text:
User:Eml4500.f08.FEABBQ/HW7

=Class Notes November 17, 2008=

Model Frame Problem with 2 Elements:
In this lecture, the two element system below was used to introduce the concept of frames.

Frames include beam elements which involve both translational and rotational degrees of freedom (dofs). Figure 2 below shows the free-body diagrams of both elements.

Element FBD's:

Note that each dof has a corresponding force. Also to be noted is that $$d^{(e)}_3$$ and $$d^{(e)}_6$$ correspond to rotational dofs. Similarly,$$f^{(e)}_3$$ and $$f^{(e)}_6$$ correspond to moments.

2-D Frame Showing Global dofs:
Figure 3 below shows the global FBD for the system. Notice the three dofs at each node compared to two dofs at each node for the similar two bar truss example.

For this example, the stiffness matrices for each element are square matrices with six rows and six columns: $$ \underline{K}^{(e)}_{6x6}$$, e=1,2

Similarly, the global stiffness matrix is a square matrix with nine rows and nine columns: $$ \underline{K}_{9x9}= {A^{e=2}_{e=1}}$$    $$\underline{K}^{(e)}_{6x6}$$

Assembly of Global Stiffness Matrix
The final portion of this lecture included the "recipe" for assembling the global stiffness matrix and is shown in figure 4 below.

=Class Notes November 19, 2008=

$${\tilde{k}}^{(e)}_{6x6} {\tilde{d}}^{(e)}_{6x1}={\tilde{f}}^{(e)}_{6x1}$$

$${\tilde{d}}^{(e)}_{6x1}=\begin{Bmatrix} {\tilde{d}}^{(e)}_{1} \\ {\tilde{d}}^{(e)}_{2}\\ {\tilde{d}}^{(e)}_{3}\\ {\tilde{d}}^{(e)}_{4}\\ {\tilde{d}}^{(e)}_{5}\\ {\tilde{d}}^{(e)}_{6} \end{Bmatrix} $$   ,    $${\tilde{f}}^{(e)}_{6x1}=\begin{Bmatrix} {\tilde{f}}^{(e)}_{1} \\ {\tilde{f}}^{(e)}_{2}\\ {\tilde{f}}^{(e)}_{3}\\ {\tilde{f}}^{(e)}_{4}\\ {\tilde{f}}^{(e)}_{5}\\ {\tilde{f}}^{(e)}_{6} \end{Bmatrix}$$

$$\tilde{z}=z $$ ( Rotation about Z-axis)

Moments about the Z-axis are equal in both coordinates: $$\tilde{f}^{(e)}_{3}=f^{(e)}_{3}, \tilde{f}^{(e)}_{6}=f^{(e)}_{6}$$

$$

\tilde{k}=\begin{bmatrix} \frac{EA}{L} &0 &0  &\frac{-EA}{L}  &0  &0 \\ 0 &\frac{12EI}{L^{2}} &\frac{6EI}{L^2} &0 &\frac{-12EI}{L^3}  &\frac{6EI}{L^2} \\ 0&0 &\frac{4EI}{L}  &0  & \frac{-6EI}{L^2} &\frac{2EI}{L} \\ \frac{-EA}{L}& 0 & 0 & \frac{EI}{L}&  0& 0\\ 0&\frac{-12EI}{L^3} &\frac{-6EI}{L^2}  & 0 &  \frac{12EI}{L^3}& \frac{-6EI}{L^2}\\ 0&\frac{6EI}{L^2} & \frac{2EI}{L} &  0& \frac{-6EI}{L^2} &\frac{4EI}{L} \end{bmatrix}_{(6X6)}$$

 * Dimension Analysis:

For Displacement the dimension $$\begin{bmatrix} \tilde{d_1} \end{bmatrix}=L= \begin{bmatrix} \tilde{d_i} \end{bmatrix} \rightarrow   i= 1,2,4,5 $$

$$L\rightarrow$$ Lenght

$$\begin{bmatrix} \tilde{d_i} \end{bmatrix} \rightarrow$$ Dimension of $$\tilde{d}_i$$

The rotation has a dimension of 1 $$\begin{bmatrix} \tilde{d_3} \end{bmatrix}=1= \begin{bmatrix} \tilde{d_6} \end{bmatrix} $$

$$AB=R\cdot \theta$$

$$AB\rightarrow$$ Arc Length

$$\theta \rightarrow$$ Angle

$$ \theta =\frac{AB}{R}$$

$$\left[\theta \right]=\frac{\left[AB \right]}{\left[R \right]}=\frac{L}{L}=1$$

=Class Notes November 21, 2008= Continuation of Dimensional Analysis

$$[ \varepsilon ] = \frac{du}{dx} = \frac{L}{L}=1$$

$$[ \sigma ] = [E] = \frac{F}{L^2}$$

$$\left[ \frac{EA}{L}\right] = \left[ \tilde{k_{11}}\right]=\frac{(F/L^2)(L^2)}{L}=\frac{F}{L}$$

$$\left[ \tilde{k_{11}} \tilde{d_1} \right] = [\tilde{k_{11}}][\tilde{d_1}]=\frac{[6][E][I]}{[L^2]} = \frac{1(F/L^2)L^4}{L^2} = F$$

Element F-d relation in global coordinates from element F=d relations in local coordinates, i.e. obtain

$$\mathbf{k}_{6x6}^{(e)}\mathbf{d}_{6x1}^{(e)}=\mathbf{f}_{6x1}^{(e)}$$

$$\mathbf{k}_{6x6}^{(e)}=\tilde{\mathbf{T}}_{6x6}^{(e)T}\tilde{\mathbf{k}}_{6x6}^{(e)}\tilde{\mathbf{T}}_{6x6}^{(e)}$$

=Class Notes November 24, 2008=

Deform Shape of Truss Element:Interpolation of Transverse Displacement $$ v(\tilde{x})$$ element
PVW for beams

$$ \int_{0}^{L}{w(\tilde{x}) \left[-\frac{\delta ^2}{\delta x^2} \left((EI)\frac{\delta^2 v}{\delta x^2} \right)  f_{t} - m\ddot{v}     \right] dx} = 0 $$ for all possible $$ w(x) $$

Now, Integrate by parts of 1st Term:

$$ \int_{0}^{L}{w(\tilde{x}) \left[-\frac{\delta ^2}{\delta x^2} \left((EI)\frac{\delta^2 v}{\delta x^2} \right)\right] dx} $$

where, $$ r(x) = \left( \frac{\partial }{\partial x}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\}\right) $$

$$ r^2(x) = \frac{\partial }{\partial x} \left( \frac{\partial }{\partial x}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\}\right) $$

$$ = \left[w \frac{\partial }{\partial x} \left\{(EI) \frac{\delta ^2v}{\delta x^2} \right\} \right]^{L}_{0} - \int_{0}^{L}{\frac{dw}{dx}\left( \frac{\partial }{\partial x}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\}\right)dx } $$

for simplicity, $$ \beta _{1} = \left[w \frac{\partial }{\partial x} \left\{(EI) \frac{\delta ^2v}{\delta x^2} \right\} \right]^{L}_{0} $$

Integrate again,

$$ = \beta _{1} - \left[\frac{dw}{dx}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\} \right]^{L}_{0} + \int_{0}^{L}{\frac{\delta ^2w}{\delta x^2} (EI) \frac{\delta^2v }{\delta x^2}dx} $$

to simplify further,

$$ \beta _2 = \left[\frac{dw}{dx}\left\{ (EI)\frac{\delta^2 v}{\delta x^2}\right\} \right]^{L}_{0} $$

$$ \gamma = \int_{0}^{L}{\frac{\delta ^2w}{\delta x^2} (EI) \frac{\delta^2v }{\delta x^2}dx} $$

Therefore the base equation (the boxed equation above) becomes,

$$ = - \beta _1 + \beta _2 - \gamma + \int_{0}^{L}{wf_t dx} - \int_{0}^{L}{wm\ddot{v}dx} $$

Now for further understanding focus on the stiffness term $$ \gamma $$ to dervie the beam stiffness matrix and also identify the beam beam shape functions.

 * Picture****

$$ v(\tilde{x}) = N_2(\tilde{x} )\tilde{d_2} +N_3(\tilde{x} )\tilde{d_3} + N_5(\tilde{x} )\tilde{d_5} + N_6(\tilde{x} )\tilde{d_6} $$

Recall: $$ u(\tilde{x}) = N_1(\tilde{x} )\tilde{d_1} N_4(\tilde{x} )\tilde{d_4} $$

 * Picture Set ****

HMWK ??? pg 246

=Class Notes December 1, 2008=

Computing $$U(\tilde{x}), V(\tilde{x)}$$ :

Local system: $$U(\tilde{x})=U(\tilde{x})\vec{\tilde{i}}+V(\tilde{x})\vec{\tilde{j}}$$

Global System:$$ U(\tilde{x})=U_{x}(\tilde{x})\vec+V_{y}(\tilde{x})\vec $$

Computing $$U(\tilde{x}), V(\tilde{x)}$$ using
 * $$V(\tilde{x)}=N_{2}(\tilde{x})\tilde{d_2}+N_{3}(\tilde{x})\tilde{d_3} +N_{5}(\tilde{x})\tilde{d_5}+N_{6}(\tilde{x})\tilde{d_6}$$
 * $$U(\tilde{x)}=N_{1}(\tilde{x})\tilde{d_1}+N_{4}(\tilde{x})\tilde{d_4}$$

Compute $$U_x(\tilde{x}), V_y(\tilde{x)}$$ from $$U(\tilde{x}), V(\tilde{x)}$$

$$\begin{Bmatrix} U_x(\tilde{x)}\\ U_y(\tilde{x)} \end{Bmatrix}= R^T \begin{Bmatrix} U(\tilde{x})\\V(\tilde{x})

\end{Bmatrix}$$

$$\begin{Bmatrix} U(\tilde{x})\\V(\tilde{x})

\end{Bmatrix}=\begin{bmatrix} N_1 &0 & 0 &N_4  &0  & 0\\ 0 &N_2 &N_3  & 0 & N_5 & N_6 \end{bmatrix}\begin{Bmatrix} \tilde{d_1^{(e)}}\\ \tilde{d_2^{(e)}}\\ \tilde{d_3^{(e)}}\\ \tilde{d_4^{(e)}}\\ \tilde{d_5^{(e)}}\\ \tilde{d_6^{(e)}} \end{Bmatrix}$$

let: $$ \begin{bmatrix} N_1 &0 & 0 &N_4  &0  & 0\\ 0 &N_2 &N_3  & 0 & N_5 & N_6 \end{bmatrix} =\mathbb{N(\tilde{\mathrm{x})}}$$

$$\begin{Bmatrix} U_x(\tilde{x)}\\ U_y(\tilde{x)} \end{Bmatrix}=R^T \mathbb{N(\tilde{\mathrm{x})}}\tilde{T}^{(e)}d^{(e)}$$

=Class Notes December 5, 2008=

$$\begin{Bmatrix} u_x(\tilde{x})\\ u_y(\tilde{x}) \end{Bmatrix} = \mathbf{R}^T\begin{Bmatrix} u(\tilde{x})\\ v(\tilde{x}) \end{Bmatrix}$$  (1)

$$\begin{Bmatrix} u_x(\tilde{x})\\ u_y(\tilde{x}) \end{Bmatrix} = \begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0\\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix} \begin{Bmatrix} \tilde{d_1}^{(e)}\\ \vdots\\ \tilde{d_1}^{(e)} \end{Bmatrix}$$  (2)

Perform dimensional analysis of equation (2):

$$[u]=L$$
 * First, a look at the normal linear degrees of freedom:

$$[N_1]=[N_4] = 1$$

$$[N_1][\tilde{d_1}],[N_4][\tilde{d_4}]$$

$$1\cdot L,1\cdot L = L$$

The dimensional analysis for DOF's 2 and 5

 * Now a look at the dimensional analysis of a rotational DOF:

$$[v]=L$$

$$[\tilde{d_3}]=[1]$$

$$[N_3][\tilde{d_3}],[N_6][\tilde{d_6}]$$

$$L\cdot 1,L\cdot 1 = L$$

Derivation of Shape Functions

Recall: Governing PDE for beams. Look at governing equations without internal forces of transverse forces or distributed truss loads. This is the static case:

$$\frac{\delta ^2}{\delta x^2} \left\{(EI) \frac{\delta v^2}{\delta x^2}\right\}=0$$

If EI is constant along the beam: $$\frac{\delta ^2}{\delta x^2}v=0$$

After integrating this equation four times, you arrive at: $$\Rightarrow v(x)=C_0+C_1x^1+C_2x^2+C_3x^3$$

Now the boundary conditions can be applied to find $$C_0, C_1, C_2, C_3$$ for each shape function.

 * $$\mathbf{N_2}$$

$$v(0)=1=C_0$$

$$v(L)=0=1+C_1L+C_2L^2+C_3L^3$$

$$v'(x)=C_1+2C_2x+C_3x^2$$

$$v'(0)=0=C_1$$

$$v'(L)=0=2C_2L+3C_3L^2$$

Now solve for $$C_2, C_3$$:

$$C_3={-2C_2}/{3L}$$

plugging $$C_3$$ back into the equation for v(L): $$0=1+C_2L^2+{-2C_2}/{3L}L^3$$

$$C_2=-3/L, C_3=2/L^3$$

$$N_2(x)=1-\left( \frac{3}{L^2}\right)x+\left( \frac{2}{L^3}\right)x^3$$

 * $$\mathbf{N_3}$$

change... $$v(0)=0=C_0$$

$$v'(0)=1=C_1$$

$$v(L)=0=L+C_2L^2+C_3L^3$$

solving for $$C_3=-1/L^2-C_2/L$$

$$v'(L)=0=1+2C_2L+3C_3L^2$$

solving for $$C_3=-1/3L^2-2C_2/3L$$

Now solve for $$C_2, C_3$$ by first setting $$C_3$$ equal to itself.

$$C_3=-1/3L^2-2C_2/3L=-1/L^2-C_2/L$$

$$C_2=-2/L$$

$$C_3=1/L^2$$

$$N_2(x)=x-\left( \frac{2}{L}\right)x^2+\left( \frac{1}{L^2}\right)x^3$$

 * $$\mathbf{N_5}$$

$$v(0)=0=C_0$$

$$v'(0)=0=C_1$$

$$v(L)=1=0+C_2L^2+C_3L^3$$

$$v'(x)=C_1+2C_2x+C_3x^2$$

$$v'(L)=0=2C_2L+3C_3L^2$$

Now solve for $$C_2, C_3$$:

$$C_3={1-C_2L^2}/{L^3}$$ from v(L)

$$C_3={-2C_2}/{3L}$$ from v'(L)

Setting $$C_3$$ equal to itself: $${(1-C_2L^2)}/{L^3}={-2C_2}/{3L}$$

$$C_2={3}/{L^2}, C_3={-2}/{L^3}-2L/3$$

$$N_5(x)=\left( \frac{3}{L^2}\right)x^2-\left( \frac{2}{L^3}\right)x^3$$

 * $$\mathbf{N_6}$$

$$v(0)=0=C_0$$

$$v'(0)=0=C_1$$

$$v(L)=0=C_2L^2+C_3L^3$$

Solve for $$C_3$$... $$C_3=-C_2/L$$

$$v'(L)=1-2C_2L+3C_3L^2$$

Solving again for $$C_3$$... $$C_3 = 1/3L^2-{2C_2}/{3L}$$

Setting $$C_3$$ equal to itself yields $$C_2$$: $$C_2=-1/L$$

$$C_3=1/L^2$$

$$N_6(x)=-\left( \frac{1}{L}\right)x^2+\left( \frac{1}{L^2}\right)x^3$$

=Matlab Matters=

2 Bar Truss Problem With Original Conditions
The code used to calculate the deformations for the original two bar truss problem is shown below. Please note that code was appended to the end to create the plot of the deformed elements.

The figure below is a plot of the undeformed and deformed elements.

2 Bar Truss Problem With Modified Conditions
For this homework it was also necessary to modify the 2 bar truss code to accommodate varying cross-sectional areas and elasticities. To do this, two different formulas were used to find the stiffness matrices of each element:

$$ \hat{k}_{gen}^{(e)}=\frac{(2A_{1}E_{1}+A_{2}E_{1}+A_{1}E_{2}+2A_{2}E_{2})}{6L}\begin{bmatrix} 1 &-1 \\ -1 &1 \end{bmatrix} $$

$$ \hat{k}_{avg}^{(e)}=\frac{(A_{1}+A_{2})(E_{1}+E_{2}))}{4L}\begin{bmatrix} 1 &-1 \\ -1 &1 \end{bmatrix} $$

The following Matlab code was used to calculate the deformations for the given values.

The figure below is a plot comparing the results of the two methods. Notice that due to the choice of given values, the results of from using the average values is the same as the original two bar truss.

Frame-Bar Truss
This function was used in the frame-bar truss code above

The following code is provided for solving problems that are purely frames (no bars as above)

=Wikiversity vs. E-learning=

=Contributing members:=
 * Stark 21:14, 9 December 2008 (UTC)
 * Esmail Hadman 23:07, 8 December 2008 (UTC)
 * Nik Vitt 00:04, 9 December 2008 (UTC)
 * Andy Koby 18:58, 9 December 2008 (UTC)
 * J.Jayma 19:48, 9 December 2008 (UTC)

title: User:Eml4500.f08.FEABBQ/HW7


text: File:Frame jamama.GIF

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

title: File:Frame jamama.GIF


text: File:Interpolations.jpg

title: File:Interpolations.jpg


text: File:BeamDisplacements.jpg

title: File:BeamDisplacements.jpg


text: File:FBD1 jamama.GIF

title: File:FBD1 jamama.GIF


text: File:FBD2 jamama.GIF

title: File:FBD2 jamama.GIF


text: File:Global jamama.GIF

title: File:Global jamama.GIF


title: File:Stiffness(9x9).GIF


text:
User talk:Thom Hastings

Welcome
Hello Thom Hastings, 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
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Draft of a Working Paper for Education at Brown University
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Automation (disambiguation)

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User talk:Stress9/HW6

Templates: $$\displaystyle \clubsuit$$ Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Artistic aspect: Comments: Overall rating:

 Note: The Overall rating: here is more of a "gut feeling" than a formal quantitative grading. Follow our list of criteria for formal grading and commenting. In case of a tie, let's favor new teams for the top-3 positions. Specific comments are given for a few teams, with a focus on the top teams to serve as a standard that other teams are measured against. The best of HW5 wiki page for each course is the grading standard; I also add some annotations in the HW reports that I looked through so to warn you of errors, incompleteness, etc. As you grade and see the best features in the HW reports, do not hesitate to add the teams with these best features to this wiki page. AERO: Ready for grading. Eml4500.f08 14:49, 26 November 2008 (UTC) FEAD: Ready for grading. Stress9 22:13, 29 November 2008 (UTC)

=EAS 4200C=

User:eas4200c.f08/HW_report_table

General comments for HW6

 * The The best of HW6 is the grading standard, which is also accessible at the bottom of the HW report table.

 * classnotes 50% + matlab problem 50% = 100%

 * contributing team members: I noticed that not all team members contributed to the HW reports; see e.g., Team Aero6 HW4. Non-contributing team members get zero HW grade. Please revisit HW1, HW2, HW3, HW4 to check on contributing team members; non-contributing team members should get zero HW grade.

Specific comments on team HW6 reports
=EML 4500=

User:Eml4500.f08/HW_report_table

General comments for HW6

 * The The best of HW6 is the grading standard, which is also accessible at the bottom of the HW report table.

 * classnotes 50% + matlab problem 50% = 100%

 * contributing team members: I noticed that not all team members contributed to the HW reports; see e.g., Team Aero6 HW4. Non-contributing team members get zero HW grade. Please revisit HW1, HW2, HW3, HW4 to check on contributing team members; non-contributing team members should get zero HW grade.

title: User talk:Stress9/HW6


text:
User:Eml4500.f08/HW report table/The best of HW6

 Under construction; not final (but close to being stable after a couple of weeks). The intention here is to document the best features in any HW report for the readers (including you); if you see excellent features in any HW report (including your team's) that I may have missed noticing, don't hesitate to let me know. I don't have time to read all HW reports in detail. In fact, due to time constraint, I only selectively looked at a few important features. On the other hand, I also added some annotations (e.g., related to errors) in the HW reports that I looked through for the benefit of everyone. Eml4500.f08 21:57, 26 November 2008 (UTC)

 * classnotes graded over 100%
 * principle of virtual work (PVW). Complete (not necessarily the "most" complete report since I did not read all reports): Team Bike (with annotations), with some errors, Team ATeam(Sean), with errors.
 * model problem: elastodynamics of a bar with varying cross section and varying Young's modulus
 * motivation: discrete equation of motion
 * multi-dofs system
 * single dof system
 * continuous case
 * integration by parts
 * axial forces at the boundary of the bar
 * selection of weighting function: discrete case (motivation), continuous case
 * comparison between continuous PVW and discrete PVW. Best table: Team Wiki1
 * interpolation
 * motivation
 * detailed expression
 * element stiffness matrix $$\displaystyle \mathbf k^{(e)}$$. Best overall solution and explanation: Team Wiki1
 * linearly tapered bar with linearly varying Young's modulus
 * particularization to case with $$\displaystyle E={\rm const} $$, $$\displaystyle A={\rm const} $$
 * element stiffness matrix $$\displaystyle \mathbf k^{(e)}_{ave}$$ by averaging $$\displaystyle E$$ and $$\displaystyle A$$ separately.
 * difference $$\displaystyle \mathbf k^{(e)} - \mathbf k^{(e)}_{ave}$$. NOTE: Several teams make an error here; Team Bike (with annotations), Team ATeam(Sean), Team FEABBQ (with annotations), etc.

 * detection of mechanisms by eigenvalue problem: Redo. Complete (not necessarily the "most" complete report since I did not read all reports): Team Bike (with annotations). See also Team ATeam(Sean).
 * square truss system
 * without diagonal brace
 * with diagonal brace

 * matlab problem graded over 100%
 * electric pylon problem
 * static analysis
 * element with highest compressive stress
 * element with highest tensile stress
 * plot of undeformed and deformed shapes
 * results: highest compressive and tensile stresses
 * Team Bike (with annotations): -8.6957 MPa, 9.0511 MPa. See also Team ATeam(Sean).
 * vibrational frequency analysis
 * lumped mass matrix
 * generalized eigenvalue problem
 * results: 3 highest vibration periods (3 lowest frequencies)
 * Team Bike (with annotations): 0.54654 s (1st mode, lowest eigenvalue), 0.12647 s (2nd mode), 0.11587 s (3rd mode). See also Team ATeam(Sean) with the same comments as for Team Bike.

EAS 4200C, The best of HW6


text:
User talk:Ariosto~enwikiversity

Hi!

Welcome
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User:Ariosto~enwikiversity

Hi!

title: User:Ariosto~enwikiversity


text:
President of the United States/Syllabus

 * Units:
 * Unit 1: The Founders
 * Washington
 * John Adams
 * Jefferson
 * Madison
 * Monroe
 * Unit 2: Change and Uncertainty
 * John Quincy Adams
 * Jackson
 * Van Buren
 * William Henry Harrison
 * Tyler
 * Unit 3: Trouble Brews in Dixie
 * Polk
 * Zach Taylor
 * Filmore
 * Pierce
 * Buchanan
 * Unit 4: From the front lines to the frontier to the slums
 * Lincoln
 * Andrew Johnson
 * Grant
 * Hayes
 * Garfield
 * Cleveland
 * Ben Harrison
 * Unit 5: Building Foreign Policy and The Modern War
 * McKinley
 * Teddy Roosevelt
 * Taft
 * Wilson
 * Harding
 * Coolidge
 * Hoover
 * FDR
 * Truman
 * Unit 6: A New World Power
 * Eisenhower
 * JFK
 * LBJ
 * Nixon
 * Ford
 * Carter
 * Reagan
 * George Hebert Walker Bush
 * Clinton
 * George W. Bush
 * Barack Obama

 * Assessment
 * Worksheets for bonus points on tests
 * Unit Quizes - TBD% (TBD% each)
 * Cumulative Test - TBD% (TBD% each)
 * Final Exam: Essay- TBD% (TBD% each)

title: President of the United States/Syllabus


text:
User:Eml4500.f08.delta 6.krueger/Notes Monday November 24th

Motivation: deformed shape of truss element and Interpolation of transverse displacement v(s), where s=$$tilde{x}$$. Recall the governing partial differential equation for the bar element is: $$ +\frac{d^{2}}{dx^{2}}[ (EA)\frac{\partial u}{\partial x} ] + \mathit{f_{a}(x,t)} = m(x)\ddot{u} $$ The positive sign in front of the equation is to emphasize the difference between the above equation and the governing equation for a distributed transversal load on a bar element. The first term, $$\frac{d}{dx^{2}}[ (EA)\frac{\partial u}{\partial x} ]$$, is a second order differential equation with respect to x. The second term ,$$\mathit{f_{a}(x,t)}$$, refers to an axial distributed load. Where the subscript a is to distinguish the axial load equation from the transversal load equation. The load may be changing with time. Therefore, the load may be a function of time as well as distance (x).

 Principle of Virtual Work for Beams:

The derivation is the same as the derivation of the principle of virtual work for the dynamics of elastic bar (refer to homework 6, PVW of dynamics of elastic bar) with one difference. The difference is that integration by parts has to be performed twice instead of once. Eqn (1)

$$ \int_{0}^{L}{w(\tilde{x})\left\{\frac{d^{2}}{dx^{2}} [ (EI) \frac{\partial^{2} u}{\partial x^{2}}] + f_{t} - m\ddot{v}\right\}dx = 0} $$

For all possible w(x), where $$f_{t}$$ is the transversal distributed load. First the first term will be integrated by parts. For simplicity the minus sign will be dropped and the first term will be defined as $$\alpha$$:

$$\alpha:= \int_{0}^{L}{w(\tilde{x})\frac{d^{2}}{dx^{2}} [ (EI) \frac{\partial^{2} u}{\partial x^{2}}]dx} $$ Where $$\tilde{x}=s(\tilde{x})$$. The position (r) is: $$r = \frac{d}{dx} [ (EI) \frac{\partial^{2} v}{\partial x^{2}}] $$ and the velocity ($$ \dot{r}$$) is: $$\dot{r} = \frac{d}{dx}(\frac{d}{dx} [ (EI) \frac{\partial^{2} v}{\partial x^{2}}]) $$

Now the argument x is omitted for simplicity and the equation is integrated.

$$=\left[w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}\right]_{0}^{L} - \int_{0}^{L}\frac{dw}{dx}\frac{\partial }{\partial x}\left\{(EI)\frac{\partial^{2} v}{\partial x^{2}}\right\}dx$$

Now again for simplicity $$\beta_{1}$$ is defined as:

$$\beta_{1} = w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}$$

Now Integration by parts once more.

$$=\beta_{1} - \left[w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}\right]_{0}^{L} + \int_{0}^{L}\frac{d^{2}w}{dx^{2}}\frac{\partial }{\partial x}\left\{(EI)\frac{\partial^{2} v}{\partial x^{2}}\right\}dx$$

Now $$\beta_{2}$$ and $$\gamma$$ are defined as:

$$\beta_{2} = \left[w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}\right]_{0}^{L}$$ and  $$\gamma = \int_{0}^{L}\frac{d^{2}w}{dx^{2}}\frac{\partial }{\partial x}\left\{(EI)\frac{\partial^{2} v}{\partial x^{2}}\right\}dx$$

Eqn (1) now becomes:

$$-\beta_{1}+\beta_{2}-\gamma+\int_{0}^{L}wf_{t}dx-\int_{0}^{L}wm\ddot{v}dx=0$$

For all possible $$w(\tilde{x})$$

Now look at the stiffness term $$\gamma$$ to derive the beam stiffness matrix and to identify the beam shape functions.

The axial displacements $$\tilde{d_{1}}$$ and $$\tilde{d_{4}}$$ are ignored because the focus is on the beam.

The transverse displacement ($$v(\tilde{x})$$) is: $$v(\tilde{x})=N_{2}\tilde{d_{2}} + N_{3}\tilde{d_{3}} + N_{5}\tilde{d_{5}} + N_{6}\tilde{d_{6}}$$. Recall: $$u(\tilde{x})= N_{1}(\tilde{x})\tilde{d_{1}} + N_{4}(tilde{x})\tilde{d_{4}}$$

The rotation of $$N_{2}(\tilde{x})$$ is equal to zero. The slope is therefore equal to zero. This is true at the ends of the beam. The expressions for the shape functions are:

$$N_{2}(\tilde{x})=1-\frac{3\tilde{x^{2}}}{L^{2}}+\frac{2\tilde{x^{3}}}{L^{3}}$$

$$N_{3}(\tilde{x})=\tilde{x}-\frac{2\tilde{x^{2}}}{L}+\frac{\tilde{x^{3}}}{L^{2}}$$

$$N_{5}(\tilde{x})=\frac{3\tilde{x^{2}}}{L^{2}}-\frac{2\tilde{x^{3}}}{L^{3}}$$

$$N_{6}(\tilde{x})=-\frac{\tilde{x^{2}}}{L}+\frac{\tilde{x^{3}}}{L^{2}}$$

Observe that when the shape function is multiplied by the displacement (d) it must give a displacement.


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Talk:Russian/Introduction/Alphabet

I suggest that all the lettres be shown with their origin. Currently, all that shows up is lettres similar to Latin lettres. However, being a linguist, I am familiar with both the Greek and Hebrew alphabets, from which many Cyrillic lettres come. Threfore, it is my suggestion that this be edited to show each lettre with its language or alphabet of origin, rather than only those derived from Latin.

title: Talk:Russian/Introduction/Alphabet


text:
Talk:Wiki Campus Radio/Active audio stream/Winter Festivals

Per IRC / #wikimedia-radio (Freenode)
A thought has been discussed about the performance of a Mystery Play on WCR over the Christmas Period. We have been looking in co-operation with Wikisource for a PD Version of the Mystery Play cycles, and this link seems to be a good bet - Its adapted from the Wakefield cycle and is completely out of copyright, making it a perfect piece for use.

Star of Bethlehem (Archive.org)

We will continue to look into this for use.

Iceflow 01:18, 27 November 2008 (UTC)


text: Talk:Educational Media Awareness Campaign/Geography: Atlases/POTD 3


text: File:Radsam hw6 profile1.png

title: File:Radsam hw6 profile1.png


text: File:Radsam hw6 profile2.png

title: File:Radsam hw6 profile2.png


text: File:Radsam hw6 clampprofile.png

title: File:Radsam hw6 clampprofile.png


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User:Mikhailov Kusserow~enwikiversity

My name is Mikhailov Kusserow. Call me Mikhailov.

title: User:Mikhailov Kusserow~enwikiversity


text:
User talk:Mikhailov Kusserow~enwikiversity

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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 Mikhailov Kusserow. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Mikhailov Kusserow~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 23:43, 17 March 2015 (UTC)

Renamed
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text:
Category talk:Learning projects

Could this cat use some cleanup? Emesee

Yes. I Real  Think  So. 173.178.93.250 13:28, 27 March 2011 (UTC)

Project Page

You Can Do  One About  Adolf  Hitler. A Suggestion  Only. 173.178.93.250 13:29, 27 March 2011 (UTC)

I WIll  Start A  WIki  Project  About  Adolf  Okay. 173.178.93.250 13:31, 27 March 2011 (UTC)

title: Category talk:Learning projects


text:
User:Amarcelo

Alvin B. Marcelo, MD

Director, International Open Source Network ASEAN+3

www.alvinmarcelo.com

title: User:Amarcelo


text:
User:Lfsciolis

Leonardo Francesco Sciolis

this is my 'userpage' and i am going to use it to experiment and familiarize myself with wikipedia. Specifically I will use this page to learn how to create and edit wiki articles.

title: User:Lfsciolis


text: Category:Political history

title: Category:Political history


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User talk:SB Johnny/Reflections

"the extreme tensions he had created through demonizing certain users in order to divide the community have almost completely relaxed."
 * Almost, but not totally. See, for example:  Jolie's analogy of JWSchmidt's remarks.  The Jade Knight (d'viser) 12:42, 27 November 2008 (UTC)

 * Well, I think it's become more of a tangential thing, rather than dominating the community discussions, though some recent comments hint that he might go on the offensive again during your (JK's) nomination discussion. The point is that he's less able to cause serious disruption because he's not in the position to bully. --SB_Johnny talk 15:13, 27 November 2008 (UTC)
 * This is true. The Jade Knight (d'viser) 12:21, 28 November 2008 (UTC)

title: User talk:SB Johnny/Reflections


text:
History/Categorization

This is a unified categorization guide for the School of History.

Introduction
The categorization of History pages is very important, and should be considered a high priority for the School—the Browse feature at Wikiversity relies 100% on proper categorization, as does the History Portal. Unless projects and Departments are properly categorized, new participants are much less likely to find the content they're looking for, and are less likely to contribute to the School. It's worth noting that Portal:History gets significantly more traffic than School:History does.

General Guidelines

 * No History pages should be placed in Category:History except for major portal pages. All other History pages should be filed under the History sub-categories (as follow).
 * Where appropriate, all projects should be categorized under each of the following three categories (in a relevant sub-category):
 * Category:Historical Subjects by Area (for the region the project applies to)
 * Category:Historical Subjects by Period (for the time period the project took place in)
 * Category:Historical Subjects by Topic (for any topical categories which might apply)
 * Not all of these categories will always apply—a Period category is unlikely to be helpful for a History project spanning centuries, for example.
 * It is helpful to browse the history categories in the areas you are interested in, so that you will be most familiar with where things should go.

Further Tips

 * If you would like your learning project to be listed at Browse, you should add it to Category:Learning projects.
 * For additional exposure, it would be good to place your learning projects in the appropriate subcategory of Category:Resources by type. This will especially help other instructors who are thinking of developing similar sorts of materials for their own courses.
 * It may also be useful to list your projects under the appropriate learning level: Category:Pre-school Education, Category:Primary education, Category:Secondary Education, Category:Tertiary Education, Category:Non-formal Education, or Category:Research.

National Histories

 * Examples: US History, History of Germany, etc.

 * National histories should be filed under Category:History by Nation in addition to the subject-specific categories listed above.
 * Remember when doing so to use the format " ]", as many national history project pages start with "history of" instead of the nation's name. Example:  " ", for History of Germany.  Doing this ensures that the page will be categorized in a way that will make it easier to find.

Departments

 * In order to be listed at Browse, Category:Departments should be added to all departments. Additionally, the School of History has its own internal department category: Category:History departments.

title: History/Categorization


text:
Talk:Epistemology: Agnosticism or Gnosticism?

It would be helpful if you linked to the Wikisource pages for the readings you would have your students do. The Jade Knight (d'viser) 14:29, 27 November 2008 (UTC)


text:
Talk:Data sizes and speeds

= Data sizes and speeds = The whole document would benefit from a clear description of decimal and binary prefixes. It would also help to use the SI standard units (http://physics.nist.gov/cuu/Units/binary.html) or at least mention that for readebilty these are expicitly not used. Theking2 (discuss • contribs) 06:55, 2 September 2014 (UTC)

 * I agree that the treatment of decimal vs binary multipliers is unclear. New prefixes and new standards initially make the situation still more confusing. I've been initially confused, then skeptical, regarding the new prefixes, but contrary to my expectations they seem to be getting adopted. More surprisingly, there seems to be some support for the principle of limiting the use of the binary quantities to situations where they are relevant. For example my file manager reports the size of a 14,355,162 byte file as 14.4 MB rather than 13.7 MB (or MiB). I'm considering having a go at fixing up the whole page, but I would value some feedback on whether to go ahead with representing the IEC/ISO standard as the norm.--Alkhowarizmi (discuss • contribs) 11:01, 8 April 2016 (UTC)

1024kB squared
1024kB squared would be weird, I'm thinking 1T(B2)...

Word
The size of a word has conventionally been set at 32 bits (4 bytes) not two bytes (16 bits).

KB
The section on Kilobyte is badly worded and does not truly explain the difference between KB (1000) and KiB (1024) in an understandable and correct way.

 * Actually, 1000 bytes is 1 kB. KB is either plain wrong, or follows a convention (due to, or propagated by O'Reilly?) of using KB for the binary kilobyte. However, awareness that prefixes are case-sensitive is not widespread, and it's not uncommon to see brag sheets claiming lots of mB. (What would a millibyte look like - anything like a square byte?)--Alkhowarizmi (discuss • contribs) 11:01, 8 April 2016 (UTC)

Byte
It's just not true that a Byte has to be 8 bits. 8 bits is an Octet, but there are 12 bits/Byte and 13 bits/byte (and way more) other machines out there.

GigaByte
The wording in this part is quit confusing. There is only one "actual" size of a harddisk. Theking2 (discuss • contribs) 06:52, 2 September 2014 (UTC)

 * Well, that's quite confusing! On my reading, the wording (then as now) doesn't imply more than one "actual" size. However it does say (absurdly) that manufacturers' sales pitch understates the "true" capacity, where it should say overstates. I was going to correct this, but wouldn't want to do that without doing something about that "true" problem, and I haven't decided how. I'm more inclined to leave it for a major revision of the whole page.--Alkhowarizmi (discuss • contribs) 11:01, 8 April 2016 (UTC)

Provide everyday size references
It may be helpful to provide "everyday" size references to provide the student with some sense of scale. How many bytes is: a number, a natural language word, a natural language sentence, a book, or encyclopedia? How big is a CD, a DVD, photo image, mp3 song, or a YouTube video? How much data does Google store? How many bits is a DNA molecule? Thanks!

title: Talk:Data sizes and speeds


text:
User:Eml4500.f08.delta 6.krueger/HW dimensions of ktilde dtilde

Verify the dimensions of all terms of $$\tilde{k_{ij}}\tilde{d_{j}}$$

Start with the matrices of $$\tilde{k_{ij}}$$ and $$\tilde{d_{j}}$$.

$$\tilde{k_{ij}}\tilde{d_{j}}=\begin{bmatrix} \frac{EA}{L} & 0 & 0 & -\frac{EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & -\frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L} \\ -\frac{EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & -\frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{6EI}{L^{2}} & -\frac{4EI}{L} \end{bmatrix} \begin{pmatrix} \tilde{d_{1}} \\ \tilde{d_{2}} \\ \tilde{d_{3}} \\ \tilde{d_{4}} \\ \tilde{d_{5}} \\ \tilde{d_{6}} \end{pmatrix}$$

The displacements $$\tilde{d_{j}}$$ where i=1,2,4 and 5 are transverse displacements. The dimensions of transverse displacements are length. Shown below is the dimensional analysis. $$\left[\tilde{d_{i}}\right] = \left[\tilde{d_{1}}\right] = \left[\tilde{d_{2}}\right] = \left[\tilde{d_{4}}\right] = \left[\tilde{d_{5}}\right] = L$$ The square brackets, in dimensional analysis, indicate that the dimension of the enclosed are being found. The displacements $$\tilde{d_{j}}$$ where i=3 and 6 are rotational displacements. The dimensions of rotational displacements are 1. Shown below is the dimensional analysis. $$\left[\tilde{d_{3}}\right] = \left[\tilde{d_{6}}\right] = 1$$ From the class notes, it is known that: $$\left[E\right] = \frac{F}{L^{2}}$$ and $$\left[A\right] = L^{2}$$ and $$\left[I\right] = L^{4}$$ Therefore: $$\left[\frac{EA}{L}\right] = \frac{F}{L}$$ and $$\left[\frac{EI}{L}\right] = FL$$ and $$\left[\frac{EI}{L^{2}}\right] = F$$ and $$\left[\frac{EI}{L^{3}}\right] = \frac{F}{L}$$ Since zero is dimensionless, it is known that the dimensions of all $$\tilde{k_{ij}}\tilde{d_{j}}$$ that equal to zero have dimensions of 1. Zero is obviously dimensionless, and a dimensionless parameter is 1. Below are all $$\tilde{k_{ij}}\tilde{d_{i}}$$ that have a dimension of 1.

$$\left[\tilde{k_{12}}\tilde{d_{2}}\right] = \left[\tilde{k_{13}}\tilde{d_{3}}\right] = \left[\tilde{k_{15}}\tilde{d_{5}}\right] = \left[\tilde{k_{16}}\tilde{d_{6}}\right] = \left[\tilde{k_{21}}\tilde{d_{1}}\right] = \left[\tilde{k_{24}}\tilde{d_{4}}\right] = \left[\tilde{k_{31}}\tilde{d_{1}}\right] = \left[\tilde{k_{34}}\tilde{d_{4}}\right] = \left[\tilde{k_{42}}\tilde{d_{2}}\right] = \left[\tilde{k_{43}}\tilde{d_{3}}\right] = \left[\tilde{k_{45}}\tilde{d_{5}}\right] = \left[\tilde{k_{46}}\tilde{d_{6}}\right]$$ $$ = \left[\tilde{k_{51}}\tilde{d_{1}}\right] = \left[\tilde{k_{54}}\tilde{d_{4}}\right] = \left[\tilde{k_{61}}\tilde{d_{1}}\right] =  \left[\tilde{k_{64}}\tilde{d_{4}}\right] = 1 $$

When the rest of the stiffness parameters are multiplied out with the corresponding displacements the following dimensional parameters are found:

$$\left[\tilde{k_{11}}\tilde{d_{1}}\right] = \left[\tilde{k_{14}}\tilde{d_{4}}\right] = \left[\tilde{k_{22}}\tilde{d_{2}}\right] = \left[\tilde{k_{25}}\tilde{d_{5}}\right] = \left[\tilde{k_{41}}\tilde{d_{1}}\right] = \left[\tilde{k_{44}}\tilde{d_{4}}\right] = \left[\tilde{k_{52}}\tilde{d_{2}}\right] = \left[\tilde{k_{55}}\tilde{d_{5}}\right] = F$$

$$\left[\tilde{k_{23}}\tilde{d_{3}}\right] = \left[\tilde{k_{26}}\tilde{d_{6}}\right] = \left[\tilde{k_{32}}\tilde{d_{2}}\right] = \left[\tilde{k_{33}}\tilde{d_{3}}\right] = \left[\tilde{k_{35}}\tilde{d_{5}}\right] = \left[\tilde{k_{36}}\tilde{d_{6}}\right] = \left[\tilde{k_{53}}\tilde{d_{3}}\right] = \left[\tilde{k_{56}}\tilde{d_{6}}\right] = \left[\tilde{k_{62}}\tilde{d_{2}}\right] = \left[\tilde{k_{63}}\tilde{d_{3}}\right] = \left[\tilde{k_{65}}\tilde{d_{5}}\right] = \left[\tilde{k_{66}}\tilde{d_{6}}\right] = FL$$


text:
User:Benja/It is easy for Bayesians to sell information

Suppose that in a group of perfect Bayesian rational actors with a common prior and different information partitions, one actor wants to sponsor a prediction market, paying the other actors to reveal their private information to the group. Suppose further -- this is an important limitation of the following, as it stands -- that the full true state of the world, not just the value of one particular random variable, will be revealed to the group at a later time $$t'$$. Can the sponsor make an offer to the other agents that makes it rational for them to share all their information, truthfully? I.e., an offer such that agents cannot try to gain more by manipulating the market, for example?

If the agents have no reasons exogenous to the sponsor's offer for wanting the other agents to be misinformed, and the common prior assigns non-zero probability to every possible state of the world, this is easy. Suppose that for every $$\omega \in \Omega$$, the sponsor is willing to pay $$f(\omega)$$ for being informed that $$\omega$$ is not the true state of the world. Then, for every $$\omega$$, the sponsor offers to the first taker a contract that will pay the taker $$f(\omega)$$ at $$t'$$ if $$\omega$$ is revealed not to be the true state of the world, and will cost the taker $$f(\omega) / p(\omega)$$ at $$t'$$ if $$\omega$$ is revealed to be the true state of the world, where $$p$$ is the prior common to the agents.

Let $$I$$ be the information of a particular agent. Then, the agent's expected gain from taking the contract for a particular $$\omega$$ is

 * $$(1-p(\omega|I))f(\omega) - p(\omega|I)f(\omega)/p(\omega)$$.

Now, if $$\omega \notin I$$, then $$p(\omega|I)=0$$, so the agent has an incentive to take the contracts for all $$\omega$$, $$f(\omega)>0$$, that they definitely know not to be the true state of the world.

If $$\omega \in I$$, then $$p(\omega|I) \geq p(\omega)$$, and therefore, $$p(\omega|I)f(\omega)p(\omega) \geq p(\omega)f(\omega)/p(\omega) = f(\omega) > (1-p(\omega|I))f(\omega)$$; the last inequality is strict because we have required that $$\forall \omega \in \Omega : p(\omega) > 0$$ (which we need because we divide by $$p(\omega)$$). Therefore, if $$\omega \in I$$, the expected value of taking the contract for $$\omega$$ is negative, so the actor does not have an incentive to take the contracts for $$\omega$$'s that they do not definitely know not to be the true state of the world.

Finally, revealing an $$\omega$$ to be false does not let other agents to rule out any $$\omega'$$ that the revealing agent doesn't already know to be false also (XXX verify that claim).

Question: Can any of this be extended if only partial information about $$\omega$$ is revealed at $$t'$$?


text:
User:Eml4500.f08.delta 6.krueger/HW Prove FD-rel

Prove Element FD-rel

Start with: Eqn 1$$\tilde{f}^{(e)}_{6x1} = \tilde{k}_{6x6} ^{(e)} \tilde{d}_{6x1}^{(e)}$$

Then plug the two following equations into Eqn 1:

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

The following equation is found:

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

The transpose of the $$\tilde{T}$$ is taken to move the matrix from the right side to the left and the following equation is found: Eqn 2$$f^{(e)} = \begin{bmatrix} \tilde{T}^{(e)^{T}} \tilde{k}^{(e)}  \tilde{T}^{(e)}*d^{(e)} \end{bmatrix}$$

Now recall the following equation:

$$k_{6x6}^{(e)}=\tilde{T}_{6x6}^{(e)T}\tilde{k}_{6x6}^{(e)}\tilde{T}_{6x6}^{(e)}$$

This equation is substituted into Eqn 2 to get:

$$f_{6x1}^{(e)}=k_{6x6}^{(e)}d_{6x1}^{(e)}$$


text: File:FDrel.jpg

title: File:FDrel.jpg


text:
User:Vinhax

Vinhax aka Vincent Pafumy III was born September 22, 1987. Vincent was diagnosed in 2004 with Spinal Meningitis after he was completely blind and paralyzed. He recovered a few months later, learning how to walk again. Vincent dropped out of highschool that year, but obtained his G.E.D. in December, 2004.

more soon

title: User:Vinhax


text: Talk:Educational Media Awareness Campaign/History/POTD 6


text:
President of the United States/Worksheets

Here are a set of Worksheets.
 * /Worksheet 1/ - Leap Years
 * /Worksheet 1.A/ - Answer Sheet

 * /Worksheet 2/ - Presidential Terms
 * /Worksheet 2.A/ - Answer Sheet

 * /Worksheet 3/ - Succeeding the President
 * /Worksheet 3.A/ - Answer Sheet

 * /Worksheet 4/ - Selective Memory
 * /Worksheet 4.A/ - Beginning, Middle, & End
 * /Worksheet 4.B/ - Succession


text:
President of the United States/Worksheets/Worksheet 1

Return to Election Years are Leap Years.

Leap Years
In this exercise, we begin to create a timeline, which is initially built on Leap Years. Using the table below, begin with the year 1776 and write down the date of every fourth year until a date is in every cell.

On the surface, we simply have a long list of numbers, but with a few basic facts, we can add meaning and relevance to a list of leap years.

 * 1) What year was the Declaration of Independence signed?
 * 2) What year was the US Constitution ratified?
 * 3) What year was the first US Presidential Election?
 * 4) Who was the first US President?
 * 5) Which past election years are not actually leap years?
 * 6) What election year occurred during the Civil War?
 * 7) Who was elected President during the Civil War?
 * 8) What year did the Civil War end?
 * 9) Who was President when final hostilities ceased?

The Timeline
../Worksheet 1.A/


text:
President of the United States/Worksheets/Worksheet 1.A

Worksheet 1: Leap Years (Answer Sheet)
In this exercise, we begin to create a timeline, which is initially built on Leap Years. Using the table below, begin with the year 1776 and write down the date of every fourth year until a date is in every cell.

On the surface, we simply have a long list of numbers, but with a few basic facts, we can add meaning and relevance to a list of leap years.

 * 1) What year was the Declaration of Independence signed? -1776
 * 2) What year was the US Constitution ratified? -1788.
 * 3) What year was the first US Presidential Election? -1789.
 * 4) Who was the first US President? -George Washington.
 * 5) Which past election years are not actually leap years? -1800 & 1900. And 1789, too.
 * 6) What election year occurred during the Civil War? -1864.
 * 7) Who was elected President during the Civil War? -Abraham Lincoln.
 * 8) What year did the Civil War end? -1865.
 * 9) Who was President when final hostilities ceased? -Andrew Johnson.


title: User:Mlester711


text:
Talk:Introduction to Metaphysics: Cosmology & Ontology

What Pythagorean corpus? My understanding is that none of Pythagoras' works survived? The Jade Knight (d'viser) 12:38, 28 November 2008 (UTC)

 * I said "Pythagorean," not "Pythagoras'" (just as "Hermetic" not "Hermes'" corpus ;) many of his ideas were recorded by Pythagoreans, so I mean that and other texts like Classical biographies.--Dchmelik 03:24, 8 December 2008 (UTC)

 * I see. The Jade Knight (d'viser) 03:34, 8 December 2008 (UTC)


text: File:Winmerge.png

title: File:Winmerge.png


text: File:Wireshark.png

title: File:Wireshark.png


text:
Linux/Basic commands/ln

ln is a standard Unix command used to create links (link) to files.

Link files
Links allow more than one file name to refer to the same file, elsewhere.

There are two types of links, both of which are created by ln :
 * symbolic links, which refer to a symbolic path indicating the abstract location of another file, and
 * hard links, which refer to the specific location of physical data.

These links behave differently when the source of the link (what is being linked to) is moved or removed. Symbolic links are not updated (they merely contain a string which is the pathname of its target); hard links always refer to the source, even if moved or removed.

Specification
The Single Unix Specification (SUS) specifies the behaviour that a link or links (either symbolic or hard) will be created where specified that will link to the target file (or directory) specified.

More precisely, ln can be invoked in one of two ways: two arguments -- first an argument specifying the source file then the target, or multiple (greater than two) arguments, specifying firstly a number of source files, then a directory in which all the links are to be created. In the latter invocation, the names of the links will be that of the source file. This invocation will be assumed if the last argument is a directory. If invoked in the latter form, ln 's behaviour is not specified (it is implementation-defined).

ln is specified to use behaviour identical to that of the standard unlink and link functions.

Usage
The SUS mandates for ln two options: -f will force removal of existing files to allow the link to be created; and -s will create symbolic links. Therefore, ln with no options creates a hard link, ln -f forces a hard link, ln -s creates a symbolic link, and ln -fs forces a symbolic link.

Other Unix and Unix-like operating systems may add extra options. GNU ln adds options such as -b to back up files before creating links, -v to print out the name of each file before links are created, and others. BSD ln adds -h, preventing ln from descending into targets whose symlinks point to directories

ln file_name link_name

would have the effect of creating a hard link called link_name that points to the same data as the existing file file_name.

Symbolic link creation and deletion:

To begin with, enter the directory called "test": alex@ubuntu:~/playground/test$ ls -ali total 8 969797 drwxr-xr-x 2 alex alex 4096 Dec 9 09:10. 1036602 drwxr-xr-x 3 alex alex 4096 Dec 9 09:10 .. alex@ubuntu:~/playground/test$ There is nothing in it. Let's create a small text file called data.txt: alex@ubuntu:~/playground/test$ echo "some data" > data.txt alex@ubuntu:~/playground/test$ ls -ali total 12 969797 drwxr-xr-x 2 alex alex 4096 Dec 9 09:11. 1036602 drwxr-xr-x 3 alex alex 4096 Dec 9 09:10 .. 969768 -rw-r--r-- 1 alex alex  10 Dec  9 09:11 data.txt alex@ubuntu:~/playground/test$ Notice that the reference count of the directory (total 8) just increased (total 12). The new file data.txt is stored in inode 969768. Let's create a symbolic (soft) link to data.txt. We will name our symbolic link slink.txt : alex@ubuntu:~/playground/test$ ln -s data.txt slink.txt alex@ubuntu:~/playground/test$ ls -ali total 12 969797 drwxr-xr-x 2 alex alex 4096 Dec 9 09:11. 1036602 drwxr-xr-x 3 alex alex 4096 Dec 9 09:10 .. 969768 -rw-r--r-- 1 alex alex  10 Dec  9 09:11 data.txt 969817 lrwxrwxrwx 1 alex alex   8 Dec  9 09:11 slink.txt -> data.txt alex@ubuntu:~/playground/test$ The reference count of the directory has not changed (total 12). Our symbolic (soft) link is stored in a different inode than the text file (969817). The information stored in data.txt is accessible through the slink.txt : alex@ubuntu:~/playground/test$ file slink.txt slink.txt: symbolic link to `data.txt' alex@ubuntu:~/playground/test$ cat slink.txt some data alex@ubuntu:~/playground/test$ If we delete the text file data.txt, slink.txt becomes a broken link and our data is lost. alex@ubuntu:~/playground/test$ rm data.txt alex@ubuntu:~/playground/test$ ls -ali total 8 969797 drwxr-xr-x 2 alex alex 4096 Dec 9 09:36. 1036602 drwxr-xr-x 3 alex alex 4096 Dec 9 09:32 .. 969817 lrwxrwxrwx 1 alex alex   8 Dec  9 09:11 slink.txt -> data.txt alex@ubuntu:~/playground/test$ file slink.txt slink.txt: broken symbolic link to `data.txt' alex@ubuntu:~/playground/test$ cat slink.txt cat: slink.txt: No such file or directory alex@ubuntu:~/playground/test$ If slink.txt was a hard link, our data would still be accessible through slink.txt. Also, if you delete the soft link itself, the data would still be there:

$ ls -ali

63500 drwxr-xr-x  2 sc69876  support      96 Aug 29 18:14. 43038 drwxr-xr-x 12 sc69876  support    1024 Aug 29 18:13 .. 104690 -rw-r--r--  1 sc69876  support      10 Aug 29 18:13 data.txt 104825 lrwxrwxrwx  1 sc69876  support       8 Aug 29 18:14 slink.txt -> data.txt $ rm slink.txt

$ ls -ali

63500 drwxr-xr-x  2 sc69876  support      96 Aug 29 18:15. 43038 drwxr-xr-x 12 sc69876  support    1024 Aug 29 18:13 .. 104690 -rw-r--r--  1 sc69876  support      10 Aug 29 18:13 data.txt

Manual pages

 * ln -- manual page from GNU coreutils
 * ln -- manual page from OpenBSD

title: Linux/Basic commands/ln


text:
User talk:Eas4200c.f08.aero6.ortega/HW4

User:eas4200c.f08.aero6.ortega/HW7

title: User talk:Eas4200c.f08.aero6.ortega/HW4


text:
User:Eas4200c.f08.aero6.ortega/HW7

Vu-Quoc example of open cross-section with varying stringer areas

mean value theorem:

$$Q_{z} = \int_{A} ydA = \bar{y}\int_{A}da = \bar{y}A$$

$$Q_{y} = \int_{z}dA = \bar{z}A$$

$$A = \sum^{4}_{i=1}A_{i}$$(neglecting the skin and the spar webs)

Recall from chapter 5.5 of C.T. Sun's Mechanics of Aircraft Structures $$q(s) = (k_{yz}Q_{z} - k_{z}Q_{y})V_{z}\!$$

Notice the shear flow is constant along the webs because the area is restricted to the stringers. Also:

1)$$V_{z}\!$$

2)$$k_{yz}, k_{z}\!$$

3)$$Q_{z}, Q_{y}\!$$

Are all independent of the surface contour. However, $$q(s)$$ would increment or "jump" when crossing a stringer.

Step 1. find ($$\bar{y}_{c}, \bar{z}_{c}$$)

Step 2. find $$I_{y}, I_{z}, I_{yz}\!$$

Step 3. find $$k_{y}, k_{z}, k_{yz}\!$$

Step 4. follow the contour path (s) to find $$q_{12}, q_{23}, q_{34}$$

where

$$q_{12} = (k_{yz}Q_{z}^{12} - k_{z}Q_{y}^{12})V_{z}$$

$$Q_{z}^{12} = y_{1}A_{1}$$ where $$y_{1}$$ is the y-coordinate for stringer 1.

$$Q_{y}^{12} = z_{1}A_{1}$$

note$$Q_{z}^{23} = y_{1}A_{1} + y_{2}A_{2}$$

Note that the contour path is not fixed. If beginning the contour from the opposite direction, the terms are as follows:

to be continued...

Mini-plan
S) Single-cell section
 * S.1) without stringers
 * S.2) with stringers

M) multi-cell sections
 * M.1) without stringers
 * M.2) with stringers

S.1) without stringers
$$q = c = $$constant shear flow

Question: can this set up resist $$V_{z}$$ (transverse shear flow)?-No.

$$R^{z} = R_{AB}^{z} + R_{BA}^{z}$$ where R is the resultant of q

$$R_{AB}^{z} = -q\bar{A'B'} = -R_{BA}^{z} ==R^{z} = 0$$

S.2) with stringers
Neglecting the contribution to web bending:

When you cross a stringer, you have a jump. This is a somewhat step-wise function.$$q_{12} \ne q_{23} \ne q_{31}$$. However, $$q_{ij}(s) = c = \!$$constant within each panel. Therefore q(s) is piece-wise constant throughout the contour.

$$R^{z} = V_{z} \ne  0\!$$. Thus this is a non-constant shear flow.

The principle of Super Position is a powerful tool for solving these problems and is readily available due to the linearity of the components.

$$q_{12} = q + \bar{q_{12}}$$

$$q_{23} = q + \bar{q_{23}}$$

$$q_{31} = q + \bar{q_{31}}$$

$$q_{ij} = q + \bar{q_{ij}}$$

One "q" ($$\bar{q_{ij}}$$ are known after solving the case without stringers) and one equation for each unknown.

Method
Data:$$V_{y}, V_{z}$$

1) Solve "2" for $$\bar{q_{12}}, \bar{q_{23}} (\bar{q_{31}} = 0)$$

2) Moment equation: take the moment about any point in the y-z plane
 * 2.1) Super-position:$$q_{ij} = q + \bar{q_{ij}}$$
 * 2.2) Select a point ( denoted $$\bar{o}$$) in plane y-z
 * $$\sum_{\bar{o}}Moments = \sum_{\bar{o}} of q_{ij}$$

3) Back to Super position

$$q_{ij} = q + \bar{q_{ij}}$$

where $$q_{ij}$$ is the true shear flow, $$q$$ is the closed-cell constant shear flow and $$\bar{q_{ij}}$$ is the open-cell piece-wise constant shear flow.

M.1) Multi-cell without stringers
q = 0 at the cuts and is constant through the contour of the skin, thus q = 0 as shown.

$$R^{z} = R^{z1} + R^{z2} = V_{z}\!$$ where $$R^{z1} = 0\!$$

Looking at stringer 3

$$\sum F_{x} = 0 = \int_{A_{3}} [\theta_{xx}(x + dx) - \theta_{xx}(x)]dA_{3} = [-\bar{q_{23}} -\bar{q_{43}} + \bar{q_{31}}]dx$$. Notice the integrand is simply a Taylor series expansion.

$$\bar{q_{31}} = \bar{q_{23}} + \bar{q_{43}} + q^{(3)}$$

$$q^{(3)} = -\int_{A_{3}}\frac{d\theta_{xx}}{dx}dA_{3}$$ This is the contribution to shear flow by stringer 3.

Recall $$V_{y} = \frac{dM_{z}}{dx}, V_{x} = \frac{dM_{y}}{dx}$$

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

$$q^{(3)} = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z}^{(3)} - (k_{z}V_{z} - k_{yz}V_{y})Q_{y}^{(3)}$$ where $$Q_{z}^{(3)} = \int_{A_{3}}ydA_{3}$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

and $$Q_{y}^{(3)} = \int_{A_{3}}zdA_{3}$$

Now the process is repeated for the other stringers.

Stringer 2
$$\bar{q_{23}} = \bar{q_{12}} - \bar{q_{24}} + q^{(2)}$$ where $$\bar{q_{12}} = \bar{q_{24}} =0$$ and $$q^{(2)}$$ is the shear flow due to bending at stringer 2. Thus

$$\bar{q_{23}} = q^{(2)}$$ which is solved by the same structure to that of stringer 3.

$$Q_{z}^{(2)} = y_{2}A_{2}$$ where $$y_{2}$$ is the y-coordinate of stringer 2

$$Q_{y}^{2} = z_{2}A_{2}$$

Stringer 4
$$\bar{q_{43}} = \bar{q_{24}} - \bar{q_{41}} + q^{(4)}$$ again $$\bar{q_{24}} =\bar{q_{41}} =0$$ and $$q^{(4)}$$ is solved just like that of stringer 3. Due to $$\bar{q_{31}}$$:

Super-position (again)

$$q_{ij} = \bar{q_{ij}} + q_{k}$$

$$q_{12} = \bar{q_{12}} + q_{1}$$

$$q_{23} = \bar{q_{23}} + q_{1}$$

$$q_{31} = \bar{q_{31}} + q_{1}$$

$$q_{24} = \bar{q_{24}} + q_{2}$$

$$q_{43} = \bar{q_{43}} + q_{2}$$

$$q_{41} = \bar{q_{41}} + q_{2}$$

Hence there are 3 unknowns ($$q_{1}, q_{2}, q_{3}$$) and 3 equations:

1) Moment equation (usually take at a convenient point such as to cancel the contribution of $$V_{y}, V_{z}$$

2) $$\theta_{1} = \theta_{2}\!$$ where $$\theta\!$$ must be computed with the true shear flow!

3)$$\theta_{2} = \theta_{3}\!$$

Stringers continued (lecture on 12/08/08)
$$ 0 = q^{(1)} + q^{(2)} + q^{(4)} \!$$

This not true or possible so...

$$ 0 = q^{(1)} + q^{(2)} + q^{(3)} + q^{(4)} = \sum_{e=1}^{4}{q^{(e)}}$$

$$ q^{(e)}= n_{z}Q_{z}^{(e)} +n_{y}Q_{y}^{(e)} $$

$$ n_{z} := -(k_{y}V_{y} - k_{yz}V_{z}) \!$$

$$n_{y} := -(k_{z}V_{z} - k_{yz}V_{y}) \!$$

Therefore

$$\sum_{e=1}^4 q^{(e)} = n_z \sum_{e=1}^4 Q_z^{(e)} + n_y \sum_{e=1}^4 Q_y^{(e)} $$

$$Q_{y}(\hat{z})=\int zdA = 0 = z_{c}dA \!$$

NACA airfoil back of envelope verification

Given a rectangle with points $$\bar{B}, \bar{E}, \bar{F}, \bar{H} \!$$

$$ \bar{B}\bar{E} = \bar{F}\bar{H} = .5(BE + FH) $$

Bidirectional bending Continuation (from Lecture 12/8/2008)
$$M_y=-EY_{yz}\frac{d^2v_o}{dx^2}-EI_y\frac{d^2w_o}{dx^2}$$ $$M_y=-EY_{z}\frac{d^2v_o}{dx^2}-EI_{yz}\frac{d^2w_o}{dx^2}$$

$$ \begin{Bmatrix} M_y\\ M_z \end{Bmatrix} = \begin{vmatrix} I_y & I_{yz}\\ I_{yz} & Iz \end{vmatrix}

\begin{Bmatrix} -Ex_z\\ -Ex_y \end{Bmatrix}

$$

$$\epsilon_{xx} = -yx_y - zx_z = \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -x_z\\ -x_y \end{Bmatrix}$$

$$ \sigma_{xx} = E\epsilon_{xx} = E \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -x_z \\ -x_y \end{Bmatrix} = E \begin{bmatrix} z & y \end{bmatrix} \frac{1}{E} I^{-1} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} = \begin{bmatrix} z & y \end{bmatrix} I^{-1} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} $$

where we know

$$ I^{-1} = \frac{1}{D} \begin{bmatrix} I_z & -I_{yz} \\ -I_{yz} & I_y \end{bmatrix} $$ and $$ D = I_y I_z - \left(I_{yz}\right)^2 \,$$

For the shear flow equation in matrix notation:

$$ q = -\int_A \frac{d\sigma_{xx}}{dx}dA $$

where we know

$$ \frac{d\sigma_{xx}}{dx} = \begin{bmatrix} z & y \end{bmatrix} I^{-1} \begin{Bmatrix} \frac{dM_y}{dx} \\ \\ \frac{dM_z}{dx} \end{Bmatrix} $$

and we also know

$$ V_y = \frac{dM_y}{dx} $$

and

$$ V_z = \frac{dM_z}{dx} $$

and additionally, we know

$$ Q_y = \int_A z \quad dA \,$$

and

$$ Q_z = \int_A y \quad dA \,$$

Back to the NACA Airfoil
The following MATLab Code calculates the shear flow in each segment of a single-cell and three-celled NACA 4-digit series airfoil. The results for the NACA 2415 (with spar webs at quarter and three-quarter chord in the multicell airfoil) are as follows:

For the single cell,

The $$q^{(1)}\!$$, $$q^{(2)}\!$$, $$q^{(3)}\!$$ for the three-celled airfoil are,

And the $$\sigma_{BF}\!$$ and $$\sigma_{EH}\!$$ values are,

Plate Buckling
Using the following equation from Dr. Vu-Quoc's plate buckling website, [Plate Buckling], we calculate $$\displaystyle\lambda$$ using an aspect ratio, $$\vartheta$$, of 1.5

$$  \displaystyle \lambda =  \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{ 1 	 + 	 \frac{81}{625} +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{1 + 9 \vartheta^2}	 \right)^2 +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{9 + \vartheta^2}	 \right)^2 \right\} \right]^{1/2} = 0.0739 $$

Using these two values we can complete the following matrix which is also found in the Plate Buckling website:

$$  \displaystyle \left[ \begin{array}{lllll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} &	 0	 &	 0	 &	 0	 \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 0	 &	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 0	 &	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 0	 &	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

$$  \displaystyle \left[ \begin{array}{lllll} 0.3471 	 &	 0.4444	 &	 0	 &	 0	 &	 0	 \\	 0.4444	 &	 5.5533  	 &	 -0.8000	 &	 -0.8000	 &	 1.4400	 \\	 0	 &	 -0.8000	 &	 14.8382  	 &	 0	 &	 0	 \\	 0	 &	 -0.8000	 &	 0	 &	 4.1588  	 &	 0	 \\	 0	 &	 1.4400	 &	 0	 &	 0	 &	 28.1135        \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

We can rearrange this matrix as follows:

$$  \displaystyle \left[ \begin{array}{lllll} 5.5533 	 &	 -0.8000	 &	 -0.8000	 &	 1.4400	 \\	 -0.8000	 &	 14.8382  	 &	 0	 &	 0	 \\	 -0.8000	 &	 0	 &	 4.1588  	 &	 0	 \\	 1.4400	 &	 0	 &	 0	 &	 28.1135        \end{array} \right] \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0.4444C_{11} \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

Using Matlab we can then express $$C_{22}, C_{13}, C_{31}, C_{33}$$ in terms of $$C_{11}$$:

$$\left\{ \begin{array}{l} C_{22} = 0.07630C_{11} \\	 C_{13} = -0.00411C_{11} \\	 C_{31} = -0.01468C_{11} \\	 C_{33} = 0.00391C_{11} \end{array} \right\}$$

We can also derive the transverse displacement of the beam using the following formulas:

$$  \displaystyle \psi (x,y) =   c_{mn} \sin \left(     \frac      {m \pi x}      {a}   \right) \sin \left(     \frac      {n \pi y}      {b}   \right) \, \ {\rm for} \ m,n = 1, 2, 3, \ldots $$

$$  \displaystyle u_z (x,y) =  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \psi_{mn} (x,y) $$

These equations and the above 'C' terms yield the following specific bending:

$$u_z (x,y) = C_{11}sin(\frac{\pi x}{a})sin(\frac{\pi y}{b})+$$ $$           C_{22}sin(\frac{2\pi x}{a})sin(\frac{2 \pi y}{b})+$$ $$           C_{13}sin(\frac{\pi x}{a})sin(\frac{3 \pi y}{b})+$$ $$           C_{31}sin(\frac{3 \pi x}{a})sin(\frac{\pi y}{b})+$$ $$           C_{33}sin(\frac{3 \pi x}{a})sin(\frac{3 \pi y}{b})$$

$$         = C_{11}sin(\frac{\pi x}{a})sin(\frac{\pi y}{b})+$$ $$           0.07630C_{11}sin(\frac{2\pi x}{a})sin(\frac{2 \pi y}{b})+$$ $$           -0.00411C_{11}sin(\frac{\pi x}{a})sin(\frac{3 \pi y}{b})+$$ $$           -0.01468C_{11}sin(\frac{3 \pi x}{a})sin(\frac{\pi y}{b})+$$ $$           0.00391C_{11}sin(\frac{3 \pi x}{a})sin(\frac{3 \pi y}{b})$$

If we set $$C_{11} = 1, a = 3, abd b = 2$$ we are left with:

$$u_z (x,y) = sin(\frac{\pi x}{3})sin(\frac{\pi y}{2})+$$ $$           0.07630sin(\frac{2\pi x}{3})sin(\frac{\pi y}{1})+$$ $$           -0.00411sin(\frac{\pi x}{3})sin(\frac{3 \pi y}{2})+$$ $$           -0.01468sin(\frac{\pi x}{1})sin(\frac{\pi y}{2})+$$ $$           0.00391sin(\frac{\pi x}{1})sin(\frac{3 \pi y}{2})$$

The following figure shows the amount of deflection the beam undergoes at specific (x,y) locations. Where the x location or the y location is zero the bar undergoes zero deflection. It undergoes its greatest deflection near the center of the beam which makes sense if we assume that the beam is fixed on all four edges.

Recommended Software
Writing Wiki Articles:

 * MediaWiki is similar in many ways to HTML, so having some knowledge in that language is very helpful. Now that so many web companies use this language for their websites, having the basic knowhow and ability to design in MediaWiki is a real advantage in the workplace.

Writing LaTex Equations:

 * Knowing how to use the LaTex Equation Editor is vital for programming in MediaWiki, and is useful for displaying equations in MATLab, however Microsoft Word 2007 has an equation editor which is much easier and far more user friendly. The learning curve is virtually non-existent and the equations can be saved as mpegs or jpegs and then uploaded.  For novices, this is probably not a bad alternative.

Drawing Figures:

 * Again, Microsoft Word 2007 provided our group with the best figure drawing capabilities, whether it was drawing simple boxes and circles or complex airfoils or curved surfaces. MS Paint and Adobe Photoshop were also a convenient tools.

Contributing Team Members
Kris Loper Eas4200c.f08.aero6.loper 17:52, 9 December 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 17:56, 9 December 2008 (UTC)

Eduardo Villalba Eas4200c.f08.aero6.villalba 20:30, 9 December 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 20:44, 9 December 2008 (UTC)

Felipe Ortega Eas4200c.f08.aero6.ortega 21:33, 9 December 2008 (UTC)

Marlana Behnke Eas4200c.f08.aero6.behnke 21:46, 9 December 2008 (UTC)

title: User:Eas4200c.f08.aero6.ortega/HW7


text:
User talk:Crux~enwikiversity

Editing other people's comments
Hi Crux, I saw you hadn't yet been officially welcomed to Wikiversity, so I added the welcome above. I also noticed you changed my word "Jade" to "the Jade Knight":. While this doesn't cause a problem, in this case, it's generally not allowed to edit other people's comments, since this can make it look like they said something they didn't. You also appear to have fixed an accidental deletion of my comments by Jade, which is fine. It's a good idea to add an edit summary (below the edit window), when you do something unusual like this, though, so we can understand what happened. Happy editing ! StuRat 20:27, 28 November 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 Crux. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Crux~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 23:30, 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) 06:05, 19 April 2015 (UTC)

title: User talk:Crux~enwikiversity


text:
Spanish 2/Chapter 4 (Fashion)

Shopping

 * entrada - entrance
 * ganga - bargain
 * letrero - sign
 * liquidación - sale
 * mercado - market
 * salida - exit

Note: Like all words that end with -ción, liquidación is feminine.

Types of colors

 * claro(a) - light
 * de sólo un color - solid-colored
 * oscuro(a) - dark
 * pastel - pastel
 * vivo(a) - bright

Clothing material

 * ¿De qué está hecho(a)? - What is it made of?
 * Está hecho(a) de... - It is made of...
 * algodón - cotton
 * cuero - leather
 * lana - wool
 * seda - silk
 * tela sintética - synthetic fabric

Paying for purchases

 * alto(a) - high
 * bajo(a) - low
 * caja - cash register
 * cajero(a) - cashier
 * cheque - check
 * cheque de viajero - traveler's check
 * cupón de regalo - gift certificate
 * en efectivo - cash
 * gastar - to spend
 * precio - price
 * tan (+ adjective) - so
 * tarjeta de crédito - credit card

Clothing purchases

 * apretado(a) - tight
 * escoger - to choose
 * estar de moda - to be in fashion
 * estilo - style
 * exagerado(a) - outrageous
 * flojo(a) - loose
 * marca - brand
 * mediano(a) - medium
 * número - shoe size
 * probarse - to try on
 * talla - size

Note: Probarse is a stem-changing verb that changes from o to ue when conjugated.

Other useful words and expressions

 * anunciar - to announce
 * encontrar - to find
 * en realidad - really
 * me/te importa(n) - it matters (it's important)/they matter to me/to you
 * inmediatamente - immediately
 * me parece que - it seems to me that
 * ¿Qué te parece? - What do you think?/How does it seem to you?
 * recientemente - recently

Note: Encontrar is a stem-changing verb that changes from o to ue when conjugated.

Preterite of regular verbs
To talk about actions that occured in the past, use the preterite tense. To form the preterite tense of a regular verb, add the preterit endings to the verb's stem.

Preterite endings for -ar verbs:

 * yo: é
 * tú: aste
 * Ud./él/ella: ó
 * nosotros(as): amos
 * vosotros(as): asteis
 * Uds./ellos/ellas: aron

Preterite endings for -er and -ir verbs:

 * yo: í
 * tú: iste
 * Ud./él/ella: ió
 * nosotros(as): imos
 * vosotros(as): isteis
 * Uds./ellos/ellas: ieron

Example conjugation: mirar
 * miré - I looked
 * miraste - you looked (singular)
 * miró - he/she looked
 * miramos - we looked
 * mirasteis - you looked (plural)
 * miraron - they looked

Example conjugation: aprender
 * aprendí - I learned
 * aprendiste - you learned (singular)
 * aprendió - he/she learned
 * aprendimos - we learned
 * aprendisteis - you learned (plural)
 * aprendieron - they learned

Example conjugarion: escribir
 * escribí - I wrote
 * escribiste - you wrote (singular)
 * escribió - he/she wrote
 * escribimos - we wrote
 * escribisteis - you wrote (plural)
 * escribieron - they wrote

Demonstrative adjectives
To point out something or someone that is far from both you and the person you are speaking to, you use demonstrative adjectives. All demonstrative adjectives come before the noun and must agree with the noun in gender and number.

 * este, esta - this
 * ese, esa - that
 * aquel, aquella - that one over there
 * estos, estas - these
 * esos, esas - those
 * aquellos, aquellas - those over there

Vocabulario adicional

 * estar pasado(a) de moda - to be out of style
 * accesible - affordable

Cultural Insight (Fiestas)
In Spanish-speaking countries, fiestas feature food, drinks, music, and dancing like other cultures throughout the world. However, fiestas usually celebrate events like a wedding or holiday. Parties usually revolve around the family and friends. Parties are usually held in living rooms, which are the biggest rooms in a typical Spanish-speaking country's home.

title: Spanish 2/Chapter 4 (Fashion)


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President of the United States/Election Years are Leap Years

What is a Leap Year?
In simplest terms, a Leap Year is a "calendar correction". According to the Gregorian Calendar, the current standard calendar in most of the world, leap day falls on February 29th which occurs for those years evenly divisible by four, unless it marks the century and isn't divisible by 400. For example, 1900 was not a leap year, even though it is divisible by four.

So, for instance, the year 1776, when the United States Declaration of Independence was adopted by the Second Continental Congress of the original Thirteen Colonies, was a Leap Year. Another Leap Year, certainly less well known than 1776, but quite important for our purposes, occurred just 12 years later in 1788.

The First Presidential Election
The United States Constitution was adopted on September 17, 1787, by the Constitutional Convention. The Convention submitted the Constitution to the Congress of the Confederation, where it received approval according to Article 13 of the Articles of Confederation.

While Delaware was the first to sign the Constitution on December 7, 1787, the 9th signature, provided by New Hampshire on June 21, 1788 determined the ratification. Once ratified, the Constitution provided definition for a new Office of the President to head the Executive Branch of the US government. Before the first elections were held, the 10th & 11th colonies signed the Constitution, Virginia on June 25, 1788, and New York on July 26, 1788. The election polls opened on December 15, 1788 and closed on January 10, 1789. Because of the end date, the first election is generally referred to as the Election of 1789.

George Washington, the 1st President was inaugurated April 30, 1789, and served until March 4, 1797, having been reelected to a second term in 1792.

Leap Year Revisited
From 1792 to the present day, the United States has held a Presidential Election every four years, even when at war with itself. During the Civil War, Abraham Lincoln, the 16th President was elected to a second term in 1864.

Oddly enough, though an election occurs every four years, not every election year falls on a Leap Year. The reason is simple enough and is once again based on the Gregorian Calendar, which dictates that years marking the end of a century (ie: multiples of 100) are only leap years if also divisible by 400. As a result, 1600, 2000, & 2400 are leap years, but 1800, 1900, & 2100 are not.

However, this triviality is a minor blemish to the value otherwise provided by the simple generalization that Election Years are Leap Years. Not even the larger discrepancy that the very first election was completed in 1789 is sufficient deterrent to employing this convenient tool.

Assignment
Complete Worksheet 1.


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Talk:President of the United States

For what it's worth, I think this is a clever idea for a US History course. The Jade Knight (d'viser) 09:35, 29 November 2008 (UTC)

title: Talk:President of the United States


text:
Talk:President of the United States/Worksheets/Worksheet 1

Something to consider: Do you want students modifying this page directly? How do you want them to fill out this worksheet? Will this worksheet be somehow evaluated? Will they get feedback? If feedback is slow in coming, what should they do? Etc. The Jade Knight (d'viser) 09:39, 29 November 2008 (UTC)


text:
User talk:Chagfeh

 Hello Chagfeh, 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 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! --The Jade Knight (d'viser) 09:40, 29 November 2008 (UTC)

Social psychology (psychology)
Hi Chagfeh,

Thanks for your message and you are very welcome to participate in social psychology. I have just been finishing up final marking and grades for the on-campus version of this unit, so I'm sorry for my slow reply. We probably should have a little bit of a chat first about what interests you, your background etc. I'm not sure that doing the whole unit as it stands is ideal for online learning - it is so far really a face-to-face unit taught at a traditional university which has been "dumped" on Wikiversity. But I am keen to adapt and can burn a DVD of lectures, etc. for you (or find somewhere to upload them). I would like to develo and use the social psychology wikibook (Social psychology) but its rather incomplete, so we also used a traditional social psychology textbook - but many others could also be used. Hope this helps.

Sincerely, James -- Jtneill - Talk - c 04:59, 2 December 2008 (UTC)

 * I will also teach a different unit next year - and this unit will be taught by someone else (but not on Wikiversity). So, what's available on Wikiversity should also be adapted further for online participation. -- Jtneill - Talk - c 05:01, 2 December 2008 (UTC)

title: User talk:Chagfeh


text:
User:Hatashe

I, Hasanuzzaman Talukdar Shemul am a Wikiholic|Editor|Writer.

Click on Hatashe

title: User:Hatashe


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User talk:Shshwiki

فلسفه و اقتصاد
right text در زبان فارسی مطالب اندکی درباره‌ی فلسفه و اقتصاد وجود دارد. بنابراین ما تکمیل دانش ایرانیان در این زمینه نیازمند تشریک مساعی همگی ماست

title: User talk:Shshwiki


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President of the United States/Presidential Terms are Four Years

What is a Term of Office?
As defined by Article 2 of the Constitution, a Term of office for the US President is four years. After the election, which is held in a leap year, the President Elect takes office by reciting the Oath of Office during an Inaugeration ceremony. If reelected to another term of office, another ceremony is performed and the President recites the Oath again. Only Franklin D. Roosevelt, the 32nd President, has been inaugerated for more than two terms.

When do Terms Start and Stop?
With a few notable exceptions, inaugeration of the early Presidents was performed on March 4th of the year following the election. Presidential terms therefore start and stop the years after leap year. This simple rule always applies, so its fairly easy to build up a short chain of Presidents and their terms based on a leap year. For instance:
 * 1788 was a Leap Year.
 * George Washington was 1st inaugerated in 1789.
 * This 1st of all Presidential terms expired in 1793.
 * In 1792, George Washington was reelected to a 2nd term.
 * The inaugeration for the 2nd term occurred in 1793, just as the 1st expired.
 * The 2nd term expired four years later in 1797.

Long before March 4, 1797, George Washington and the new Nation knew that he would not serve a 3rd term, for the simple fact that he had choosen not to run for the office. Succeeding his position by the election of 1796, John Adams became the 2nd President and served only one term, 1797-1801.

The First Eight Presidents
Let's take a quick look at the available List of Presidents and notice that, with one previously stated exception, the first eight all took office on March 4th following an election year and held office for a full four or eight years. The exception is George Washington, who was inaugerated for his first term on April 30, 1789.

Eight names is a fairly long list to remember, but there are ways to group the information. For instance, take a look at Unit 1. All five of these Presidents, George Washington, John Adams, Thomas Jefferson, James Madison, and James Monroe, were founding members of the United States. With one exception, they all served two terms, for a total of eight years, while John Adams served only one term. With these bits of information, we can build ourselves a little table, like so:

 * 1) Number first five Presidents from 1st to 5th.
 * 2) Enter their names.
 * 3) Enter the number of terms. (Who held only one term?)
 * 4) List the years they were elected.
 * 5) What years were they in office?

Well, that's the first five, but we were looking for eight. For the rest, let's take a look at Unit 2, where we can see that the first three of these Presidents, John Quincy Adams, Andrew Jackson, and Martin Van Buren, all served full terms of office. Like his father before him, John Quincy Adams only served one term, while Andrew Jackson served two, and Martin Van Buren only one. With this information, we are sufficiently armed to add three more rows to our table, just as we did before:

The names and terms for the first eight Presidents aren't too difficult to remember, however this is less than a fifth of the total 44 Presidents, so the problem with memorizing will get progressively more difficult as the list grows. Still, let's take advantage of what we know so far.

Assignment
Complete Worksheet 2.


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User:Eml4500.f08.gravy.mmm/HW7

Element 1: E1(1)=2, E2(1)=4, A1(1)=.5, A2(1)=1.5

Element 2: E1(2)=3, E2(2)=7, A1(2)=1, A2(2)=3

A comparison of the solution for 2-bar truss with tapered elements are shown in a previous report. Modification of 2-bar truss code to accommodate general k(i)

Frame element is a truss bar element(which is axial deformation) and a beam element (which is transverse deformation).

Model frame problem with two elements

Free Body diagrams for the elements

$$d_1^{(1)} \longrightarrow f_1^{(1)}$$

$$ d_2^{(1)} \longrightarrow f_2^{(1)}$$

$$d_3^{(1)} \longrightarrow f_3^{(1)}$$

$$ d_4^{(1)} \longrightarrow f_4^{(1)}$$

$$d_5^{(1)} \longrightarrow f_5^{(1)}$$

$$ d_6^{(1)} \longrightarrow f_6^{(1)}$$

$$d_1^{(2)} \longrightarrow f_1^{(2)}$$

$$d_2^{(2)} \longrightarrow f_2^{(2)}$$

$$d_3^{(2)} \longrightarrow f_3^{(2)}$$

$$d_4^{(2)} \longrightarrow f_4^{(2)}$$

$$ d_5^{(2)} \longrightarrow f_5^{(2)}$$

$$d_6^{(2)} \longrightarrow f_6^{(2)}$$

$$ d_2^{(1)} \longrightarrow f_2^{(1)}$$

$$d_3^{(1)} \longrightarrow f_3^{(1)}$$

$$ d_4^{(1)} \longrightarrow f_4^{(1)}$$

$$d_5^{(1)} \longrightarrow f_5^{(1)}$$

$$d_6^{(1)} \longrightarrow f_6^{(1)}$$

In general, $$d_i^{(e)} \longrightarrow \underbrace{f_i^{(e)}}_{general forces}$$

e=1,2  i=1,...,6

$$\begin{matrix} d_3^{(e)}\\ d_6^{(e)} \end{matrix} \}$$ rotational degrees of freedom $$\longrightarrow \begin{matrix} f_3^{(e)}\\ f_6^{(e)} \end{matrix} \} $$ bending moments

2-D Frame global dofs;

Two element stiffness matrices with $$ \mathbf{k}_{6x6}^{(e)}, e=1,2$$

and the Global stiffness matrix becomes $$ \mathbf{k}_{9x9}= A_{e=1}^{e=2}\mathbf{k}_{6x6}^{(e)}$$

$$ \tilde{\mathbf{k}_{6x6}^{(e)}}\tilde{\mathbf{d}_{6x1}^{(e)}}= \tilde{\mathbf{f}_{6x1}^{(e)}}$$

$$ \tilde{\mathbf{d}^{(e)}}= \begin{bmatrix} d_1^{e} \\ d_2^{e} \\ d_3^{e} \\ d_4^{e} \\ d_5^{e} \\ d_6^{e} \end{bmatrix}$$, $$\tilde{\mathbf{f}^{(e)}}= \begin{bmatrix} f_1^{e} \\ f_2^{e} \\ f_3^{e} \\ f_4^{e} \\ f_5^{e} \\ f_6^{e} \end{bmatrix}$$

Note: $$\tilde{\mathbf{f}_3^{(e)}}= \mathbf{f}_3^{(e)}, \tilde{\mathbf{f}_6^{(e)}}= \mathbf{f}_6^{(e)}$$ moments about $$\tilde{x}=z$$

Dimensional Analysis
$$ \underbrace{\left[\tilde{d_1}\right]}_{dimension of} = \underbrace{L}_{Length}= \underbrace{\left[\tilde{d_1}\right]}_{displacement}$$ i=1,2,4,5

$$\left[\tilde{d_3}\right]= \underbrace{1}_{No dimension}=\underbrace{\left[\tilde{d_6}\right]}_{rotational}$$

$$\underbrace{\widehat{AB}}_{arc lenth}=R \underbrace{\theta}_{radians}$$

$$ \theta=\frac{\widehat{AB}}{R}$$

$$\left[\theta\right]=\frac{\widehat{\left[AB\right]}}{\left[R\right]}=\frac{L}{L}=1$$

$$\sigma=E \epsilon \Rightarrow \left[\sigma\right]=\left[E \right]\underbrace{\left[\epsilon\right]}_1$$

monday 12/8
$$ \tilde{k}_{23}= \frac{6EI}{L^2}= \int_0^L \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2},dx= \frac{dN_2}{dx}(EI)\frac{d^2N_3}{dx^2} \frac{d^2N_2}{dx^2}(EIx)\frac{d^2N_3}{dx^2} \frac{d^2N_2}{dx^2}(EI)\frac{dN_3}{dx} \Bigg|_0^L $$

In general, $$\tilde{k}_{ij}=\int_0^L \frac{d^2N_i}{dx^2}(EI)\frac{d^2N_3}{dx^2}dx$$ i,j=2,3,5,6

Elastodynamics (trusses, frames, 2-D, 3-D, elasticity)
A model problem can be found in a previous report Here

Discrete Principle of Virtual work
$$ \mathbf{\bar{w}} \cdot [ \mathbf{\bar{M}} \mathbf{\ddot{\bar{d}}} + \mathbf{\bar{K}} \mathbf{\bar{d}}- \mathbf{\bar{F}} ] =0$$

for all $$\mathbf{\bar{w}}$$

The bars mean the matrix is reduced which is to say the boundary conditions have already been applied

$$\Rightarrow $$ $$\mathbf{\bar{M}} \mathbf{\ddot{\bar{d}}} + \mathbf{\bar{K}} \mathbf{\bar{d}}=  \mathbf{\bar{F}}(t)$$ equation 1 $$\mathbf{\bar{d}}(0) = \mathbf{\bar{d}}_0$$ $$\mathbf{\dot{\bar{d}}}(0) = \mathbf{\bar{v}}_0$$

Complete ordinary differential equation (ODEs) (2nd order in time)+ initial conditions governing the elastodynamics of discretized continuous problem (MDOF)

Solving equation 1:

1) Consider unforced vibration problem

$$ \mathbf{\bar{M}}_{nxn} \mathbf{\ddot{v}}_{nx1} + \mathbf{\bar{K}}_{nxn} \mathbf{v}_{nx1}= \mathbf{0}_{nx1}$$   unforced equation 2

assume $$v(t)=(sin \omega t) \mathbf{\phi}_{nx1}$$ where $$\mathbf{\phi}$$ is not time dependent

$$ \mathbf{\ddot{\bar{v}}}=-\omega^2sin\omega t \phi$$

$$=\omega^2sin\omega t \mathbf{\bar{M}} \phi + sin\omega t \mathbf{\bar{K}} \phi = \mathbf{0}$$

$$\Rightarrow \mathbf{\bar{K}} \phi=\omega^2\mathbf{\bar{M}}\phi$$ equation 3 Which is the Generalized eval. Problem

$$ \mathbf{Ax}=\lambda \mathbf{Bx}$$ with $$\lambda$$ as the eval. Which is the General Form

Standard eval. problem: $$ \mathbf{Ax}=\lambda \mathbf{x}$$ with B = I with I as the Identity Matrix

The Identity Matrix I=$$\begin{bmatrix} 1 & 0\\0&1 \end{bmatrix}$$ which has ones on the diagonal and zeros everywhere else.

$$\lambda =\omega^2$$ and ($$\lambda_i,\phi_i$$) are eigenpairs with i=1,...,n

mode i $$\Rightarrow \begin{cases} v_1(t)=(sin\omega_1t)\phi \\ i=1,\cdots,n \end{cases}$$

2)Model superposition method:

Orthogonal properties of eignepairs:

$$\underbrace{\mathbf{\phi}_i^T}_{1xn} \underbrace{\mathbf{\bar{M}}}_{nxn} \underbrace{\mathbf{\phi}_i}_{nx1} = \underbrace{\delta_{ij}}_{Kronecker \delta}= \begin{cases} 1,& \mbox{if } i=j \\  0,  &\mbox{if } i=j \end{cases}$$

Mass orthogonal of eigenvectors

Using equation 2 and equation 3

$$\mathbf{\bar{M}}\omega_j=\lambda_j^{-1}\mathbf{\bar{K}}\omega_j$$

$$

\underbrace{\phi_i^T\mathbf{\bar{M}}}_{\delta_{ij}}\phi_j=\lambda_j^{-1}\phi_i^T\mathbf{\bar{K}}\phi_j$$

$$\Rightarrow \phi_i^T\mathbf{\bar{K}}\phi_j=\lambda_j \delta_{ij}$$

From equation 1 we get: $$\mathbf{\bar{d}}=\sum_{i=1}^n \xi_i(t)\phi_i$$

$$ \mathbf{\bar{M}} \underbrace{(\sum_{j}\ddot{\xi}_j \phi_j)}_{\mathbf{\ddot{d}}} + \mathbf{\bar{K}}\underbrace{(\sum_{j}\xi_j \phi_j)}_{\mathbf{\bar{d}}}= \mathbf{F}$$

$$\sum_j\ddot{\xi}_j\underbrace{(\phi_i^T\mathbf{\bar{M}}\phi_j)}_{\delta_{ij}} + \sum_j\xi_j \underbrace{(\phi_i^T\mathbf{\bar{K}}\phi_j)}_{\lambda_i\delta_{ij}}= \phi_i^T \mathbf{F}$$

$$\Rightarrow \ddot{\xi}_i+\lambda_i\xi_i=\phi_i^T F$$ i=1,...n

Pictures

title: User:Eml4500.f08.gravy.mmm/HW7


text: File:Schematic.svg

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text: File:AUA-Pipeline.png

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text: Talk:Educational Media Awareness Campaign/Geography: Climate/POTD 7


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Introduction to Judaism/Lashon Hara

Overview
Lashon Harah (לשון הרע,gossip) is the act of speaking about someone behind their back. A common misconception about Lashon Harah is that it is when a person speaks wrongly about someone, but Lashon Harah can be speaking about someone behind their back in any way, even in a positive manner, as it is considered the speaker is drawing unwanted attention to their subject, and this attention even if positive in content could portray the subject - and the speaker - in a negative light.

title: Introduction to Judaism/Lashon Hara


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President of the United States/Worksheets/Worksheet 2

Return to Presidential Terms are Four Years.

Presidential Terms
In this exercise, we continue creating our timeline by moving from Election Years to Presidential Terms. Using the table below, begin with the first Election Year and write down the beginning and end dates of every Presidential Term of Office.

On the surface, we again have a long list of numbers, but with the few facts we've so far gathered, we can add some meaning and relevance to a long list of four-year ranges.

The First Eight Presidents

 * 1) Who served the 1st term as the US President?
 * 2) When did his term start?
 * 3) How many terms did he serve?
 * 4) Who was the 2nd President?
 * 5) Of the first eight presidents, who served only one term?
 * 6) Who was inaugerated at the beginning of the new century?
 * 7) Who was the first father & son group of Presidents?
 * 8) What number President was the son?
 * 9) What years did the son serve as President?
 * 10) Which President left office in 1841?

Adjusting the Timeline
In the table provided below, also notice that the first column of sequential numbers has been moved down so that 1. starts at 1788.
 * 1) Why do you think this change has been made?
 * 2) How can this help us remember dates?
 * 3) What are the drawbacks?

The Civil War Era

 * 1) Which election was held during the Civil War?
 * 2) Who was elected during the Civil War?
 * 3) When was Abraham Lincoln first elected?
 * 4) Which years were Abraham Lincoln's inaugerations?
 * 5) What year was Abraham Lincoln assassinated?
 * 6) Who finished out his term as President?
 * 7) When did this term finish?
 * 8) Who was elected President in 1868?
 * 9) How many terms did he serve?
 * 10) What year did he leave office?

The Timeline
../Worksheet 2.A/