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Wikiversity:Community Review/Moulton's block/Archive

Procedural Matter
Closed The start of this subsection has been struck out by the user who brought it up. There have been no further comments in one week, and since the original nom considers the issue addressed, I'll archive it now. If there any further concerns, please open a new section. --mikeu talk 14:11, 15 October 2008 (UTC)

I would like to amend the above matter and focus on a broader issue, the role of Jimbo. What power should he have. He is not an active member of this community, and arbitrary actions seem to make us subject to his will, a violation of academic freedom. Geo.plrd 01:38, 7 October 2008 (UTC) The community has addressed the issue of Jimbo making the block. Geo.plrd 15:08, 8 October 2008 (UTC)

 * Could you detail your involvement with "Arbcab," your multiple rejected bids for admin/MedCab on en.wp, your sanction for providing legal advice re. MyWikiBiz, what Zoe, Quiddity and ^demon have determined your motiviations were for your various en.wp edits, along with your involvement in "Esperanza" and the disruptive recreation of such? Inquring minds want to know exactly how much drama you intend to bring here before someone makes you stop. Why aren't you out powertripping at Conservapedia, exactly? Salmon of Doubt 16:36, 7 October 2008 (UTC)
 * Excuse me? I was under the impression that this is Wikiversity? If I am mistaken, please let me know. Geo.plrd 19:52, 7 October 2008 (UTC)
 * Could you also comment on your invasion of User:H's privacy (I believe you were banned from en.wp over this), as it's especially relevent to this matter. Salmon of Doubt 16:42, 7 October 2008 (UTC)
 * You mean the mistake Daniel made? He thought another user was me and made an honest mistake. Geo.plrd 19:52, 7 October 2008 (UTC)
 * And how is any of that relevant, Salmon? If you have a problem with what Geo.plrd is doing, why not address that, rather than attempt to poison the well? Sχeptomaniacχαιρετε 17:02, 7 October 2008 (UTC)

 * It's only poisoning the well if it's unrelated. Geoff's consistant actions to create disruptive bureaucracy and then attempt join it in a position of power is exactly what we have here. It's laughable. Salmon of Doubt 17:06, 7 October 2008 (UTC)
 * Salmon, you are spinning discussion away from the pressing issue of a bad precedent. For general information, here is what I have done here. I revamped a dormant department and am working on a survey course of US history? What is so bureaucratic about that? Geo.plrd 19:52, 7 October 2008 (UTC)
 * Blocking disruptive influences with a minimum of your inflamed bureaucracy is not a bad precedent. You are spimming discussion away from the pressing issue of your three year long abuse of process and sockpuppetry. Salmon of Doubt 20:42, 7 October 2008 (UTC)
 * Salmon, while Geo may hold a minority view here at Community Review, his contributions at Wikiversity which I have noticed have been thus far constructive. I do not think it is appropriate to be attacking him for his actions at other pages here.  The Jade Knight (d'viser) 20:19, 7 October 2008 (UTC)
 * I disagree. If Geo, a consumate bureaucratic warrior, were arguing for anything but more bureaucratic nonsense, that would be true. Salmon of Doubt 20:42, 7 October 2008 (UTC)
 * Are you ignoring his other constructive contributions at Wikiversity recently? The Jade Knight (d'viser) 21:04, 7 October 2008 (UTC)
 * No. Beyond the damage he's causing here, I am deeply concerned that a high school junior (16 years old, for those not in the US) whose most substantial contribution to the public discourse is at "conservapedia," widely seen as anti-historical and a polemical far-rightist institution of disinformation intends to be the "INSTRUCTOR" in a "survey of the history of the United States of America, taught at the college level." Luckily, Wikiversity is the university that everyone can edit - when he attempts to inform us that Uncle Roy Cohn was a GREAT MAN WHO DID NOT DIE OF AIDS AND WAS CERTAINLY NO HOMOSEXUAL, I'll be right there to remind everyone that that's nonsense. Salmon of Doubt 21:09, 7 October 2008 (UTC)
 * It seems to me that you are ignoring his other constructive contributions. Please be aware, Salmon, that right now you appear to be engaging in outing, and you are without question being very rude.  I know a few teenagers with enough understanding of American History to be able to teach many others about elements of it, including myself, and I hold a degree in History.  As you are not involved at the School of History at all, I do not see how you consider yourself to be in any position to judge who should or should not contribute there.  I do not see Geo doing "any damage" on Wikiversity.  Apart from his disagreement with you on this page, can you point out any other specific examples of "damage" he has done on Wikiversity?  The Jade Knight (d'viser) 21:14, 7 October 2008 (UTC)
 * I would be engaging in "outing" if I revealed anything at all about Geoff that was non public or even remotely hidden. It is, however, not. Geoff reveals his name here. He talks about Conservapedia here. He reveals his age here. This is all public. I look foward to Geoff to not damaging wikiversity any further. Salmon of Doubt 21:22, 7 October 2008 (UTC)
 * Your actions are still very hostile and rude. And you have ignored my other comments and concerns.  The Jade Knight (d'viser) 21:26, 7 October 2008 (UTC)
 * Agreed. Unnecessarily rude and hostile behavior is certainly not any improvement over advocating unnecessary bureaucracy.  A simple argument would have been far better than dragging up what I still consider irrelevant history. Sχeptomaniacχαιρετε 23:20, 7 October 2008 (UTC)
 * Salmon, please back off. We'll deal with issues and activity that happen here in Wikiversity - I've tried to address the issue Geoff brings up, as have others. Cormaggio talk 09:58, 8 October 2008 (UTC)
 * Fine. I'm done here. Salmon of Doubt 12:48, 8 October 2008 (UTC)
 * Geoff: Jimbo acted as he very occasionally does, and in the same way as all stewards occasionally do: intervening in (often) unprecedented circumstances, and where the community is either non-existent, or damaged, or where there could be a danger of serious community damage. I think the 'precedent' we should be discussing here is action which had not been explicitly discussed in a community forum - not whether Jimbo has any authority here. The key issue is whether we, as a community, have policies and procedures in place to deal with problems when they arise - which, when working within Wikimedia's broad mission, would mean that Jimbo, or any steward, or Wikimedia agent would never have to intervene. Cormaggio talk 20:53, 7 October 2008 (UTC)

 * Rather than attempt to tell Jimbo what to do, it would be wiser and more effective just to make his intervention unnecessary in the future. This page seems to be a move to create a more appropriate place to discuss the less cut-and-dried issues that will arise now and then. Sχeptomaniacχαιρετε 23:20, 7 October 2008 (UTC)


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User:Eml4500.f08.ateam.didomenico/hw 4

Five-Bar Truss System: MATLAB Dissection
Analysis of the five-bar truss system MATLAB code is important before attempting to write code for the three-bar truss system. The following code is from the  file found at the website for the class textbook, Fundamental Finite Element Analysis and Applications.

Additional Functions
The following functions are essential for the MATLAB program above to run. They are also found at the course textbook website. These functions include,  , and. The details of these functions are outlined below.

This function generates the element stiffness matrix for each element. When the function is called it calculates the X and Y coordinate positions of the element ends from the coordinates passed in. It produces the element length from this information. Next it calculates the direction cosines of the element. Using the Young's modulus and cross-sectional area of the element, the element stiffness matrix, k, is assembled. The stiffness matrix is passed back to the.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

This function outputs the axial strain, axial stress, and axial force of each element.

MATLAB Results
The following is a printout of the exact results of the MATLAB analysis. It includes the complete Global Stiffness matrix (K), the Applied Force matrix (R), the Displacement matrix (d), the reaction forces, and the resulting stresses and strains.

K = Columns 1 through 6 32600       76067       -32600       -76067            0            0        76067  2.9749e+005       -76067 -1.7749e+005            0    -1.2e+005 -32600      -76067  2.4309e+005  1.1914e+005       -32998        32998 -76067 -1.7749e+005 1.1914e+005  2.4309e+005        32998       -32998 0           0       -32998        32998    1.53e+005       -32998 0   -1.2e+005        32998       -32998       -32998    1.53e+005 0           0 -1.7749e+005       -76067    -1.2e+005            0 0           0       -76067       -32600            0            0 Columns 7 through 8 0           0            0            0 -1.7749e+005       -76067 -76067      -32600    -1.2e+005            0 0           0  2.9749e+005        76067 76067       32600 R = 0          0           0     -150000           0           0           0           0 d = 0           0      0.53895     -0.95306       0.2647      -0.2647            0            0 reactions = 54927 1.5993e+005 -54927     -9926.7 results = -0.0001743     -34.859 -1.3944e+005 -3.15e-005     -6.2999       -25200 -5.2941e-005     -10.588       -31764 -5.2941e-005     -10.588       -31764 0.00032087      22.461        44922

The d matrix represents the displacements at each node.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\ 0 \\ 0.53895 \\ -0.95306 \\ 0.2647 \\ -0.2647 \\ 0 \\ 0 \\ \end{Bmatrix}$$

Reaction forces at Global Node 1 and Global Node 4:

Resulting Stresses and Strains:

Three-Bar Truss System: MATLAB Dissection
The following code solves the three-bar truss system presented in class. The two-bar truss system code provided by Dr. Vu-Quoc on the EML4500 website was modified to account for the additional element. The original code can be found here. After analysis, the resulting deformation was scaled by a factor of 1/500 for presentation.

MATLAB Results
n_nodes = 4 n_elem = 3 k = 0.9     0.51962         -0.9     -0.51962      0.51962          0.3     -0.51962         -0.3         -0.9     -0.51962          0.9      0.51962     -0.51962         -0.3      0.51962          0.3 k = 0.6     0.34641         -0.6     -0.34641      0.34641          0.2     -0.34641         -0.2         -0.6     -0.34641          0.6      0.34641     -0.34641         -0.2      0.34641          0.2 k = 0.3         0.3         -0.3         -0.3          0.3          0.3         -0.3         -0.3         -0.3         -0.3          0.3          0.3         -0.3         -0.3          0.3          0.3 K = Columns 1 through 6 0.9     0.51962         -0.9     -0.51962            0            0      0.51962          0.3     -0.51962         -0.3            0            0         -0.9     -0.51962          1.8        1.166         -0.6     -0.34641     -0.51962         -0.3        1.166          0.8     -0.34641         -0.2            0            0         -0.6     -0.34641          0.6      0.34641            0            0     -0.34641         -0.2      0.34641          0.2            0            0         -0.3         -0.3            0            0            0            0         -0.3         -0.3            0            0  Columns 7 through 8 0           0            0            0         -0.3         -0.3         -0.3         -0.3            0            0            0            0          0.3          0.3          0.3          0.3 R = 0    0     0    30     0     0     0     0 d = 0           0      -435.17       671.77            0            0            0            0 reactions = 42.588      24.588       28.392       16.392      -70.981      -70.981

Unstable Three-Bar Truss: MATLAB Dissection
The following code is the three-bar truss system code modified for the unstable three-bar truss system shown at the left of the figure below. To stabilize this truss, an additional member is added at shown at the right of the figure. A force, P, of magnitude 0.5 is applied as shown to evaluate the truss system stability.

Breakdown of FEM
Running this MATLAB code results in a breakdown of the FEM. The system above is unstable and results in a singular stiffness matrix. FEM cannot be used to solve for the displacements and forces at each node. This breakdown is shown below.

EDU>> three_unstable n_nodes = 4 n_elem = 3 k = 0    0     0     0     0     6     0    -6     0     0     0     0     0    -6     0     6 k = 6    0    -6     0     0     0     0     0    -6     0     6     0     0     0     0     0 k = 0    0     0     0      0     6     0    -6      0     0     0     0      0    -6     0     6 K = 0    0     0     0     0     0     0     0      0     6     0    -6     0     0     0     0      0     0     6     0    -6     0     0     0      0    -6     0     6     0     0     0     0      0     0    -6     0     6     0     0     0      0     0     0     0     0     6     0    -6      0     0     0     0     0     0     0     0      0     0     0     0     0    -6     0     6 R = 0     0      0      0      1      0      0      0 Warning: Matrix is singular to working precision. > In NodalSoln at 12 In three_unstable at 67 d = 0     0    Inf NaN NaN NaN 0     0 reactions = NaN NaN NaN NaN

Stabilization Due to Additional Member
To add the cross-bar member to the truss the MATLAB code is modified to the following.

This code produces the following results.

EDU>> three_stable n_nodes = 4 n_elem = 4 k = 2.2496e-032 3.6739e-016 -2.2496e-032 -3.6739e-016 3.6739e-016           6 -3.6739e-016           -6 -2.2496e-032 -3.6739e-016 2.2496e-032  3.6739e-016 -3.6739e-016          -6  3.6739e-016            6 k = 6           0           -6            0            0            0            0            0           -6            0            6            0            0            0            0            0 k = 0    0     0     0     0     6     0    -6     0     0     0     0     0    -6     0     6 k = 2.1213      2.1213      -2.1213      -2.1213       2.1213       2.1213      -2.1213      -2.1213      -2.1213      -2.1213       2.1213       2.1213      -2.1213      -2.1213       2.1213       2.1213 K = Columns 1 through 7 2.1213      2.1213 -2.2496e-032 -3.6739e-016      -2.1213      -2.1213            0 2.1213      8.1213 -3.6739e-016           -6      -2.1213      -2.1213            0 -2.2496e-032 -3.6739e-016           6  3.6739e-016           -6            0            0 -3.6739e-016          -6  3.6739e-016            6            0            0            0 -2.1213     -2.1213           -6            0       8.1213       2.1213            0      -2.1213      -2.1213            0            0       2.1213       8.1213            0            0            0            0            0            0            0            0            0            0            0            0            0           -6            0  Column 8 0           0            0            0            0           -6            0            6 R = 0           0          0.5            0            0            0            0            0 d = 0           0      0.40237 -2.4638e-017 0.31904   -0.083333            0            0 reactions = -0.5        -0.5            0          0.5

title: User:Eml4500.f08.ateam.didomenico/hw 4


text: User:Roux

title: User:Roux


text: User talk:Roux

title: User talk:Roux


text:
User:EAS4200c.f08.Blue.E/Lecture 22

How to relate strains to $$E,\nu$$

first is to define Hooke's law in three dimensions which is

for normal strains:

$$\epsilon_{xx}=\frac{\sigma_{xx}}{E}-\frac{\nu*\sigma_{yy}}{E}-\frac{\nu*\sigma_{zz}}{E}$$

$$\epsilon_{yy}=\frac{\sigma_{yy}}{E}-\frac{\nu*\sigma_{xx}}{E}-\frac{\nu*\sigma_{zz}}{E}$$

$$\epsilon_{zz}=\frac{\sigma_{zz}}{E}-\frac{\nu*\sigma_{xx}}{E}-\frac{\nu*\sigma_{yy}}{E}$$

for shear strains:

$$\gamma_{xy}=2*\epsilon_{xy}=\frac{\tau_{xy}}{G}$$

$$\gamma_{yz}=2*\epsilon_{yz}=\frac{\tau_{yz}}{G}$$

$$\gamma_{xz}=2*\epsilon_{xz}=\frac{\tau_{xz}}{G}$$

from the symmetry of the strain tensor

$$   \mathbf{\epsilon} = \begin{bmatrix} \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{bmatrix} ~. $$

it is clear that $$\epsilon=[\epsilon_{ij}], \sigma=[\sigma_{ij}]$$ which are both three by three matrices.

next arrange the strain components $$\epsilon_{ij}$$ and the stress components in column matrices. this is also called the voigt notation.

$$ \mathbf{\epsilon_{ij}} = \begin{bmatrix} \epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33}\\ \epsilon_{23}\\ \epsilon_{31}\\ \epsilon_{12}\\ \end{bmatrix} $$

and

$$ \mathbf{\sigma_{ij}} = \begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{23}\\ \sigma_{31}\\ \sigma_{12}\\ \end{bmatrix}

$$

next hookes law for an isotropic material can be written which is

$$                     \begin{bmatrix} \epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33}\\ \epsilon_{23}\\ \epsilon_{31}\\ \epsilon_{12}\\ \end{bmatrix}~= $$                            $$                      \begin{bmatrix} \frac{1}{E}&\frac{-\nu}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{1}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{-\nu}{E}&\frac{1}{E}&0&0&0\\ 0&0&0&\frac{1}{2*G}&0&0\\ 0&0&0&0&\frac{1}{2*G}&0\\ 0&0&0&0&0&\frac{1}{2*G}\\ \end{bmatrix}~* $$ $$                     \begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{23}\\ \sigma_{31}\\ \sigma_{12}\\ \end{bmatrix}

$$

which can also be written as

$$                     \begin{bmatrix} \epsilon_{xx}\\ \epsilon_{yy}\\ \epsilon_{zz}\\ \gamma_{yz}\\ \gamma_{zx}\\ \gamma_{xy}\\ \end{bmatrix}~= $$                            $$                      \begin{bmatrix} \frac{1}{E}&\frac{-\nu}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{1}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{-\nu}{E}&\frac{1}{E}&0&0&0\\ 0&0&0&\frac{1}{G}&0&0\\ 0&0&0&0&\frac{1}{G}&0\\ 0&0&0&0&0&\frac{1}{G}\\ \end{bmatrix}~* $$ $$                     \begin{bmatrix} \sigma_{xx}\\ \sigma_{yy}\\ \sigma_{zz}\\ \tau_{yz}\\ \tau_{zx}\\ \tau_{xy}\\ \end{bmatrix}

$$

The poisson ratio $$\nu$$ for certain materials was also mentioned in class such as cork which is unique for the fact that is has a zero poissons ratio which means that when it is compressed or put under tension there is no change in volume in the directions other than that of the force applied. for example if there was a cylinder of cork of some radius R and it was put under compression the radius would not change. More can be found about the poisson ratio here

title: User:EAS4200c.f08.Blue.E/Lecture 22


text:
User:Eas4200c.f08.blue.a/Lecture 22

How to relate strains to $$E,\nu$$

first is to define Hooke's law in three dimensions which is

for normal strains:

$$\epsilon_{xx}=\frac{\sigma_{xx}}{E}-\frac{\nu*\sigma_{yy}}{E}-\frac{\nu*\sigma_{zz}}{E}$$

$$\epsilon_{yy}=\frac{\sigma_{yy}}{E}-\frac{\nu*\sigma_{xx}}{E}-\frac{\nu*\sigma_{zz}}{E}$$

$$\epsilon_{zz}=\frac{\sigma_{zz}}{E}-\frac{\nu*\sigma_{xx}}{E}-\frac{\nu*\sigma_{yy}}{E}$$

for shear strains:

$$\gamma_{xy}=2*\epsilon_{xy}=\frac{\tau_{xy}}{G}$$

$$\gamma_{yz}=2*\epsilon_{yz}=\frac{\tau_{yz}}{G}$$

$$\gamma_{xz}=2*\epsilon_{xz}=\frac{\tau_{xz}}{G}$$

from the symmetry of the strain tensor

$$   \mathbf{\epsilon} = \begin{bmatrix} \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{bmatrix} ~. $$

it is clear that $$\epsilon=[\epsilon_{ij}], \sigma=[\sigma_{ij}]$$ which are both three by three matrices.

next arrange the strain components $$\epsilon_{ij}$$ and the stress components in column matrices. this is also called the voigt notation.

$$ \mathbf{\epsilon_{ij}} = \begin{bmatrix} \epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33}\\ \epsilon_{23}\\ \epsilon_{31}\\ \epsilon_{12}\\ \end{bmatrix} $$

and

$$ \mathbf{\sigma_{ij}} = \begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{23}\\ \sigma_{31}\\ \sigma_{12}\\ \end{bmatrix}

$$

next hookes law for an isotropic material can be written which is

$$                     \begin{bmatrix} \epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33}\\ \epsilon_{23}\\ \epsilon_{31}\\ \epsilon_{12}\\ \end{bmatrix}~= $$                            $$                      \begin{bmatrix} \frac{1}{E}&\frac{-\nu}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{1}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{-\nu}{E}&\frac{1}{E}&0&0&0\\ 0&0&0&\frac{1}{2*G}&0&0\\ 0&0&0&0&\frac{1}{2*G}&0\\ 0&0&0&0&0&\frac{1}{2*G}\\ \end{bmatrix}~* $$ $$                     \begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{23}\\ \sigma_{31}\\ \sigma_{12}\\ \end{bmatrix}

$$

which can also be written as

$$                     \begin{bmatrix} \epsilon_{xx}\\ \epsilon_{yy}\\ \epsilon_{zz}\\ \gamma_{yz}\\ \gamma_{zx}\\ \gamma_{xy}\\ \end{bmatrix}~= $$                            $$                      \begin{bmatrix} \frac{1}{E}&\frac{-\nu}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{1}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{-\nu}{E}&\frac{1}{E}&0&0&0\\ 0&0&0&\frac{1}{G}&0&0\\ 0&0&0&0&\frac{1}{G}&0\\ 0&0&0&0&0&\frac{1}{G}\\ \end{bmatrix}~* $$ $$                     \begin{bmatrix} \sigma_{xx}\\ \sigma_{yy}\\ \sigma_{zz}\\ \tau_{yz}\\ \tau_{zx}\\ \tau_{xy}\\ \end{bmatrix}

$$

The poisson ratio $$\nu$$ for certain materials was also mentioned in class such as cork which is unique for the fact that is has a zero poissons ratio which means that when it is compressed or put under tension there is no change in volume in the directions other than that of the force applied. for example if there was a cylinder of cork of some radius R and it was put under compression the radius would not change. More can be found about the poisson ratio here

title: User:Eas4200c.f08.blue.a/Lecture 22


text:
User:Eml4500.f08.wiki1.ambrosio/hw4

Assignment (10/8): Derive $$\tilde{d}^{(e)}_2$$
(NOTE: $$\tilde{d}^{(e)}_1$$ was derived on Page 12-4.)

(the component of $$\overrightarrow{d}^{(e)}_{local node 1}$$ along $$\overrightarrow{\tilde{j}}$$, i.e. the $$\tilde{y}$$ axis)

$$ \begin{array}{lcl} \begin{align}

\tilde{d}^{(e)}_2 & = \overrightarrow{d}^{(e)}_{local node 1} \cdot \overrightarrow{\tilde{j}} \\

&=(d^{(e)}_1 \overrightarrow{i} + d^{(e)}_2 \overrightarrow{j}) \cdot \overrightarrow{\tilde{j}} \\

&= d^{(e)}_1 (\overrightarrow{i} \cdot \overrightarrow{\tilde{j}}) + d^{(e)}_2(\overrightarrow{j} \cdot \overrightarrow{\tilde{j}}) \\

&= d^{(e)}_1 (-\sin \theta ^{(e)}) + d^{(e)}_2(\cos \theta^{(e)}) \\

&= d^{(e)}_1 (-m^{(e)}) + d^{(e)}_2(l^{(e)}) \\

\end{align} \end{array} $$

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

Where $$\tilde{d}^{(e)}_2 = q^{(e)}_2$$

Assignment (10/10): Interpretation of Transverse DOFs $$ \mathbf{\tilde{f}}^{(e)} = \mathbf{\tilde{T}}^{(e)} \mathbf{f}^{(e)} $$
First we note that $$\mathbf{\tilde{f}}^{(e)}_x$$ and $$\mathbf{\tilde{d}}^{(e)}_x$$ point in the same directions. Since $$\mathbf{\tilde{T}}^{(e)} $$ is only depenedent on $$\theta$$, we can also use it to transform from $$\mathbf{f}^{(e)}$$ to $$\mathbf{\tilde{f}}^{(e)}$$.

We can easily convert between $$\mathbf{\tilde{f}}^{(e)} $$ and $$\mathbf{f}^{(e)} $$ by using this relation.
 * $$ \mathbf{\tilde{f}}^{(e)} = \mathbf{\tilde{T}}^{(e)} \mathbf{f}^{(e)} $$

When expanded, the relation looks like:

\begin{pmatrix} \mathbf{\tilde{f}}^{(e)}_1\\ \mathbf{\tilde{f}}^{(e)}_2

\end{pmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} \begin{pmatrix} \mathbf{f}^{(e)}_1\\ \mathbf{f}^{(e)}_2 \end{pmatrix} $$

Assignment (10/10): Verify $$\mathbf{k}^{(e)} = \mathbf{\tilde{T}}^{(e)\mathbf{T}} \mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)}$$
So far we know that $$\mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$. If we multiply both sides by $$\mathbf{\tilde{T}}^{(e)}$$, we get
 * $$\mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)} \mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{T}}^{(e)} \mathbf{f}^{(e)}$$

We can then obtain
 * $$\mathbf{\tilde{T}}^{(e)-1} \mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)} \mathbf{d}^{(e)} = \mathbf{f}^{(e)}$$

It has been proven that $$\mathbf{\tilde{T}}^{(e)-1} = \mathbf{\tilde{T}}^{(e)T}$$, so

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

Now notice the relation between the two equations.
 * $$[\mathbf{\tilde{T}}^{T} \mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)}] \mathbf{d}^{(e)} = \mathbf{f}^{(e)} \Leftrightarrow  \mathbf{k}^{(e)}  \mathbf{d}^{(e)} =  \mathbf{f}^{(e)}  $$

We can now see,
 * $$ \mathbf{k}^{(e)} = \mathbf{\tilde{T}}^{T} \mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)}$$

title: User:Eml4500.f08.wiki1.ambrosio/hw4


text:
User:Eml4500.f08.Ateam.philbin/HW4

Homework 4

Mtg. 20 10-10-08

Changing global degrees of freedom to a local coordinate system aligned along the truss
The tilde vectors represent the axial and normal forces from the axis of the bar. The same equation is used to start the derivation.

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

A transformation vector will be used to get $$ \tilde\mathbf {d} $$ as well as $$ \tilde {f} $$.

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

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

By substituting the values for $$ \tilde \mathbf{T} $$ it can be proven.

$$ \tilde \mathbf{T} = \begin{bmatrix} cos (\Theta) & sin (\Theta) & 0 & 0\\ -sin (\Theta) & cos (\Theta) & 0 & 0\\0 & 0 & cos (\Theta) & sin (\Theta)\\0 & 0 & -sin (\Theta) & cos (\Theta)\end{bmatrix}=\begin{bmatrix} \mathbf{R}^{(e)} & \mathbf{0}\\\mathbf{0} & \mathbf{R}^{(e)}\end{bmatrix}$$

$$ \tilde \mathbf{T}^{(1)} \mathbf{f}^{(1)}= \begin{bmatrix} cos (30) & sin (30) & 0 & 0\\ -sin (30) & cos (30) & 0 & 0\\0 & 0 & cos (30) & sin (30)\\0 & 0 & -sin (30) & cos (30)\end{bmatrix}\begin{bmatrix}-4.378\\-2.5622\\4.4378\\2.5622\end{bmatrix} = \tilde\mathbf{f}^{(1)}$$

$$\tilde \mathbf{f}^{(1)} = \begin{bmatrix} -5.1243\\0\\5.1243\\0\end{bmatrix} $$

This confirms the previous statics work for $$ \tilde \mathbf{f}^{(1)} $$.

By substituting the previous relationships into $$ \tilde \mathbf{f}^{(e)}= \tilde \mathbf{k}^{(e)} \tilde \mathbf{d}^{(e)}$$, this equation can be rewritten as follows.

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

Verify transformation matrix transpose theorem
If the transformation matrix, $$\tilde \mathbf{T}^{(e)}$$, is invertible then the equation can be changed into the following form.

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

Now since only a square matrix can be inverted, the following process was investigated. Since $$\tilde \mathbf{T}^{(e)}$$ is composed of R matrices which are sin and cos functions, when the transpose of $$\tilde \mathbf{R}^{(e)}$$ is multiplied by $$\tilde \mathbf{R}^{(e)}$$, the result is the identity matrix.

$$ \mathbf {R}^{(e)T} \ \mathbf {R}^{(e)} = \begin{bmatrix} \sin^2 {\Theta} + \cos^2 {\Theta} & 0 \\ 0 & \sin^2 {\Theta} + \cos^2 {\Theta} \end{bmatrix}=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$$

Therefore if both sides are divided by $$ \mathbf {R}^{(e)}$$, then the following relationship can be made.

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

This simplifies the math involved significantly and allows for rectangular matrices to be used instead of solely square matrices. Since $$\tilde \mathbf{T}^{(e)}$$ is comprised of a diagonal $$\tilde \mathbf{R}^{(e)}$$ matrix, the previous relationship between the transpose and inverse can be applied to $$\tilde \mathbf{T}^{(e)}$$ as well.

To verify that $$ [\tilde \mathbf{T}^{(e)-1} \ \tilde \mathbf{k}^{(e)} \ \tilde\mathbf {T}^{(e)}] $$, the values from the 2-bar truss will be inserted. The first element's values will be used.

$$ [\tilde \mathbf{T}^{(e)-1} \ \tilde \mathbf{k}^{(e)} \ \tilde\mathbf {T}^{(e)}] = \begin{bmatrix} cos (30) & -sin (30) & 0 & 0\\ sin (30) & cos (30) & 0 & 0\\0 & 0 & cos (30) & -sin (30)\\0 & 0 & sin (30) & cos (30)\end{bmatrix}\begin{bmatrix}0\\0\\4.4352\\6.1271\end{bmatrix} \begin{bmatrix} cos (30) & sin (30) & 0 & 0\\ -sin (30) & cos (30) & 0 & 0\\0 & 0 & cos (30) & sin (30)\\0 & 0 & -sin (30) & cos (30)\end{bmatrix}$$

$$= \begin{bmatrix} 0.5625 & 0.325&-0.5625&-0.325\\0.325&0.1875&-0.325&-0.1875\\-0.5625&-0.325&0.5625&0.325\\-0.325&-0.1875&0.325&0.1875\end{bmatrix}= \mathbf{k}^{(1)}$$

This verifies the derivation made earlier.

title: User:Eml4500.f08.Ateam.philbin/HW4


text:
User:Eml4500.f08.delta 6.krueger/Notes Friday October 17th

When equations one and two, from the equilibrium of node two, are rewritten they become: $$f^{(1)}_{3} + f^{(2)}_{1}= 0$$ $$f^{(1)}_{4} + f^{(2)}_{2}= 0$$ The element forces can then be written in terms of the element stiffness and displacements. $$f_{3}^{(1)} = \begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix}$$ $$f_{1}^{(2)} = \begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}$$ The forces written in terms of the elements stiffness and displacements are then substituted into equation one and equation one becomes. $$ \begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix} + \begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}=0$$ Now recall the relationship between the global degrees of freedom and the element degrees of freedom: $$d_{1}=d_{1}^{(1)},d_{2}=d_{2}^{(1)},d_{3}=d_{3}^{(1)}=d_{1}^{(2)},d_{4}=d_{4}^{(1)}=d_{2}^{(2)},d_{5}=d_{3}^{(2)},d_{6}=d_{4}^{(2)}$$ The global degrees of freedom can now be substituted into equation one for the element degrees of freedom and equation one becomes. $$ \begin{bmatrix} k_{31}^{(1)}d_{1} + k_{32}^{(1)}d_{2} + k_{33}^{(1)}d_{3} + k_{34}^{(1)}d_{4} \end{bmatrix} + \begin{bmatrix} k_{11}^{(2)}d_{3} + k_{12}^{(2)}d_{4} + k_{13}^{(2)}d_{5} + k_{14}^{(2)}d_{6} \end{bmatrix}=0$$ Now the common terms are grouped and factored (i.e. The terms that include $$d_{3}$$ and $$d_{4}$$). The same process is repeated for equation two. Equation one results in the third row of the global stiffness matrix and equation two results in the fourth row of the global stiffness matrix. The work is shown in the homework for this section. =Assembly of the global stiffness matrix= Before starting the assembly of the global stiffness matrix a quick review of the goals and the end that is being worked towards is necessary. The first goal is the use of The Principle of Virtual Work (PVW) which allows the formation of the global stiffness matrixK by eliminating rows that correspond to the boundary conditions. The ultimate goal is to use partial differential equations (PDE's) to derive the Finite Element Method (FEM). Recall now two equations that have already been derived in the process of the FEM. $$q^{(e)}_{2x1}$$ = $$T^{(e)}_{2x4}$$ $$d^{(e)}_{4x1}$$ $$K^{(e)}= T^{(e)^(T)} * \hat{K}^{(e)} * T^{(e)}$$

The assembly of the global stiffness matrix (K) is accomplished using the element stiffness matrices ($$k^{(e)}$$) and an assemble operator (A). Where e= 1,....,nel, and nel is the number of elements. The assembly operator is introduced because the sigma operator is not correct. The global stiffness matrix is assembled using a combination of element stiffness matrices. The element stiffness matrices are small when compared to the global stiffness matrix. The global stiffness matrix could be orders of magnitude larger than the element stiffness matrices. The sum of $$k_{nedxned}^{(e)}$$ doesn't equal to a $$K_{nxn}$$ matrix. Therefore the assembly operator (A) is necessary for the notation of the process to be correct. $$K=Ak^{(e)}$$ With this, new notation needs to be introduced. n is equal to the number of global degrees of freedom before eliminating the boundary conditions. ned is the number of element degrees of freedom. Where ned is very small when compared to n. Recall that the force displacement relationship for a bar or spring is kd=F. After some algebra: $$ kd-F=0$$ This will be called equation three and is equivalent to: w(kd-F)=0 This will be call equation four. This holds true for all values of w. Therefore it is trivial to imply that equation three is equal to equation four but is extremely important to understand that equation four is equal to equation three. This is true because equation four is valid for all w. So letting w=1, equation four is converted back into equation three.


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User:Eas4200c.f08.vqcrew.f/hw3

HW 3

 Torsion of a Uniform Non-Circular Bar

A uniform non-circular bar leads to warping of the cross-section. Warping is defined as the axial displacement along the x-axis (i.e. along bar length) of a point on the deformed rotated cross-section.

$$U_y =$$ y component of displacement vector $$\vec{PP}'$$

$$U_z =$$ z component of displacement vector $$\vec{PP}'$$

Important Definition : $$\displaystyle \theta = \frac{\displaystyle \alpha}{x}$$       =   Rate of Twist

Referring to Figure-4, assuming that the cross-section at x=0 remains stationary, and if the rotation angle $$\displaystyle \alpha$$ is small, then the displacement components at P are given by the following:

$$u = -r \displaystyle \alpha sin\displaystyle \beta = -\alpha y$$, which can be simplified to Equation (3.11):

(1) $$U_y = - \displaystyle \theta x z$$

Now the we have the displacement in the y direction, we can solve for the displacement $$U_z$$ which yields:

$$U_z = + PP' cos\displaystyle \beta = +\displaystyle \alpha y_p$$, this can be simplified to Equation (3.12):

(2) $$U_z = \displaystyle \theta x y$$

Now that we have the vertical and horizontal displacements the warping displacement in the x-axis can be found as a function of the y and z directions.

(3) $$U_x = \displaystyle \theta \displaystyle \psi (y, z)$$

Based on these three equations come the Kinematic Assumptions.

Small Angle Approximation

Road Map for Torsional Analysis of Aircraft Wing:

A) Kinematic Assumption

B) Strain Displacement

C) Equilibrium Equation (stresses)

D) Prandtl Stress Function $$\displaystyle \phi$$

E) Strain Compatibility Equation

F) Equation for $$\displaystyle \phi$$

G) Boundary Conditions for $$\displaystyle \phi$$

H) $$T = 2 \iint_A \displaystyle \phi \,dA$$

$$T = G J \displaystyle \theta$$

$$ J =\frac{-4}{\nabla^2 \phi}\iint_A \displaystyle \phi \,dA$$

I) Thin-Walled Cross-Section, $$T = 2 q A $$ (3.48) This equation was also derived in Problem 1.1.

J) Twist Angle $\displaystyle \theta$ Method 1.

$$\displaystyle \theta = \frac{1}{2 G A} \oint \frac{q}{t}\,ds$$

title: User:Eas4200c.f08.vqcrew.f/hw3


text:
User talk:Mhemraj

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title: User talk:Mhemraj


text:
User:Eml4500.f08.ramrod.c/10-8-08

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

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

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

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

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

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

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

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

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

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

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

title: User:Eml4500.f08.ramrod.c/10-8-08


text:
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text: File:5 bar truss U and D.jpg

title: File:5 bar truss U and D.jpg


text:
User:Eml4500.f08.qwiki.bishop/Lecture 22

=Meeting 22, Wednesday October 15, 2008=

Formal Justification of Assembly Process of Element Stiffness Matrix into Global Stiffness Matrix
For a formal justification, we will consider the two-bar truss discussed in meeting 9. We must also recall that the Force-Displacement relationship for each element is $$\mathbf{k}^{(e)}_{(4x4)}\mathbf{d}^{(e)}_{(4x1)}=\mathbf{f}^{(e)}_{(4x1)}$$. This relationship allows us to compute the nodal displacements and forces, unlike statics which only considers element displacements and forces. Therefore, if we follow the Finite Element Method we can isolate and analyze each node and element in the system according to the Euler Cut Principle.

In terms of the DOFs and their naming for this justification, we refer to the Free Body Diagrams of Element 1 and 2 from Meeting 4. Notice that global node 2 DOFs correspond to node 2 of element 1 and node 1 of element 2.

 * Global Node 1:
 * $$d_1=d_1^{(1)}$$
 * $$d_2=d_2^{(1)}$$

 * Global Node 2:
 * $$d_3=d_3^{(1)}=d_1^{(2)}$$
 * $$d_4=d_4^{(1)}=d_2^{(2)}$$

 * Global Node 3:
 * $$d_5=d_3^{(2)}$$
 * $$d_6=d_4^{(2)}$$

Since global node 2 is the intersecting node of element 1 and 2, we can examine the equilibrium of global node 2

The next step of the justification process is to invoke the element FD relationship $$\mathbf{k}_{(4x4)}^{(e)}\mathbf{d}_{(4x1)}^{(e)} = \mathbf{f}_{(4x1)}^{(e)}$$ for each element. We will replace each internal element force in equations 1 and 2 shown above with its respective internal displacements, where $$f_i^{(e)} = \sum_{j=1}^{4}{k_{ij}^{(e)}d_{j}^{(e)}}$$
 * $$\sum{F}_x = -f_3^{(1)} - f_1^{(2)} = 0$$ Eqn. 1
 * $$\sum{F}_y = \mathbf{P} - f_1^{(4)} - f_2^{(2)} = 0$$   Eqn. 2


text:
User:EML4500.f08.JAMAMA/Homework4

Methods to Assemble the Global Stiffness Matrix $$K$$

Connecting array "conn," consider the 2-bar truss system (p. 5-6 $$ \rightarrow $$ shown above)

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

Location matrix master array: "lmm" (the same 2-bar truss system (p. 5-6) that is shown above)

lmm(i,j) = equation number (global dof number) for the element stiffness coefficients corresponding to the j -th local dof number.

Alternative Method for Transforming the Local Degrees of Freedom
Method 2 to derive $$ \mathbf{d}_{4 \times 4}^{(e)}: $$

Transform a system with $$ \mathbf{4} $$ dofs into a system with also $$ \mathbf{4} $$ dofs (instead of only $$ \mathbf{2} $$ dofs), so that the transformatin matix is now $$ \mathbf{4 \times 4} $$ and hopefully invertible (local-to-local transformation).

The objective is to transform the set of local degrees of freedom $$\tilde{d}^{(e)}$$ to a different set degrees of freedom $$\tilde{d}^{(e)}_{4X1}$$ using an invertible matrix $$\tilde{T}^{(e)}{_{4X4}}$$.

$$\tilde{d}^{(e)}_{4X1} = \tilde{T}^{(e)}_{4X4}d^{(e)}_{4X1}$$

Lets define equation (1) bellow:

$$\tilde{d}^{(e)}_{1} = [l^{(e)}m^{(e)}]\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{Bmatrix}$$

$$\tilde{d}_{2}^{(e)} = \overrightarrow{d}_{1}^{(e)}\cdot \overrightarrow{j}$$ (comp. of $$ \overrightarrow{d}_{1}^{(e)}$$ along $$\vec\tilde{j}$$ i.e. y-axis)
 * $$= sin(\theta^{(e)}) + cos(\theta^{(e)}) d_{2}^{(e)}$$

And lets define equation (2) bellow:

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

By putting (1) and (2) in matrix form:

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

$$\underbrace{\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \tilde{d}_{3}^{(e)}\\ \tilde{d}_{4}^{(e)}\\ \end{Bmatrix}}_{\tilde{d}_{4X1}^{(e)}} = \underbrace{\begin{bmatrix} R_{2X2}^{(e)}&0_{2X2}\\ 0_{2X2}&R_{2X2}^{(e)}\\ \end{bmatrix}}_{\tilde{T}_{4X4}^{(e)}}\underbrace{\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ \end{Bmatrix}}_{d_{4X1}^{(e)}}$$

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

$$\tilde{f}^{(e)}_{4X1} = \tilde{k}^{(e)}_{4X4}\tilde{d}^{(e)}_{4X1}$$

Note: Consider the case: $$\tilde{d}{_{4}}^{(e)}\neq 0$$

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

$$\tilde{f}^{(e)}=\tilde{k}^{(e)}\tilde{d}^{(e)}=0$$ 4X1  4X4  4X1   4X1

Interp of transverse DOF's.

Recall from Meeting 19: $$\tilde{d}^{(e)}=\tilde{T}^{(e)}d^{(e)}$$

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

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

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

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

$$\left[ \tilde{T}^{(e)-1}\hat{k}^{(e)}\tilde{T}^{(e)}\right]d^{(e)}=f^{(e)}$$ (K matrix)

$$\tilde{T}^{(e)}$$=block diagonal matrix (P.19.3)

Consider a general block-diagram matrix:

$$A=\begin{bmatrix} D1 & &  &  &0 \\ & ..&  &  & \\ &  &  ..&  & \\ &  &  &  ..& \\ 0&  &  &  & D5 \end{bmatrix}$$

Question: What is $$A^{-1}$$

Answer using a simpler example: Using a diagonal matrix:

$$B=\begin{bmatrix} d_{11} & &  &  &0 \\ & d_{22}&  &  & \\ & &  d_{33}&  & \\ & &  &  d_{44}& \\ 0& &  &  &d_{mn} \end{bmatrix}= diag\left[d_{11},d_{22},d_{33}....d_{mn} \right]$$

assuming that $$d_{ij}\neq 0 $$ for i=1,....n then $$B^{-1}=diag\left[\frac{1}{d_{11}},\frac{1}{d_{22}},\frac{1}{d_{33}}....\frac{1}{d_{mn}} \right]$$

For a block diagonal matrix A:

$$A=diag\left[D_{1},....D_{5} \right]$$

$$A^{-1}=diag\left[D_{1}^{-1},....D_{5}^{-1} \right]$$

$$\tilde{T}^{(e)-1}=diag\left[R^{(e)-1},R^{(e)-1} \right]$$

P.19-2 $$R^{(e)T}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)}&l^{(e)} \end{bmatrix}$$

$$R^{(e)T}R^{(e)}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)}&l^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)}&l^{(e)} \end{bmatrix}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}=I $$ (2x2 identity matrix)

Proof of the Identity Matrix Above

$$R^{(e)T}R^{(e)}=\left[ \begin{matrix} l^{2}+m^{2} & 0 \\ 0 & l^{2}+m^{2} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\   0 & 1  \\ \end{matrix} \right]=I_{_{2x2}}$$ NOTE: $$\left. \begin{align} & l^{(e)}=\cos \theta \\ & m^{(e)}=\sin \theta \\ \end{align} \right\}\therefore l^{2}+m^{2}=\cos ^{2}\theta +\sin ^{2}\theta =1$$

... $$\Rightarrow R^{(e)-1}=R^{(e)T}$$

$$\Rightarrow \tilde{T}^{(e)-1}=diag\left[R^{(e)T},R^{(e)T} \right] =\left( diag\left[R^{(e)},R^{(e)} \right]\right)^{T}$$ ...  $$\tilde{T}^{(e)-1}=\tilde{T}^{(e)T}$$

so from earlier it can be shown that: $$\left[ \tilde{T}^{(e)T}\hat{k}^{(e)}\tilde{T}^{(e)}\right]d^{(e)}=f^{(e)}$$

The Meaning of Eigenvectors and Eigenvalues
An eigenvector of an n x n matrix A is a nonzero vector v such that K v =λ v for some scalar λ.

Eigenvalue problem: K v =λ v ( M is the identity matrix)

Let {u1, u2, u3, u4} be the pure eigenvectors corresponding to the 4 zero eigenvalues.

$$K_{_{6x6}}u_{i_{6x1}}=\underbrace{0\cdot u_{i}}_{0_{6x1}},\text{ }i=1,2,3,4$$

The linear combination of {ui, i=1,2,3,4}

$$\sum\limits_{i=1}^{4}{\alpha _{i_{1x1}}u_{i_{6x1}}=:W_{_{6x1}}}=0_{_{1x1}}\cdot W_{_{6x1}}$$

Where $$=:$$ means "equal by definition"

$$\alpha _{i}\text{ is a real number (scalar)}$$

A nonzero solution for d e is possible only if K cannot be inverted. (Because the $$\det K=0$$, the inverse will include a scalar divided by 0, which is an undefined value, therefore making the inverse undefined.) The eigenvalues represent the buckling loads. Substituting each eigenvalue into the global equations, you can compute the nodal displacements. These nodal displacements are the eigenvectors, which are the buckling modes, or mode shapes.

 These mode shapes (corresponding to the zero eigenvalues) may come out as a linear combination of the pure mode shapes. (3 pure rigid body motions and 1 pure mechanism)

For (a), there will be only one zero eigenvalue because the set up is unstable, for instance, you would simply have to push on one of the top nodes to force it out of equilibrium, however, (b) is more stable, and will have more than just one zero eigenvalue.

Justification of the Assembly of the Element Stiffness Matrix $$k^{(e)}$$ into the Global Matrix $$K$$
Consider the 2 bar truss matrix again.

The FD relation for each element is

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

Then using the Euler cut method 2 and look at node 2 with elements 1 and 2.

The sum of the forces at node 2 is zero so

$$\sum{F_{x}}=0=-f_{3}^{1}-f_{1}^{2}=0$$

$$\sum{F_{y}}=0=-f_{4}^{1}-f_{2}^{2}=0$$

The next step is to utilize the knowledge that the sum of the forces at the global node 2 are equal to zero and substituting each element force with the appropriate force displacement relation: $$k_{4x4}^{(e)}d_{4x1}^{(e)}=f_{4x1}^{(e)}$$

 * For the equations $$f_{3}^{(1)}+f_{1}^{(2)}=0$$ and $$f_{4}^{(1)}+f_{2}^{(2)}=P$$, the substitutions are as follows:

1) Replace $$f_{3}^{(1)}$$, $$f_{1}^{(2)}$$, $$f_{4}^{(1)}$$, and $$f_{2}^{(2)}$$ with the following:

 $$\begin{align} & \text{Row 3: }\underbrace{\left[ k_{31}^{(1)}d_{1}^{(1)}+k_{32}^{(1)}d_{2}^{(1)}+k_{33}^{(1)}d_{3}^{(1)}+k_{34}^{(1)}d_{4}^{(1)} \right]}_{f_{3}^{(1)}}+\underbrace{\left[ k_{11}^{(2)}d_{1}^{(2)}+k_{12}^{(2)}d_{2}^{(2)}+k_{13}^{(2)}d_{3}^{(2)}+k_{14}^{(2)}d_{4}^{(2)} \right]}_{f_{1}^{(2)}}=0 \\ & \\ &  \\ \end{align}$$

$$\text{Row 4: }\underbrace{\left[ k_{41}^{(1)}d_{1}^{(1)}+k_{42}^{(1)}d_{2}^{(1)}+k_{43}^{(1)}d_{3}^{(1)}+k_{44}^{(1)}d_{4}^{(1)} \right]}_{f_{4}^{(1)}}+\underbrace{\left[ k_{21}^{(2)}d_{1}^{(2)}+k_{22}^{(2)}d_{2}^{(2)}+k_{23}^{(2)}d_{3}^{(2)}+k_{24}^{(2)}d_{4}^{(2)} \right]}_{f_{2}^{(2)}}=P$$

2) Now replace all of the local degrees of freedom with their global equivalents.

The equivalent local and global degrees of freedom:

$$\begin{align} & d_{1}=d_{1}^{(1)} \\ & d_{2}=d_{2}^{(1)} \\ & d_{3}=d_{3}^{(1)}=d_{1}^{(2)} \\ & d_{4}=d_{4}^{(1)}=d_{2}^{(2)} \\ & d_{5}=d_{3}^{(2)} \\ & d_{6}=d_{4}^{(2)} \\ \end{align}$$

$$\begin{align} & \text{Row 3: }\underbrace{\left[ k_{31}^{(1)}d_{1}+k_{32}^{(1)}d_{2}+\underbrace{k_{33}^{(1)}d_{3}+k_{34}^{(1)}d_{4}}_{\text{common displacements}} \right]}_{f_{3}^{(1)}}+\underbrace{\left[ \underbrace{k_{11}^{(2)}d_{3}+k_{12}^{(2)}d_{4}}_{\text{common displacements}}+k_{13}^{(2)}d_{5}+k_{14}^{(2)}d_{6} \right]}_{f_{1}^{(2)}}=0 \\ & \\ \end{align}$$

$$\text{Row 4: }\underbrace{\left[ k_{41}^{(1)}d_{1}+k_{42}^{(1)}d_{2}+\underbrace{k_{43}^{(1)}d_{3}+k_{44}^{(1)}d_{4}}_{\text{common displacements}} \right]}_{f_{4}^{(1)}}+\underbrace{\left[ \underbrace{k_{21}^{(2)}d_{3}+k_{22}^{(2)}d_{4}}_{\text{common displacements}}+k_{23}^{(2)}d_{5}+k_{24}^{(2)}d_{6} \right]}_{f_{2}^{(2)}}=P$$

3) Now factor out the common global displacements (noted above):

$$\begin{align} & \text{Row 3: }k_{31}^{(1)}d_{1}+k_{32}^{(1)}d_{2}+\underbrace{\left( k_{33}^{(1)}+k_{11}^{(2)} \right)}_{\begin{matrix} {} & \text{Addends for Row 3 } \\ {} & \text{ of the } \\ {} & \text{Global Stiffness Matrix} \\ {} & K \\ \end{matrix}}\left( d_{3} \right)+\underbrace{\left( k_{34}^{(1)}+k_{12}^{(2)} \right)}_{\begin{matrix} {} & \text{Addends for Row 4 } \\ {} & \text{of the } \\ {} & \text{Global Stiffness Matrix} \\ {} & K \\ \end{matrix}}\left( d_{4} \right)+k_{13}^{(2)}d_{5}+k_{14}^{(2)}d_{6}=0 \\ & \\ &  \\ \end{align}$$

$$\text{Row 4: }k_{41}^{(1)}d_{1}+k_{42}^{(1)}d_{2}+\underbrace{\left( k_{43}^{(1)}+k_{21}^{(2)} \right)}_{\begin{matrix} {} & \text{Addends for Row 3 } \\ {} & \text{          of the }  \\ {} & \text{Global Stiffness Matrix} \\ {} & \text{             }K  \\ \end{matrix}}\left( d_{3} \right)+\underbrace{\left( k_{44}^{(1)}+k_{22}^{(2)} \right)}_{\begin{matrix} {} & \text{Addends for Row 4 } \\ {} & \text{          of the }  \\ {} & \text{Global Stiffness Matrix} \\ {} & \text{             }K  \\ \end{matrix}}\left( d_{4} \right)+k_{23}^{(2)}d_{5}+k_{24}^{(2)}d_{6}=P$$

By factoring these out, the necessity of adding the $$3^{rd}\text{ and }4^{th}$$ rows of the global stiffness matrix $$K$$ has been proven for both equilibrium equations derived above. For a visual of both rows, the global stiffness matrix is reproduced below:

 $$K=\begin{matrix} {} \\   {}  \\   Row\text{ }3  \\ Row\text{ }4 \\ {} \\   {}  \\ \end{matrix}\left[ \begin{matrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 \\ k_{31}^{(1)} & k_{32}^{(1)} & \left( k_{33}^{(1)}+k_{11}^{(2)} \right) & \left( k_{24}^{(1)}+k_{12}^{(2)} \right) & k_{13}^{(2)} & k_{14}^{(2)} \\ k_{41}^{(1)} & k_{42}^{(1)} & \left( k_{43}^{(1)}+k_{21}^{(2)} \right) & \left( k_{44}^{(1)}+k_{22}^{(2)} \right) & k_{23}^{(2)} & k_{24}^{(2)} \\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} \\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \\ \end{matrix} \right]$$

Assembly of the Element Stiffness Matrix
$$k^{(\text{e})}:\text{ e}=1,...,nel$$, into global stiffness matrix $$K$$.

$$K_{_{n\text{ x }n}}=\underset{\text{e}=1}{\overset{nel}{\mathop{A}}}\,k_{_{ned\text{ x }ned}}^{(\text{e})}$$

Where A is the assembly operator. Note: $$\sum – $$ can be used because it implies that $$k^{(\text{e})}$$ has the same dimensions and $$K$$. For example, if each element in a truss has 4 degrees of freedom, and there are 1 million global degrees of freedom, $$k^{(\text{e})}$$ is still a $$4x4$$ matrix.

It is now important to introduce:  $$\begin{array}{*{35}l} {} & n:\text{ the total number of global degrees of freedom before eliminating the boundary conditions}  \\ {} & {} \\   {} & \begin{align} & ned\text{: the number of element degrees of freedom} \\ & \\ & nel:\text{ the number of elements} \\ \end{align} \\ \end{array}$$

In general,

$$ned\ll \text{ }n$$, where $$\ll $$ means "very small compared to".

Principle of Virtual Work (PVW)

The PVW justifies how to eliminate the rows corresponding to the boundary conditions to obtain the global matrix $$K$$. (The columns are eliminated by the boundary conditions alone.)

Recall,

$$q_{_{2x1}}^{(e)}=T_{_{2x4}}^{(e)}d_{_{4x1}}^{(e)}$$

and

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

 * We now begin: Deriving the Finite Element Method for Partial Differential Equations

The force displacement relationship for a bar or spring: $$kd=F$$

$$kd=F\underbrace{\Rightarrow }_{implies}kd-F=0$$

However,

$$kd=F\underbrace{\Leftrightarrow }_{equivalent}W(kd-F)=0,\text{ }\underbrace{\text{for all }W}_{\begin{smallmatrix} **\text{Extremely } \\ \text{    Important**} \end{smallmatrix}}$$

Where $$W$$ is the weighting coefficient, and this is also known as the "weak form" $$\equiv $$ PVW

We begin the proof of why $$kd=F\Rightarrow kd-F=0$$ is trivial, while $$kd=F\Leftrightarrow W(kd-F)=0$$ is not.

Because $$W(kd-F)=0,\text{ for all }W$$, you can select any $$W$$:

For example, if $$W=1$$, then $$W(kd-F)=0$$ becomes $$\left[ (1)(kd-F)=0 \right]\Rightarrow \left[ kd-F=0 \right]$$, which is not trivial, because it is specific for $$W=1$$.

Contributing Team Members
Megan Alvarez --EML4500.f08.jamama.megan 23:19, 23 October 2008 (UTC)

Arunas Janulevicius --Eml4500.f08.jamama.jan 09:24, 24 October 2008 (UTC)

Chris Alford --Eml4500.f08.jamama.chris 16:50, 24 October 2008 (UTC)

Cedric Adam --Eml4500.f08.jamama.adam 18:20, 24 October 2008 (UTC)

William Mueller Eml4500.f08.jamama.mueller 19:02, 24 October 2008 (UTC)

Justin McIntire --Eml4500.f08.jamama.justin 19:29, 24 October 2008 (UTC)

title: User:EML4500.f08.JAMAMA/Homework4


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User:Eas4200c.f08.blue.a/Homework4

Solving for Areas using the Method of Quadrature

Cubature - a method used for integrating volumes by breaking up the object into cubes and adding the individual cubes to find the total volume. To make the method more accurate, use smaller cubes to better approximate the volume.

Squaring (quadrature) of the circle - wikipedia.

Algebraic sums of the areas as the airfoil is swept counterclockwise is the NACA airfoil area.

A = A1 + A2 = A1 - |A2|

NOTE: When writing the code plotting the NACA arifoil, we do not need if statements to calculate the area if the arifoil. As long as the cross product is taken between the 2 sweeping vectors, the signs will add up correctly to give the appropriate area.

How about using trapezoids to integrate??

The method of trapezoids is easier to integrate over the surface and gives a more precise answer than Riemann Sums however it has a distinct disadvantage over the method of quadrature. The disadvantage being, changes in curvature are not accurately taken into account. For example, a segment with a lot of curvature is not approximated very well with a straight line. By using the method of quadrature, as long as there are a large number of segments, the area can be approximated better. Not to mention, THE TRAPEZOID METHOD IS NOT AS ELEGANT AS THE METHOD OF QUADRATURE (TRIANGLES)!!!!!!!

Returning to the single cell airfoil:

Shear flow along the airfoil is constant: q = q1 = q2 = q3

The angular rate of twist is:

$$ \theta =\frac{1}{2GA}q\sum_{j=1}^3 \frac{l}{t} $$

$$ \theta =\frac{1}{2GA}q [\frac{\pi b}{t1} + \frac{a}{t2} + \frac{\sqrt{a^2 + b^2}}{t3}] $$

$$ \theta =(factor)q $$

Homework: Computing the factor for the rate of twist.

Max shear stress = tau_max

If tau_max = tau_all (which is given), and since q = T/2A then,

Tall = 2Atauall[min{t1,t2,t3}]

Homework: If tau_all = 100 Gpa, fint T_all.

Quadrature allows an observer outside of a NACA airfoil to find the cross sectional area. An example of this is shown below in Figure 1.

Figure 1

$$\bar{A}=\bar{A_{1}}+\bar{A_{2}} =\bar{A_{1}}\left|\bar{A_{2}} \right|$$

In this close up, $$d\vec{A}$$ is defined as:

$$d\vec{A}=\frac{1}{2}\vec{r}+\vec{PQ}=-\left|dA \right|\vec{i}$$

Dividing the area into trapezoids and adding them is another method used to find the area under a curve called the (Trapezium rule), but it is not as elegant or effective as triangular quadrature in more complicated applications such as finding the area of the oddly shaped airfoil in Figure 2 below.

Figure 2

Single Cell Airfoil Calculations - Twist Angle
Returning to the idealized single cell airfoil example from above, where the shear flow is constant, the twist angle can be found using the formula found in step J of the road-map for torsional analysis of an aircraft wing.

$$\Theta =\frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds}$$

Because $$\left(q \right)$$ is constant,

$$\left(q \right)=\left( q_{1}\right)=\left( q_{2}\right)=\left( q_{3}\right)$$

$$\Theta =\frac{1}{2G\bar{A}}q\sum_{j=1}^{3}{\frac{l_{j}}{t_{j}}}$$ $$=\frac{1}{2G\bar{A}}q\left[\frac{2\pi \left(\frac{b}{2} \right)}{t_{1}}+\frac{a}{t_{2}}+\frac{\sqrt{a^{2}+b^{2}}}{t_{3}} \right]=q\left(\frac{1}{2\left(5.57m^{2} \right)G} \right)\left(\frac{\pi \left( 1m\right)}{0.008m}+\frac{4m}{0.01m} +\frac{\sqrt{16m^{2}+4m^{2}}}{0.01m} \right)=\frac{q}{\left( 11.14m^{2}\right)G}\left(1239.91 \right)=111.24m^{-2}\frac{q}{G}$$

The Max Shear Stress $$\left( \tau _{max}\right)$$ can also be found now using the formula:

$$ \tau _{max}=\frac{q}{min\left\{t_{1},t_{2},t_{3} \right\}}$$

If the maximum shear stress is equal to the allowable shear stress and from the formula to find shear flow, the allowable torque $$\left[ T_{allow}\right]$$ can be found.

$$\tau _{max}=\left( \tau _{allow}\right)$$

$$q=\frac{T}{2\bar{A}}$$

$$T_{allow}=2\bar{A}\tau _{allow}\left(min\left\{ t_{1},t_{2},t_{3}\right\} \right)$$

Therefore, if $$\tau _{allow}=100GPa$$, and because $$t_{1}=0.008m$$, $$t_{2}=t_{3}=0.01m$$

$$T_{allow}=2\bar{A}\tau _{allow}t_{1}=2\left(5.57m^{2} \right)\left(100GPa \right)\left(0.008m \right)=8.91\left(10^{9} \right)Nm$$

Example of an Ad Hoc method of proving the relation:

$$\theta =\frac{1}{2G\bar{A}}\oint_{}^{}{}\frac{q}{t}ds$$

Uniform bar with non Circular Cross section subjectred to twist:

Here the displacement of PP' due to $$\alpha$$ is:

$$ \frac{PP'}{OP}=tan\alpha \cong \alpha$$

making a small angle assumption for $$\alpha$$. Now project the displacement orthogonal to OP'.

$$PP''=PP'cos\alpha $$

$$PP''=(OPcos\alpha )tan\alpha $$

$$PP=OPtan\alpha$$

Recall: OP = r and OP'' = $$\rho$$

Therefore:

$$PP''=(rcos\alpha)tan\alpha$$

Where PP" is the displacement of P in the direction of the tangent to the surface of the bar, $$rcos\alpha$$ is $$\rho$$ and $$tan\alpha$$ is approximated to be $$\alpha$$ with a small angle assumption.

Strain can be defined as:

$$\gamma =\frac{PP''}{dx}=\frac{\rho\alpha }{dx}=\rho \theta $$

with

$$d\theta =\frac{\alpha }{dx}$$ as the rate of twist.

Hooke's Law shows that $$\tau = G\gamma =G\rho \theta $$

In this case: $$\tau (s)=G\rho (s)\theta (s)$$

and integration along the contour yields:

$$\oint_{}^{}{\tau (s)}=G\theta (x)\oint_{}^{}{\rho (s)}$$

Where $$\tau (s)=\frac{q(s)}{t(s)}$$ and$$\oint_{}^{}{\rho (s)}=2\bar{A}$$

Discussion

Why is this method ad hoc? *to get PP' we assume $$\alpha$$ is small *to get PP" we assume finite $$\alpha$$
 * 1) PP' is projected tangent to the surface but, it is not necessarally on the surface (it is only close)
 * 2) $$\tau$$ is assumed to be uniform across the wall thickness
 * 3) Inconsistancy in assuming the size of $$\alpha$$

Formal Justification of Elasticity Theory
$$u_{x}(y,z)=\theta \psi (y,z)$$

$$u_{y}(x,z)=-\theta \psi (x,z)$$

$$u_{z}(x,y)=\theta \psi (x,y)$$

We assume, given an applied torque abouyt the x-axis that:

$$\varepsilon _{xx}=\varepsilon _{yy}=\varepsilon_{zz} =\gamma _{yz}=0$$

Where each strain is computed as:

$$\varepsilon _{xx}=\frac{du_{x}}{dx}(y,z)$$

$$\varepsilon _{yy}=\frac{du_{y}}{dy}(x,z)$$

$$\varepsilon _{zz}=\frac{du_{z}}{dz}(x,y)$$

$$\gamma _{yz}=\frac{du_{y}}{dz}+\frac{du_{z}}{dy}$$

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

Returning back to the formal derivation:

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

This matrix can also be represented by

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

i is the row index and j is the column index

How to relate strains to $$E,\nu$$

First is to define Hooke's law in three dimensions which is

For normal strains:

$$\epsilon_{xx}=\frac{\sigma_{xx}}{E}-\frac{\nu*\sigma_{yy}}{E}-\frac{\nu*\sigma_{zz}}{E}$$

$$\epsilon_{yy}=\frac{\sigma_{yy}}{E}-\frac{\nu*\sigma_{xx}}{E}-\frac{\nu*\sigma_{zz}}{E}$$

$$\epsilon_{zz}=\frac{\sigma_{zz}}{E}-\frac{\nu*\sigma_{xx}}{E}-\frac{\nu*\sigma_{yy}}{E}$$

For shear strains:

$$\gamma_{xy}=2*\epsilon_{xy}=\frac{\tau_{xy}}{G}$$

$$\gamma_{yz}=2*\epsilon_{yz}=\frac{\tau_{yz}}{G}$$

$$\gamma_{xz}=2*\epsilon_{xz}=\frac{\tau_{xz}}{G}$$

From the symmetry of the strain tensor

$$   \mathbf{\epsilon} = \begin{bmatrix} \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{bmatrix} ~. $$

It is clear that $$\epsilon=[\epsilon_{ij}], \sigma=[\sigma_{ij}]$$ which are both three by three matrices.

Next arrange the strain components $$\epsilon_{ij}$$ and the stress components in column matrices. this is also called the voigt notation.

$$ \mathbf{\epsilon_{ij}} = \begin{bmatrix} \epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33}\\ \epsilon_{23}\\ \epsilon_{31}\\ \epsilon_{12}\\ \end{bmatrix} $$

and

$$ \mathbf{\sigma_{ij}} = \begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{23}\\ \sigma_{31}\\ \sigma_{12}\\ \end{bmatrix}

$$

Next hookes law for an isotropic material can be written which is

$$                     \begin{bmatrix} \epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33}\\ \epsilon_{23}\\ \epsilon_{31}\\ \epsilon_{12}\\ \end{bmatrix}~= $$                            $$                      \begin{bmatrix} \frac{1}{E}&\frac{-\nu}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{1}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{-\nu}{E}&\frac{1}{E}&0&0&0\\ 0&0&0&\frac{1}{2*G}&0&0\\ 0&0&0&0&\frac{1}{2*G}&0\\ 0&0&0&0&0&\frac{1}{2*G}\\ \end{bmatrix}~* $$ $$                     \begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{23}\\ \sigma_{31}\\ \sigma_{12}\\ \end{bmatrix}

$$

which can also be written as

$$                     \begin{bmatrix} \epsilon_{xx}\\ \epsilon_{yy}\\ \epsilon_{zz}\\ \gamma_{yz}\\ \gamma_{zx}\\ \gamma_{xy}\\ \end{bmatrix}~= $$                            $$                      \begin{bmatrix} \frac{1}{E}&\frac{-\nu}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{1}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{-\nu}{E}&\frac{1}{E}&0&0&0\\ 0&0&0&\frac{1}{G}&0&0\\ 0&0&0&0&\frac{1}{G}&0\\ 0&0&0&0&0&\frac{1}{G}\\ \end{bmatrix}~* $$ $$                     \begin{bmatrix} \sigma_{xx}\\ \sigma_{yy}\\ \sigma_{zz}\\ \tau_{yz}\\ \tau_{zx}\\ \tau_{xy}\\ \end{bmatrix}

$$

Poisson Ratio
The poisson ratio $$\nu$$ for certain materials was also mentioned in class such as cork which is unique for the fact that is has a zero poissons ratio which means that when it is compressed or put under tension there is no change in volume in the directions other than that of the force applied. for example if there was a cylinder of cork of some radius R and it was put under compression the radius would not change. More can be found about the poisson ratio here

Suggested Reading For This Report
MIT OpenCourseWare Structural Mechanics Units 4 - 12

Matlab Code Certification
I, the undersigned, certify that I can read, understand, and write matlab codes, and can thus contribute effectively to my team.

Mark Barry Eas4200c.f08.blue.a 20:31, 19 October 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 16:58, 21 October 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 01:54, 22 October 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 02:04, 22 October 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 03:08, 22 October 2008 (UTC)

Clay Robertson Eas4200c.f08.blue.f 16:46, 23 October 2008 (UTC)

Team Members To Contributed To This Report
Mark Barry Eas4200c.f08.blue.a 20:31, 19 October 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 04:16, 20 October 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 16:58, 21 October 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 01:54, 22 October 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 02:03, 22 October 2008 (UTC)

Clay Robertson Eas4200c.f08.blue.f 16:46, 23 October 2008 (UTC)

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C Sharp/Variables

Lesson Goals
This lesson will focus on the creation and use of different variables in C#. At the end of this lesson you should be able to:
 * Understand what a variable is.
 * Understand the basic C# types.
 * Create variables of different types.
 * Assign values to those variables.
 * Cast from one variable type to another.

Variables
A variable is a place in memory where the application can store some type of information. This information can be a number, a string, a class, a delegate, or even a custom data type.

The primary concept of variables is the naming conventions: For example names like "Hello", "a1234", "_variable", "is_good", "zXYzz" are valid, but "1bac", "C#", "hello world", "invalid.var" are invalid.
 * The only valid characters of a variable is English alphabet (A to Z and a to z), digits (0 to 9) and underscore character (_)
 * The name cannot start by a digit

Types
In the above algebraic example, we have taken for granted that x can actually store a number. Programming languages such as C# have types associated with their variables. Here is a list of all the types in C#.

Variable Instantiation
C# has a common method for the creation of variables. Where type is the "type" of variable to initialize. "name" is a custom name that the programmer assigns the variable. Lets declare our variable x from above. The above code not only creates a variable x that can store a number (as specified by int), but it also stores a value of 5 into that variable. So now whenever we call the variable x the computer will see that as a value of 5.

You can also assign values to your variables at any point in the program. This can be done with the = operator. The above code will create a variable named x that can store an integer number (as specified by int). The initial value of x is zero. However we change the value of x to 5 at the second line of code.

Output
The variable x contains 0 The variable x contains 5 The variable y contains 2 The variable z contains 2.4 The integer cast of z is 2 Hit any key to end...

Remarks
Lets step through this code one line at a time. This line initialized a new integer named x. The initial value of x is set to zero. If we did not define this initial value to zero, then the program would not compile properly. This line outputs some data to the console. This line in particular prints the value of x on the screen. The operator + performs a string concatenation. This means the string value of x is appended to the end of the text and then displayed in the console. Next we've reassigned the value of x. We no longer need the keyword int because we've already told the compiler that we'd like a variable named x. After reassigning the value of x to 5, we then output this to the screen. Next we've created a new variable named y. This variable is initialized to 2. The purpose of this code is to show that your variable can be initialized to whatever you'd like. The result is then output to the console. Here we've created a new variable type. A float can store non-integer numbers. We've initialized this float to 2.4f. The f is to inform the compiler that 2.4f is a floating point value. If we'd used 2.4 instead of 2.4f then the compiler would have assumed that 2.4 represented a double. It would have then thrown an error, because the compiler is not sure how to convert a double to a float without help from the programmer. Here we've casted the floating point value of 2.4f to an integer. This will remove any precision after the decimal place (as we can see on the output). These lines of code pause the execution of the program. Without these lines the program would terminate very quickly, and we wouldn't be able to view the program output. Console.ReadyKey waits until the user presses a key.

Casting
It is often important to be able to convert between different types of data. For example, the user may request a double precision value from your program, but your program stores them as floats. The requires a conversion method between types. This conversion is accomplished via casting. The syntax for casting is as follows: Where type is the type to convert to, and name is the name of the variable to convert.

Output
Original area: 304.12 Truncated area: 304 Hit any key to end...

Remarks
This line of code performs a type cast on the floating point variable named area. It is important to note that a cast does not change the contents of the variable area. Instead, a new integer value of area is created and stored in the variable truncated. The name truncated comes from the fact that this type of conversion (from float to int) removes any precision past the decimal place. This is called truncating.

String
String is any sequence of characters, without limitation of character. String can be empty. Some examples are: In order to convert string to a number (int or double) we use these expressions: For the opposite usage, we use ToString(n). For example:

var keyword
If a variable's type is clear according to the code, we can simply use "var" for its type. For example is these statements:

But this statement is wrong: because the compiler cannot understand what type of variable it is. Also, you cannot define a field (variable of a class) by var.

Summary
(type)name
 * Variables are used to store different types of data in C#.
 * C# has many types. Commonly used types include int, float and bool.
 * Each variable is given a unique name by the programmer. The name should reflect the type of data that is being stored.  Words such as Area and IsValid are more descriptive than t_var1e.  Using descriptive variable names can help with code comprehension.
 * Casting between basic types is possible using the syntax
 * Casting can introduce problems if the destination data type causes a loss of information. An example is when converting from a float to an int.  The float will be truncated to remove any precision after the decimal place.  This causes a loss of data and it is important that the program be robust enough to deal with the loss of data accordingly.

Practice Exercises

 * Create a program that can convert a double to an int and then back to a double. What is the final value of your variable?  Did you expect this outcome?  Make lots of explanatory console outputs so that you can better understand what is going on at each step of your code.
 * Create a program that converts from an int to a char. What happens when you output this char to the screen using Console.WriteLine?

title: C Sharp/Variables


text: File:3 bar plot U and D.jpg

title: File:3 bar plot U and D.jpg


text:
User:JWSchmidt/Blog/19 October 2008

This is a continuation from User:JWSchmidt/Blog/15 September 2008.

case 8
case 8 details. McCormack was not happy that I prevented him from destroying Wikiversity's long-standing cultural practice of welcoming new visitors who self-identify as students and that I foiled his attempt to destroy the system we had for inviting students to participate at Wikiversity as editors. McCormack was not happy that I created a learning resource that described his effort to delete and otherwise disrupt the main Wikiversity student portal page. It was entirely proper for me to defend the long-standing invitation extended by Wikiversity to new student participants. The learning resource I created was a very useful learning resource for student participants at Wikiversity. Because I have prevented McCormack from undermining the mission of Wikiversity he called me a troll and made a large number of false and distorted charges against me. He used those false and distorted charges against me to get me banned, blocked and de-sysoped. I will continue to refute all of the false and distorted charges McCormack against me. It makes me physically ill to read the false and distorted charges McCormack manufactured and to think about the damage he has done to Wikiversity. I hope that by answering all these false and distorted charges I can help the Wikiversity community avoid this kind of witch hunt in the future. --JWSchmidt 23:26, 19 October 2008 (UTC)

case 9
case 9 details. My edit provided useful balance to what was an existing deceptive claim about the Browse page. The Browse page originally functioned to facilitate the development of Wikiversity content, but it had been systematically "repurposed" to subvert that important function. I made a small edit in an attempt to correct the bias of the page and for my trouble I was charged with making an edit that looks like vandalism. Background: McCormack called me a troll and told me that he refuses to talk to me. If I were dealing with a normal member of the community who was willing to discuss matters, then talk page discussions would be a viable option. However, McCormack is part of a special team of custodians that does not have to be civil and follow policy, so I feel free to defend Wikiversity from his disruption by using the direct method of editing Wikiversity pages so as to correct problems that McCormack has created. In order to game the system, he makes false charges about my editing such as implying that my edit to the Brows page was apparent vandalism. Making false charges against an editor and using those false charges to "justify" blocking and de-sysoping is a clear abuse of custodial power. --JWSchmidt 16:11, 23 October 2008 (UTC)

Policy manipulation
Four Wikiversity custodians have charged me with "policy manipulation" because I have participated in the development of Wikiversity and its policies. This absurd and sickening charge comes from people who openly violated Wikiversity policy and worked to prevent the development of critically needed policies. I call upon the Wikiversity community to carefully examine the actions of these four custodians and how their actions have damaged and continue to damage the Wikiversity community. I stand by my efforts to help develop Wikiversity policy; those efforts are a normal part of developing a wiki community. The charge that my efforts the build Wikiversity are some kind of disruptive "policy manipulation" are absurd and groundless (see also this general reply from JWSchmidt for the charge of "policy manipulation"). Each specific charge is addressed below.

case 21
case 21 details. The Wikiversity Cite sources policy has long had a special place in the heart of User:SB Johnny. Citing sources for one's work is a fundamental part of the educational process, but for some reason SB Johnny blocked the designation of "cite sources" as official Wikiversity policy. Other fundamental Wikiversity policies, including the research policy that was explicitly requested by the Wikimedia Foundation Trustees, refer to, and rely on, the Cite sources policy. Because I have worked through the normal collaborative editing process to develop Wikiversity policy as directed by the Foundation, the four custodians who published all these false and distorted charges against me have charged me with "policy manipulation". I stand by the importance of Cite sources policy and my effort to have it designated as official Wikiversity policy. I ask the community: why is Cite sources not official policy? I ask the community: why is Privacy policy not official policy? I ask the community: who has stood in the way of development of critically needed policies like Cite sources and Privacy policy and in so doing done great harm to the community?

case 22
case 22 detailed reply. As for "case 21", Verifiability is one of a group of policies that is important for Wikiversity and the Wikiversity research policy that was explicitly requested by the Wikimedia Foundation Trustees. I make no apologies for my efforts to develop Wikiversity and the research policy and other related policies. The charge that I "manipulated policy" by my good faith efforts to find collaborators to develop these policies, hold community discussions about these policies and finally to designate them as official is absurd and sickening. That such absurd charges have been used to "justify" blocking, banning and desysoping me is outrageous.

case 23
case 23 detailed reply. Reliable sources is an interlocking policy with Verifiability and Cite sources. These three policies work together to make sure that reliable and verifiable sources are cited to support information on Wikiversity webpages. That normal good faith wiki editing and development of the Wikiversity project is called "manipulation" is truly sickening.

case 24
case 24 detailed reply. There have always been wiki participants who are tempted to use the vandalism reverting tools for purposes other than reverting vandalism, so it is important for wikis to have policies that say only use these tools for reverting vandalism. Tools for quickly reverting vandalism are a useful feature of MediaWiki. I am proud to take a stand against those who treat good faith edits as if they were vandalism. I find it truly sickening that four Wikiversity custodians with charge me with "policy manipulation" because I have worked to make sure that the anti-vandalism revert tools are only used for their intended purpose. --JWSchmidt 17:54, 23 November 2008 (UTC)

case 25
case 25 detailed reply. The rules for custodians were worked out by means of a large number of discussions and collaborative editing. The claim that marking a needed policy as "official" is in some way "policy manipulation" is absurd and this charge was only manufactured in an attempt to assemble a large number of false and distorted charges that were then used to "justify" blocking, banning and desysoping me. I think the Wikiversity community should closely examine the actions of Wikiversity participants who, rather than help develop needed policies, have obstructed their development, tried to prevent them from being marked "official" and have even ignored them and violated them. --JWSchmidt 18:59, 23 November 2008 (UTC)

case 36
case 36 detailed reply. User:McCormack has worked hard to ignore, undermine and replace the Wikimedia Foundation-approved Wikiversity project proposal. I prevented him from writing the "learn by doing model" out of Wikiversity. It is truly sickening that misguided custodians have accused me of "policy manipulation" for defending the Foundation-approved Wikiversity project proposal and the idea that Wikiversity is a wiki where learners can learn by editing. --JWSchmidt 18:59, 23 November 2008 (UTC)

to be continued
I intend to respond to all of the false and distorted charges that were made against me. It is a sickening process to have to read these false and distorted charges. Worse is the fact that a few Wikiversity participants continue to use these false and distorted charges to try to 'justify" the removal of my custodial status, the bad block that was made against me and keeping me banned from participating in #wikiversity-en.

title: User:JWSchmidt/Blog/19 October 2008


text:
User:Eml4500.f08.team.Bengtson/teamHW4

 HW Redo: The derivation for the transverse displacement degrees of freedom is added. The figure of the elemental subject to axial and transverse degrees of freedom and forces is also added.

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

Eml4500.f08.team.foskey.ckf 00:30, 3 November 2008 (UTC)

Transforming Local Degrees of Freedom
Goal: Want to find $$ \tilde{\mathbf{T}}^{(e)}$$ that frames fom the set of local (element) definions $$\boldsymbol{d}^{(e)}$$ to another set of local (element) defs $$\tilde{\boldsymbol{d}}^{(e)}$$ that $$ \tilde{\mathbf{T}}^{(e)}$$ is invertible.

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

This image shows that one can show displacements can be made in the x and y axis but the axises can also be aligned along the length of the beam. This causes the x~ axis displacement to be either tension or compression.

$$\tilde{\boldsymbol{d}}^{(e)}_{1}=\boldsymbol{q}^{(e)}_{1}$$ (comp of $$\tilde{\boldsymbol{d}}^{(e)}_{[1]} $$ along the $$\jmath $$ also known as $$\tilde{y}$$ axis

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

$$\tilde{\boldsymbol{d}^{(e)}_{2}=\boldsymbol{d}}^{(e)}_{[1]}\jmath$$

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

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

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

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

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

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

$$\boldsymbol{R}^{(e)}$$ is a 2x2 matrix $$\boldsymbol{0}^{(e)}$$ is a 2x2 0 matrix.

$$\tilde{\boldsymbol{f}}^{(e)} = K^{(e)}\begin{bmatrix} 1 & 0& -1 &0 \\ 0 &0  & 0 & 0\\ -1 &0  & 1 & 0\\ 0 & 0 & 0 &0 \end{bmatrix} \tilde{\boldsymbol{d}}^{(e)} $$

This means only the forces that act on the beam are the ones along beam. This means the force either acts a compression or tension.

Transverse forces do not stretch the springs.

title: User:Eml4500.f08.team.Bengtson/teamHW4


text:
User:Kristenmaher

Hello, my name is Kristen. I'm a fashion marketing major at a small school in Seattle, WA. [IMG]http://i33.tinypic.com/2s8fz2f.jpg[/IMG]

title: User:Kristenmaher


text: File:Element displacement bar.png

title: File:Element displacement bar.png


text:
User talk:Kristenmaher

Welcome
Hello Kristenmaher, 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;
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 * Browse our portals, schools, and research activities;
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Emesee 23:53, 19 October 2008 (UTC)

Knitting
You may find Knitting helpful. Consider sharing what you've learned here after you've learned more about it. Good luck! The Jade Knight (d'viser) 09:19, 30 October 2008 (UTC)

title: User talk:Kristenmaher


text: File:Displacement bar.png

title: File:Displacement bar.png


text: File:Equilibrium of node 2.png

title: File:Equilibrium of node 2.png


text:
User:Georgekalogiros

I' am 26 year old culinary student at the art institute of Seattle and this is my first quarter and I look forward to meeting new and interesting people. I was instantly impressed with Seattle when I first came out to visit my sister and made my mind up that I was going to move here. From the musical and food culture's to the different views on life and government the people of Seattle have a lot to offer me. Coming from Edison New Jersey where I was born and raised now I find myself in a new city, living on my own and very eager to get to know this city and it's people.

title: User:Georgekalogiros


text: File:3 bar truss setup.jpg

title: File:3 bar truss setup.jpg


text: File:Hw 4 1.png

title: File:Hw 4 1.png


text:
User:Eml4500.f08.ateam.carr/HW4

Eigenvalue Analysis - Continuation
The links below illustrate the application of Eigenvalue Modal Analysis to real world examples:

 * Frequency analysis of shell structures with contact
 * Frequency Solutions with Contact

The first link demonstrates various modes of the oscillations in a shell structure. The second link is an example of rigid body modes with zero eigenvalues.

In order to generate the plots of modal analysis, the Eigenvalue is simply multiplied by a sinusoidal function. For example:

$$d(t)=\lambda cos(\omega t)$$

where $$ \omega= \frac{2\pi}{f}$$

f is the frequency of vibration.undefined

Assembly of Element Stiffness Matrix into Global Stiffness Matrix
Consider the previous example of the two bar truss and recall the element force-displacement relationship:

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

We have previously proved the equilibrium of Node 2 from the free body diagrams of Element 1 and Element 2.

$$d_{3}=d_3^{(1)}=d_1^{(2)}$$

$$d_{4}=d_4^{(1)}=d_2^{(2)}$$

For Node 2, identify the global degree of freedom to element degree of freedom for both Element 1 and 2.

Equilibrium of Node 2

$$ \sum F_x=0=-f_{3}^{(1)}-f_{1}^{(2)} $$

$$ \sum F_y=0=P-f_{4}^{(1)}-f_{2}^{(2)} $$

Next replace each f above with the individual element kd relationship.

title: User:Eml4500.f08.ateam.carr/HW4


text: File:FEA Drawing.JPG

title: File:FEA Drawing.JPG


text: File:FEA Drawing Revised.JPG

title: File:FEA Drawing Revised.JPG


text:
User talk:Anmarlow

Welcome
Hello Anmarlow, 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;
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 * Read and help develop our community policies, or
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Emesee 01:59, 20 October 2008 (UTC)

title: User talk:Anmarlow


text:
User talk:Georgekalogiros

Welcome
Hello Georgekalogiros, 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;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * [ Ask] a question, discuss Wikiversity, etc. at the Colloquium;
 * Read and help develop our community policies, or
 * Chat with other Wikiversians on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Emesee 02:00, 20 October 2008 (UTC)

title: User talk:Georgekalogiros


text: File:Fea drawing 2.jpg

title: File:Fea drawing 2.jpg


text: File:Me.JPG

title: File:Me.JPG