User:Eml4500.f08.ateam/hw 7

A-Team Teampage

Homework Report #7

This page is intended for Homework Report #7 of the A-Team.

Due date: Tuesday, 09 December 08, 5 pm EST = 22:00 UTC.

Two-Bar Frame System
In this section frame structure is introduced. Analysis of frame structures is very similar to analysis of truss structures but there are a few key differences outline below. The properties of each element are shown below.

Element 1: $$E_1^{(1)} = 2.0$$ $$ E_2^{(1)} = 4.0$$ $$A_1^{(1)} = 0.5$$ $$ A_2^{(1)} = 1.5$$

Element 2: $$E_1^{(2)} = 3.0$$ $$ E_2^{(2)} = 7.0$$ $$A_1^{(2)} = 1.0$$ $$ A_2^{(1)} = 3.0$$

Model Frame With Two Elements:

Element 1 is the on the left while Element 2 is on the right.



In the previous picture, the frame has a rigid connection between the two elements. This means that when the frame is deformed, there is no change in the angle between the elements. This leads to bending moments at either end of each element. We will now consider the free body diagrams of each element of the frame:





It must be noted that in the diagrams, $$d_i^{(e)}$$ and $$f_i^{(e)}$$ are generalized forces:    $$d_3^{(e)}$$ and $$d_6^{(e)}$$ are the rotational degrees of freedom and $$f_3^{(e)}$$ and $$f_6^{(e)}$$ are the bending moments.

Now we can consider the global degrees of freedom for the frame.



With the two $$6x6$$ element stiffness matrices, we can develop the $$9x9$$ global stiffness matrix $$\underline{K}_{9x9} = A\underline{k}_{6x6}^{(e)}$$



The element force displacement equation in local coordinates is represented by:

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

Where the local element stiffness matrix is:

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

It is important to note that $$\tilde{f_3}^{(e)} = f_3^{(e)}$$ and $$\tilde{f_6}^{(e)} = f_6^{(e)}$$ because they are moments about the z axis and are unaffected by the differences in local and global coordinate systems.

Dimensional Analysis
When performing dimensional analysis, the following is the notation that will be used to represent the dimension of:

$$\left[\tilde{d_i}\right]$$ where i = 1,2,4,5

Where $$\left[\tilde{d_1}\right] = L$$ and $$\left[\tilde{d_3}\right] = 1 = \left[\tilde{d_6}\right]$$



From the previous figure, R can be found to help calculate the arc length, AB : $$\widehat{AB} = R\theta$$

Note that $$\theta$$ should be in radians for proper use in the following equation.

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

Now stress and strain can be found which will help verify the values of the stiffness matrix.

$$[\sigma] = [E][\epsilon]$$

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

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

$$[A] = L^2$$

$$[I] = L^4$$

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

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

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



Using the principle of virtual work and focusing only on the bending effect, the expression for the matrix $$\tilde{k}$$ becomes:

$$\frac{\partial^2 }{\partial x^2}[(EI)\frac{\partial^2v }{\partial x^2}] - f_t(x) = m(x)\ddot{v}$$

Principle of Virtual Work for Beams
Motivation: The deformed shape of the tress element and the interpolation of the transverse displacements v(s), where $$s = \tilde{x}$$.

The PVW for beams can be defined as:

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

Begin the integration by parts for the first term:

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

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

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

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

where $$s'(x) = \frac{\partial w}{\partial x}$$

and $$\beta_1 = \left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right]^L_0$$.

After integrating by parts again, $$\alpha$$ becomes:

$$ = \beta_1 - \left[\frac{dw}{dx}\left(EI\right)\frac{\partial ^2v}{\partial x^2}\right]^L_0 + \int_{0}^{L}{\frac{\partial^2w}{\partial x^2}\left(EI\right)\frac{\partial ^2v}{\partial x^2}}dx$$

where $$\beta_2 = \left[\frac{dw}{dx}(EI)\frac{\partial ^2v}{\partial x^2}\right]^L_0$$

and the last integral can be replaced with the symbol $$\gamma$$. Note that this integral is symmetric. Equation (1) can now be rewritten and shortened to:

$$ - \beta_1 + \beta_2 - \gamma + \int_{0}^{L}{wf_tdx} - \int_{0}^{L}{wm\ddot{v}dx} = 0$$ for all possible w(x).

Stiffness Term $$\gamma$$
The focus is now on the stiffness term $$\gamma$$ to derive the beam stiffness matrix and to identify the beam shape function.



For this beam, the equation for $$v(\tilde{x})$$ can be seen below.

$$ v(\tilde{x}) = N_2(\tilde{x})\tilde{d}_2 + N_3(\tilde{x})\tilde{d}_3 + N_5(\tilde{x})\tilde{d}_5 + N_6(\tilde{x})\tilde{d}_6$$

Recall that $$u(\tilde{x}) = N_1(\tilde{x})\tilde{d}_1 + N_4(\tilde{x})\tilde{d}_4$$

$$N_2(\tilde{x}) = 1 - \frac{3\tilde{x}^2}{L^2} + \frac{2\tilde{x}^3}{L^3}\,\,\,\,\,\,\,\tilde{d}_2 $$

$$N_3(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\,\,\,\,\,\,\,\tilde{d}_3 $$

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

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

Beam Shape Functions
The beam shape functions for the six forces are as follows:







Note the similarities between the corresponding horizontal and vertical forces as well as the moments. The vertical sides of the first four shapes have a magnitude of 1. The initial slope of mode 3 and the final slope of mode 6 both have a magnitude of 1. The final displacement vector can be written in the following form.

$$v(\tilde{x}) = N_2(\tilde{x})\hat{d}_2 + N_3(\tilde{x})\hat{d}_3 + N_5(\tilde{x})\hat{d}_5 + N_6(\tilde{x})\hat{d}_6$$

$$u(\tilde{x}) = N_2(\tilde{x})\hat{d}_1 + N_2(\tilde{x})\hat{d}_4 $$

In order to compute these components, the relationship developed in a previous lecture will be used.

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

This relationship is used to reorient the axial coordinate system to the global coordinate system.

Computing the Displacement Vector
The displacement vector can be written in terms of its components, both in the axial and global coordinate systems.

$$\mathbf{u}(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+v(\tilde{x})\vec{\tilde{j}}\quad (axial\ coordinates)$$

$$\mathbf{u}(\tilde{x})=u_x(\tilde{x})\vec{i}+v_y(\tilde{x})\vec{j}\quad (global\ coordinates) $$

In order to find $$u_x(\tilde{x})\ and\ v_y(\tilde{x})$$ a transformation matrix $$R^T$$ will be used.

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

Now the equation for $$\begin{Bmatrix} u(\tilde{x}) \\ v(\tilde{x})  \end{Bmatrix}$$ must be found. It is related to the displacements and the equation derived from the mode shapes.

$$\begin{Bmatrix} u(\tilde{x})\\v(\tilde{x})\end{Bmatrix} = \begin{bmatrix} N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix} \begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)}\\\tilde{d}_3^{(e)}\\\tilde{d}_4^{(e)}\\\tilde{d}_5^{(e)}\\\tilde{d}_6^{(e)}  \end{Bmatrix} $$

Using these equations, they can be rewritten to give the final form of the global displacement vector.

$$\begin{Bmatrix} u_x(\tilde{x})\\v_y(\tilde{x})\end{Bmatrix} = R^T \begin{bmatrix} N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix} \tilde{T}^{(e)} d^{(e)}$$

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:

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

Discrete Principal of Virtual Work
$$\bar{w}\cdot[\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}-\bar{F}]=0 \quad\ for\ all\ \bar{w}$$

The bar notation denotes that the boundary conditions have already been applied. The ordinary differential equations (second order in time) and the initial conditions governing the elastodynamics of the discrete continuity problem are as follows.

$$\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}=\bar{F}(t)\quad\ (1)$$

$$\bar{d}(0)=\bar{d}_0\quad \quad (initial\ position)$$

$$\dot{\bar{d}}(o)=\bar{v}_0 \quad \quad (initial\ velocity)$$

Eigenvalue Solution
1) Consider the unforced vibration problem:

$$\bar{M}\ddot{v}+ \bar{k} v = 0 \quad \ (2)$$

Assume $$v(t)=(sin {wt})\mathbf{\phi}$$. Here $$\phi$$ is a column matrix of size (nx1) that is not time dependent.

$$\ddot{v}=-w^2sin {wt}\phi$$

Plug the previous equation back into equation (1):

$$-w^2 sin{wt} \bar{M}\phi + sin {wt} \bar{k} \phi = 0 $$

This can be simplified to the generalized eigenvalue problem:

$$\bar{k}\phi = w^2 \bar{M} \phi$$

An even more generalized version can be written in terms of matrices A and B.

$$\mathbf{A} x=\lambda \mathbf{B} x$$

Once evaluated, the following relations are made.

$$\mathbf{\lambda} = w^2$$

$$Mode\ i=\begin{cases} V_i(t)=(sin {w_it})\phi_i \\ i=1,...,n\end{cases}$$

2) Model superposition method:

Orthogonal property of eigenvalues is shown in the following equation:

$$\phi_{i} \bar{M} \phi_{j} = \delta _{i j} = \begin{cases} 1, & \mbox{if }\ {i=j} \\ 0, & \mbox{if }\ {i \ne j} \end{cases}$$

Here $$ \mathbf{\delta}_{ij}$$ is the Kronecker delta which has a variety of uses including digital processing.

$$\bar{M}\phi_j=\lambda^{-1}_j\phi^T_i\bar{k}\phi_j$$

$$\phi_i^T\bar{M}\phi_j=\delta_{ij}\lambda^{-1}_j\phi^T_i\bar{k}\phi_j$$

$$\phi^T_i\bar{k}\phi_j=\lambda_j \delta_{ij}$$

Next look at:

$$\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}=\bar{F}\quad \ (3)$$

$$\bar{d}(0)=\bar{d}_0$$

$$\dot{\bar{d}}(0)=\bar{v}_0$$

then

$$\bar{d}(t)=\sum_{i=1}^n \xi_i(t)\phi_i $$

Now plug the previous equation into equation (3):

$$\bar{M}(\sum_j \xi_j\phi_j)+\bar{k}(\sum_j\xi_j\phi_j)=\bar{F}$$

where

$$\ddot{\bar{d}}=\sum_j \xi_j\mathbf{\phi_j}\quad and\ \bar{d}=sum_j\xi_j\mathbf{\phi_j}$$

Finally,

$$\sum_j \ddot{\xi}_j(\mathbf{\phi_i^T}\mathbf{\bar{M}}\mathbf{\phi_j})+ \sum_j \ddot{\xi}_j(\mathbf{\phi_i^T}\mathbf{\bar{k}}\mathbf{\phi_j})=\mathbf{\phi^T_i} \mathbf{F}$$

$$\ddot{\xi}_i+\lambda_i\xi_i=\phi^T_i\mathbf{F}$$

Team View of Mediawiki
Mediawiki is an interesting medium for our class to use to complete our homework. Its learning curve is hard at first but once the proper syntax is learned, very professional work can be created. Mediawiki has one main advantage, as well as disadvantage, in the fact that anybody can access it. A password is not needed to access the information, unlike in the e-Learning system where only members are able to view it. This is a disadvantage because now anyone can edit the information to something that could be false. The math function in particular can create very precise and systematic functions that are easy to understand. Since all members of the group are using the same editors, a cohesive presentation can be achieved throughout each homework assignment. Since there are multiple groups working on each assignment, once turned in, each student has access to many different views and approaches to each problem.

With the new method of teaching with MediaWiki, there is a dissenting opinion about using it for coursework. In using the MediaWiki software, there are several issues that make it not appealing for classes. The first reason is the difficulty and time it takes to create the Wiki entries. The Wiki pages contain figures and equations, each of which must be entered using a special Wiki format. This takes an enormous amount of time to do when first exposed to it. Even after mastering the MediaWiki software, it still takes time to enter all the equations. In doing so, the focus of the course shifts from learning the Finite Element Method to creating a nice-looking Wiki homework report. FEM is an extremely powerful method and we, as students at UF, have a unique opportunity to take a class entirely devoted to it. Not many universities require this as an undergraduate course. Unfortunately, the implementation of Wiki homework reports greatly hampers the learning experience for the course.

The time spent formatting the Wiki pages takes away from the time that could be spent learning and working practice problems. For students that are taking many classes, time spent making wiki pages is time that could be better used on coursework.

The final reason the MediaWiki software is not acceptable for coursework is the concern for cheating. Because the entries are stored on the public domain, it is accessible to anyone, including students in the same class. This allows for students to copy other students work and submit it as their own.

Comparison of MediaWiki and E-Learning
MediaWiki has many advantages over E-Learning (ELS). MediaWiki allows for collaboration to produce professional-looking reports and projects. There is currently no way in ELS to collaborate in a group project. The editing ability of MediaWiki is far greater than that of ELS. Pages can be organized with many parameters and many types of media can be uploaded. A table of contents can be generated and linked to sections within a page.

MediaWiki also has the ability to store a greater amount of information for an indefinite time. Anything posted using MediaWiki is permanent and can always be accessed unless deletion is requested. An ELS account can be closed by the professor when the course has been complete. Additionally, the resources of MediaWiki are superior to those of ELS. During peak usage times, ELS is rendered useless while Wikiversity has the capability to handle a large server load.

The advantage ELS has over MediaWiki is privacy. MediaWiki is public domain, and anything posted or uploaded can be seen and sometimes used by others. This creates a large opportunity for plagiarism and cheating. In addition, grades and evaluations can be posted on ELS so that only the student can view them. MediaWiki does not have this capability.

Suggested Software to Improve Productivity
Most of the aspects of each homework assignment are similar and flow throughout the assignment. However, it was noticed that the programs that each student used for creating images were not the same. Therefore, some images were more plain that the others. In order to create a more uniform look, each group should choose an image software program at the beginning of the class and stick with it throughout the course.

Besides the Paint program that comes on every Windows PC, there are several programs available online that have enhanced features. Some such as Paint.NET and e-Paint are available for free and offers more advanced tools that the basic paint program. Another program that can easily be found online is GIMP (GNU Image Manipulation Program) which is more designed to modify images. Inkscape is a vector based graphics editor which can create Scalable Vector Graphics (SVG) file format. This is different that the usual scalar approach and allows the size of images to be easily changed with little loss of quality.

Other common programs useful for creating neat looking images include those in the Microsoft Office Suite. If one has the latest version, MS Word is excellent for the easy production of images. If the 2003 version must be used, MS PowerPoint is also good for generating clean, higher resolution images.

Anyone of these programs would be effective at creating the images that were needed for this class. In future courses, if a group were to decide on one, it would increase the overall continuity between individual contributions.

Additionally, if it is decided that the course must have an online component, it may be advisable to explore Google Documents. While Google Docs does not have some of the editing capability of MediaWiki, it may be a more viable alternative. The learning curve is not nearly as high as that of WikiMedia. Users can assemble documents, spreadsheets, presentations, and forms to be shared with others and even published. These forms are similar to MS Suite applications which are familiar to almost everyone. Like Wikiversity, Google Docs also allows collaboration for projects. An advantage of Google Docs in this area is it has the capability for multiple users to edit the same file at the same exact moment. Google's software saves the file almost constantly, storing edits as users make them.

Much like Wikiversity, each revision of a file is saved and the file can be reverted to a selected revision at anytime. Google Docs also displays the revision history in a more informative script, displaying exactly what changes were made, and their author. A document behaves much like MS Word in that images can be added and a table of contents can be generated. Users could still use the Latex Equation Editor and paste in the images of equations instead of the code. Also, tables can be generated within a document or spreadsheet.

A great advantage of Google Documents is privacy. A team member can create a file and choose to share it with members of the group, the TA's, and the professor. No one else can access or see the team's work. This would prevent a large amount of copied work. There would also be no need to link the homework report to a homework table before a deadline. The grader simply has to access the file that has been shared with them.

Using this method simply requires the creation of a Google Account much like creating a Wikiversity account. Creating a Wikiversity account had no other use outside classwork as it was strictly recommended not to edit other articles using the Wiki account. A Google account can have a Gmail account and calendar in addition to other benefits. The Gmail account can be linked to access UF Webmail - like WebCT, a somewhat outdated interface - and provide the student an email address upon graduation when their  account is closed. Google's services have proven to be highly useful and reliable and would possibly be of benefit to this course.

MATLAB: Two-Bar Truss System - Varying Cross-sectional Areas
The set-up and data for this two-bar truss system are seen below.



In this problem, two different methods of assembling the element stiffness matrices were explored. In one method, the varying cross-sectional areas are accounted for using the following relation general stiffness matrix formula.

$$\hat{k}^{(i)}_{gen}=\frac{1}{6\cdot L}\left(2E_{1}A_{1}+(E_{1}A_{2} + E_{2}A_1)+2E_2A_2\right)\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} $$

Additionally, the areas were averaged and assembled into the average stiffness matrix. This relation is as follows.

$$\hat{k}^{(i)}_{avg}=\frac{(E_1+E_2)\cdot (A_1 +A_2)}{4L}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

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

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

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

Plot of Deformed Shape and Deformed Shape of Two-Bar Truss System Using Average Areas
This plot depicts the deformation of the two-bar truss system using the average stiffness matrix in red and the deformation of the system using the general stiffness matrix in green. The difference is clearly shown in the plot and the table below.



MATLAB: Two-Bar Frame System
This two-bar frame system is very similar to the one analyzed in | Homework Report #2 and further in | Homework Report #5. It uses the same data as before but, for this case, Element 1 is changed to a frame element, creating a frame system. The cross-section for this system is assumed to be square. The applied load, P, has a value of 7.



This functions generates the element stiffness matrices an the element R matrix. It acts similarly to the  seen in previous problems.

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

Plot of Undeformed and Deformed Shapes
This plot depicts the undeformed and deformed shapes of the two-bar frame system. Additionally, the deformed shape of the two-bar truss system is plotted in green for comparison. This deformation is scaled by a factor of 0.5 for viewing purposes.

From the plot, the difference in frame elements and truss elements is apparent. The frame element was divided into 20 segments to show its complete deformation as a curve. The fixed boundary conditions cause an additional degree of freedom at each node, resulting in rotation or bending of the bar.



MATLAB: Electric Pylon Frame System
The structure in this problem is a electric-line pylon consisting of 91 elements. The MATLAB code for accurately assembling this system was obtained from Dr. Loc Vu-Quoc and can be found here or seen below.

The objective of this problem is to expand upon analysis performed in the previous homework report, | Homework Report #6. For this problem, the electric pylon truss system was converted to a frame system with each element having a square cross-section. The applied load P is applied vertically downward at the far right tip of the arm of the pylon. Clamped boundary conditions are applied at the base of the frame model, replacing the simply supported conditions of the previous problem. All elements are constructed of 300M steel and have identical cross-sectional areas. Properties for this material were given by Dr. Vu-Quoc and verified online here. They are displayed in the table below.



Analysis Requirements

 * Convert the electric pylon truss system to a frame system
 * Redo the static analysis using the same geometric, material and loading data as for Homework Report #6
 * Compare the nodal displacements of the node under the applied force for the frame model to those of the truss model
 * Identify the frame element or elements with the highest nodal bending moment and transverse shear force
 * Plot the undeformed shape, deformed shape of the truss model and "incomplete" deformed shape of the frame model using the same magnification factor
 * Determine if the problem is statically determinant and provide justification through argument or MATLAB programming
 * Compare the reactions obtained from the truss model to those from the frame model
 * Construct the lumped mass matrix of the electric pylon
 * Solve the generalized eigenvalue problem
 * Find the lowest three eigenpairs and, in 3 separate plots, plot the eigenvectors as deformed shapes superposed onto the undeformed shape
 * Find the three lowest vibrational periods of the electric pylon
 * Generate an animation of the three lowest vibrational periods of the electric pylon

MATLAB Analysis Code
The code below is the  file written to solve the problem as assigned. This program requires additional MATLAB functions to run. These functions are also detailed below.

This functions generates the element stiffness matrices an the element R matrix. It acts similarly to the PlaneTrussElement.m seen in previous problems.

This function returns three matrices consisting of the Axial Force Matrix,, the Bending Moment Matrix,  , and the Shear Force Matrix,. Each of these matrices lists the X and Y coordinates and the respective value for each element.

This function returns the length of an element based on the element coordinates it receives. It was independently written and cannot be found with the preceding functions.

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

This function was written to output the reduced matrices, $$\bar{M}$$ and $$\bar{K}$$. It was independently written and cannot be found with the preceding functions.

Running Analysis Code
The code below is the complete results from running the MATLAB program. Each part is detailed in later sections of this homework report.

Plot of Deformed and Undeformed Shapes
The extended MATLAB code produces the following plot of the deformation of the truss system. The frame system is detailed in red while the truss system is shown in green. Since the systems overlap, it is difficult to view. The red frame system is plotted on top of the green truss system. To scale the electric pylon truss to be 60m tall, the nodal positions were multiplied by a factor of 8.0753701211306. It is important to note that the deformation has been magnified by a factor of 20 for viewing purposes.



Statically Determinant?
The electric pylon structure is not statically determinate. While two nodes of the system are fixed, these two nodes act as the termination point of four elements. Thus, there are a total of four unknown reaction forces in a system with only three equations with which to attempt to solve them. This cannot be solved using the principles of statics.

It is important to note that, while similar in number of fixed nodes to previous systems analyzed, this system differs in that the fixed nodes contain more than one element.

Comparison of Reactions
The reactions for the truss system and the frame system are almost identical. They are displayed in the following table.

Comparison of Nodal Displacements of Truss and Frame Systems
The nodal displacements of the frame and truss systems vary by minuscule amounts. The following picture magnifies the displacements for viewing. To better understand the scale, the distance between the two nodes is approximately one ten-thousandth of a meter.



The exact displacements are presented in the table below. The values for the truss and frame systems are almost identical.

Highest Nodal Bending Moment
As shown on the plot and results from running, the greatest nodal bending moment occurs in Element 81. The value of the moment is 6.2874 N-m.

Highest Transverse Shear Force
As shown on the plot and results from running, the highest transverse shear force occurs in Element 81. The value of the force is -5.4827 N.

Lumped Mass Matrix
The lumped mass matrix was assembled as a diagonal matrix of the masses of both degrees of freedom for each node. To determine the masses, each elements length was multiplied by its cross-sectional area and its density. Once the global nodes of each end of the element were known, the half of the element's mass was assigned to both nodes. Continuing this for each element created the lumped mass matrix.

Eigenpair Analysis
The three smallest eigenvalues for the truss system were determined using the generalized eigenvalue problem shown below. In this relation $$\bar{K}$$ is the reduced stiffness matrix and $$\bar{M}$$ is the reduced lumped mass matrix.

$$\bar{K}\nu =\lambda \bar{M}\nu$$

By plotting the columns of the eigenvector matrix that correspond to the smallest eigenvalues, modes of the truss can be depicted. In this case, the smallest eigenvalues were 132.86, 2471.2 and 2942.6. Therefore, the columns of the eigenvector matrix, V, that correspond to these values are plotted similarly to plotting displacements. These values are simply added to the original nodal positions to give the mode shape.

The following code was appended to the original truss problem MATLAB file to plot the eigenvectors. It is important to note that the eigenvectors were multiplied by a factor of 4 for viewing purposes.

Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the smallest eigenvalue.

Second Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the second smallest eigenvalue.

Third Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the third smallest eigenvalue.

Vibrational Period Analysis
To find the three lowest vibration periods of the electric pylon, the eigenvalues are set equal to the angular frequencies as shown in the equation below.

$$\lambda = \omega^2$$

The angular frequencies are converted to vibrational frequencies as follows.

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

From the vibrational frequencies, the vibrational periods can be determined using the following equation.

$$T = \frac{1}{f} = \frac{2\cdot \pi}{\omega}$$

Following these relations reveals the three lowest vibrational periods, shown in the table below. In comparison to the values obtained from the electric pylon truss system, the vibrational periods are similar but shorter. This is due to the added rigidity of the frame system. The values do not seem to vary much at all but it is important to remember that for both the frame and truss systems, the actual deformation was extremely small, so this difference in vibrational period is significant.

Eigenmode Analysis of the Electric Pylon Truss System
movies were created by adding the following MATLAB code segment to the code from Homework Report #6. These files were converted to  format for compatibility with Wikimedia.

Creating Videos for Upload to Wikimedia Commons
A more detailed explanation of how to generate and convert video files for upload to Wikimedia Commons can be found here or seen in the collapsible table below.

Animation of Lowest Eigenmode
This animation represents the lowest eigenmode of the electric pylon system.



Animation of Second Lowest Eigenmode
This animation represents the second lowest eigenmode of the electric pylon system.



Animation of Third Lowest Eigenmode
This animation represents the third lowest eigenmode of the electric pylon system.



Contributing Team Members
The following students contributed to this report:

Mark Barry Eml4500.f08.Ateam.barry 03:46, 9 December 2008 (UTC)

Daniel DiDomenico Eml4500.f08.ateam.didomenico 14:31, 25 November 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 18:31, 9 December 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 01:35, 9 December 2008 (UTC)

Matthew Philbin Eml4500.f08.Ateam.philbin 12:30, 07 December 2008 (UTC)

Ian Slevinski Eml4500.f08.Ateam.Slevinski 05:34, 09 December 2008 (UTC)