User:Eml4500.f08.gravy.jad/Notes7.g

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

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

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

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

Model frame problem with two elements


Free Body diagrams for the elements



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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

2-D Frame global dofs;



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

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



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

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

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





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

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



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

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

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

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

$$\ \left \lbrack \epsilon \right \rbrack = \frac{\left \lbrack du \right \rbrack}{\left \lbrack dx \right \rbrack} = \frac{L}{L}= 1 \,$$

$$\ \left \lbrack \sigma \right \rbrack = \left \lbrack E \right \rbrack = \frac{F}{L^{2}} \,$$

$$\ \left \lbrack A \right \rbrack = L^{2}, \left \lbrack I \right \rbrack = L^{4} \,$$

$$\ \left \lbrack \frac{EA}{L} \right \rbrack = \left \lbrack \tilde{k}_{11} \right \rbrack = \frac {\left ( F/L^2 \right ) \left ( L^2 \right )}{L} = \frac{F}{L} \,$$

$$\ \left \lbrack \tilde{k}_{11} \tilde{d}_{1} \right \rbrack = \left \lbrack \tilde{k}_{11} \right \rbrack \left \lbrack \tilde{d}_{1} \right \rbrack = F \,$$

$$\ \left \lbrack \tilde{k}_{23} \tilde{d}_{3} \right \rbrack = \left \lbrack \tilde{k}_{23} \right \rbrack \left \lbrack \tilde{d}_{2} \right \rbrack = \frac {\left \lbrack 6 \right \rbrack \left \lbrack E \right \rbrack \left \lbrack I \right \rbrack}{\left \lbrack L^2 \right \rbrack} = \frac{1.(F/L^2).L^4}{L^2} = F \,$$

Verifying the dimensions of all the terms in the stiffness matrix multiplied by the displacement matrix
Elem FD rel. in golbal coordinate from element FD rel. in local coordinate:

ie. obtain

$$\ \mathbf{k}^e_{6x6} \mathbf{d}^e_{6x1} = \mathbf{f}^e_{6x1} \,$$ $$\ \mathbf{k}^e_{6x6} = \mathbf{\tilde{T}}^{e^{T}}_{6x6} \mathbf{\tilde{k}}^e_{6x6} \mathbf{\tilde{T}}^{e}_{6x6} \,$$ from $$\ \tilde{\mathbf{k}}^e_{6x6} \tilde{\mathbf{d}}^e_{6x1} = \tilde{\mathbf{f}}^e_{6x1} \,$$

Obtaining Element FD relation in global coord. From element FD relation in local coord.
$$\ \begin{Bmatrix} \tilde{d}_{1} \\ \tilde{d}_{2} \\ \underline{\tilde{d}}_{3} \\ \tilde{d}_{4} \\ \tilde{d}_{5} \\ \underline{\tilde{d}}_{6} \end{Bmatrix}_{6x1} = \underbrace {\begin{bmatrix} \mathbf{\underline{R}}& &0&0&0\\ & & 0&0&0&0\\ 0&0&\mathbf{1}&0&0&0\\ 0&0&0& \mathbf{\underline{R}}& &0\\ 0&0&0& & &0\\ 0&0&0&0&0&\mathbf{1} \end{bmatrix}_{6x6}}_{\tilde{\mathbf{T}}} \begin{Bmatrix} d_1\\d_2\\ \underline{d}_3\\d_4\\d_5\\ \underline{d}_6 \end{Bmatrix}_{6x1} \,$$ where $$\ \underline{d}_3 \,$$ and $$\ \underline{d}_6 \,$$ are rotational components, not displacement components

Two Bar Frame Element in Matlab
The two bar frame system is the same one analyzed in HW2 and HW5, with the same lengths, areas and elasticities for both elements, but for this case, the elements are considered to be frame elements with square cross sections. This requires adding the moment of inertia to the previous code and using PlaneFrameElement.m, taken from here, to build the global stiffness matrix.

Plotted deformed shape on top of previous figure


Derivation of $$\ \mathbf{\tilde{k}}^{(e)} \,$$ from PVW, focusing only on bending effect;

$$\ - \frac{\partial ^2}{\partial x^2} \left ( \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right ) + f_{t} \left ( x \right ) = m \left ( x \right ) \ddot{V} \,$$

$$\ + \frac{\partial ^2}{\partial x^2} \left ( \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right ) + f_{a} \left ( x,t \right ) = m \left ( x \right ) \ddot{u} \,$$

PVW for beams:

$$\ \int_{0}^{L} W\left ( x \right ) \left \lbrack - \frac{\partial^2}{\partial x^2} \left ( \left ( EI \right ) \frac{\partial^2 v}{\partial x} \right ) + f_t - m \ddot v \right \rbrack dx = 0 \,$$ for all possible W(x) (Eq 1)

Integrate by Parts of 1 st term:

$$\ \alpha = \int_{0}^{L} \underbrace{W \left ( x \right )}_{s \left ( x \right )} \frac{\partial^2}{\partial x^2} \left \lbrace \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right \rbrace dx \,$$ where $$\ \underbrace{\frac{\partial}{\partial x} \left ( \frac{\partial}{\partial x} \left \lbrace \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right \rbrace \right )}_{r'\left ( x \right )} \,$$

$$\ = \left \lbrack W \frac{\partial}{\partial x} \left \lbrace \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right \rbrace \right \rbrack _{0}^{L} - \int _{0}^{L} \underbrace{\frac{dW}{dx}}_{s' \left ( x \right )} \underbrace{ \frac{\partial}{\partial x} \left \lbrace \left ( EI \right ) \frac{\partial ^2 v}{\partial x^2} \right \rbrace}_{r \left ( x \right )} dx \,$$

$$\ = \beta_1 - \underbrace{\left \lbrack \frac{dW}{dx} \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right \rbrack_{0}^{L}}_{\beta_2} + \underbrace{\int_{0}^{L} \frac{d^2 W}{dx^2} \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} dx}_{\gamma} \,$$

Equation (1) becomes: $$\ \beta_1 - \beta_2 - \gamma + \int_{0}^{L} W f_t dx - \int_{0}^{L} W m \ddot{v} dx = 0 \,$$ for all possible W(x)

Now focusing on the stiffness term $$\ \gamma \,$$ for the time being to derive the beam stiffness matrix and to identify the beam shape functions:



$$\ v \left ( \tilde{x} \right ) = N_2 \left ( \tilde{x} \right ) \tilde{d}_2 + N_3 \left ( \tilde{x} \right ) \tilde{d}_3 + N_5 \left ( \tilde{x} \right ) \tilde{d}_5 + N_6 \left ( \tilde{x} \right ) \tilde{d}_6 \,$$ Equation 2

Recall: $$\ u \left ( \tilde{x} \right ) = N_1 \left ( \tilde{x} \right ) \tilde{d}_1 + N_4 \left ( \tilde{x} \right ) \tilde{d}_4 \,$$ Equation 4



In accordance with the book, the following equations represent the above picure:

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

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

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

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



As determined previously in an above section:

$$\ \tilde{\mathbf{d}}^{(e)}_{6x1} = \tilde{\mathbf{T}}^{(e)}_{6x6} \mathbf{d}^{(e)}_{6x1} \,$$ where $$\ \mathbf{d}^{(e)}_{6x1} \,$$ is known after solving the Finite Element Systems.

The only thing left is to compute $$\ u \left ( \tilde{x} \right ) \,$$ and $$\ v \left ( \tilde{x} \right ) \,$$



where $$\ \mathbf{u} \left ( \tilde{x} \right ) = u \left ( \tilde{x} \right ) \vec{\tilde{i}} + v \left ( \tilde{x} \right ) \vec{\tilde{j}} \,$$

Now we compute $$\ u \left ( \tilde{x} \right ) \,$$, $$\ v \left ( \tilde{x} \right ) \,$$ using Equations (1) and (2) as stated previously.

Computing $$\ u_x \left ( \tilde{x} \right ), u_y \left ( \tilde{x} \right ) \,$$ from $$\ u \left ( \tilde{x} \right ), v \left ( \tilde{x} \right ) \,$$

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

$$\ \begin{Bmatrix} u_x \left ( \tilde{x} \right ) \\ u_y \left ( \tilde{x} \right ) \end{Bmatrix} = \underbrace_{\mathbb{N} \left ( \tilde{x} \right )} \underbrace{\begin{Bmatrix} \tilde{d}_1^{(e)} \\ \vdots \\ \tilde{d}_6^{(e)} \end{Bmatrix}}_{\mathbf{\tilde{d}}^{(e)}} \,$$ Equation (2)

$$\ \begin{Bmatrix} u_x \left ( \tilde{x} \right ) \\ u_y \left ( \tilde{x} \right ) \end{Bmatrix} = \underbrace{\mathbf{R}^T}_{2x2} \mathbb{N} \underbrace{\left ( \tilde{x} \right )}_{2x6} \underbrace{\mathbf{\tilde{T}}^{(e)}}_{6x6} \underbrace{\mathbf{d}^{(e)}}_{6x1} \,$$ Equation (3)

Now lets note: how Equ (2), Dimensional Analysis

$$\ \lbrack u \rbrack = L \,$$, previously we also noted that $$\ \lbrack N_1 \rbrack = \lbrack N_4 \rbrack = 1 \,$$, and $$\ \underbrace{\lbrack N_1 \rbrack}_{=1} \underbrace{\lbrack \tilde{d}_1 \rbrack}_{= L} + \underbrace{\lbrack N_4 \rbrack}_{= 1} \underbrace{\lbrack \tilde{d}_4 \rbrack}_{= L} \,$$

also

$$\ \lbrack v \rbrack = L, \underbrace{\lbrack N_2 \rbrack}_{=1} \underbrace{\lbrack \tilde{d}_2 \rbrack}_{= L} = L \,$$(displacement, transverse)

The same can be said with the third component: $$\ \underbrace{\lbrack N_2 \rbrack}_{=1} \underbrace{\lbrack \tilde{d}_2 \rbrack}_{= L} = L \,$$

Recall: Governing PDE for beams

Equ (1), without ft (distributed transversal load) and without inertia force, $$\ m \ddot{v} \,$$ (static case = 0) :

$$\ \frac{\partial ^4}{\partial x^2} \begin{Bmatrix} \left ( EI \right ) \frac{\partial ^2 v}{\partial x^2} \end{Bmatrix} = 0 \,$$

Further consider constant EI: $$\ \frac{\partial ^4}{\partial x^4} v = 0 \,$$ integrate fout times and get C0, ..., C3.

Thus $$\ \Rightarrow v \left ( x \right ) = C_0 + C_1 x^1 + C_2 x^2 + C_3 x^3 \,$$

To obtain $$\ N_2 \left ( x \right ) \,$$ ( $$\ \tilde{x} \cong x \,$$ ) for simplicity

$$\ v \left ( 0 \right ) = 1, v \left ( L \right ) = 0, v' \left ( L \right ) = 0, v' \left ( L \right ) = 0 \,$$

These are the boundary conditions to use to solve for C0, ..., C3.

$$\ v \left ( 0 \right ) = 1 = C_0 \,$$

$$\ v \left ( L \right ) = 1 + C_1 \cdot L + C_2 \cdot L^2 + C_3 \cdot L^3 = 0 \,$$ Equation (1)

$$\ v' \left ( x \right ) = C_1 + 2 C_2 \cdot x + 3 C_3 \cdot L^2 = 0 \,$$

$$\ v' \left ( 0 \right ) = C_1 = 0 \,$$

$$\ v' \left ( L \right ) = 2 C_2 L^2 + 3 C_3 L^3 = 0 \,$$

$$\ (2) \Rightarrow C_3 = - \frac{2}{3} \frac{C_2}{L} \,$$

$$\ (1) \Rightarrow 1 + C_2 L^2 - \frac{2}{3} \frac{C_2}{L} (L^3) = 1 + \frac{1}{3} C_2 L^2 = 0 \,$$

$$\ C_2 = \frac{-3}{L^2} \,$$ and $$\ C_3 = \frac{-2}{3} \frac{1}{L} ( \frac{-3}{L^2} ) = \frac{2}{L^3} \,$$ Compare with expression for N2 previously.

For N3: Boundary Conditions

$$\ v \left ( 0 \right ) = 0, v \left ( L \right ) = 0, v' \left ( L \right ) = 1, v' \left ( L \right ) = 0 \,$$

For N5: Boundary Conditions

$$\ v \left ( 0 \right ) = 0, v \left ( L \right ) = 1, v' \left ( L \right ) = 1, v' \left ( L \right ) = 0 \,$$

For N6: Boundary Conditions

$$\ v \left ( 0 \right ) = 0, v \left ( L \right ) = 0, v' \left ( L \right ) = 1, v' \left ( L \right ) = 1 \,$$

(See previous plots for N5 and N6.)

Derive coefficients in $$\ \tilde{\mathbf{k}} \,$$ (element stiffness matrix) coefficients with EA: Done

Derive coefficients in $$\ \tilde{\mathbf{k}} \,$$ (element stiffness matrix) coefficients with EI: To be done

$$\ \tilde{k}_{22} = \frac{12EI}{L^3} = \int_{0}^{L} \frac{d^2 N_2}{dx^2} (EI) \frac{d^2 N_2}{dx^2} dx \,$$

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

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

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

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

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

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

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

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

Solving equation 1:

1) Consider unforced vibration problem

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

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

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

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

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

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

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

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

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

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

2)Model superposition method:

Orthogonal properties of eignepairs:

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

Mass orthogonal of eigenvectors

Using equation 2 and equation 3

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

$$

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

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

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

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

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

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

Frame Pylon
The following matlab code was used to solve the electric pylon. The functions: PlaneFrameElement, PlaneFrameResults,and NodalSoln were obtained from the Fundamental Finite Element Analysis and Applications book website. Here

The element number for the element with the highest tensile stress is represented by Tbn and equals 91. The highest tensile stress is represented by Ts and equals 33.763

The element number for the element with the highest compressive stress is represented by Cbn and equals 0 so all of the elements have the same compressive stress.

This problem is statically indeterminate. One reaction was given but there are too many reactions remaining to solve using the static equations.

Recommended software to improve productivity
We used the tool bar for Wikipedia and firefox powered by Greasemonkey to write our equations. We used the formula help page to find the comands for the different formula parts. This was easier than typing it in word because if the equation didn't parse when the page was previewed it normaly would give a hint as to the problem.

After Professor Vu-Quoc suggested using Inkscape we switched from using paint to using Inkscape. The images became easier to do in Inscape the more we used the program but it was hard at first considering we were all acustomed to using paint. Learning Inscape was worth the time becasue we found our images looked nicer than if they had been done in paint.

We recommend the Wikipedia and firefox powered by Greasemonkey toolbar and Inkscape program for completing work on Wikipedia.

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 20:06, 9 December 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 04:12, 30 November 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 16:57, 9 December 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 19:15, 9 December 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 19:40, 9 December 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 20:46, 9 December 2008 (UTC)