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User:Eml4500.f08.wiki1.aguilar/hw7

Wikiversity vs. E-Learning - Wiki1 Group Verdict
While both the Wikiversity and the E-Learning systems are very useful, each was designed for a specific purpose in mind. While E-Learning was designed to facilitate the communication between instructors and students, wikis are more designed to allow groups and organizations to easily post, edit, and share information. Because they are used with different purposes in mind, they are difficult to compare.

While E-Learning is in no way perfect, it does facilitate communication between professors and students when used correctly. By using E-learning as the standard and trying to familiarize all the professors with it, it bypasses the need for teachers to create a website to post information that needs to be accessible to all the students. We are not familiar with the back end of E-Learning, but we are sure that using E-learning to post files and information is easier than teaching professors how to build their own websites with attached files and documents. E-Learning does allows easy methods for professors to do a variety of different things. For example, E-Learning allows professors and TA's to upload grades and then for the various assignments and exams, students can view their grades at any time throughout the semester. Professors can also allow it to provide statistics on the assignment or exam like the average, the high and low scores, as well as histograms showing the grade distributions. E-Learning allows the professors to easily post announcements and documents as well, like homework solution PDFs and course syllabi. There is a feature for a discussion board which could easily be used to allow students to ask other students and TAs various questions, and there is also a calendar to allow various important dates to be posted. This are just some of the features provided for by the E-Learning system, many of which cannot be adequately provided for through Wikiversity.

It becomes more of a question what is it that we are trying to do? If, for example, the end goal is to provide a large, multimedia report that is to be worked on by an entire group, then Wikiversity does indeed provide a very good solution for this. But E-Learning is currently not meant to serve this function. For the purposes of this class, "E-Learning" is more of an indirect method for group collaboration. If you had to use E-Learning to do this, the process would vary group by group. Groups could come up with various ways to create these reports. Some groups might use sharing features of Google documents to write up the report and allow all members of the group to contribute text, images and various other elements to the report. This could then be saved as either a PDF or a Word document which could then be submitted on E-Learning. The advantage to this method would be that there isn't any "coding" to learn so that would minimize some of the learning curve, and it would also protect reports from being viewed by others outside the group. The downside to this would be that reports from group to group would be much less streamlined, and having to save equations as images could become quite a hassle. Another alternative would be to use Word to allow groups to create the report. Word 2007 does contain a much improved equation editor and almost everyone uses Microsoft Word already. E-Learning could be set up to allow groups members to upload their parts, and then someone at the end would combine into a final report. This method would have similar pros and cons to the Google Shared documents method.

Clearly, for a group to put together an entire report like what was required of us, Wikiversity does seem to provide a very good solution, keeping in mind that no solution will be perfect. Wikiversity allows for students to be introduced to wikis, something that is becoming much more widespread throughout the "real world." These skills are much more likely to becomes valuable later on down the road. Wikiversity is major project and it's performance will not be affected by the activity of our groups even under peak conditions. The downside is that the Wikiversity approach does, however, lend itself to the possibility of being vandalized or copied by members from other groups.

As a group, we think that for our class purposes, the Wikiversity method could be improved. This could be done by cutting out much of the tedious busy work that tends to be more "filler" within our reports. While each homework report usually looked impressive when it was turned in, by essentially having each of the groups rewrite all the notes from the course lectures was of little added benefit to each student. Much time was spent tediously formatting the lengthy equations that time was taken away from more beneficial areas of the homework. Each student should be responsible for taking his or her own notes thus emphasizing the importance of class attendance. The reports should focus on actual homework with each group's solutions to various problems and MATLAB assignments with included images, videos, and coding from all the members of the group. In our opinion, this approach to the Wikiversity homework reports would be more beneficial to the students, and should be used in conjunction with many of the features of E-Learning, such as grade statistics and feedback, as well as others.

Frame Analysis
A frame element is a combination of a truss (bar element) with a beam element.

The differences between a truss and a beam is that because of it's pin connections, a truss only incurs only axial deformation while the fixed nature of the nodes of a beam experience transverse deformation.

2-D Frame Local DOFs

In general, $$d_{(i)}^{(e)}\Rightarrow f_{(i)}^{(e)}$$, where $$f_{(i)}^{(e)}$$ are the generalized forces. Here, e=1,2 and i=1,...,6

$$\left.\begin{matrix} d_{(3)}^{(e)}\\d_{(6)}^{(e)} \end{matrix}\right\}$$ rotational degrees of freedom

$$\left.\begin{matrix} f_{(3)}^{(e)}\\f_{(6)}^{(e)} \end{matrix}\right\}$$ bending moments

2-D Frame Global DOFs

Therefore, there are two element stiffness matrices, $$k_{6\times6}^{(e)}$$ where e=1,2.

This means the global stiffness matrix is $$K_{9\times9}^{(e)}=Ak_{6\times6}^{(e)}$$.

This is assembled by combining the two element stiffness matrices as shown below.

Transformed Coordinate System

$$\mathbf{\tilde{k}}_{(6\times 6)}^{(e)} \mathbf{\tilde{d}}_{(6\times 1)}^{(e)}=\mathbf{\tilde{f}}_{(6\times 1)}^{(e)}$$, where each is defined below.

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

Note: The rotational displacements and moments remain unchanged. $$\tilde{d}_{3}^{(e)}={d}_{3}^{(e)}$$ $$\tilde{d}_{6}^{(e)}={d}_{6}^{(e)}$$ $$\tilde{f}_{3}^{(e)}={f}_{3}^{(e)}$$ $$\tilde{f}_{6}^{(e)}={f}_{6}^{(e)}$$

and

$$ \mathbf{\tilde{k}}_{(6\times 6)}^{(e)}= \overbrace{

\begin{bmatrix} \frac{EI}{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{4EI}{L} & 0 & -\frac{6EI}{L^2} & \frac{2EI}{L}\\ -\frac{EI}{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{4EI}{L} & 0 & -\frac{6EI}{L^2} & \frac{4EI}{L} \end{bmatrix}}

^{\begin{matrix} \;\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{matrix}}

\left.\begin{matrix} \\ \\ \\ \\ \\ \end{matrix}\right\}

\begin{matrix} \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{matrix} $$

Dimensional Analysis
The square bracket notation around a variable means "dimension of".

$$ \begin{bmatrix} \tilde{d}_1 \end{bmatrix}=L = \begin{bmatrix} \tilde{d}_i \end{bmatrix} $$, where i=1,2,4,5 and L signifies that it has a dimension of length.

Note: There is a difference between 'units' and 'dimensions'. Displacement has a dimension of length Mass, length, time, charge and temperature are all fundamental dimensions. They cannot be broken down any further, while a dimension such as force can be broken down even further as shown below.

$$Force=mass \times acceleration = \frac {mass \times distance}{{time}^2}=\frac{m\times L}{t^2}$$

$$\begin{bmatrix} \tilde{d}_3 \end{bmatrix}=1$$, meaning it has no dimension.

$$\sigma = E\varepsilon \Rightarrow \begin{bmatrix}\sigma \end{bmatrix}= \begin{bmatrix}E\end{bmatrix} \begin{bmatrix}\varepsilon   \end{bmatrix} =1$$

$$\left[\varepsilon \right]={[du] \over [dx]}={L \over L}=1$$

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User:Eml4500.f08.jamama.adam/homework7

Meeting 38
To get motivated lets consider the deformed shape of a truss element interp. of transverse displacement $$ V(s) with s = \tilde{x}$$.

Principal of Virtual Work (PVW): Eq. (1)

$$\int_{0}^{1}{w(x)}\left[-\frac{\partial{}^{2}}{\partial{x^{2}}}(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}} + f_{T} - m\ddot{v}\right]dx = 0$$ for all possible $$w(x) $$

For the first order term we will integrate by parts:

Let us recall the integration by parts method:

$$ r(x), s(x) $$

$$ (rs)' = r's + rs'$$

$$ r' = \frac{dr}{dx}  ;   s' = \frac{ds}{dx}$$

$$ \underbrace{\int {(rs)'}}_{(rs)}=\int{r's} + \int{rs'}$$

$$ =>\int{r's} = rs - \int{rs'}$$

Therefore if we let (α) = $$\int_{0}^{1}{\underbrace{w(x)}_{s(x)}}\frac{\partial{}^{2}}{\partial{x^{2}}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]dx $$

then we can write that $$\underbrace{\frac{\partial{}}{\partial{x}}\left(\underbrace{\frac{\partial{}}{\partial{x}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]}_{r(x)}\right)}_{r^'(x)}$$

$$= \left[\underbrace{W\frac{\partial{}}{\partial{x}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]}_{\beta^1}\right]_{0}^{L} - \int_{0}^{L}\underbrace{\frac{dw}{dx}}_{s'(x)}\underbrace{\frac{\partial{}}{\partial{x}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]dx}_{r(x)}$$

$$= \beta^{1} - \underbrace{\left[\frac{dw}{dx}(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]_{0}^{L}}_{\beta^{2}} + \underbrace{\int_{0}^{L}\frac{d^{2}w}{dx^{2}}(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}dx}_{\gamma}$$ The symmetry should be noted!

Then Eq. (1) becomes:

$$- \beta^{1} + \beta^{2} - \gamma + \int_{0}^{L}{wf_{t}}dx - \int_{0}^{L}{wm\ddot{v}}dx = 0$$ for all possible $$w(x)$$

We will now focus on the stiffness term $$\gamma$$ to derive the beam stiffness matrix and to identify the beam shape functions.

$$V(\tilde{x}) = N_{2}(\tilde{x})\tilde{d}_{2} + N_{3}(\tilde{x})\tilde{d}_{3} + N_{5}(\tilde{x})\tilde{d}_{5} + N_{6}(\tilde{x})\tilde{d}_{6}$$

Let us recall: $$U(\tilde{x}) = N_{1}(\tilde{x})\tilde{d}_{1} + N_{4}(\tilde{x})\tilde{d}_{4}$$

From the book p.246, the following functions can be found:

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

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

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

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

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User:Eas4200c.f08.wiki.d/HW7

Single Cell Airfoil
EDU>> NewAirfoil(2415,.5,[0 0 0])

Single Cell

q_0 =

2.9437e+004

q_hf =

2.9437e+004

q_fb =

-2.2512e+005

q_be =

3.2195e+004

q_eh =

2.8555e+005

Note: All values in N/m

Multi Cell Airfoil
EDU>> NewAirfoil(2415,.5,[0 0 0])

Multi Cell

q_1 =

3.2721e+004

q_2 =

2.6482e+004

q_3 =

3.7939e+004

q_hf =

3.7939e+004

q_fb =

-2.2807e+005

q_be =

3.5480e+004

q_eh =

2.8260e+005

q_w1 =

-6.2391e+003

q_w2 =

1.1456e+004

Note: All values in N/m

Also, it can be seen here that the shear flow values in the skin panels are similar between the single and multi cell cases. This verifies the validity of the two cases.

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User:Eml4500.f08.delta 6.guzman/MATLAB Code and Plot of Undeformed and Deformed Shapes for the Two Bar Truss System

Data

MATLAB Code

In order to run this MATLAB code the following functions are needed:

Note: this MATLAB functions were originally obtained from the MATLAB.zip folder provided by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=2256&itemId=0471648086&resourceId=4981 and later modified in order to applied to the MATLAB code presented in this section.

New Results

Previous Results (HW#2)

Plot


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User:Eas4200c.f08.nine.s/Lecture 7

Homework 1

Homework 2

Homework 3

Homework 4

Homework 5

Homework 6

Homework 7

Group nine - Homework 7

Without Stringers
Question: Can this setup resist $V_z$? Answer: No

$R^{z} = R_{AB}^{z} + R_{BA}^{z}$

where $R^{z}$ is the total resultant in the z-direction and $R_{AB}^{z}$ is the resultant of q in AB.

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

With Stringers
We can neglect the contribution of web to bending and we will look at a cell with three stringers.

$q_{12} \neq q_{23} \neq q_{31}$ $q_{ij}(s) = const.$ within each panel. $\Rightarrow q(s)$ is piecewise constant with respect to s.

Analysis Algorithm : 1) Solve $P2$ for $\tilde{q_{12}}, \tilde{q_{23}}, \left(\tilde{q_{31}} = 0\right)$|undefined 2)Take moment about any point in plane (y,z) and use superposition.

Without Stringers
$R^z = \sum_{i=1}^{n cells}{} R^{z}_i$ $\Rightarrow R^z = 0$

With Stringers
$R^z = R^{z1} + R^{z2}$

Stringer Equilibrium
$$\sum F_{x}=0=\int_{A_{3}} [\sigma_{xx}(x+dx)-\sigma_{xx}(x)]dA_{3} $$

Where we can exapand these terms into a taylor series:

$$[\sigma_{xx}(x+dx)-\sigma_{xx}(x)]dA_{3} \rightarrow

\frac{\partial \sigma_{xx}}{dx}dx+h.o.t \times [-\tilde{q}_{23} -\tilde{q}_{43} +\tilde{q}_{31}]dx $$ Contribution of shear flow by stringer 3 is given by:

$$q^{3}=-\int_{A_{3}} \frac{\partial \sigma_{xx}}{dx} A_{3}$$

Where:

$$V_{y}=\frac{\partial d M_{z}}{dx}$$

And:

$$V_{z}=\frac{\partial d M_{y}}{dx}$$

So we have:

$$ q^{3}=-(KyVy-KyzVz)Qz^{3}-(KzVz-KyzVy)Qy^{3} $$

And:

$$Qz^{3}=\int_{A_{3}} ydA_{3}$$

$$Qy^{3}=\int_{A_{3}} zdA_{3}$$

Similarly for stringer 2:

$$\tilde {q}_{23}=\tilde {q}_{12} - \tilde {q}_{24} + {q}^{2}$$

$$Qy^{2}=\int_{A_{2}} zdA_{2}\,$$

And for stringer 4:

$$\tilde {q}_{43}=\tilde {q}_{24} - \tilde {q}_{41} + {q}^{4}$$

$$Qy^{4}=\int_{A_{4}} zdA_{4}$$

______________________________________________  Using superposition principle:

$$q_{ij}=\tilde {q}_{ij} + q_{k}$$

First term is known, second term is unknown:

$$q_{12}=\tilde {q}_{12} + q_{1}$$

$$q_{23}=\tilde {q}_{23} + q_{1}$$

$$q_{31}=\tilde {q}_{31} + q_{1}$$

$$q_{24}=\tilde {q}_{24} + q_{2}$$

$$q_{43}=\tilde {q}_{43} + q_{4}$$

$$q_{41}=\tilde {q}_{41} + q_{1}$$

3 unknowns:

$$q_{1}, q_{2}, q_{3}$$

3 equations:

1. Moment equations of $$V_{y} V_{z} $$ and $$[q_{12}, ... q_{41}]$$ about any convenient point, usually where lines of action of $$V_{y} V_{z}\,$$ intersect. 2. $$\theta_{1}=\theta_{2}$$ 3. $$\theta_{2}=\theta_{3}$$

Plotting Buckling Shape under Shear Stress
$$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ in terms of $$C_{11}$$.

1. Find $$\lambda$$ for $$\vartheta = 1.5$$

Using equation (30) found on website:

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

Plugging known values: $$\displaystyle \lambda = .02875$$

2. Evaluate numerically $$k_{sys}$$ using equation (26) found on website:

Equation 26:

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

3. Calculate: $$[k_{ij}]$$

$$\begin{bmatrix} k_{12} & k_{23} & \cdots & k_{25} \\ \vdots & \ddots & & \vdots \\ \vdots & & \ddots & \vdots \\k_{32} & \cdots & \cdots & k_{55}\end{bmatrix} \begin{bmatrix} C_{22} \\ \vdots \\ \vdots \\ C_{33}\end{bmatrix} = \begin{bmatrix} -\frac{u_z}{a}C_{11} \\ 0 \\ 0 \\ 0\end{bmatrix}$$

Solving for: $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ intersect with $$C_{11}$$.

Since the $$K$$ matrix can be written as: $$\underline{\overline{K}}$$

Then its inverse is: $$\underline{\overline{K}}^{-1}$$

$$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix} = \underline{\overline{K}}^{-1}\ \begin{bmatrix} -\frac{u_z}{a}C_{11} \\ 0 \\ 0 \\ 0\end{bmatrix}$$

Finally:

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

Bidirectional Bending
Under bidirectional bending, the longitudinal displacement expansions are given by the following equations.

$u = u_0(x) + z\phi_y(x) + y\phi_z(x)$ $v = v_0(x)$ $w = w_0(x)$

$\phi_y$ and  $\phi_z$ are the rotations of the cross section about the y and z axes respectively. The corresponding strains are given below.

$\epsilon_{xx} = \frac{\delta u}{\delta x} = \frac{du_0}{dx} + z\frac{d\phi_y}{dx} + y\frac{d\phi_z}{dx}$ $\gamma_{xy} = \frac{\delta v}{\delta x} + \frac{\delta u}{\delta y} = \frac{dv_0}{dx} + \phi_z$ $\gamma_{xy} = \frac{\delta w}{\delta x} + \frac{\delta u}{\delta z} = \frac{dw_0}{dx} + \phi_y$

Assuming $\gamma_{xy} = \gamma{xz} =0 $ we yield the following relations,

$\phi_z = -\frac{dv_0}{dx}$ $\phi_y = -\frac{dw_0}{dx}$

and the we obtain the following by substitution.

$\epsilon_{xx} = \frac{du_0}{dx} - y\frac{d^{2}v_0}{dx^{2}} - z\frac{d^{2}w_0}{dx^{2}}$|undefined

Noting that $du_0/dx = 0$ if $N_x = 0$ and thus, the bending strain is reduced to,

$\epsilon_{xx} = - y\frac{d^{2}v_0}{dx^{2}} - z\frac{d^{2}w_0}{dx^{2}}$|undefined

The bending moments about the y and z axes are as follows,

$M_y = \int{}\int_{A}z\sigma_{xx}dA = -EI_{yx}\frac{d^{2}v_0}{dx^{2}} - EI_y\frac{d^{2}w_0}{dx^{2}}$|undefined $M_z = \int{}\int_{A}y\sigma_{xx}dA = -EI_{z}\frac{d^{2}v_0}{dx^{2}} - EI_yz\frac{d^{2}w_0}{dx^{2}}$|undefined

where

$I_y = \int\int_A z^{2}dA$ $I_z = \int\int_A z^{2}dA$ $I_{yz} = \int\int_A yzdA$

Comparing MediaWiki to E-learning
After using MediaWiki for the entire semester, it is this groups opinion that MediaWiki is better than E-learning. With MediaWiki, it is possible to work with your group without ever having to meet with your group. It is very easy to access, and can be used by others for future reference. MediaWiki also has a history that saves and keeps track of all changes made to its page. The history can be used to undo any unwanted changes by simply selecting a saved version of the article in the history that is prior to the change. E-learning does not allow any manipulation outside of posting text. There is no way to do any coding. It does however allow a teacher to easily post grades and course work.

Contributing Team Members
The following students contributed to this report:

David Phillips Eas4200C.f08.nine.d 7:40, 9 December 2008 (UTC)

Stephen Featherman Eas4200c.f08.nine.s 7:45, 9 December 2008 (UTC)

Ricardo Albuquerque Eas4200c.f08.nine.r 6:55, 9 December October 2008 (UTC)

Felix Izquierdo Eas4200c.f08.nine.F 9:45, 9 December 2008 (UTC)

title: User:Eas4200c.f08.nine.s/Lecture 7


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User:EML4500.f08.Ateam.Miller/homework7

Dimensional Analysis of Displacements
Axial Displacements

$$\left[u \right]=L$$

$$\left[N_{1} \right]= \left[N_{4} \right]=1$$

$$\left[N_{1} \right]\left[\tilde{d_{1}} \right]+\left[N_{4} \right]\left[\tilde{d_{4}} \right]$$

Transverse Displacements

$$\left[v \right]=L$$

$$\left[N_{2} \right]\left[\tilde{d_{2}} \right]=L$$

$$\left[N_{3} \right]\left[\tilde{d_{3}} \right]=L$$

Derivation of Beam Shape Functions
$$N_2,\ N_3,\ N_5,\ N_6$$

Refer to plots shown above to derive the beam shape functions.

The governing partial differential equation (PDE) for beams is as follows.

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

Considering EI to be constant reveals the following.

$$\frac{\delta ^{4}}{\delta x^{4}}v=0$$

By integrating this four times we obtain four constants in a relation for V(x).

$$V\,(x) = c_0+c_1x^1+c_2x^2+c_3x^3$$

In order to obtain $$N_2(x)$$ where $$\tilde{x}\equiv x$$ the following boundary conditions must be considered to solve for the four constants determined above.

$$V\,(0)=1$$

$$V\,(L)=0$$

$$V\,'(0)=V\,'(L)=0$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=1=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=1$$

$$V'(0)=0=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=0$$

$$V(L)=0=1+0+c_2(L)^2+c_3(L)^3=1+c_2L^2+c_3L^3\Rightarrow (1)$$

$$V'(L)=0=0+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2\Rightarrow (2)$$

From equation (2),

$$c_3=-\frac{2}{3}\frac{c_2}{L}$$

Now, inserting c3 from above into equation (1) yields

$$0=1+c_2L^2+\left( -\frac{2}{3}\frac{c_2}{L}\right)L^3\Rightarrow c_2=-\frac{3}{L^2}$$

and $$c_3=-\frac{2}{3}\frac{1}{L}\left({-\frac{3}{L^2}} \right)=\frac{2}{L^3}$$

Now, inserting c0, c1, c2, and c3 into V(x),

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

at the N2 boundary conditions. This expression is the same as the N2(x) found previously.

Likewise for $$N_3$$ the following boundary conditions are considered.

$$V\,(0)=0$$

$$V\,(L)=0$$

$$V\,'(0)=1$$

$$V\,'(L)=0$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=0=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=0$$

$$V'(0)=1=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=1$$

$$V(L)=0=1+0+c_2(L)^2+c_3(L)^3=1+c_2L^2+c_3L^3\Rightarrow (1)$$

$$V'(L)=0=1+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2\Rightarrow (2)$$

Using MATLAB to solve the system of 2 equations and 2 unknowns for equations (1) and (2),

$$c_2=\frac{-2}{L}$$

$$c_3=\frac{1}{L^2}$$

Now, inserting c0, c1, c2, and c3 into V(x),

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

at the N3 boundary conditions. This expression is the same as the N3(x) found previously.

Likewise for $$N_5$$ the following boundary conditions are considered.

$$V\,(0)=0$$

$$V\,(L)=1$$

$$V\,'(0)=0$$

$$V\,'(L)=0$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=0=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=0$$

$$V'(0)=0=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=0$$

$$V(L)=1=0+0+c_2(L)^2+c_3(L)^3=c_2L^2+c_3L^3=1\Rightarrow (1)$$

$$V'(L)=0=0+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2=0\Rightarrow (2)$$

Using MATLAB to solve the system of 2 equations and 2 unknowns for equations (1) and (2),

$$c_2=\frac{3}{L^2}$$

$$c_3=\frac{-2}{L^3}$$

Now, inserting c0, c1, c2, and c3 into V(x),

$$V(x)=\frac{3}{L^2}x^2-\frac{2}{L^3}x^3$$

at the N5 boundary conditions. This expression is the same as the N5(x) found previously.

Likewise for $$N_6$$ the following boundary conditions are considered.

$$V\,(0)=0$$

$$V\,(L)=0$$

$$V\,'(0)=0$$

$$V\,'(L)=1$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=0=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=0$$

$$V'(0)=0=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=0$$

$$V(L)=0=0+0+c_2(L)^2+c_3(L)^3=c_2L^2+c_3L^3=0\Rightarrow (1)$$

$$V'(L)=0=0+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2=0\Rightarrow (2)$$

Using MATLAB to solve the system of 2 equations and 2 unknowns for equations (1) and (2),

$$c_2=\frac{-1}{L}$$

$$c_3=\frac{1}{L^2}$$

Now, inserting c0, c1, c2, and c3 into V(x),

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

at the N6 boundary conditions. This expression is the same as the N6(x) found previously.

Element Stiffness Matrix Coefficient Derivation
The coefficients containing EA have already been derived. This leaves the terms containing an EI component. These coefficients are derived here.

$$\tilde{k_{22}}=\frac{12EI}{L^{3}}=\int_{0}^{L}{\frac{d^{2}N_{2}}{dx^{2}}(EI)\frac{d^{2}N_{2}}{dx^{2}}dx}$$

$$\tilde{k}_{23}=\frac{6EI}{L^2}= \int_{0}^{L} \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}\, dx$$

Or to write this in a more generalized form:

$$\tilde{k}_{ij}= \int_{0}^{L} \frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}\, dx\quad i,j=2,3,5,6$$

title: User:EML4500.f08.Ateam.Miller/homework7


text:
User:Eas4200c.f08.vqcrew.d/hw7

Solving P2 - Euler Cut Principle
The Euler Cut Principle involves finding the equilibrium of each stringer. This is basically a statics problem where the sum of the forces, which are given through the shear flows, are equal to zero. Stringer 3 is shown in Figure # in two dimensions and again in three dimensions in Figure #, with the stringer expanded into the x-direction. The sum of the forces in the x-direction can be expressed as follows:

 * $$\sum{F_x}=0=\int_{A_3}^{}{\left[\sigma _{xx}(x+dx) -\sigma _{xx}(x)\right]}dA_3=\int_{A_3}^{}{\left[\frac{d\sigma _{xx}}{dx}dx + HOT\right]}dA_3=\left[ \tilde{q_{31}}-\tilde{q_{23}}-\tilde{q_{43}}\right]dx$$

Thus we solve for $$\tilde{q_{31}}$$ as:

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

Where $$q^{(3)} \;$$ is the contribution to shear flow by Stringer 3:

 * $$q^{(3)}=-\int_{A_3}^{}{\frac{d\sigma _{xx}}{dx}}dA_3$$

The integral can be solved for $$q^{(3)} \;$$ as:

 * $$q^{(3)}=-\left(k_yV_y-k_{yz}V_z \right)Q_z^{(3)}-\left(k_zV_z-k_{yz}V_y \right)Q_y^{(3)}$$

where:

 * $$V_y=\frac{dM_z}{dx}$$
 * $$V_z=\frac{dM_y}{dx}$$
 * $$Q_z^{(3)}=\int_{A_3}^{}{ydA_3}$$
 * $$Q_y^{(3)}=\int_{A_3}^{}{zdA_3}$$

The same procedure can be used to solve for Stringer 2:

 * $$\tilde{q_{23}}= \underbrace{\tilde{q_{12}}}_{flow in}-\underbrace{\tilde{q_{24}}}_{flow out}+q^{(2)}$$

The value of $$q^{(2)} \;$$ can be computed as done with Stringer 3 with the values of $$Q_z^{(2)}$$ and $$Q_y^{(2)}$$ as:

 * $$Q_z^{(2)}=y_2A_2$$
 * $$Q_y^{(2)}=z_2A_2$$

where $$y_2 \;$$ and $$z_2 \;$$ are the y and z coordinates of Stringer 2, and $$A_2 \;$$ is the area of Stringer 2.

The same procedure can be followed on Stringer 4 to find $$q^{(4)} \;$$ and $$\tilde{q_{24}}$$. Hence, we can also deduce $$\tilde{q_{31}}$$.

Using the principles of superposition, the values of q can be determined.

 * $$q_{ij}=\tilde {q}_{ij} + q_{k}$$

On the right side of the equations, the tilde terms are the knowns and the non-tilde terms are unknowns.

 * $$q_{12}=\tilde {q}_{12} + q_{1}$$
 * $$q_{23}=\tilde {q}_{23} + q_{1}$$
 * $$q_{31}=\tilde {q}_{31} + q_{1}$$
 * $$q_{24}=\tilde {q}_{24} + q_{2}$$
 * $$q_{43}=\tilde {q}_{43} + q_{2} - q_{3}$$
 * $$q_{41}=\tilde {q}_{41} + q_{3}$$

These equations reveal that there are three unknowns, $$q_1, q_2, q_3 \;$$, therefore three equations are needed to solve for these values to complete the problem. The three equations to be use are: 1) Moment equation: Take the moment of $$V_y \;$$, $$V_z \;$$, and $$\left\{q_{12},...,q_{41} \right\}$$ about any convenient point (usually where the lines of action of $$V_y \;$$ and $$V_z \;$$ intersect). 2) $$\theta _1=\theta _2 \;$$ 3) $$\theta _2=\theta _3 \;$$ Equations 2 and 3 are the compatibility equations.

To recap, in order to solve P2 for each cell you must follow path "$$s_i \;$$" and then determine the equilibrium of each stringer on the path. There are two ways to accomplish this: 1) Complete method using FBD as shown above 2) A consequence of the first method, as exemplified in Figure #:

 * $$\tilde{q_{j6}} = \tilde{q_{j2}} - \tilde{q_{j5}} - \tilde{q_{j8}} + q^{(j)}$$

Also, the left side of the equation simply becomes negative if the direction of $$\tilde{q_{j6}} \;$$ changes (and becomes $$\tilde{q_{6j}}$$), as shown in Figure #.

 * $$\tilde{q_{6j}} = \tilde{q_{j2}} - \tilde{q_{j5}} - \tilde{q_{j8}} + q^{(j)}$$

Using the Euler Cut Principle brings forth a question, what happens if the object is cut such that one stringer is isolated as shown in Figure #? Due to the cuts made, $$\tilde{q_{23}}=\tilde{q_{31}}=\tilde{q_{34}}=0$$. We then sum the forces acting on the FBD shown in Figure #.

 * $$\underbrace{\tilde{q_{31}}}_{0} = \underbrace{\tilde{q_{23}}}_{0} - \underbrace{\tilde{q_{34}}}_{0} + q^{(3)}$$

This shows that $$q^{(3)}=0 \;$$, which is not possible. Therefore, when using the Euler Cut method a stringer must not be isolated from the rest of the body.

title: User:Eas4200c.f08.vqcrew.d/hw7


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text:
User:Eml4500.f08.team.Bengtson/teamHW7

November 21st, 2008: Meeting 38
Motivation: The deformed shape of the tress element and the interpolation of the transverse displacements v(s)and s = \tilde{x} elements.

The PVW for Beams is as follows: $$\int_{0}^{L}{w(\tilde{x})}\left[-\frac{\partial ^2}{\partial x^2} \left((EI)\frac{\partial ^2v}{\partial x^2}\right) + f_t - m\ddot{v}\right]dx = 0 $$

This is true for all possible w(x), $$ x=\tilde{x}$$

Integration by parts of the first term.

$$(\alpha) = \int_{0}^{L}{w(\tilde{x})}\frac{\partial^2}{\partial x^2}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}dx$$

To do integration by parts, one must separate the different parts of the integration to do them. $$ r(x) = \left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}$$ $$r'(x) = \frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right)$$ $$ s(x)= w(x) $$ $$s'(x) = \frac{\partial w}{\partial x}$$ Putting these together into an integration by parts, you get:

$$ (\alpha) = \left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right]^L_0 - \int_{0}^{L}\frac{dw}{dx}\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}dx$$

Where $$(B_1)=\left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right]^L_0$$

Continue Integrating $$= (B_1) - \left[\frac{dw}{dx}\left(EI\right)\frac{\partial ^2v}{\partial x^2}\right]^L_0 + \int_{0}^{L}{\frac{\partial^2w}{\partial x^2}\left(EI\right)\frac{\partial ^2v}{\partial x^2}}dx$$

This gives us: $$(B_2) = \left[\frac{dw}{dx}(EI)\frac{\partial ^2v}{\partial x^2}\right]^L_0$$ and $$v_\gamma =\int_{0}^{L}{\frac{\partial^2w}{\partial x^2}\left(EI\right)\frac{\partial ^2v}{\partial x^2}}dx$$

Please note $$v_\gamma $$ is symmetrical

$$-(B)_1 + (B)_2 - v_\gamma + \int_{0}^{L}{wf_tdx} - \int_{0}^{L}{wm\ddot{v}dx} = 0 $$

For all possible w(x)

Stiffness Term Lets focus on the stiffness term for $$ v_\gamma$$ to derive the beam stiffness matrix and to identify the beam shape function. $$v(\tilde{x}) = N_2(\tilde{x})\tilde{d}_2 + N_3(\tilde{x})\tilde{d}_3 + N_5(\tilde{x})\tilde{d}_5 + N_6(\tilde{x})\tilde{d}_6$$ Recall that $$u(\tilde{x}) = N_1(\tilde{x})\tilde{d}_1 + N_4(\tilde{x})\tilde{d}_4$$ $$N_2(\tilde{x}) = 1 - \frac{3\tilde{x}^2}{L^2} + \frac{2\tilde{x}^3}{L^3}\,\,\,\tilde{d}_2 $$ $$N_3(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\,\,\,\tilde{d}_3 $$

$$N_5(\tilde{x}) = \frac{3\tilde{x}^2}{L^2} - \frac{2\tilde{x}^3}{L^3}\,\,\,\tilde{d}_5$$ $$N_6(\tilde{x}) = -\frac{\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\,\,\,\tilde{d}_6$$

title: User:Eml4500.f08.team.Bengtson/teamHW7


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User:Eml4500.f08.team.Bengtson/teamHW7b

November 21st, 2008 Meeting 37
$$[\varepsilon ] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$

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

$$ [\mathbb{A}]=L^2$$

$$[\mathbb{I}]=L^4$$

$$ \frac{EA}{L}=[\tilde{k}_{11}] = \frac{(\frac{F}{L^2})L^2}{L} = \frac{F}{L} $$

$$[\tilde{k}_{11}\tilde{d}_1] = [\tilde{k}_{11}][\tilde{d}_1] = F$$

$$ [\tilde{k}_{23}\tilde{d}_3] = \frac{6[E][I]}{L^2} = \frac{(\frac{F}{L^2})L^4}{L^2} = F $$

$$ \begin{Bmatrix} \tilde{d}_1\\ \tilde{d}_2\\ \tilde{d}_3\\ \tilde{d}_4\\ \tilde{d}_5\\ \tilde{d}_6 \end{Bmatrix} =\begin{bmatrix} R& & 0 & 0 &0  &0 \\ & & 0 & 0 & 0 &0 \\ 0 & 0 & 1 &R  &  &0 \\ 0& 0 & 0 & &  &0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{Bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6 \end{Bmatrix}$$ See MATLAB portion for two bar system. Derivative of $$\hat{K}$$$$ \frac{\partial^2 }{\partial x^2}[(EI)\frac{\partial^2v }{\partial x^2}] - f_t(x) = m(x)\ddot{v}$$

title: User:Eml4500.f08.team.Bengtson/teamHW7b


text:
User:Eml4500.f08.team.figuerrez/hw7

Meeting 39 (12/1/08)

Meeting 40 (12/5/08)

Meeting 41 (12/8/08)

title: User:Eml4500.f08.team.figuerrez/hw7


text:
User:Eml4500.f08.team.figuerrez/hw7/meeting39

Meeting 39 - Principle Axial & Transverse Dislocations Of A Beam (Shape Functions)
From Meeting 37: $$\tilde{\boldsymbol{d}}_{6x1}^{(e)}=\tilde{\boldsymbol{T}}_{6x6}^{(e)}*{\boldsymbol{d}}_{6x1}^{(e)}$$, where $${\boldsymbol{d}}_{6x1}^{(e)}$$ is known after solving the finite element system.

Now we will discuss how to compute $${\boldsymbol{u}}\tilde{\boldsymbol{(x)}},{\boldsymbol{v}}\tilde{\boldsymbol{(x)}}$$

We compute $${\boldsymbol{u}}\tilde{\boldsymbol{(x)}},{\boldsymbol{v}}\tilde{\boldsymbol{(x)}}$$ using equations (1) & and (2) from meeting (38-3).

We Compute $${\boldsymbol{U_{x}}}\tilde{\boldsymbol{(x)}},{\boldsymbol{U_{y}}}\tilde{\boldsymbol{(x)}}$$ from $${\boldsymbol{u}}\tilde{\boldsymbol{(x)}},{\boldsymbol{v}}\tilde{\boldsymbol{(x)}}$$

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

$$\begin{Bmatrix}u\tilde{(x)} \\ v\tilde{(x)}\end{Bmatrix}=\begin{bmatrix}N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6\end{bmatrix} \begin{Bmatrix}\tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)}\end{Bmatrix} $$     (2)

We will denote $$\begin{bmatrix}N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6\end{bmatrix}$$ by N, and the matrix $$\begin{Bmatrix}\tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)}\end{Bmatrix}$$ as $$\tilde{d}^{(e)}$$ therefore:

$$\begin{Bmatrix}U_{x}\tilde{(x)} \\ U_{x}\tilde{(x)}\end{Bmatrix}=R^{T}*{\boldsymbol{N}}\tilde{(x)}*\tilde{T}^{(e)}d^{(e)}$$    (3)


text: Category:Social psychology (psychology)/Admin/Tutor notes


text: Social psychology (psychology)/Tutorials/Assessment workshop/Tutor notes


text: Social psychology (psychology)/Tutorials/Cross-cultural training/Tutor notes


text: Social psychology (psychology)/Tutorials/Prejudice, aggression, and prosocial behaviour/Tutor notes


text:
Social psychology (psychology)/Tutorials/Communication/Tutor notes

 * Also at the start include - how to "images" with Wikiversity


text:
Social psychology (psychology)/Lectures/Introduction/Instructor notes

There were several issues in 2007 which were meant to be considered for 2008, but weren't (due to unavailability of files due to computer rebuild):

Thus, most issues from 2007 remain to be done:

2007

 * The slides were the first prepared in OOImpress, based on:
 * 2005 Lecture 1
 * The new Textbook - Baumeister, including the demo PPT text slides
 * 2008 - Small/minimal improvements.
 * added design template
 * removed ~10-15 slides to make is shorter
 * lightly edited/reviewed
 * added/improved some text/images

To do

 * Remove animation
 * Check layout - some overlapping text
 * Add attributions for all images
 * Shorten (it was far too long): Culture & Nature should be shifted into the next week with Social Self
 * Add more examples of “social psychology” situations (e.g., as per p. 4-5 in Baumeister & Bushman, 2008.
 * Take account of (and incorporate 2005 Lecture 2 History and Research) (major oversight)
 * Incorporate textbook digital images (if available)
 * Listen to 2005 audio & 2007 & 2008 audio/video & review/write-up accordingly
 * Add age of universe
 * Add quote from book about being on planet earth
 * Add quote from Buckminster Fuller about spaceship earth
 * Animations needs to be fixed/stripped
 * Sometimes multiple clicks are needed to go to the next slide
 * Add page numbers

Strengths

 * If well presented, students can find this introduction 'engaging' in part due to the style (aka Dick Hardt's Identity 2.0 presentation)


text: Category:Social psychology (psychology)/Admin/Instructor notes


text:
Social psychology (psychology)/Lectures/Social self/Instructor notes

2008

 * Good, but too long
 * Check what slide number I got up to.
 * Replaced title image: Instead of M. C. Escher's "Hand with Reflecting Sphere" (http://en.wikipedia.org/wiki/Hand_with_Reflecting_Sphere), a public domain image was found and used.

2007

 * Admin
 * Blog Feeds
 * Blog Posts
 * Active Blogs
 * Online Engagement - meaning

 * Self
 * Maybe add a list of self-constructs and improve definitions
 * Relate self to social ? This was a bit weak.


text:
Social psychology (psychology)/Lectures/Social thinking/Instructor notes

 * At the start, summarise about essay signup
 * The 3 sections seems to work OK
 * Still went over time - didn't finish last section (only had 18 mins)
 * Part 1: 11.30 - 12.37
 * Part 2: 12:43 to 1:08?
 * Part 3: 1:08 to 1:33!!
 * 2009 draft already started - make sure to use this one (from laptop)
 * Warn students re starting/end time of next lecture - and inform about DVD in library
 * Include Cross-cultural cognitive dissonance material from Jenny Owen's 2008 essay.


text: Social psychology (psychology)/Lectures/Aggression/Instructor notes


text:
Social psychology (psychology)/Lectures/Prejudice/Instructor notes

 * 1) Sample exam questions (Wilderdom)


text:
Social psychology (psychology)/Lectures/Relationships/Instructor notes

What was done in 2008

 * Removed animations
 * Replaced many proprietary images (there are still several to be worked on) and added many free culture images
 * Rationalised some slides (e.g., several are now hidden)
 * Kept the Social Exclusion video in
 * Adjusted master slides to aid visual clarity
 * Didn't read chapter – near 0 pre-knowledge – this was poor

Comments

 * Timing was better than 2007; approximately 30 in attendence
 * Tended to read “down into the slides” a bit too much; rather than look at audience
 * Skipped some slides back and forth; some re-ordering helpful; several slides skipped
 * Somewhat mellowish/tiredish/stayed low key; 6.5/10
 * The audio contains several additional comments
 * Beauty activities worked well – add some more
 * Add heat map of beauty marketing (available on web)
 * Breaking before social exclusion
 * Showing the social exclusion video at the start of Pt2
 * It could be good to add a briefer summary slide of the consequences of social exclusion and to find /show relevant references (and effect sizes)
 * Add slide/reference to Gottman's studies - “thin slices” (did provide a useful summary)
 * Lienke's Law mentioned
 * Showed 4 relationship adds at the end; worked; maybe 3 enough though (e.g. leave out plane add)

What's to be done

 * Not all 2007 comments were addressed
 * Not all animations were removed – grrrr (mostly the remaining ones are towards the end).
 * If possible, reduce some image file sizes – the handout is almost 3MB

Summary

 * This was "OK" e.g., 7/10 - no more.
 * Intrinsically interesting topic
 * I was a bit mellow/flat, but OK.

Slides

 * Several slides were skipped - see video
 * Took too long on initial material, so Social Exclusion was skipped
 * Also, video on social exclusion wasn't shown bec. no audio from laptop - need additional cord.
 * Basic structure of material is pretty good.
 * Add slide back in about Harlow's Monkeys.
 * Some proprietary images from the textbook used; replace these.
 * What is love? Some more romantic, widespread, definitions?
 * More activities (e.g., attractiveness) - these invariably worked nicely - build more of these.
 * Not all animations removed
 * Summary to be developed / relate to assessment items

Assessment

 * Add some assessment items around slides only presented in the lecture (i.e., around attractiveness)
 * Need to generate assessment items for additional lecture content e.g., Hatfield's model

2005
(note that the trailer needed to be downloaded and buffered before the lecture started)
 * 6 topics - 3 in each lecture half - overall, good scope and efficient - aiming to give 15-20 minute insight into each
 * Time was about right; finished a few minutes early and had time to indulge a bit with each section
 * At lecture break time, I showed the trailer for "An Inconvenient Truth" - could replace with trailer for Blue Eyes / Brown Eyes and Rwanda 3rd year videos.
 * Use document projector to show / pupil / faces from Manwatching and the Manwatching cover


text:
Social psychology (psychology)/Lectures/Groups and leadership/Instructor notes

To do

 * Further reduce proprietary images
 * Improve/plan timings
 * Any activities – e.g., what makes a group, what makes a leader? Identify e.g., key questions for blog/e-portfolios

2008

 * Get video - Taylor, R. (2003). The confidence zone (Programme 2: Leadership). Videorecording. 15 mins. BBC Worldwide: Sydney, NSW. UC Library Call No: BF575.S39.C66 2003 3 hour loan.
 * Reduced by ~15 slides, with a few added
 * Something got out of sequence, so all NOTES need checking and all SLIDE TITLES titles need rechecking
 * Several images were replaced with OERs – several still to go

2007

 * No notes
 * Leadership video wasn't shown
 * Audio trimmed and converted


text:
Social psychology (psychology)/Lectures/Prosocial/Instructor notes

Generic
Needs audio cable to show gorilla video

To do

 * 1) Most slides titles need names or renamed to lowercase
 * 2) Add:from Teall McQueen 2008 essay on Bystander Effect: Recent studies have demonstrated the effect of diffusion of responsibility on peoples’ helping behaviour in impersonal settings, such as electronic communication. In a study conducted by Blair, Thompson, and Wuensch (2005), 400 undergraduates were sent a bogus email from a new student, seeking assistance about how to access an on-line journal database. The number of other students, who participants were led to believe had also received the email was manipulated, resulting in four conditions (0, 1, 14, or 49 other recipients). Consistent with diffusion of responsibility, participants were less likely to respond to the email when they believed other people had also received the email. Similar findings were found by Barron and Yechiam (2002). In this study, an email generated more responses when it was sent to a single recipient, and responses were more detailed, compared to when it was sent to multiple recipients.
 * 3) Blair, C. A., Thompson, L. F., & Wuensch, K. L. (2005). Electronic helping behavior:
 * 4) The virtual presence of others makes a difference. Basic and Applied Social
 * 5) Psychology, 27, 171-178.

Report

 * Altruisim – does it exist (“debate” - good)
 * I was a bit underdone in terms of reading preparation (overall) – it would be good to keep finding more references/studies to be able to refer to in more depth.
 * Overall: 6.5/10
 * Ran out of time by ~ 10-15%
 * The lecture recording failed :( - best to write up notes in more detail to accompany the lecture.
 * Include Kohlberg's stage of moral development

Done

 * 1) Most slide headings renamed to lowercase

To do

 * 1) Print out Parable of the Good Samaritan
 * 2) Work as prosocial behaviour? -> Volunteering
 * 3) Blink – Sorry
 * 4) Bystander effect – Indian boy begging
 * 5) Gorilla rescues kid video
 * 6) Kohlberg's stages of moral development
 * 7) Asch or other comformity videos

2007
Material to maybe to consider for next time

Oskar Schindler – Holocaust rescues

Research out some videos for this topic, e.g.,

 * Stanley Milgram (Youtube Search)

Teaching Tip: The novel (and later film) Pay It Forward offers an interesting variation on the reciprocity norm. A 12-year-old boy does favors for three individuals and, when asked how they can repay him, tells them to “pay it forward” by doing favors for three others, who then are to be instructed to pay it forward to three more recipients. (Book: http://www.amazon.com/Pay-Forward-Catherine-Ryan-Hyde/dp/0743412028/sr=1-1/qid=1164504990/ref=pd_bbs_sr_1/104-0317742-5903166?ie=UTF8&s=books)  (Film:  http://www.imdb.com/title/tt0223897/)

Teaching Tip: The short story “The Ones Who Walk Away From Omelas” by Ursula K. Le Guin provides an interesting perspective on happiness, or lack thereof, when equality is not achieved. (http://www.cbe.wwu.edu/dunn/rprnts.omelas.pdf)

Teaching Tip: The play (and later film) Death of a Salesman forwards the idea that some think they are a burden on other people, that others do things for them that they cannot reciprocate; and so the others would be better off if they were dead as the main reason for the suicide of the central character, Willy Loman.

Prisoner's Dilemma

Technology Tip: Check out the special issue of Science magazine devoted to the idea of the tragedy of the commons. (http://dieoff.org/page95.htm)

Technology Tip: A computer demonstration of the tragedy of the commons is available at “Tragedy of the Commons HubNet” (http://ccl.northwestern.edu/netlogo/models/CompHubNetTragedyoftheCommonsHubNet).

Technology Tip: Various online versions of the prisoner’s dilemma are available on the web, including the relatively simple version “You Have Found the Prisoner’s Dilemma” (http://serendip.brynmawr.edu/playground/pd.html) and a more complex version, “The Prisoner’s Dilemma,” (http://www.princeton.edu/~mdaniels/PD/PD.html)  that allows for the manipulation of variables.

Forgiveness

Technology Tip: An interesting article available from Science Blog, “New Scientific Study Finds Forgiveness a Factor in Decreasing Spread of AIDS.” found that AIDS victims who forgave were less likely to endanger others through risky sex practices (http://www.scienceblog.com/community/older/2003/D/20031547.html).

Obediance

Technology Tip: The “Soup Nazi” episode of Seinfeld (Season 7) provides a humorous look at the customer’s obedience.

Technology Tip: NPR produced a segment, “Father Cares: The Last of Jonestown,” chronicling life during the final months of Jonestown. Details and entire audio are available online (http://www.npr.org/programs/specials/jonestown.html).

Food for Thought - Restaurants, Rules, and the Bad Taste of Nonconformity

In restaurants we tend to be nonconforming

People tend to order different menu items

People who order the same item, when it is their first choice, enjoy it more than those who switch to avoid conformity

 * De Ropp's Master Game
 * Sorry in Blink by Gladwell

Is Bad Stronger Than Good? #Moral and Immoral

Establishing a good reputation can be shattered by one scandal.

A good impression, established by moral behavior, can be ruined by the slightest immoral behaviour.

Very difficult to reverse an initial bad impression with subsequent good behaviours.

The Social Side of Sex #Helping, Sex, and Friends

Many people are introduced to sex partners by friends, co-workers, and relatives

Spring break sex

Men will support and help each other to engage in casual sex

Women will support and help each other to avoid casual sex

What Makes Us Human?

Humans that help individuals and society

Humans frequently exhibit prosocial behavior toward others who are not family

Rule following, obedience, and conformity are prosocial acts

Reciprocity and cooperation with strangers


text:
Social psychology (psychology)/Lectures/Environmental/Instructor notes

History

 * 2005, James Neill, in Social Psychology
 * 2008, James Neill, in Social Psychology

2008

 * A reasonably smooth effort, despite minimal sleep
 * Well-paced, good length, reasonable coverage
 * Did not include/use Application Module D from the textbook because I announced three other PDF readings to students, then later found the Module at the back of the book (I suspect in the 2007 pre-release this module may not yet have been available?). However, I may be better anyway to use the other readings because the content appears to be quite American and because APS requires that (3rd year?) units do not soley rely on textbook readings, but this should have further review/reconsideration in future.
 * Updated some images from 2005
 * 2008 Environmental Psychology
 * Went surprisingly smoothly - 60 slides fitted in well, timing-wise

To do

 * Find/add Greenwich Emotion Map video
 * Add outdoor child play video from youtube? (I haven't watched this yet)
 * Add top environmental concerns (find research on this – e.g., Morgan)
 * See note about the textbook readings in 2008
 * Add authorship/website details to notes on first slide

 * Add activities??
 * Add Greenwich Emotion Map
 * Add Defensible Space
 * Add Top environmental concerns
 * Add Risk perceptions
 * planning and design solutions that incorporate green spaces within urban designs that offer those in high density living arrangements a place of escape and restoration (Van den Berg, Hartig & Staats, 2007)?.
 * Healey (2000) quotes that the most densely populated 1% of Australia contains 84% of the population.
 * Healey, J. (2000). Population growth: Issues in society. Balmain: Spinney Press.
 * Van den Berg, A.E., Hartig, T., & Staats, H. (2007). Preference for nature in
 * urbanized societies: Stress, restoration, and the pursuit of sustainability.
 * Journal of Social Issues, 63, 79-96.

Activities?

 * Favourite place? Describe it.
 * Sensory pleasures? List them.

Louv video exercept?

~10 minutes for approx. 3 minutes

~15 minutes for research


text:
Social psychology (psychology)/Lectures/Review/Instructor notes

2005

 * This was the year with no textbook – so the slides were based on the readings; these could be reconsidered, particularly the Psychology and Everyday Life reading, and incorporated/brought in

2006-2007

 * A summary lecture wasn't given, in part because the final lecture slot was used for the final exam.

2008

 * Overall – 7/10, but, as usual, was on the run, not as well prepared as desirable, etc. But given that the slides were created quickly, a solid review effort.
 * This was effectively a new lecture – 69 slides cobbled together quickly based on Tricia's 2003 lecture and outline/overview/summary slides from each of the previous 9 lectures
 * In general, I think this lecture should be continued to give the unit some sense of returning to its original goals and reviewing those, marking out signposts along the way
 * Preamble was longer than for other lectures, but I think this was OK – it told the bigger picture story before diving in; suggest transcribing and edited and put on the web too
 * Timing was OK – arrived/started late – up to 10 mins; no break; finished with 2 mins to go
 * Performance: 7/10 – underprepared, particularly on latter topics
 * Several current, contemporary, spontaneous examples were given during the lecture – could use even more – transcribe what these were:
 * Looking-glass self: Weight loss with full-length mirror on fridges
 * John McCain's choice of Sarah Palin as VP running mate in 2008 USA presidential election – was he smitten and irrational?
 * Hot drink / attraction study
 * It helped that I'd prepared half the exam beforehand – some concepts were salient
 * Slides were not put up beforehand :(
 * Slides were revised afterwards :)
 * Add exam questions – say 1 per topic from this lecture
 * Slides were not put up beforehand :(
 * Slides were revised afterwards :)
 * Add exam questions – say 1 per topic from this lecture

To do

 * Listen to the 2008 previous lecture – would be useful preparation both beforehand, but particularly also on the way in
 * Add some more slides on the latter topics
 * Review and increase correspondence between given lecture key points, review content, and exam.


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af:user:Zenman ang:user:Zenman an:user:Zenman ar:user:Zenman bat-smg:user:Zenman diq:user:Zenman eo:user:Zenman et:user:Zenman gl:user:Zenman jv:user:Zenman ln:user:Zenman pl:user:Zenman sv:user:Zenman tr:user:Zenman uz:user:Zenman zh-min-nan:user:Zenman

title: User:Zenman


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User:Eml4500.f08.team.figuerrez/hw7/meeting40

Meeting 40 - Derivation & Application Of Shape Functions
Equation (2) from meeting (39), we will be using dimensional analysis:

$$\left[ u \right]=Length=L$$

From meeting (31): $$\left[ N_1 \right]=\left[ N_4 \right]=1$$

Therefore equation (2) from meeting (39) becomes:

$$\left[ N_1 \right]\left[ \tilde{d}_1 \right]+\left[ N_4 \right]\left[ \tilde{d}_4 \right]$$

Converting the given values equates to:

$${\boldsymbol{1 * L}} + {\boldsymbol{1 * L}}$$

$$\left[ V \right]=Length=L$$

$$\left[ N_2 \right]\left[ \tilde{d}_2 \right]=L$$

Converting the given values equates to:

$${\boldsymbol{1 * L}} = {\boldsymbol{L}}$$    (displacement, transverse)

$$\left[ N_3 \right]\left[ \tilde{d}_3 \right]=L$$

Converting the given values equates to:

$${\boldsymbol{L * L}} = {\boldsymbol{L}}$$    (rotational)

Now we solve the derivation of beam shape functions: N2, N3, N5, N6  (from meeting 38)

Recall: Governing PDE for beams

From meeting (37), equation (1), without ft (distributed trans. load) and without inertia force $$m\ddot{v}$$ (static case):

$$\frac{\partial^{2}}{\partial x^{2}}\left[(EA)\frac{\partial^{2}v}{\partial x^{2}}\right]=0$$

Further more, consider the constant EI:

$$\frac{\partial^{4}}{\partial x^{4}}V=0$$, where you integrate four times to get four constants: C0, C1, C2, C3

Solving for v equates to: $${\boldsymbol{v(x)=C_{0}+C_{1}x^{1}+C_{2}x^{2}+C_{3}x^{3}}}$$

To obtain N2(x)    ($$\tilde{x}=x$$ for simplicity), use the boundary conditions:

$${\boldsymbol{v(0)=1, v(L)=0}}$$

$${\boldsymbol{v'(0)=v'(L)=0}}$$

Use the above boundary conditions to solve for: C0, C1, C2, C3

$${\boldsymbol{v(0)=C_{0}}}$$

$${\boldsymbol{v(L)=1+C_{1}L+C_{2}L^{2}+C_{3}L^{3}}}$$    (1)

$${\boldsymbol{v'(L)=C_{1}+2C_{2}x+3C_{3}x^{2}}}$$

$${\boldsymbol{v'(L)=C_{1}=0}}$$

$${\boldsymbol{v'(L)=2C_{2}L+3C_{3}L^{2}=0}}$$    (2)

Solving for C3 from equation (2) yields:

$$C_{3}=-\frac{2}{3}\frac{C_{2}}{L}$$

Plugging C3 back into equation (1) yields:

$$0=1+C_{2}L^{2}+\left[-\frac{2}{3}\frac{C_{2}}{L}\right]L^{3}$$

$$C_{2}=-\frac{3}{L^{2}}$$

$$C_{3}=-\frac{2}{3}\frac{1}{L}\frac{-3}{L^{2}}=\frac{2}{L^{3}}$$

To obtain N3(x), use the boundary conditions:

$${\boldsymbol{v(0)=v(L)=0}}$$

$${\boldsymbol{v'(0)=1, v'(L)=0}}$$

To obtain N5(x), use the boundary conditions:

$$\tilde{d}_{5}=displacement$$

$${\boldsymbol{v(0)=0, v(L)=L}}$$

$${\boldsymbol{v'(0)=0, v'(L)=0}}$$

To obtain N6(x), use the boundary conditions:

$$\tilde{d}_{6}=rotationals$$

$${\boldsymbol{v(0)=v(L)=0}}$$

$${\boldsymbol{v'(0)=0, v'(L)=1}}$$

Derivation of coefficient in $$\tilde{K}$$ (element stiffness matrix)

Coefficient with EA: D

Coefficient with EI yields:

$$K_{22}=\frac{12EI}{L^{3}}=\int_{0}^{L}\frac{\partial ^2N_{2}}{\partial x^{2}}(EI)\frac{\partial ^2N_{2}}{\partial x^{2}}dx$$


text:
Introduction to social psychology/Lecture/Instructor notes

2005
(measure attitude first) (measure attitudes after)
 * create an 'include' (single file) for the first slide for each lecture - with contact details
 * About the right amount of content for 100 mins (with 30 min vid), plus 10 min break
 * Check for some duplicates slides
 * Cull some detail
 * Refine the amount of text on each slide to prevent excessive zooming
 * Includes 27 min VHS video
 * Use overheads?
 * Include exercises?
 * 3 attitudes - What's your attitude towards guns? george bush?  john howard?  (choose object (gun), person (john howard/george bush), and an event (boxing))
 * I am... (10 items / what # are group based)
 * 3 reasons why bin laden / john howard is good
 * Add a history slide
 * Social relatively recent development from nexus b/w psychology and sociology
 * sociology focuses on groups; psychology on individuals; social psychology on the dynamic relationship between individuals and others
 * Figure 1 can be redrawn in a non propietry form and include:
 * Political psychology
 * Environmental psychology
 * Developmental psychology
 * Cognitive psychology
 * Clinical psychology
 * Organisational / Industrial psychology
 * Marketing psychology
 * Consider/try finding an image for each of these themes and updating/changing the gallery so its more like a flickr slideshow
 * Content on 'self' at the end (last 5 mins) was a bit weak - check textbook more carefully and consider moving it to next week (bec. it can lead into relationships) - maybe include a self-test
 * Add diagrams showing circles and relations b/w groups and individuals (see Intro 3rd year SP lecture) - important to show/emphasize the kinds of relations
 * Circles
 * I (Individual)
 * G (Group
 * depict possible relations / interactions
 * Add 'unit of analysis' as a feature
 * Some references incomplete


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User:KTucker/My Sandbox/use cases

=Use Case Template=

Use Cases Template
Based on Alastair's work

Template:use case

title: User:KTucker/My Sandbox/use cases


text:
Social psychology (psychology)/Lectures/Instructor notes

2005

 * Could bring some of the video out of week 1, bust it up, and present each part in conjunction with its respective lecture presentation
 * Include more activities (~30 mins) - would be desirable still to have an interactive or take-away activity to accompany each of these important topics.
 * Each topic needs a take-away message / summary to be emphasized and summarised at the end of the two lectures.
 * Need to read the textbook chapters more thoroughly - only partially read / skimmed in 2005


text: Social psychology (psychology)/Tutorials/Tutor notes


text:
Social psychology (psychology)/Assessment/Exam/Instructor notes

 * An open-book, 3-hour exam, 150-item exam was reasonable successful
 * Average mark was 71%
 * In previous years the exam had been too hard (not enough time; 2 hours); added one hour (2 → 3 hours)
 * Reduced draft from 165 to 150 items
 * Redrafted and reselected the items in 2008.
 * Approximately 2 to 3 days was put in – quite a lot – to set this up even better for this year and the future.
 * There is still work to be done in this area
 * In 2009, prepare the items more continuously and systematically as part of tackling each topic.
 * Become more clear about whether the items are to be stored in Respondus or other format – currently Respondus files are the master, but this is somewhat inefficient in selecting items.
 * The 2007 idea was to create a large database with randomly selected items by topic area, using Respondus
 * In part, this is to allow the creation of random quizzes, etc. e.g., in Moodle
 * The 2008 idea is create a reasonably standard exam, and use the other items for a test bank which could be on Wikiversity and/or Moodle for each chapter
 * Moderated by: Melisah Feeney (see email comments)
 * Faculty review by: Laurie Grealish (see .doc comments)
 * Item length in many cases was reduced.
 * Item style was made more consistent e.g., with punctuation.
 * American spelling changed to Australian.
 * Manual formatting applied
 * Keep with next
 * Hanging indent for answers
 * Reviewed item response frequency analysis from 2007 and redrafted accordingly; some items were also further redrafted further away from the textbooks items.
 * Added items to cover extra Environmental and tutorial topics
 * Ch9 – Selection didn't match the Respondus allocation – re-allocate the Respondus allocation accordingly.
 * Ch10 – Attraction For items 104-110 no selection was made – adjust Respondus; instead selected 2 from 35-46
 * Ch11 – Selections didn't match Respondus; adjust Respondus
 * Ch12 – Selections didn't match Respondus; adjust Respondus
 * Process
 * Master databases by chapter/topic in Respondus
 * Planned number of items per topic (170)
 * Manually selected items per topic, based on Sets in Respondus
 * Manually merged and reordered items for exam draft
 * Removed 5 items to bring to 165 (didn't record which ones)
 * To do
 * This exam could/should be used as basis for future exams (build and improve on it)
 * Eventually further reduced to 150 items
 * Retrospectively create a 2008 Ch plan for the 150 items as the new Master. (see notes page in Ch plan)
 * Suggest re-using with relatively minor modification, based on:
 * Should get some up-front planning thought and on-going preparation during Semester
 * Check/concentrate more on lecture content (as I prepare each lecture).
 * Conduct item response analysis. Proof-reading could further reveal useful copy-edits.
 * Most efficient would probably be to revise the word processing version; making master adjustments in the Respondus database.
 * To find item numbers, text search the 165 item with codes draft from 2008 (could make a new one of these up).
 * Request to schedule this at a similar time as 2008 - preferably Fri am in the first week of exams in 6B4*. Alternatively it could be Wed or Thu am of exams.
 * Alternatively, this could be an ideal unit to deliver an online Moodle exam. However, to be fair, this would require either regular Moodle quizzed (by topic) or a practice exam.


text:
Social psychology (psychology)/Assessment/Essay/Instructor notes

 * Involved unique topic per person in order to reduce redundancy and to encourage individual, unique effort
 * This required some instructor expertise and experience to "pull-off"
 * Students had the option of putting the essay into the public domain, generating content which could be used to help build a social psychology wiki-book (Social psychology)
 * To do
 * Reduce penalty for over word count to 2.5% per 100 words
 * Clarify how words are counted e.g., one student noted that the website said that “references were not counted. I thought that this included in-text references.”


text:
Social psychology (psychology)/Review

E-portfolio

 * Transcluded from: E-portfolio instructor notes

Essay

 * Transcluded from: Essay instructor notes

Exam

 * Transcluded from: Exam instructor notes

Lectures

 * A good balance/combination was struck in 2008 (9 lectures + 1 review lecture)
 * Lectures were prepared in Open Document Presentation (.odp) format and hosted on http://slideshare.net.
 * It is perhaps still an issue that the Aggression lecture is only presented electronically (in favour of showing Ghosts of Rwanda. This documentary provides an excellent point of reference and many students comment on its impact. The Aggression lecture could instead be pre-recorded and made available electronically.
 * Environmental psychology was surprisingly popular, based on e-portfolio comments - recommend keeping.
 * See also: Instructor notes on each specific lecture

Tutorials

 * A good balance/combination was struck in 2008 (5 tutorials + 1 tutorial as an Assessment Workshop)
 * Students appreciated the novelty and intrinsic interest level of the topic matter
 * Attendance was moderate (attendance was strongly recommended but not compulsory) - the e-portfolio was designed to assess attendance
 * These tutorials could continue to be worked on as engaging, dynamic, experiential, but also substantial content-worthy, practical-skill tutorials.

E-reserve

 * Selected e-reserve readings could be better incorporated into lectures, tutorials, and assessment

Textbook

 * Baumeister & Bushman (2008) was generally popular and well-received
 * See also: B&B2008 textbook

Websites

 * Ongoing efforts were made towards making the unit materials available electronically using free and open educational resources.
 * The unit was taught using Wikiversity as an organising framework, and ucspace for uploading and embedding materials which couldn't easily be done with Wikiversity.
 * Students were required to create Wikiversity accounts in order to prepare e-portfolios

title: Social psychology (psychology)/Review


text:
Social psychology (psychology)/Assessment/E-portfolio/Instructor notes

 * An experiment; it 'worked' in general as a way for students to document their learning journey through the activities of the unit
 * It served as an appropriate and more engaging/dynamic alernative to "tutorial participation"
 * It tended (as desired) to help reward those students engaged throughout the unit
 * It involved extra support (e.g., via technical assistance and ongoing feedback) and final marking on the part of the instructor - and this required some technical expertise to "pull off" (i.e., successful mentor)
 * Additional handouts, demos (e.g., demo videos), online tutorials etc could be offered to minimise the technical barriers.


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Poetry/Practice/You Disturbed Me

I was in the corridor when I heard a thud… Minutes later I picked up a note …it read:

Why did you have to disturb me as I read my poems? Why did you just come into my room and disrupt my thoughts? I was seated on the bed – in the room – alone! Why did you come? I was reading silently – absorbing thoughts – seeing minds at work and you came with your mouth – humming and shouting. Why did you come?

I needed space. I wanted time to be alone. I didn’t want visitors. You came. Why?

You disturbed me. It was your fault. I wouldn’t have harmed you if you respected my privacy. I wouldn’t even have touched you! You went around touching things in the room – without permission! And you made a lot of noise.

I wasn’t even angry when I killed you. I was just a little irritated. And I didn’t hit you badly – just one simple blow. My aim was to silence you – but not silence you forever. Forgive me! I ask for forgiveness!

It was just a single blow then you cried… then you stopped…

Why did you have to die? Forgive me… but you... you are dead …you can’t even forgive me – I just hope guys will understand me:

I killed you because you disturbed me …you silly wasp!

title: Poetry/Practice/You Disturbed Me


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Wikiversity:Deletions/Working copy

Merged into Deletion policy. Adambro 13:09, 25 March 2010 (UTC)

title: Wikiversity:Deletions/Working copy


text:
Wikiversity talk:Deletions/Working copy

WV:RFD has become horribly bogged down because it's been impossible to reach consensus on the deletion of essentially blank pages. part of the goal for this working copy will be to establish a speedy deletion criteria for such pages so that each of these pages do not require a drawn-out vote (in many cases the deletion discussion takes far more effort and content than does the creation of the page. I think there should be a set of speedy deletion criteria much like what's on the deletion policy page, as well as delayed deletion for pages of this type. --SB_Johnny talk 13:22, 9 December 2008 (UTC)

No praise?
A question on the "no praise" criteria that is on the main copy page: if someone has genuinely helped you out a lot and you would like to thank them or make references to the help they've given you, what is the policy here? Would it be acceptable to say "a special thank you to" or something along those lines? Exactly what sort of praise is this referring to? Trinity507talk 17:56, 28 August 2009 (UTC)
 * Where do you find "no praise"? The Jade Knight (d'viser) 19:07, 28 August 2009 (UTC)

 * See Deletion_policy. "This includes pages with only... praises (e.g. "this guy is great!")..."  Trinity507, I think the keyword there is only.  That is, an empty page with nothing but "this guy is great!"  --mikeu talk 19:12, 28 August 2009 (UTC)
 * In other words, there's no problem with thanking or praising other users on talk pages. Go right ahead!  The Jade Knight (d'viser) 19:16, 28 August 2009 (UTC)
 * Oh, lol. Keep in mind we also had issues a couple years ago with a spambot that randomly created pages with complimentary messages. That might be where that came from. --SB_Johnny talk 19:26, 28 August 2009 (UTC)
 * Makes sense! Thanks for clearing that up - maybe we should specify that on the policy page. Trinity507 05:14, 29 August 2009 (UTC)

title: Wikiversity talk:Deletions/Working copy


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User:Eml4500.f08.ateam.carr/HW7

Electric Pylon as a frame
The electric pylon structure from before is now analyzed as a frame. The following procedures are to be completed:

 * 1) Update Matlab code to accommodate frame elements
 * 2) Plot the undeformed and deformed structure
 * 3) Identify the frame element(s) with the highest nodal bending moment and shear force
 * 4) Is the frame model statically determinate?
 * 5) Compare the reactions obtained from the truss model to those from the frame model
 * 6) Find the lowest 3 eigenpairs
 * 7) Plot the eigenvectors

Recall the properties of the pylon:

Young's Modulus: 200 GPa

Moment of inertia: 1.33 x 10-8m4

Cross-sectional area: 4 x 10-4m2

Additional Code Required
Length.m ReduceM.m ReduceS.m PlaneTrussElement.m NodalSoln.m PlaneTrussResults.m

PlaneFrameResults.m

2) Plot of Undeformed and Deformed Truss and Frame Structures

The red represents the Pylon as a frame approximation while the cyan is representative of the truss modeling.

Code output
Per the Matlab output:

max_moment = 6.2872 max_momentele = 81 max_shear = 5.4826 max_shearele = 81

The highest moment is 6.29 N-m and occurs in element 81. The highest shear stress (transverse) is 5.48Pa and also occurs in element 81.

4) Statically determinate?
This problem is not statically determinate. The two-force member assumption does not hold and therefore there are more unknowns than available equations.

6) Lowest Theee Eigenpairs
Per the Matlab output, the lowest three eigenpairs are:

lam = 132.86      2471.2       2942.6

7) Plot of the Three Corresponding Eigenvectors

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User:Eml4500.f08.team.figuerrez/hw7/meeting41

Meeting 41 - Derivation Elastodynamics (Combining FEA & Vibrations)
$$\tilde{K_{23}}=\frac{6EI}{L^{2}}=\int_{0}^{L}\frac{\partial ^2N_{2}}{\partial x^{2}}(EI)\frac{\partial ^2N_{3}}{\partial x^{2}}dx$$

In general, $$\tilde{K_{ij}}=\int_{0}^{L}\frac{\partial ^2N_{i}}{\partial x^{2}}(EI)\frac{\partial ^2N_{j}}{\partial x^{2}}dx$$, where $${\boldsymbol{i,j=2,3,5,6}}$$

Elastodynamics (trusses, frames, 2-d, 3-d elasticity)

From meeting 31: Model problem

Discrete PVW (Principle of Virtual Work):

$${\boldsymbol{w\left[M\ddot{d}+KJ-F\right]=0}}$$

Note: The boundary conditions are already applied to w, and K is the reduced stiffness matrix. The equation then becomes:

$${\boldsymbol{M\ddot{d}+Kd=F(t)}}$$

$${\boldsymbol{d(0)=d_{0}}}$$, where 0 is unforced

$${\boldsymbol{\dot{d}(0)=v_{0}}}$$

Complete ordinary differential equations (ODEs)

2nd order in time plus inital conditions governing the elastodynamics of discrete-fixed cont. problems (MDoF)

Solving Equation (1):

1.) Consider unforced vibration problem.

$$M_{nxn}\ddot{v}_{nx1}+K_{nxn}v_{nx1}=0_{nx1}$$

Now, assume that $${\boldsymbol{v(t)_{nx1}=(sinwt)_{nx1}\phi}}$$, where $$\phi$$ is not time dependent.

Then, taking the double derivative of the function yields:

$${\boldsymbol{\ddot{v}=-w^{2}sinwt\phi}}$$

$${\boldsymbol{-w^{2}sinwtM\phi+sinwtK\phi=0}}$$

$${\boldsymbol{K\phi=w^{2}M\phi}}$$

Generalized Eigen value problem:

$${\boldsymbol{Ax = \lambda Bx}}$$, where lambda is an eigenvalue.

standard eigenvalue problem: $${\boldsymbol{Ax = \lambda x}}$$

This means that:

$${\boldsymbol{\lambda = \omega ^{2}}}$$ is an eigenvalue and $${\boldsymbol{(\lambda _i,\mathbf{\phi _i})}}$$ is the eigenpairs with i = 1 through n.

The ith mode animation can be represented as:

$${\boldsymbol{v_i(t)=(sinw_it)\phi _i}}$$, where i = 1 through n

2.) Model superposition method (using orthogonal properties of the eigenpairs):

$$\phi _{i}^{T} M \phi = \delta _{ij} = 1$$, if $$i=j$$

$$\phi _{i}^{T} M \phi = \delta _{ij} = 0$$, if $$i\neq j$$

Where $${\boldsymbol{\delta _ij}}$$ is referred to as the Kronecker delta.

This reduction is possible due to the mass orthogonality of the eigenvector.

Now, applying this to Eq (1) yields:

$${\boldsymbol{M \phi _{j} = \lambda _{j}^{-1} \cdot K \phi _{j}}}$$

$${\boldsymbol{\phi _{i}^{T} M \phi _{j} = \lambda _{j}^{-1} \phi _{i}^{T} K \phi _{j}}}$$, where $${\boldsymbol{\phi _{i}^{T} M \phi _{j}}}$$ is equal to the Kronecker delta, therefore:

$${\boldsymbol{\phi _{i}^{T} K \phi _{j} = \lambda _{j} \delta _{ij}}}$$

$${\boldsymbol{d_{nx1}(t) = \sum{\zeta _{i}(t) \phi _{inx1}}}}$$

Equation (1) can then be written as:

$${\boldsymbol{M(\sum_{j}^{} \ddot{\zeta }_{j} \phi _{j}) + K(\sum_{j}^{} \zeta _{j} \phi _{j}) = F}}$$

$${\boldsymbol{\sum_{j}^{} \ddot{\zeta}_{j}(\phi _{i}^{T} M \phi _{j}) + \sum_{j}^{} \zeta _{j} \phi _{i}^{T} K \phi _{j} = \phi _{i}^{T} F}}$$

$${\boldsymbol{\ddot{\zeta } + \lambda _{i} \zeta _{i} = \phi _{i}^{T} F}}$$, with i = 1 through n.


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text: User:EAS4200C.F08.WIKI.A/My Contribution: 7


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User:Eml4500.f08.team.guan/hw7

Derivation of the Stiffness Term and Beam Shaping Functions
The motivation of the following derivation is to describe and model the deformed shape of a truss element by using the interpolation of the transverse displacement along with the axial displacement. To begin the derivation, let's recall the second order differential equation that describes the dynamics of an elastic bar. This equation was displayed in the 29.1 lecture. $$\frac{\partial}{\partial{x}}\left[A(x)E(x)\frac{\partial{u}} {\partial{x}}\right]+f(x,t)=m(x)\ddot{u}$$ Here, the term A(x)E(x) is the stiffness matrix, the $$\frac{\partial{u}} {\partial{x}}$$ corresponds to the displacement matrix, and f is the external force exerted on the bar. Applying the above equation to beams and using the principle of virtual work, we get the following equation. $$\int_0^L{w(x)\left[-\frac{\partial^2}{\partial{x^2}}(EI \frac{\partial^2{V}}{\partial{x^2}})+f_t-m\ddot{u}\right]}dx=0$$ for all possible w(x) ....... (1) Next, let's integrate by parts the 1st term. We will call this term alpha. $$\alpha=\int_0^L{w(x)\left[\frac{\partial^2}{\partial{x^2}}(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]dx}$$ where $$\displaystyle s(x)=w(x)$$ and $$r(x)=\frac{\partial}{\partial{x}}\left[(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]$$ and $$r^'(x)=\frac{\partial^2}{\partial{x^2}}\left[(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]$$ Integrating by parts the first time we will get the following. $$=\left[w\frac{\partial}{\partial{x}}((EI) \frac{\partial^2{V}}{\partial{x^2}})\right]\bigg|^L_0-\int_0^L{\frac{dw}{dx}\left[\frac{\partial}{\partial{x}}(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]dx}=\beta_1-\int_0^L{\frac{dw}{dx}\left[\frac{\partial}{\partial{x}}(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]dx}$$ Integrating again by integration by parts, we get the following. $$=\beta_1-\left[\frac{dw}{dx}(EI)\frac{\partial^2{V}}{\partial{x^2}}\right]\bigg|^L_0+ \int_0^L{\frac{d^2w}{dx^2}(EI) \frac{\partial^2{V}}{\partial{x^2}}dx}=\beta_1-\beta_2+ \gamma$$

Thus, in all simplified notation.  $$\displaystyle \alpha=\beta_1-\beta_2+ \gamma$$ Equation 1 above can be written as follows. $$-\beta_1+\beta_2-\gamma+\int_0^L{wf_tdx}-\int_0^L{wm\ddot{V}dx}=0$$ for all possible w(x) The boundary conditions will be evaluated later. Now, let's focus on the stiffness term gamma to derive the beam stiffness matrix and to identify the shape functions for beams. Consider the free body diagram of a beam below. The shaping functions at each node combined with the displacements of the nodes will allow for an equation of the transverse displacement to be written.

 The transverse displacement of the beam can be described by the following equation. $$V(\tilde{x})=N_2(\tilde{x})\tilde{d}_2+N_3(\tilde{x})\tilde{d}_3+N_5(\tilde{x})\tilde{d}_5+N_6(\tilde{x})\tilde{d}_6$$ where the shaping functions are the following. $$N_2(\tilde{x})=1-\frac{3\tilde{x^2}}{L^2}+\frac{2\tilde{x^3}}{L^3}$$ $$N_3(\tilde{x})=\tilde{x}-\frac{2\tilde{x^2}}{L}+\frac{\tilde{x^3}}{L^2}$$ $$N_5(\tilde{x})=\frac{3\tilde{x^2}}{L^2}-\frac{2\tilde{x^3}}{L^3}$$ $$N_6(\tilde{x})=-\frac{\tilde{x^2}}{L}+\frac{\tilde{x^3}}{L^2}$$ Graphically, the shaping functions are represented by the figure below.  Also, recall that the axial displacement described by its corresponding shaping functions and displacements is the following. $$\displaystyle u(\tilde{x})=N_1(\tilde{x})\tilde{d_1}+N_4(\tilde{x})\tilde{d_4}$$

title: User:Eml4500.f08.team.guan/hw7