 * How can precious metals be mined in ethical ways?
 * How can precious metals be most efficiently extracted from junk electronics?

title: Extraction of metals


text:
User:Eas4200c.f08.radsam/Lecturenotes

= Lecture Notes=

= Homework 1= {| class="toccolours collapsible collapsed" style="width:100%" ! Homework 1

Meeting 5

 * }

title: User:Eas4200c.f08.radsam/Lecturenotes


text:
User talk:Eml4500.f08.echo.marshall.cwd

 Hello Eml4500.f08.echo.marshall.cwd, 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. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on 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.

 * 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.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 17:54, 12 September 2008 (UTC)

title: User talk:Eml4500.f08.echo.marshall.cwd


text:
User talk:Eml4500.f08.group.gaar

 Hello Eml4500.f08.group.gaar, 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. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on 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.

 * 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.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 17:55, 12 September 2008 (UTC)

title: User talk:Eml4500.f08.group.gaar


text: File:Stress distribution diagram.JPG

title: File:Stress distribution diagram.JPG


text:
User:Eml4500.f08.lulz.abcd

User:Eml4500.f08.lulz.abcd/HW1 User:Eml4500.f08.lulz.abcd/HW2 User:Eml4500.f08.lulz.abcd/HW3 User:Eml4500.f08.lulz.abcd/HW4 User:Eml4500.f08.lulz.abcd/HW7

title: User:Eml4500.f08.lulz.abcd


text:
Learning from conflict and incivility/Erkan Yilmaz

see aims of this project

User:JWSchmidt
I once asked JWS if he sees some recent actions as: "me vs them" - I think I am responsible in that now many people think badly about JWS. Probably all these incidents lead to this: JWS feels more isolated. How can I help now ?

"Welcoming: For JWSchmidt, "welcoming" means deleting or merging as little content as possible. JWSchmidt believes that deleting and merging content is rude and shows hostility to newcomers and their "seedling" projects. This is very important to JWSchmidt." (source)

User:JWSchmidt is not the only user who has a certain view on deletion:

 * my idea of WV

 * I remember when I started at WV (de + en) I was also always one of the first to delete:
 * And as of today (2008-09-13) I am still leading at en.WV + de.WV in the deletion stats. Just looked it up: from the persons who can use the delete button in all 10 WVs I have deleted the most WV pages: 1417. So, am I a "deletionist" ?
 * What made me change my behaviour ? My deeper involvement and seeing people come and go ? Seeing in the RCs the many (unsuccessful) attempts to participate ? I think at one point when you see this all you ask yourself (and everyone of us has a different answer to these questions):
 * Why is every one here at Wikiversity ?
 * How can I help you increase your happiness level ?
 * What can this all evolve to when we just wait more ?

 * 2007 October 6:
 * Normally when I see copy + paste of Wikipedia I write a comment to the author (see e.g. What Wikiversity is not, section: "A duplication of other Wikimedia projects"). But today I have seen for the first time something different. JWSchmidt has "rescued" this page to Wikiversity. While chating with him I got to know this project: Wikimedia Garbage Detail - great idea and probably one of the means to achieve the mission of Wikiversity.

 * Topic talk:Ring theory

 * copied from here:
 * how would the wiki-verse be when there is no delete button anymore ? More happy or unhappy people ?
 * How about when anyone can delete ?
 * How about when all participants in a wiki have or have not same access to tools ?
 * Who is to blame (is blaming a correct approach ?) that people think (unconscious): WV = WP ? Is it some elements at Wikipedia (which is known far more than other wikis - and that they have and are and will influence (future) participants to only think in one direction only what a wiki can be ?) ? Is it human's own thinking/need ? How can people here at WV still overcome their unconscious thinking: that WV = WP ? Diversity (in opinions) is just interesting.

 * I also think that deletion debates are a good thing, people get to know what other participants ideas of WV are. This helps WV.

User:Moulton

 * I once told Moulton: you would be a very valuable resource at WV if you would not mention WP in every sentence
 * When I joined the Teamspeak chat, I listen to Moulton's voice and can "see" his deep emotions.
 * Recently when Moulton was searching for a song topic I requested one from him: "Moulton and Wikiversity in 7 years" (no response so far :-. This brought me the urge to make this edit:
 * "Imagine: participants after being 7 years at Wikiversity (and forget their "actions" from begin - agf)"

 * User talk:Moulton
 * I should treat Moulton more like other users - but Moulton, you make it hard for me sometimes :-)

Asking myself
JWS and McCormack didn't chat (yet),
 * Why is chatting to relieve one's emotions - which worked before - not used anymore ?
 * Is it because "external" people are stressing some WV participants here until it all turns out to take forms like on other wikis now ?

 * For a conflict there is more than one person necessary and I give many people the "blame" (including myself - see above). Do I and others realize what was done wrong?
 * I just have a certain amount of time per day to do things - I started too late to "act" (whatever that means)
 * But: this is a wiki - god damn it - everybody can edit - why had it come so far ? Ask yourself ! Was your own time also more important than influencing/helping ?

Emotions I had

 * After having seen how things between User:JWSchmidt and User:McCormack turned out - with a review request - I didn't contribute much thereafter the next day. I wondered why people didn't learn from the incidents before from newcomers (who don't yet understand WV's philosophy - hopefully they will eventually) ? I had thoughts of leaving WV (I am not kidding) - I released my thoughts in chat and felt better afterwards. And as one person said there: it was good to relieve things which stress oneself. I also let my feelings out by creating Which hands do you think is best (for Wikiversity) ?
 * Sometimes Moulton's actions make me very angry, and sometimes I feel guilty towards Moulton.
 * The talk on this talk page finally helped me decide to release my thoughts to the public, too. I already feel calm now after writing this.
 * I am also disappointed in one person (won't tell name). I hope that with this edit I will overcome this when reading this previous sentence after a few days


title: File:Element1wiki1.JPG


text:
User:Eml4500.f08.echo.sell.cm/HW2

09.10.08
Global FD rel:  K d  =  F

09.12.08
In compact Notation:

$$\begin{bmatrix}K_{ij}^{}\end{bmatrix}_{6x6} \left\{d_{ij}\right\}_{6x1} = \left\{F_{i}\right\}_{6x1}$$

In the more general form (nxn):

$$\sum_{j=1}^{6} K_{ij}d_{j}= F_{i}$$; i = 1, 2, ..., 6 $$K_{nxn} = \begin{bmatrix}K_{ij}^{}\end{bmatrix}_{nxn}$$ = global stiffness matrix $$d_{nx1} = \left\{d_{j}^{}\right\}_{nx1}$$ = global displacement matrix $$F_{nx1} = \left\{F_{i}^{}\right\}_{nx1}$$ = global force matrix

Recall Element Force-Displacement relationship:

$$k_{4x4}^{(e)} d_{4x4}^{(e)} = f_{4x1}^{(e)}$$ $$k_{4x4}^{(e)} = \begin{bmatrix}k_{ij}^{(e)}\end{bmatrix}_{4x4}$$ = element stiffness matrix $$d_{4x1}^{(e)} = \left\{d_{j}^{(e)}\right\}_{4x1}$$ = element displacement matrix $$f_{4x1}^{(e)} = \left\{f_{i}^{(e)}\right\}_{4x1}$$ = element force matrix

Use an assembly process to go from element matrices(stiffness, displacement, force) to global matrices.

 * Identify the correspondence between the element displacement degrees of freedom's and the global displacement degrees of freedom's.

Global Level:

title: User:Eml4500.f08.echo.sell.cm/HW2


text:
User:Eml4500.f08.echo.mott/HW2

This is the drafting page for homework report 2 - due Friday, September 26.

The link to the FINAL SUBMITTAL PAGE is HERE
 * This time I will leave everything on this page and just copy it to the final page. Chris 15:11, 26 September 2008 (UTC)

Mon, Sept 8
2 Force body example:

Solution to this problem:

$$\sum F_x = 0; R_1cos( \theta_1) - R_2cos( \theta_2) + Pcos( \theta_P) = 0$$

$$\sum F_y = 0; R_1sin( \theta_1) - R_2sin( \theta_2) + Psin( \theta_P) = 0$$

$$\sum M_1 = 0; R_2sin( \theta_2)*(l_2cos( \theta_2) + l_1cos( \theta_1)) + Psin( \theta_P)*(l_1cos( \theta_1)) = 0$$

Three equations and three unknowns, therefore it is statically determinate.

Hey are there values for this? I know he wanted us to solve it using statics but I don't know if he gave us values - --70.185.100.27 00:28, 26 September 2008 (UTC)Eml4500.f08.echo.bruce.jp

I don't have any values either. I don't think he gave us any values, just wanted us to do it symbolically or something. I did add Step 1) below (it was from the previous lecture) Eml4500.f08.echo.sell.br 04:54, 26 September 2008 (UTC)

I thought he wanted us to solve the problem that had P just being a vertical force but that sounds like maybe it would be too easy so i just wanted to see what everyone else had. I have time today Friday 9/26 to fix it but i wasn't sure if that was what he was asking for.- --70.185.100.27 00:28, 26 September 2008 (UTC)Eml4500.f08.echo.bruce.jp

I don't really know for sure. The rest of this section should be finished now. Eml4500.f08.echo.sell.br 06:44, 26 September 2008 (UTC)

I just realized the picture is slightly wrong. I'm going to update it and add the R1, R2,theta's, etc. Eml4500.f08.echo.lacroix 15:22, 26 September 2008 (UTC)

Finite Element Method

Step 1) Draw and label Global Picture

Step 2) Draw and label Element Pictures

Step 3) Create global FD at element level

 * Drawing and Labeling the Global Picture:

 * Breaking it down into two elements:

 * Statically indeterminate examples:

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

$$	\mathbf{k}^{(e)}$$ = 4x4 element stiffness matrix for element e where e = 1,2

$$	\mathbf{d}^{(e)}$$ = 4x1 element displacement matrix of element e

$$	\mathbf{f}^{(e)}$$ = 4x1 element force matrix of element e

The element stiffness matrix can be broken down as follows:

$$\mathbf{k}^{(e)}=k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2\\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)}\\-l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2\end{bmatrix}$$

The scalar value of the axial stiffness of bar element e can be found using the following equation:

$$k^{(e)} = \frac{E^{(e)}{A_c}^{(e)}}{L^{(e)}}$$

$$l^{(e)}$$ is the x-length of the element in reference to global coordinates and $$m^{(e)}$$ is the y-length of the element from a global perspective. They can be thought of as the length of the element projected onto global coordinates from local coordinates.

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

$$m^{(e)}= \mathbf{\tilde{i}} \cdot \mathbf{j}=cos( \frac{ \pi}{2} - \theta^{(e)})=sin( \theta^{(e)})$$

The direction cosines are the components of $$\mathbf{\tilde{i}}$$ (unit vector along $$\mathbf{x}$$ axis) with respect to basis $$(\mathbf{i},mathbf{j})$$ $$\mathbf{\tilde{i}}=cos( \theta^{(e)})\mathbf{i}+sin( \theta^{(e)})\mathbf{j}=l^{(e)}+m^{(e)}$$

$$\mathbf{\tilde{i}}=cos( \theta^{(e)})\mathbf{i}+sin( \theta^{(e)})\mathbf{j}$$

$$\mathbf{\tilde{i}} \cdot \mathbf{i}=cos( \theta^{(e)})$$

$$\mathbf{\tilde{i}} \cdot \mathbf{j}=sin( \theta^{(e)})$$

Eml4500.f08.echo.sell.br 06:44, 26 September 2008 (UTC)

Eml4500.f08.echo.lacroix 23:57, 22 September 2008 (UTC)

Wed, Sept 10
I'll show how to compute the k matrix from given values of the sample problem. I might need a little more information from the last part of the lecture. *Chris 15:56, 15 September 2008 (UTC)

The element stiffness matrix takes the form shown below.

$$\mathbf{k^{(e)}}=k^{(e)} \begin{bmatrix} l^{(e)^2} & l^{(e)}m^{(e)} & -l^{(e)^2} & -l^{(e)}m^{(e)} \\l^{(e)}m^{(e)} & m^{(e)^2} & -l^{(e)}m^{(e)} & -m^{(e)^2} \\-l^{(e)^2} & -l^{(e)}m^{(e)} & l^{(e)^2} & l^{(e)}m^{(e)} \\-l^{(e)}m^{(e)} & -m^{(e)^2} & l^{(e)}m^{(e)} & m^{(e)^2}\end{bmatrix}$$

The scalar $$k^{(e)}$$ can be found from the properties of the structure where E is the Elastic Modulus, A is the cross-sectional area of the element, and L is the length of the element.

$$k^{(e)} = \frac{E^{(e)}{A_c}^{(e)}}{L^{(e)}}$$

$$l^{(e)}$$ is the x-length of the element in reference to global coordinates and $$m^{(e)}$$ is the y-length of the element from a global perspective. They can be thought of as the length of the element projected onto global coordinates from local coordinates.

The x-length can be found by using: $$l_{s}= \frac{x_{2}-x_{1}}{L}$$

The y-length of the element can be found by using: $$l_{s}= \frac{y_{2}-y_{1}}{L}$$

With the inclined angle referenced to the horizontal, the horizontal and vertical lengths can be found by:

lsx = cos θ

lsy = sin θ

After reviewing the stiffness matrix two observations can be made.

1) Of all the numbers in the matrix, only three need to be computed. The other coefficients have the same absolute values.
 * $$l_s^{} m_s$$
 * $${l_s}^2$$
 * $${m_s}^{2}$$

2) Matrix k is symmetric. That is $$k_{i j}^{1}=k_{j i}^{1}$$.

Fri, Sept 12
I'll do the global and the element matrices assembly process.--C. Sell 18:28, 16 September 2008 (UTC) Still working on this days notes decided to post what i had for now.--C. Sell 16:24, 21 September 2008 (UTC)

Process to Assemble a Global Stiffness Matrix using Element Stiffness Matrix:

Consider the Global Force Displacement Relationship for a Truss with 2 elastic bars (6 dof's in Global picture.

Written in compact notation we have the following:

$$\begin{bmatrix}K_{ij}^{}\end{bmatrix}_{6x6} \left\{d_{ij}\right\}_{6x1} = \left\{F_{i}\right\}_{6x1}$$

In the more general form (nxn):

$$\sum_{j=1}^{6} K_{ij}d_{j}= F_{i}$$; i = 1, 2, ..., 6 $$K_{nxn} = \begin{bmatrix}K_{ij}^{}\end{bmatrix}_{nxn}$$ = global stiffness matrix $$d_{nx1} = \left\{d_{j}^{}\right\}_{nx1}$$ = global displacement matrix $$F_{nx1} = \left\{F_{i}^{}\right\}_{nx1}$$ = global force matrix

Now recall the Element Force-Displacement relationship:

$$k_{4x4}^{(e)} d_{4x4}^{(e)} = f_{4x1}^{(e)}$$ $$k_{4x4}^{(e)} = \begin{bmatrix}k_{ij}^{(e)}\end{bmatrix}_{4x4}$$ = element stiffness matrix $$d_{4x1}^{(e)} = \left\{d_{j}^{(e)}\right\}_{4x1}$$ = element displacement matrix $$f_{4x1}^{(e)} = \left\{f_{i}^{(e)}\right\}_{4x1}$$ = element force matrix

Use an assembly process to go from element matrices(stiffness, displacement, force) to global matrices.

 * First, Identify the correspondence between the element displacement degrees of freedom's and the global displacement degrees of freedom's.

Global Level: $$\left\{d_1, d_2, \dots, d_6\right\}$$

Element level:
 * $$\left\{d_1^{(1)}, d_2^{(1)}, d_3^{(1)}, d_4^{(1)}\right\}$$
 * $$\left\{d_1^{(2)}, d_2^{(2)}, d_3^{(2)}, d_4^{(2)}\right\}$$

The relationship between global-local(element) dof's are as follows:

$$\text{Global Node 1} = \begin{cases} d_1 = d_1^{(1)} \\ d_2 = d_2^{(1)}\end{cases}$$

$$\text{Global Node 2} = \begin{cases} d_3 = d_3^{(1)} = d_1^{(2)} \\ d_4 = d_4^{(1)} = d_2^{(2)}\end{cases}$$

$$\text{Global Node 3} = \begin{cases} d_5 = d_3^{(2)} \\ d_6 = d_4^{(2)}\end{cases}$$

Conceptual step of the assembly:

Mon, Sept 15
Taking the global stiffness matrix and substituting in the appropriate values:

$$\mathbf{K}= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} & k_{13}^{(2)} & k_{14}^{(2)}\\k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} & k_{23}^{(2)} & k_{24}^{(2)}\\0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\end{bmatrix}= \begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16}\\K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26}\\K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36}\\K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46}\\K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56}\\K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66}\end{bmatrix}$$

$$K_{11}=k_{11}^{(1)}=\frac{9}{16}=0.5625$$

$$K_{12}=K_{21}=k_{12}^{(1)}=\frac{3\sqrt{3}}{16}=0.3248$$

$$K_{13}=K_{31}=k_{13}^{(1)}=-k_{11}^{(1)}=-0.5625$$

$$K_{14}=K_{23}=K_{32}=K_{41}=k_{14}^{(1)}=-k_{12}^{(1)}=-0.3248$$

$$K_{22}=k_{22}^{(1)}=\frac{3}{16}=0.1875$$

$$K_{24}=K_{42}=k_{24}^{(1)}=-k_{22}^{(1)}=-0.1875$$

$$K_{33}=k_{33}^{(1)}+k_{11}^{(2)}=	\begin{matrix} \frac{9}{16} \end{matrix}+\begin{matrix} \frac{5}{2} \end{matrix}=3.0625$$

$$K_{34}=k_{34}^{(1)}+k_{12}^{(2)}=	\begin{matrix} \frac{3\sqrt{3}}{16} \end{matrix}+\begin{matrix} \frac{-5}{2} \end{matrix}=-2.1752$$

$$K_{35}=K_{53}=k_{13}^{(2)}=-k_{11}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{36}=K_{45}=K_{54}=K_{63}=k_{14}^{(2)}=-k_{12}^{2}=2.5$$

$$K_{43}=K_{34}=k_{43}^{(1)}+k_{21}^{(2)}=k_{34}^{(1)}+k_{12}^{(2)}=-2.1752$$

$$K_{44}=k_{44}^{(1)}+k_{22}^{(2)}=\frac{3}{16}+\frac{5}{2}=2.6875$$

$$K_{46}=K_{64}=k_{24}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{55}=k_{33}^{(2)}=\frac{5}{2}=2.5$$

$$K_{56}=K_{65}=k_{34}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{66}=k_{44}^{(2)}=\frac{5}{2}=2.5$$

$$K_{15}^{}=K_{16}=K_{25}=K_{26}=K_{51}=K_{52}=K_{61}=K_{62}=0$$

$$\mathbf{K}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0\\0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0\\-0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5\\-0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5\\0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\0 & 0 & 2.5 & -2.5 & -2.5 & 2.5\end{bmatrix}$$

Wed, Sept 17
Next, the elimination of known degree's of freedom can create the following matrices.

$$ d_1^{} = d_2^{} = d_5^{} = d_6^{}=0 $$

$$	\mathbf{K}\begin{Bmatrix} d_1=0\\d_2=0\\d_3\\d_4\\d_5=0\\d_6=0 \end{Bmatrix}=\begin{vmatrix} k_{13} & k_{14} \\ k_{23} & k_{24}\\k_{33} & k_{34}\\k_{43} & k_{44}\\k_{53} & k_{54}\\k_{63} & k_{64}\end{vmatrix}\begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\mathbf{F}$$

Applying fixed boundary conditions allows us to delete the corresponding columns in the global stiffness matrix K

By the Principal of Virtual Work (PVW) we're able to delete the corresponding rows (in this case - 1, 2, 5, 6). Corresponding rows of F are also deleted.

The resulting FD relation is: $$ \begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \end{bmatrix} \begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}$$

$$\mathbf{F}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}=\begin{Bmatrix} 0\\P\end{Bmatrix}$$

And P =>

What is the inverse of K without using a calculator?

Taking the determinant of K:

$$det \mathbf{K}=k_{33}k_{44}-k_{k34}k_{43}$$

$$\mathbf{K}^{-1}=\frac{1}{det\mathbf{K}}\begin{bmatrix} k_{44} & -k_{34} \\ -k_{43} & k_{33} \end{bmatrix}$$

Verifying:

$$\mathbf{K}\mathbf{K}^{-1}=\mathbf{K}^{-1}\mathbf{K}=\mathbf{I}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

This is essentially:

$$\frac{1}{det(\mathbf{K})}\begin{bmatrix} det(\mathbf{K}) & 0 \\ 0 & det(\mathbf{K}) \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Note: $$\mathbf{K}^T=\begin{bmatrix} k_{33} & k_{43} \\ k_{34} & k_{44} \end{bmatrix}$$

Therefore: $$\mathbf{K}^{-1}\ne \ \frac{1}{det(\mathbf{K})}\mathbf{K}^T$$

Going back to the 2 bar truss problem...

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}=\mathbf{K}^{-1}\begin{Bmatrix} 0 \\ P \end{Bmatrix}=\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Now we must compute the reactions:

 * Method 1:

Use element FD relations.

$$\mathbf{k}^{(1)}\mathbf{d}^{(1)}=\mathbf{f}^{(1)}$$

Taking a look at element 1 again and applying the knowns:

$$d^{(1)}=\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

Taking a look at element 2:

$$\mathbf{k}^{(2)}\mathbf{d}^{(2)}=\mathbf{f}^{(2)}$$

$$d^{(1)}=\begin{Bmatrix} 4.352 \\ 6.1271 \\ 0 \\ 0 \end{Bmatrix}=\begin{Bmatrix} d_1^{(2)} \\ d_2^{(2)} \\ d_3^{(2)} \\ d_4^{(2)} \end{Bmatrix}$$

 * Method 2: Not really sure when he phased into this....

Fri, Sept 19

Useful LaTeX Link
When writing equations in Wikiversity, it is beneficial to use LaTeX code.

The following is a useful link that provides information regarding how to format equations and mathematical formulas using LaTeX code: http://en.wikibooks.org/wiki/LaTeX/Mathematics

title: User:Eml4500.f08.echo.mott/HW2


text:
User:Eml4500.f08.lulz.abcd/HW1

Part 6: For, while, if - and relations
MATLAB handles most common computer code concepts like any other modern computer programming language. This includes for and while loops, if statements, and relational conditions.

For
for loops can be used very effectively in MATLAB to create matrices.

Ex 1: If you wanted to create a series of vectors where each value in the vector is the square of the index value of the position of the value in the vector, you could use this for loop:  EDU>> x = []; for i = 1:3, x=[x,i^2], end x =

1 x =

1    4 x =

1    4     9 The expression x = []; i = 3:-1:1, x=[x,i^2], end will give the same thing, except in reverse.

Ex 2: for loops can also be used to produce the Hilbert matrix. (Coding note, the semicolon prevents an intermediate printing which would be undesirable and the final H tells MATLAB to print the resulting H matrix.)  for i = 1:3 for j = 1:3 H(i,j) = 1/(i+j-1); end end H H =

1.0000   0.5000    0.3333    0.5000    0.3333    0.2500    0.3333    0.2500    0.2000

While
A while loop executes some command repeatedly so long as the condition it checks remains true.

Ex: MATLAB can calculate for a given number a, the smallest value n such that 2^n is less than or equal to a:  n = 0; a = 12; while 2^n  if n<0 parity = 0; elseif rem(n,2) == 0 parity = 2; else parity = 1; end

Relations
MATLAB uses the following relational operators in its programming language: < less than > greater than <= less than or equal to >= greater than or equal to == equal ~= not equal & and ~ not
 * or

Part 7: Scalar functions
MATLAB functions that are designed to operate on scalars can also be applied to matrixes, they will simply do so element-wise on the matrix elements. These are some common MATLAB functions that work in such a manner: sin, cos, tan, asin, acos, atan, ex, log, rem, abs, sqrt, sign, round, floor, ceil

Part 8: Vector functions
MATLAB functions that are designed to operate on vectors can also be applied to M x N matrixes (M >= 2), they will simply do so column-by-column which will produce a row vector containing the results of their operation in each column. To get these functions to operate on a row-by-row basis, use the transpose of the matrix. Common MATLAB functions that work like this are: max, min, sort, sum, prod, median, mean, std, any, all

These functions can also be nested within each other. Ex: to get the maximum single value in matrix A, use max(max(A)).

Part 9: Matrix functions
MATLAB has many matrix functions automatically built in for ease of use. These dramatically simplify programming and creating useful results from the software. These are the most useful of the functions: eig - Eigenvalues and Eigenvectors chol - Cholesky factorization svd - single value decomposition inv - inverse lu - LU factorization qr - QR factorization hess - Hessenberg form schur - Schur decomposition rref - reduced row echelon form expm - matrix exponential sqrtm - matrix square root poly - characteristic polynomial det - determinant size - size norm - 1-norm, 2-norm, F-norm, infinity-norm cond - condition number in the 2-norm rank - rank

As in other MATLAB functions, these can be used several ways. Ex: y = eig(A) and simply inputting eig(A) produce the same results. The output can also go to another matrix, [U,D] = eig(A) will produce a matrix with columns as the eigenvalues of A and a diagonal matrix D with the eigenvalues on the diagonals. This all may seem elementary, the point is that if there is a desired form of an output in MATLAB for a function, it is likely possible given the proper syntax.

Part 10: Command line editing and recall
The command line in MATLAB can be edited like most other text editors (usin up and down arrows, home and end keys, etc.). Full details of how the command line can be edited will be printed from the help cedit command.

Previous commands can be accessed using the up and down arrows on the command line. This is frequently more efficient than producing a full M-file for small problems. Commands may also be placed sequentially on the same command line using semi-colons to separate the commands.

title: User:Eml4500.f08.lulz.abcd/HW1


text:
User:Salmon of Doubt/cust

I am requesting custodianship on this project for a number of reasons.

Primarily, I intend to help with day-to-day maintenance. I have watched powerlessly as a pow-wow of three custodians were unable to clear a cross-namespace redirect for move. There are numerous edit histories on this project that are not GFDL compliant and require mergers. Image policy here requires a strong hand to fix - specifically, there is no robotic tagging and notification system for obvious copyvios. Users with bot-proficiency are necessary for a project in a growth stage, and I have that proficiency.

Additionally, I would be extra vigilant in recent change patrolling - I have already been instrumental in dealing with one extremely vulgar troll, and would be able to do more with the ability to stop vandalism in the project space without having to constantly revert while discussing vandal's intentions on their talk pages.

Finally, and perhaps more controversially, I believe that there is a serious problem with divulged personal information that cannot be solved without tools. In the event that otherwise private personal information about a user is disclosed, I will delete specific edits to protect the privacy of our contributors. If contributors to this project are driven away with threats to their health and well-being, there will be nothing left but people who make such threats.

I will not protect pages to “win” edit wars, nor will I block established contributors without seeking broad consensus of other users, nor will I delete pages without wide consensus (unless those pages are copyvios, invasions of privacy or where the sole author requests deletion)

Thank you for your consideration.

Salmon of Doubt 19:35, 12 September 2008 (UTC)

title: User:Salmon of Doubt/cust


title: File:Example1.jpg


text: File:Example2 HW1.jpg

title: File:Example2 HW1.jpg


text: File:Globalstiffnessmatrix.jpg

title: File:Globalstiffnessmatrix.jpg


title: File:Example1Data.jpg


text:
User:Uf.team.aero/HW2

= Problem 1.1=

Provided Information
A beam with closed thin-walled section is designed to carry a bending moment M, and torque T. The total wall contour length is constant and is given by $$L=2(a+b)$$. Here a and b are the fixed width and height dimensions, respectively. For the purpose of solving this problem, the bending moment is equal to the torque, and $$ {\sigma}_{allowable}= 2{\tau}_{allowable}$$.

Using the assumption that the thickness, t, is much smaller than the height and width (b and a, respectively), we can assume that the shear stress is equally distributed in the walls of the shell. Below is a diagram visualizing the shear flow in the box beam, as well as showing approximations of the force distribution as a resultant force $${\nu}$$.

Figure 3: Shear Stress Distribution and Resultant Shear Stress in thin-walled beam

The shear stress due to torsion is given by: $${\tau}=\frac {T} {2abt}$$

Assumptions
To re-iterate and make referencing easier, the assumptions are as follows:

1) $$L=2(a+b)$$ is fixed

2) $$M=T$$

3) $${\sigma}_{allowable}= 2{\tau}_{allowable}$$

4) $${\tau}=\frac {T} {2abt}$$

Objective
With the provided information and assumptions, find the optimum b/a ratio that will achieve the most efficient thin walled-section.

Case 1
Case 1 is the calculation of the maximum allowable moment when $$ \sigma_{max}\ $$ is reached first from bending stresses. The equation for normal stress from a bending moment is as follows:

$$ \sigma\ = {My \over I} $$

Maximum normal stresses occur at the top and bottom of the cross-section when y=b/2. The equation then becomes:

$$ \sigma_{max}\ = {Mb \over 2I} $$

Since $$ \sigma_{all}\ $$ is a known quantity and we want to solve for $$ M_{max} $$, we use some simple algebra to rearrange it into the form:

$$ M_{max} = { 2I\sigma_{allow}\ \over b} $$

In order to maximize $$ M_{max} $$, we want to maximize the portion of the equation $${I \over b} $$. The next step involves finding I, the second moment of area of the cross section of the beam. I can be found with a combination of the superpositioning and parallel axes theorem by:

$$I = \sum_{i=1}^4 \frac{b_{i}\,h_{i}^3}{12} + A_{i}\,d_{i}^2 = 2\left (\frac{t\,b^3}{12} \right) + 2\left [\frac{a\,t^3}{12}+at\left (\frac{b}{2} \right)^2 \right]$$

With the assumptions $$t<<b$$ and $$t<<a$$, we can reduce the second moment of area to:

$$ I = {tb^2 \over 6} (3a+b) $$

Using the relation from the constraints of the problem that $$ a = {L \over 2}-b$$ (assumption 1), the new equation for the second moment of area becomes:

$$I = {tb^2(3L-4b) \over 12}$$

We then proceed to maximize the $${I \over b} $$ equation by defining it as a function $$f(b)$$ and applying fundamental principals of calculus.

The maximum applied moment for the allowable normal stress then finally becomes:

$$ M_{max}^{(1)} = {3tL^2 \over 32}\sigma_{allow}\ $$

From the problem statement we are given that $$\tau_{max}\ = {T_{max} \over 2abt}$$ (assumption 4). We plug in our derived value of $$M_{max}^{(1)}$$ into $$T_{max}$$ (assumption 2) and solve for $$ \tau\ $$:

$$ \tau_{max}\ = {T_{max} \over 2abt} = {M_{max} \over 2abt} = \sigma_{allow}\ $$

By assumption 3:

$$ \tau_{max}\ = \sigma_{allow}\ = 2 \tau_{allow}\ > \tau_{allow}\ $$

Which cannot happen because $$ \tau_{max}\ $$ cannot exceed $$ \tau_{allow}\ $$

Case 2
For Case 2, allow $$ \tau_{max}\ = \tau_{allow}\ $$, and then check to see that $$ \sigma_{max}\ < \sigma_{allow}\ $$ by invoking the assumption that $$ T = M $$.

First, solve for

 * $$T_{max} = 2t \tau_{allow}\ (ab)_{max} $$

Note that all parameters are fixed except the product $$(ab)_{max}$$. To maximize the product, recall that a can be defined in terms of b:

 * $$ f_{2} = ab = (\frac {L} {2} - b)b = \frac {bL} {2} - b^2 $$

Now, the maximum torque based on allowable shear stress is found as

 * $$ T_{max} = 2t \tau_{allow}\ \frac {L^2} {16} $$

Invoking the assumptions that M = T and $$ \sigma\ = 2\tau\ $$ the maximum bending moment is found to be

 * $$ M_{max} = \frac {tL^2} {16} \sigma_{allow}\ $$

This defines the maximum allowable normal stress to be

 * $$ \sigma_{allow}\ = \frac {16M_{max}} {tL^2} $$

To be sure that the actual maximum normal stress has not exceeded the allowable normal stress recall the equation

 * $$ \frac {I} {b} = {tb(3L-4b) \over 12} = \frac {tL^2} {24}$$

Such that the maximum stress is given by

 * $$ \sigma_{max} = M_{max} \frac {b} {2I} = \frac {12M_{max}} {tL^2} $$

Comparing the maximum normal stress to the allowable normal stress, it is seen that

 * $$ \sigma_{max} = \frac {3} {4} \sigma_{allow} $$

Thus, Case 2 provides a square cross section such that the acceptable optimized ratio is

 * $$ \frac {b} {a} = 1 $$

= Mohr's Circle =

To help further explain the basis of the problem statement assumption that

 * $$ \sigma = 2\tau $$

it is convenient to examine a Mohr's Circle plot of a beam in uniaxial tension.

Since it can be seen by the figure of the axial member that normal stress lies only in the x-direction, a Mohr's circle would have one plot of principal stress equal to $$ \sigma_{max} $$ and one equal to zero. The relationship of shear stress to the normal stress can be seen visually on the Mohr's Circle below. Namely,
 * $$ \tau_{max} = \frac {\sigma_{max}} {2} $$.

Therefore, it is seen that the problem statement for Problem 1.1 uses a fundamental principle in its assumption that $$ \sigma_{max} = 2\tau_{max} $$.

= Derivation of Second Area Moment of Inertia for Circular Cross Section =

Formula
Or equation for the second area moment of inertia in rectangular coordinates is:

 * $$\int\int_A \,z^2\,dy\,dz$$

The transition into polar requires the following polar transformations:
 * $$z=r\,sin(\theta)$$

 * $$dy\,dz=r\,dr\,d\theta$$

The resulting second area moment of inertia in polar coordinates is:

 * $$\int\int_A \,r^3\,sin^2(\theta)\,dr\,d\theta$$

Derivation
The second area moment of inertia in polar coordinates integrated over the entire area results in the equation:

 * $$2\,\int_{0}^{\pi}{\int_{0}^{R}{ \,r^3\,sin^2(\theta)\,dr\,d\theta}}$$

By integrating and evaluating the first integral we get:

$$2\,\int_{0}^{\pi}{\frac{R^4}{4}\,sin^2(\theta)\,d\theta}=\frac{R^4}{2}\,\int_{0}^{\pi}{\,sin^2(\theta)\,d\theta}$$

To continue integration, it is necessary to implement the half angle identity $$sin^2(\theta)=\frac{1}{2} \left[1-cos(2\,\theta) \right]$$ so that our integral becomes:

 * $$\frac{R^4}{4}\,\int_{0}^{\pi}{\,1-cos(2\,\theta)\,d\theta}$$

 * $$=\frac{R^4}{4}\,\left[\theta-\frac{sin(2\,\theta)}{2}\right]_0^\pi$$

With our final evaluation, we receive the second area moment of inerta for a circular cross section:
 * $$I=\frac {\pi\,R^4}{4}$$

= Problem 1 =

Given
The following figures below are presented, case 1 and case 2. Case 1 is a circular figure with radius R, and area, $$ A^{(1)} $$. Case 2 is a 3 bar system made out of 3 bars, each with thicknesses t, and length and height of a, and be respectively. The area of case 2 is has the notation as $$ A^{(2)} $$. See the schematic below for more information concerning the problem information.

Objective
For this problem make sure to calculate the momemnt of inertia for both cases 1 and 2 and compare there values. Be sure to also locate the position of the centroid for case 2 in relation to the parallel axis, in this case the y-axis.

Solution
Using the following formulas to derive the moment of intertia for both cases, the circle and square:

For a circle:
 * $$ I_{y} = \frac {(\pi)R^4} {4} $$

For a rectangle:
 * $$ I_{y} = \frac {bh^3} {12} $$

Parallel axis theorem:
 * $$ I_{y} = I_{y,centroid} + Ad^2 $$

Using the formula for the moment of inertia for a circle with respect to the y-axis, we know that;
 * $$ I_{y} = \frac {(\pi)R^4} {4} $$

The area for case 1 is given by:

 * $$A^{(1)}=\pi R^2$$

and the moment of inertia is given as:

 * $$I_y=\frac{\pi}{4}R^4$$

Knowing that a = b, and that t=a/10, the equation for the area of case 2 can be reduced as follows:

= Importance of Stringers =

Stringers, or longerons, are critical to aircraft structures. Strings provide rigidity to the skin of the fuselage or wing, helping to support the aerodynamic loads. Stringers can be seen riveted to the fuselage skin in the cross-section of a 747 to the right.

As can be seen from the figure, stringers have an open cross-section with the vertical members canted outwards at a slight draft angle. These features are advantageous for ease of manufacturing and assembly. The skin must be riveted to the stringer, which is more easily done with an open cross-section rather than a closed, thin-walled section. Also, when manufacturing an open-section stringer, a punch can be used on sheet metal to obtain the form. This eliminates the need to roll and form a closed-section.

The outward canted draft angle provides a means for the string to be easily removed from the punch and is also convenient for stacking stringers in storage (as shown in the figure below).

= Problem 1.7 with Modification II: Circles =

Given
A circle has radius $$ R_{1} $$ and area $$ A^{(1)} $$. A dumbell shaped figure consists of a thin web of thickness $$ t = 0.1R_{1} $$, length $$ l = 2R_{1} $$ and area $$ A^{(2)} $$. On each end of the web are connected circles of radius $$ R_{2} $$.

Find
Find and compare the second area moment of inertia, $$ I_{y} $$, for the circle and dumbell if $$ R_{1} = 10 cm $$ and $$ A^{(1)} = A^{(2)} $$.

Equations
For a circle:
 * $$ I_{y} = \frac {(\pi)R^4} {4} $$

For a rectangle:
 * $$ I_{y} = \frac {bh^3} {12} $$

Parallel axis theorem:
 * $$ I_{y} = I_{y,centroid} + Ad^2 $$

Solution
Circle:

Use the formula derived in the previous problem to determine $$ I_{y}^{(1)} $$.

 * $$ I_{y}^{(1)} = \frac {(\pi)R_{1}^4} {4} = \frac {(\pi)(10 cm)^4} {4} = 7,854 cm^4 $$

Dumbell:

First, determine $$ R_{2} $$ such that $$ A^{(1)} = A^{(2)} $$.

 * $$ A^{(1)} = A^{(2)} $$

 * $$ (\pi)R_{1}^2 = 2R_{1}t + 2(\pi)R_{2}^2 $$

 * $$ R_{2} = [\frac {R_{1}^2} {2} - \frac {R_{1}t} {\pi}]^{1/2} = 6.84 cm $$

Next, use the parallel axis theorem to determine $$ I_{y}^{(2)} $$.

 * $$ I_{y}^{(2)} = \frac {t(2R_{1})^3} {12} + 2[\frac {(\pi)R_{2}^4} {4} + (\pi)R_{2}^2(R_{1} + R_{2})^2] $$

 * $$ I_{y}^{(2)} = \frac {2R_{1}^3t} {3} + 2[\frac {(\pi)R_{2}^4} {4} + (\pi)R_{2}^2(R_{1}^2 + 2R_{1}R_{2} + R_{2}^2)] = 87,469 cm^4 $$

Finally, compare $$ I_{y}^{(1)} to I_{y}^{(2)} $$.

 * $$ \frac {I_{y}^{(2)}} {I_{y}^{(1)}} = 11.1 $$

Therefore, it is shown that by redistributing the area of the circle into the shape of the dumbell, the second area moment of inertia increases by an order of magnitude.

= Shear Stress Equations =

Ad-Hoc vs Analytical Method
From the above schematic of Problem 1.1, shear stress across a rectangular cross section is approximated at $${\tau}=\frac {T} {2abt}$$

An analytic solution, applicable across any cross section, is: $$T = \oint \rho\,q\,ds = \iint_{\bar A} 2q\,dA = 2q \bar A$$

Where q is the shear flow, described by the equation $$q=\tau t$$, and Ā is the average area of the cross section.

Shear Strain
Engineering shear strain, not to be confused with tensorial, is described by the equation:

$$\gamma\ = \frac {\delta\ u} {\delta\ y} + \frac {\delta\ v} {\delta\ x} = \frac {\delta\ u_x} {\delta\ y} + \frac {\delta\ u_y} {\delta\ x}$$

Where u is the displacement along the x-axis and v is the displacement along the y. As you can see in Figure 14, the shear strain γ is analogous to the change in the right angle that the new parallelogram creates.

$$ \epsilon_{xy}\ = Tensorial\,Shear\,Strain = \frac {1}{2} \gamma_{xy}\ $$

= References =
 * http://en.wikipedia.org/wiki/Mohr%27s_circle
 * http://en.wikipedia.org/wiki/Longeron
 * http://en.wikipedia.org/wiki/Ad_hoc
 * http://en.wikipedia.org/wiki/Shear_strain

= Contributing Members = The following Team Aero members contributed to this report.

Jared Lee --Eas4200c.f08.aero.lee 16:18, 26 September 2008 (UTC)

Oliver Oyama --Eas4200c.fo8.aero.oyama 16:32, 26 September 2008 (UTC)

Ray Strods --Eas4200c.f08.aero.strods 18:08, 26 September 2008 (UTC)

William Kurth --Eas4200c.f08.aero.kurth 18:53, 26 September 2008 (UTC)

title: User:Uf.team.aero/HW2


text:
Template:Uw-block1

in accordance with Wikiversity policy for. Once the block has expired, you are welcome to make constructive contributions. If you believe this block is unjustified, you may contest the block by adding the text below.

Usage:

title: Template:Uw-block1


text:
User:Eas4200c.f08.aero.lee/HW1

= EAS4200C: An Introduction =

The University of Florida course EAS4200C Aerospace Structures will emphasize an understanding of mechanics, the ability to properly formulate problems, the ability to judge correctness of a solution and the avoidance of old ad hoc methods of structural analysis. The Finite Element Method (FEM) will be employed in order to solve the Partial Differential Equations that govern aerospace structure mechanics. FEM is also widely used to solve other challenges in engineering and applied math. A specific example is the use of FEM to aid in the exploration of oil reservoirs.

Course Information
Throughout the semester there will be many resources implemented that will encourage learning through lecture notes, assigned readings, cooperative learning assignments as well as in class exams.

Exam Guidelines
= Aerospace Structures : Chapter One =

Materials
The primary criteria that distinguish aerospace structures are high stiffness, high strength and light weight. The emphasis on light weight is of very high importance to enable aircraft to perform more efficiently in the atmospheric environment where gravity is an issue. Reducing weight allows aircraft to fly higher, faster and with greater endurance. Often, many different materials will be utilized in different areas of the structure depending on the expected loading and functional condition of that element.

! Stiffness

If the geometry of two materials are the same, stiffness can simply be compared on the basis of the Young's Modulus, E, of the material. Young's Modulus provides a relationship between the stress, σ, and the resulting strain of the material, ε. The Young's Modulus is also the slope of a stress-strain graph, and is numerically calculated as follows:
 * $${E}=\frac {\sigma} {\epsilon}$$

A typical stress-strain curve for an elastic material is shown below.  In metals, phenomenon known as hardening and softening occur before the material will fracture. A more detailed stress-strain curve, typical of metals, is shown below.

! Material Examples

Geometry
Of equal consideration to material selection is the geometry of the structure in question. To take full advantage of the material properties, the geometry of each member must be designed to maximize its capability. Typically this means reducing the amount of material in each member to until structural limits are encountered to minimize weight while maintaining an effective margin of safety.

! Space Frame Geometry

= Problem 1.1: Getting Started =

= Contributing Members = The following Team Aero members contributed to this report.

Jared Lee --Eas4200c.f08.aero.lee 19:57, 16 September 2008 (UTC)

Ray Strods --Eas4200c.f08.aero.strods 00:53, 17 September 2008 (UTC)

Oliver Oyama --Eas4200c.fo8.aero.oyama 02:47, 17 September 2008 (UTC)

Gonzalo Barcia --Eas4200c.f08.aero.barcia 06:47, 18 September 2008 (UTC)

William Kurth --Eas4200c.f08.aero.kurth 07:15, 19 September 2008 (UTC)

title: User:Eas4200c.f08.aero.lee/HW1


text:
Template:Uw-npa1

Welcome to Wikiversity. Although everyone is welcome to contribute constructively to the Community, we would like to remind you not to attack other editors. Please comment on the contributions and not the contributors. Take a look at the welcome page to learn more about contributing to this community.

title: Template:Uw-npa1


text:
Template:Uw-npa2

Please see Wikiversity's no personal attacks policy. Comment on content, not on contributors. Personal attacks damage the community and deter users. Note that continued personal attacks will lead to blocks for disruption. Please Assume Good Faith and keep this in mind while editing.

title: Template:Uw-npa2


text:
Template:Uw-npa3

Please do not attack other editors. If you continue, you may be blocked from editing Wikiversity.

title: Template:Uw-npa3


text:
Template:Uw-pinfo

Wikiversity operates on the principle that every contributor has a right if they wish to remain completely anonymous. Wikiversity's policy on that issue is strictly enforced. Posting private information about a user with the intent to annoy, threaten or harass, specifically their (alleged) name and/or personal details, is strictly prohibited, and users who do that are often immediately blocked from editing Wikiversity. Such posting can cause offence or embarrassment to the victim of the posting, not least because it means that their name, and any personal criticism or allegations made against them can then appear on web searches.

If you have posted such information, please remove it immediately. Please then email to the following [mailto:oversight-l@lists.wikimedia.org this address]. It will then be removed from the archives of Wikiversity.

If you do not ensure that the personal information you posted is removed from this site you will be blocked from editing this site. Remember: Wikiversity's privacy policy is there to protect the privacy of every user, including you.

title: Template:Uw-pinfo


text:
User talk:Aicra~enwikiversity

 Hello Aicra, 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. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on 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.

 * 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.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! -- DarkMage 21:45, 12 September 2008 (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 Aicra. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Aicra~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 23:26, 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) 06:02, 19 April 2015 (UTC)

title: User talk:Aicra~enwikiversity


text:
User:Eml4500.f08.ateam/hw 2

A-Team Teampage

Homework Report #2

This page is intended for Homework Report #2 of the A-Team.

Due date: Friday, 26 September 08, 5 pm EDT = 17:00 EDT (or 21:00 UTC).

Statically Determinate versus Statically Indeterminate
A statically determinate problem is one that can be solved using simple statics relationships. There are three main equations, setting the sum of forces in the x equal to zero and setting the sum of the forces in the y direction equal and setting the sum of the moments about a point equal to zero. With these three equations, any problem with up to three unknowns can be solved. Problems with multiple beams or more than three reaction forces are statically indeterminate since there are insufficient equations to solve for all the unknowns.

An exception to the indeterminate rules is shown in the following example.

Step 1: Global Picture

Here there are four reaction forces, two at each support and a known applied force at the top. It can be noted that the beams in this problem are both two-force members. By looking at each element separately and summing the forces in any direction it is found that they must be both equal and opposite. Summing the moment about either end shows that both forces' line of action must go through each end. Otherwise the moments would not sum to zero.

Therefore the two forces are constrained in one direction. With these simple straight beams, the line of action is along the beam itself. If the beam were curved or had an angle in it, the same logic would apply and the line of action of the two forces will be along the invisible line going through the two attachment points.

This simplification of the problem lowers the degrees of freedom (unknowns) from four to three. Since the direction of the reaction forces are now known, only the magnitude is left to be found.

Step 2: Element Picture
Here the element degrees of freedom and element forces are shown in either the global system or in a local coordinate system. The element number is in the parentheses and the forces (and displacements) are numbered from node 1 to 3, each starting with the x and then y axes.

Step 3: Global Force-Displacement Relationship
The next step is to create a relationship between the forces and displacements in the global coordinate system. The matrices will be in the following form.

$$\underline k^{(e)}*\ \underline d^{(e)}= \underline f^{(e)}$$

Where k (e) is a 4x4 matrix and d (e) and f (e) are 4x1 matrices. In order to find the values of k (e), the formula found on page 225 of Fundamental Finite Element Analysys and Applications must be used. It states that the scalar, $$K = \frac{EA}{L}$$, is multiplied by a matrix composed of functions of the director of cosines.

The director of cosines is a function of the angle of the element and given as, $$l^{(e)}= \vec \tilde{i} \centerdot \vec i = \cos \ \Theta^{(e)}$$ and  $$m^{(e)}= \vec \tilde{i} \centerdot \vec j = \sin \ \Theta^{(e)}$$. The completed matrix for K (e) is as follows.

$$ \underline K^{(e)}= K*\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} \\ l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2 \\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \end{bmatrix} $$

The k (e) matrices for elements 1 and 2 were calculated based on the given data and are as follows.

$$ \underline K^{(1)}= \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & \frac{-9}{16} & \frac{-3\sqrt{3}}{16} \\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & \frac{-3\sqrt{3}}{16} & \frac{-3}{16} \\ \frac{-9}{16} & \frac{-3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16} \\ \frac{-3\sqrt{3}}{16} & \frac{-3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 \\ -0.5625 & -0.3248 & 0.5625 & 0.3248 \\ -0.3248 & -0.1875 & 0.3248 & 0.1875 \end{bmatrix} $$

$$ \underline K^{(2)}= \begin{bmatrix} \frac{5}{2} & \frac{-5}{2} & \frac{-5}{2} & \frac{5}{2} \\ \frac{-5}{2} & \frac{5}{2} & \frac{5}{2} & \frac{-5}{2} \\ \frac{-5}{2} & \frac{5}{2} & \frac{5}{2} & \frac{-5}{2} \\ \frac{5}{2} & \frac{-5}{2} & \frac{-5}{2} & \frac{5}{2} \end{bmatrix}=\begin{bmatrix} 2.5 & -2.5 & -2.5 & 2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ 2.5 & -2.5 & -2.5 & 2.5 \end{bmatrix} $$

We can observe that K is symmetric, meaning $$k^{(e)T} = K ^(e)$$.

$$K^(e)= $$$$\begin{bmatrix} K_{11}&K_{12}&K_{13}&K_{14} \\ &K_{22}&K_{23}&K_{24} \\ & &K_{33}&K_{34} \\ &sym. & &K_{44} \\ \end{bmatrix} $$

Element 1:
From the FBD of element 1, we see that the angle from the x-axis is 30°.

θ(1) = 30ο l(1) = cosθ(1) = cos30ο = $$\frac{\sqrt3}{2}$$ m(1) = sinθ(1) = sin30° = $$\frac{1}{2}$$

For element 1, the stiffness is given as: $$K^{(1)} = \frac{E^{(1)}A^{(1)}}{L^{(1)}}$$ = $$\frac{3}{4}$$

And the matrix of k(1) is given as:

$$ \underline K^{(1)}= K*\begin{bmatrix} k^{(1)}(l^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(l^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} \\ k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(m^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(m^{(1)})^2 \\ -k^{(1)}(l^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(l^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} \\ -k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(m^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(m^{(1)})^2 \end{bmatrix} $$

In this matrix, only three values need to be calulated. The rest only differ by being positive or negative. As you can see this matrix is symetric. Therefore,

 kij(1) = kji(1)

And in general:

 kij(e) = kji(e) and  k(e)T = k(e)

Element 2:
From the FBD of element 2, we see that the angle from the x-axis is -45°.

θ(2) = -45°

l(2) = cosθ(2) = cos(-45°) = $$\frac{\sqrt2}{2}$$

m(2) = sinθ(2) = sin(-45°) = $$\frac{-\sqrt2}{2}$$

For element 2, the stiffness is given as: $$K^{(2)} = \frac{E^{(2)}A^{(2)}}{L^{(2)}}$$ = 2.5

And the matrix of k(2) is given as:

$$ \underline K^{(2)}= K*\begin{bmatrix} k^{(2)}(l^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(l^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} \\ k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(m^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(m^{(2)})^2 \\ -k^{(2)}(l^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(l^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} \\ -k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(m^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(m^{(2)})^2 \end{bmatrix} $$

Global Force Displacement Relation
The Global Force Displacement Relation is known as K d = F. Matrix K is the element stiffness matrix in global coordinates. In our example, the global stiffness matrix is a 6 x 6 grid, the global displacement matrix is 6 x 1, and the global force matrix is 6 x 1, and is shown below:

 * $$\begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} = \begin{bmatrix}

K_{11}&K_{12}&K_{13}&K_{14}&K_{15}&K_{16} \\ K_{21}&K_{22}&K_{23}&K_{24}&K_{25}&K_{26} \\ K_{31}&K_{32}&K_{33}&K_{34}&K_{35}&K_{36} \\ K_{41}&K_{42}&K_{43}&K_{44}&K_{45}&K_{46} \\ K_{51}&K_{52}&K_{53}&K_{54}&K_{55}&K_{56} \\ K_{61}&K_{62}&K_{63}&K_{64}&K_{65}&K_{66} \\ \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}$$

Since the stiffness matrix is 6 x 6, there are six degrees of freedom in the problem. The Global Force Displacement Relation can be shown in a more compact notation:
 * $$\begin{bmatrix} K_{ij} \end{bmatrix} \begin{Bmatrix} d_{j} \end{Bmatrix} = \begin{Bmatrix} F_{i} \end{Bmatrix}$$

or    $$\sum_{j=1}^6 $$ Kij dj  = Fi , i=1,2,3,4,5,6

K or $$\begin{bmatrix} K_{ij} \end{bmatrix}$$n x n is the global stiffness matrix. d or $$\begin{Bmatrix} d_{j} \end{Bmatrix}$$n x 1 is the global displacement matrix. F or $$\begin{Bmatrix} F_{i} \end{Bmatrix}$$n x 1 is the global force matrix.

These global matrices are different than the element force displacement relation k(e) d(e) = f(e). In this relationship, recall the k(e) is a 4 x 4 matrix, called the element stiffness matrix where (e) is the element number. d(e) and f(e) are the element displacement and force matrices, both of which are 4 x 1. These element matrices are smaller than the global ones because each one pertains to an element rather than the entire picture. To go from an element matrix to a global matrix, the correspondence between the element displacement degrees of freedom and the global displacement degrees of freedom need to be identified.

At the global level, the degrees of freedom are $$d_1,\ d_2,\ d_3,\ d_4,\ d_5,\ d_6$$. At the element level, there is a set of degrees of for each element.

In Element 1, the degrees of freedom are $$d_1^{(1)},\ d_2^{(1)},\ d_3^{(1)},\ d_4^{(1)}$$. Element 2 is $$d_1^{(2)},\ d_2^{(2)},\ d_3^{(2)},\ d_4^{(2)}$$.

The schematic above shows that some of the global degrees of freedom are the same as the element. At node 1:
 * d1 = d1(1)
 * d2 = d2(1)

Node 2 on the global schematic has the same displacement degrees of freedom as on both the first and second element; therefore:
 * d3 = d3(1) = d1(2)

 * d4 = d4(1) = d2(2)

Node 3 is only:
 * d5 = d3(2)
 * d6 = d4(2)

The conceptual step of assembly has a small matrix for each element stiffness matrix and two zero matrices all within the global force displacement relation. The basic matrix is shown below:

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix}$$

Assembly of Force Displacement Matrix
The global force displacement matrix is assembled from the 2 element displacement matrices.

The Global Stiffness Matrix with symbolic representation:
 * $$\begin{bmatrix}

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

Note: The overlap in the 2 matrices comes from the fact that global node 2 is shared between both elements.

Calculating the Matrix Values:
K11 = k111 = 0.5625 K12 = k112 = 0.3247 K33 = k133 + k211 = 3.0625 K34 = k134 + k212 = -2.1752 K43 = k143 + k221 = -2.1752 K44 = k144 + k222 = 2.6875

Elimination of the known degrees of freedom reduces the global force displacement relationship. d1 = d2 = d5 = d6 = 0.

Two Bar Truss Matlab Example
Here is the Matlab code run for this example:

% Two bar truss example clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4; nodes = [0, 0; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta)]; dof=2*length(nodes); conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1); %load vector R = zeros(dof,1); R(4) = P;    % Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k;   end K    R     % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e, A, ...           nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g    results

Here are the results for the code:

k_local = 0.75       -0.75        -0.75         0.75

k = 0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local = 5   -5    -5     5

k = 2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5

K = 0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5

R = 0    0     0     7     0     0

d = 0           0        4.352       6.1271            0            0

reactions = -4.4378     -2.5622       4.4378      -4.4378

results = 1.7081      5.1244       8.5406       5.1244       17.081       0.6276       1.8828        3.138       1.8828

Step 4: Elimination of Known Degrees of Freedom
The next step in the Finite Element Method is to eliminate the known degrees of freedom by reducing the global force-displacement relationship. The displacement matrix from the previous two-truss system can be reduced by applying the fixed boundary conditions:

$$d_1=d_2=d_5=d_6=0$$

The resulting displacement matrix is:

$$d= \begin{bmatrix} 0 \\ 0 \\ d_3  \\ d_4 \\ 0 \\ 0 \end{bmatrix} $$

After applying these conditions, the corresponding columns in the global stiffness matrix can be deleted and the resulting matrix is shown below.

$$F=K \begin{bmatrix} 0 \\ 0 \\ d_3  \\ d_4 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} k_{13} & k_{14} \\ k_{23} & k_{24} \\ k_{33} & k_{34} \\ k_{43} & k_{44} \\ k_{53} & k_{54} \\ k_{63} & k_{64} \\ \end{bmatrix} \begin{bmatrix} d_3 \\ d_4 \\ \end{bmatrix} $$

The next step in reducing the force-displacement relationship is to apply the Principle of Virtual Work. This allows the corresponding rows of the global stiffness matrix and the global force matrix to be deleted. The resulting force-displacement relation is

$$ \begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \\ \end{bmatrix} \begin{bmatrix} d_3 \\ d_4 \\ \end{bmatrix}= \begin{bmatrix} F_3 \\ F_4 \\ \end{bmatrix} $$ or $$\mathbf{K}\cdot\mathbf{d} = \mathbf{F}$$

Based on the given information, the reduced force matrix is a known, as shown below.

$$\underline{\overline{F}}= \begin{bmatrix} F_3 \\ F_4 \\ \end{bmatrix}= \begin{bmatrix} 0 \\ P \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 7 \\ \end{bmatrix} $$

Finally, the reduced global displacement matrix can be solved by multiplying the reduced force matrix by the inverse of the reduced global stiffness matrix.

$$ \begin{bmatrix} d_3 \\ d_4 \\ \end{bmatrix} =\begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \\ \end{bmatrix}^{-1} \begin{bmatrix} 0 \\ P \\ \end{bmatrix}= \begin{bmatrix} 4.352 \\ 6.127 \\ \end{bmatrix} $$

Step 5: Compute Reactions
After the displacements of Node 2 are computed, then the next step is to compute the reactions. There are two methods that can be used to compute the reactions. The first method is to use the element force-displacement relations

$$k^{(e)}d^{(e)}=f^{(e)}$$

for elements 1 and 2. The element displacement matrices can be determined from the global displacement matrix.

$$\underline{d}^{(1)}= \begin{bmatrix} 0 \\ 0 \\ 4.352 \\ 6.127 \\ \end{bmatrix}

\underline{d}^{(2)}= \begin{bmatrix} 4.352 \\ 6.127 \\ 0 \\ 0 \\ \end{bmatrix} $$

Solving for element forces of Element 1
$$k^{(1)}d^{(1)} = f^{(1)}$$

$$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{bmatrix} 0.5625 & 0.32776 & -0.5625 & -0.32476 \\ 0.32476 & 0.1875 & -0.32476 & -0.1875 \\ -0.5625 & -0.32476 & 0.5625 & 0.32476 \\ -0.32476 & -0.1875 & 0.32476 & 0.1875 \end{bmatrix} $$ $$\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

This relation can be reduced to the following by the observation that the first two terms in the element displacement matrix are zero. This is due to matrix algebra causing the first two columns of the element stiffness matrix will completely drop out.

 $$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{bmatrix} -0.5625 & -0.32476 \\ -0.32476 & -0.1875 \\ 0.5625 & 0.32476 \\ 0.32476 & 0.1875 \end{bmatrix} $$ $$\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Carrying out this relation reveals the element force matrix.

 $$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \\ \end{Bmatrix}$$

In which, $$f^{(1)}_1$$ and $$f^{(1)}_2$$ are reaction forces while $$f^{(1)}_3$$ and $$f^{(1)}_4$$ are internal forces.

Element 1 is in equilibrium because the reaction forces and internal forces offset each other and the sum of the moments about any point on Element 1 is zero.

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

$$\sum F_x = f^{(1)}_2 + f^{(1)}_4 = 0$$

$$\sum M_{any} = 0$$

Solving for element forces of Element 2

$$P^{(1)}_1 = \sqrt{f^{(1)}_1 + f^{(1)}_2}$$

Accounting for the applied force
In order to proceed, the applied force P must be accounted for in the free-body diagram and the equilibrium of all nodes must be verified. The equilibrium of global nodes 1 and 3 were previously verified.

Statics method
There is a second method to solve for the reactions and internal forces of the 2-bar truss depicted here. This method uses the principles of statics, but also must employ the Euler Cut Principle. The Euler Cut Principle must be utilized because the truss is considered statically indeterminate.

Method 1

Method 2

There are multiple ways of applying the Euler Cut Principle to the truss. Each way is determined simply by where the cut is made in the truss system. Depending on this choice, there are two or three resulting free-body diagrams from which the forces are derived.

Contributing Team Members
The following students contributed to this report:

Matthew Philbin Eml4500.f08.Ateam.philbin 22:28, 24 September 2008 (UTC)

Mark Barry Eml4500.f08.Ateam.barry 13:41, 25 September 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 17:29, 25 September 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 04:00, 26 September 2008 (UTC)

Daniel DiDomenico Eml4500.f08.ateam.didomenico 04:02, 26 September 2008 (UTC)

Ian Slevinski Eml4500.f08.Ateam.Slevinski 18:02, 26 September 2008 (UTC)

title: User:Eml4500.f08.ateam/hw 2


text:
Talk:Metaphysics (Alternate View)/Creation/Fields

What's AV? Emesee 22:04, 12 September 2008 (UTC)
 * Don't know either but on the subpages at begin it is mentioned "alternate view" - perhaps that ? Will write also the user per mail. Erkan Yilmaz uses the Chat (try) 09:41, 14 September 2008 (UTC)


text: Category:Introduction to Abstract Algebra/Lectures


text: Category:Abstract Algebra

title: Category:Abstract Algebra


text:
Talk:Learning from conflict and incivility/Erkan Yilmaz

I appreciate your joining the project, and your input. Would you like me to clean up your English a little, or do you prefer leaving it as-is? The Jade Knight 00:42, 13 September 2008 (UTC)
 * Feel free to edit it, Erkan Yilmaz uses the Chat (try) 07:23, 13 September 2008 (UTC)
 * Done; hope I didn't misinterpret anything you intended. The Jade Knight 08:24, 13 September 2008 (UTC)
 * Thanks very much for the help Jade. I hope I could with this edit clear your questions ? "relief chat": chat can have many purposes in my view, e.g. just making jokes to release the stress from a hectic day - for me that is the "fun chat". But since others are not familiar with my definitions I changed it. Erkan Yilmaz uses the Chat (try) 08:50, 13 September 2008 (UTC)
 * Glad to help. Let me know if you'd like me to clean anything up in the future.  If it's short, I'll generally be glad to.  I made one last little change, which I think better reflects what you mean.  The Jade Knight 08:59, 13 September 2008 (UTC)

It seems...
"Welcoming: For JWSchmidt, "welcoming" means deleting or merging as little content as possible. JWSchmidt believes that deleting and merging content is rude and shows hostility to newcomers and their "seedling" projects. This is very important to JWSchmidt."

It seems like to lean on the side of keeping content risks off-putting certain contributors less than deleting content. I don't know how many users leave projects annoyed because someone didn't delete the page that they started and they didn't request that it be deleted. Emesee 09:45, 13 September 2008 (UTC)


text:
Template:No fairuse

Files put in Category:Files needing fair use rationales.

When applying this tag: Use {{no fairuse|month=July|day=29|year=2024}} or simply {{subst:nfu}}.

title: Template:No fairuse


text:
How to be a Wikimedia sysop/User page

''On this page, consider what kinds of content might be particularly helpful on sysop's user pages. Link to examples if possible.''
 * 1) To the extent to which the sysop feels comfortable, some personal, biographical information can help the community in getting to know about those with additional tools/priveleges. This can help to build community trust and mutual understanding. This needs to be balanced with a sense of "online safety" because sysops may also become targets of frustrated users. Examples:
 * 2) * Countrymike explains a bit about who he is, his involvement and status on other wiki projects.
 * 3) A friendly and accessible Custodian makes it easy to access multiple communication channels (ideally), e.g.,
 * 4) * Gbaor provides clearly listed contact details.
 * 5) Record of actions: A transparent custodian who is committed to accountability makes his/her actions readily available for review by others, e.g.,
 * 6) * Draicone makes it very easy for others to see what he/she has been up to!

A sysop user page on Wikiversity which seems to bring all these elements together is Cormaggio's.

title: How to be a Wikimedia sysop/User page


text:
User:Eml4500.f08.team.foskey.ckf/HW2

Statically Indeterminant?
Proving that the truss problem is not Statically Indeterminate

At first glance, the truss problem appears to be statically indeterminate. The truss has four unknown reaction forces, and there are only three equations to use to solve for those values. However, the truss can be broken up into its components. Each of the components is a two force member. To further simplify things, a new coordinate system is chosen such that the x’-axis is along each of the elements.

Solving the Statics equations
In element 1, if we take the sum of the moments around node 2, we find that force f (1)2 must equal zero. Taking the sum of the moments about node 1, we find that force f (1)4 is related to the known force P, the known angle force P acts upon, and the known length of the element.

$$ \sum M_{2}^{(1)} = 0 = f_{2}^{(1)}*L^{1} $$ therefore $$ f_{2}^{(1)} = 0$$

$$ \sum M_{1}^{(1)} = 0 = f_{4}^{(1)}*L^{1}+ Pcos \theta *L^{1} $$ therefore $$ f_{4}^{(1)} = -Pcos \theta $$

In element 2, taking the sum of node1 and node 2 results in f (2)2 and f (2)4 both being zero.

$$ \sum M_{1}^{(2)} = 0 = f_{4}^{(2)}*L^{2} $$ therefore $$ f_{4}^{(2)} = 0$$

$$ \sum M_{2}^{(2)} = 0 = -f_{2}^{(2)}*L^{2} $$ therefore $$ f_{2}^{(2)} = 0$$

Combining the elements again will show that force f (1)3 corresponds to force f (2)2, therefore force f (1)3 is zero as well. Similarly, force f (1)4 corresponds to force f (2)1, and the two forces are equal.

$$ f_{3}^{(1)}=f_{2}^{2} = 0 $$ $$ f_{1}^{(2)}=f_{4}^{1} = -Pcos \theta $$

Returning to element 2, and summing the forces in the x’ direction and setting them equal to zero gives force f (2)3 being equal to –f (2)1. Returning to element 1, and setting the sum of the forces in the x’ direction equal to zero allows us to obtain the value of force f (1)1.

$$ f_{3}^{(2)}= -f_{1}^{2} = Pcos \theta $$

$$ \sum F_{x'}^{(1)} = 0 = f_{3}^{(1)} + Pcos \theta + f_{1}^{(1)} $$

$$ = -(f_{3}^{(1)} + Psin \theta ) = -Psin \theta $$

Thus all of the forces acting on the truss can be found by using statics equations, and taking into account the two force members. The Finite Element Method is not necessary to solve this problem.

Verification of two force member forces
Since the element bars that make up the truss are two force members, the axial forces on each end of the bars should be equal and opposite. This can be shown by calculating the resultant forces for each end of each bar.

Element 1
As shown earlier, by changing the local coordinate system, the axial forces are found. These were:

$$ f^{(1)}_{1} = -Psin \theta $$ $$f^{(1)}_{3} = 0 $$

where \ theta is the angle between force P and the axis of the bar, and $$f^{(1)}_{1}$$ and $$f^{(1)}_{3}$$ are the resultant axial forces for each end of the element. Knowing that in the global figure, element 1 was at a 30 degree angle from the horizontal, and that force P acted vertically, the \theta in the element 1 figure is 60 degrees. For the elemental node 2 of element 1, since the force $$f^{(1)}_{3}$$ is zero, the axial force component is composed of only the axial component of force P, which is $$Psin \theta$$. Also knowing that P = 7, we can calculate that the axial forces at each end of the element are 5.124. Each of these forces act outward from the element.

Element 2
As previously shown, the axial forces in element 2 were:

$$f^{(2)}_{1} = -Pcos \theta$$ $$f^{(2)}_{3} = Pcos \theta$$ Again, the axial forces at each end should be equal and opposite, as the element is a two force member. Using the original problem data with P = 7, the axial forces in element 2 are found to be 6.276. These forces too act outward from the center of the element.

Director Cosines Proof
Proving that    m (e)  =  sinθ (e)

Since $$ m^{(e)} = \hat \tilde i \bullet \hat j $$

and $$ \hat \tilde i = cos \theta ^{(e)} \hat i + sin \theta^{(e)} \hat j $$ expanding $$ \hat\tilde i \bullet \hat j= (cos \theta ^{(e)} \hat i + sin \theta^{(e)} \hat j ) \bullet \hat j $$

$$ = cos \theta ^{(e)} \hat i \bullet \hat j + sin \theta^{(e)} \hat j \bullet \hat j $$

Since $$ \hat i \bullet \hat j = 0 $$ and $$ \hat j \bullet \hat j= 1 $$ we obtain $$m^{(e)} = \hat \tilde i \bullet \hat j = cos \theta ^{(e)} * 0 + sin \theta ^{(e)}*1 $$ proving that $$ m^{(e)}=sin\theta ^{(e)}$$

The Stiffness Matrices
Solving the k1 and k2 stiffness matrices

Using the data for the truss problem and the equation for calculating the elemental stiffness matrix, we can simply plug in the known values into the equations. The end results for the stiffness matrices are:

$$ k^{(1)} = \frac{3*1}{4}$$ $$ \begin{bmatrix} (\frac{\sqrt{3}}{2})^{2} & (\frac{\sqrt{3}}{2} * \frac{1}{2}) & -(\frac{\sqrt{3}}{2})^{2} & -(\frac{\sqrt{3}}{2} * \frac{1}{2}) \\

( \frac{\sqrt{3}}{2}* \frac{1}{2}) & (\frac{1}{2})^{2} & - ( \frac{\sqrt{3}}{2} * \frac{1}{2}) & - ( \frac{1}{2})^{2} \\

-(\frac{\sqrt{3}}{2})^{2} & -(\frac{\sqrt{3}}{2} * \frac{1}{2})& (\frac{\sqrt{3}}{2})^{2} & (\frac{\sqrt{3}}{2} * \frac{1}{2}) \\

-(\frac{\sqrt{3}}{2} * \frac{1}{2})& -(\frac{1}{2})^{2} & (\frac{\sqrt{3}}{2} * \frac{1}{2}) &(\frac{1}{2})^{2} \\ \end{bmatrix} $$

$$ = \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} \\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \\ \end{bmatrix} $$

and

$$ k^{(2)} = \frac{5*2}{2} \begin{bmatrix} (\frac{\sqrt{2}}{2})^{2} & (\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & -(\frac{\sqrt{2}}{2})^{2} & -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) \\ (\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & -(\frac{\sqrt{2}}{2})^{2} & -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & -(\frac{\sqrt{2}}{2})^{2} \\ -(\frac{\sqrt{2}}{2})^{2} & -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & (\frac{\sqrt{2}}{2})^{2} & (\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) \\ -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & (\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) \\ \end{bmatrix} $$

$$ = \begin{bmatrix} \frac{10}{4} & -\frac{5}{2} & -\frac{10}{4} & \frac{5}{2} \\ -\frac{5}{2} & \frac{10}{4} & \frac{5}{2} & -\frac{10}{4} \\ -\frac{10}{4} & \frac{5}{2} & \frac{10}{4} & -\frac{5}{2} \\ \frac{5}{2} & -\frac{10}{4} & -\frac{5}{2} & \frac{10}{4} \\ \end{bmatrix} $$

Equilibrium of the Node
In the Finite Element Method of solving the two bar truss problem, bring in the force P back into the problem requires looking at the equilibrium of global node 2. Using the Euler Cut Principle to isolate global node 2, we can see that in order for the node to be in equilibrium, the axial forces in each element must cancel out the force P. This can easily be shown by resolving the axial forces into their horizontal and vertical components, and then setting the sum of the forces in each direction equal to zero.

$$\sum F_{horizontal} = f^{(1)}_{1}*cos(30) + f^{(2)}_{3}*cos(-45) = -5.124cos(30) + 6.276cos(-45)=0$$

$$\sum F_{vertical} = f^{(1)}_{1}*sin(30) + f^{(2)}_{3}*cos(-45) + P =-5.124sin(30) + -6.276sin(-45) + 7 =0$$

title: User:Eml4500.f08.team.foskey.ckf/HW2