User:Eml4500.f08.bike.gravois

Homework 1
Command line editing and recall

The MATLAB command line can be edited in several ways. The cursor is moved with the left and right arrows on the keyboard, and the backspace key deletes the character to the left of the cursor. Characters can also be deleted using the Home, End or Delete keys. The up and down arrows, on the keyboard, are used to scroll through previous commands.

A command line can be changed and rerun by scrolling up to, editing, and then executing the previous command. The plots of y=sin(mx) and y=sin(nx) from [0 2π] can be compared easily by editing the (mx) to (nx) in the previous command line:

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Submatrices and colon notation

Colon notation is used in MATLAB to generate vectors and reference submatrices. Colon notation also reduces the use of loops and simplifies codes.

The expression 1:4 is a row vector [1 2 3 4]. Neither the vector numbers nor the increment have to be integers. The example,

yields the vector

To cont down, a negative sign is used in the vector notation.

yields

Colon notation is also used to recall submatrices within a matrix. An example

is a column vector made up of the first three values in the second column of the matrix A.

A lone colon calls an entire row or column:

is the entire second column of the A matrix. Vectors can be used within colon notation to recall multiple columns.

recalls the entire columns two and five of the matrix A.

Columns can be replaced in one matrix from another using colon notation for Example,

replaces columns two, three, and five in matrix A with columns two, three, and four from matrix B.

M-files: script files, function files

MATLAB can run a sequence of statements that are stored in disk files, called “M-files.” M-files end with “.m” in the filename and are usually created with a local editor. The two types of M-files are script and function files.

Script files

Script files contain normal sequences of MATLAB statements. The file  will be run in MATLAB by imputing the command. Variables in the M-file are global and will replace the values of same named variables in the MATLAB environment.

Script files also allow for large matrices to be entered and corrected easily. A matrix can be input, for example, by entering diskfile data.m

A = [ 1 2 3 4 5 6 ];

Running the MATLAB command data will cause the matrix A to be created and stored in the MATLAB environment. An M-file can call on other M-files, or it can run itself repeatedly.

Function files

Function files allow for the creation of new functions, that do not exist in MATLAB already, needed for a specific problem that. Function variables are set as local variables by default.

An example of a function file:

function a = (sin(m))^(-1) %INVSIN computes inverse sin. % invsin(m) returns inverse sin
 * A = 10*invsin(m);

This example should be in a diskfile named  due to the function name. The first line declares the function name, input arguments, and output arguments. Without the first line the function file would be a script file. A MATLAB command  will pass the value 45 into the variables m in the function file, and the output will be put into variable A.

The  symbol is used for note making in an M-file. The symbol tells MATLAB to skip the rest of the line. Comments should always be used in M-files to describe the function.

Loops can be made to run faster by preallocating output vectors or matrices. An example of this can be seen in the second line of the for loop where the matrix E is made one column larger for each iteration.

F = fun(6); D = zeros(6,50); For j = 1:50
 * D(:,j) = det(F^i);
 * End

Most of the MATLAB functions are built into the software, but some are M-files. To see a listing of any non built in M-file the MATLAB command type must be run.

Week 2 Class Notes
Element Labeling

The class begins with an introduction to the nomenclature of element labeling. The figure is first drawn with all members, forces, and constraints in a global view. Each constraint is then labeled globally by an encircled number. The structure is typically numbered from left to right. Each element of the member is then labeled with a number enclosed in a triangle. The constraints are then removed and replaced with force arrows labeled as Rnx or Rny in the x and y axes respectively. The subscript n denotes the constraint node number of the global member.

Once all of the members, nodes, and reaction forces of the global member are fully labeled, a free body diagram for each member of the structure must then be drawn. The member labeled with its global member number within a triangle. The nodes are then labeled with a local node number inscribed within a square. The local reaction forces are then drawn in the x and y axes at each node. These internal forces are then labeled as fi(e) where the subscript i represents the degree of freedom (dof) of that node and the superscript e represents the element number. The displacement of each node is drawn just like the degree of freedom and labeled as di(e) with i representing the degree of freedom of that node and e represents the element number. The labeling nomenclature is illustrated below in a global and local view of a truss system.

[image]

Force-Displacement Relation

A simple 1-D spring element fixed at one end is used to illustrate the force-displacement (FD) relation. This relation is f=kd, where f is the force applied to the spring, k is the spring constant, and d is the displacement of the spring from its natural position. The spring element can be seen below.

[image]

The relation also holds true for a spring with two free ends as seen below.

[image]

The FD relation determined by observing the spring in two ways. The first case is where the observer is sitting on node one, which appears to be stationary. The resulting equation is f2=k(d2-d1). The second case with the observer on node 2 yields the equation f1=k(d1-d2). The matrix of f=kd for the above spring is presented below.

[image]

Steps to Solve Simple Truss Systems

1.     Global Picture


 * Structure level


 * Global dofs


 * o      Known part (fixed dof, constraints)


 * o      Unknown part (solved using finite element method)


 * Global forces


 * o      Known part (applied forces)


 * o      Unknown part (reactions)

2.     Element Picture


 * Element dof


 * Element forces

3.     Global FD relations


 * Element stiffness matrices in global coordinates


 * Element force matrices in global coordinates


 * Assembly of element stiffness and force matrices into global FD relation

4.     Elimination of known dofs to reduce the global FD relations


 * Stiffness matrix non-singular→inveritble


 * To be invertible unknown displacement dofs mus be less than known dofs

5.     Compute element forces from known element stresses

6.     Compute reactions (unknown forces)

Homework 2
Element 1 :

The element stiffness matrix multiplied by the element displacement matrix is k (1) d (1)

$$

=\left[ \begin{array}{cccc} 0.5625 & 0.32476 & -0.5625 & -0.32476 \\ 0.32476 & 0.1875 & -0.32476 & -0.1875 \\ -0.5625 & -0.32476 & 0.5625 & 0.32476 \\ -0.32476 & -0.1875 & 0.32476 & 0.1875 \end{array} \right]\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.127  \end{Bmatrix}

$$

The matrices can be reduced due to the zeros in the displacement matrix and set equal to the element force matrix k (1) d (1) = f (1)

$$

=\left[ \begin{array}{cc} -0.5625 & -0.32476 \\ -0.32476 & -0.1875 \\ 0.5625 & 0.32476 \\ 0.32476 & 0.1875 \end{array} \right]\begin{Bmatrix} 4.352 \\ 6.127  \end{Bmatrix} = \left[ \begin{array}{c} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \end{array} \right]  =  \left[ \begin{array}{c} f_{1}^{(1)} \\ f_{2}^{(1)} \\ f_{3}^{(1)} \\ f_{4}^{(1)} \end{array} \right]

$$

Observation:

Element 1 is in equilibrium therefore:

(1) $$\sum F_x = f_1^{(1)} + f_3^{(1)} = 0 $$

(2) $$\sum F_y = f_2^{(1)} + f_4^{(1)} = 0 $$

(3) $$\sum M_{any pt} = 0 $$



The axial force P1 can be found using the root sum square method (RSS) below.

P1(1) = [(f1(1))2 + (f2(1))2]1/2

The RSS method can be used to verify that Element 1 is in equilibrium and the results from f (1)

-P1(1) = [(f1(1))2 + (f2(1))2]1/2 P2(1) = [(f3(1))2 + (f4(1))2]1/2

-P1(1) = [(-4.4378)2 + (-2.5622)2]1/2 = 5.124 P2(1) = [(4.4378)2 + (2.5622)2]1/2 = 5.124

P2(1) = -P1(1)

Element 2 :

The figure below shows element 2 labeled with axial forces P.



Method 2 to solve for a 2-bar truss system using statics method:



P can be brought back into the big picture by performing an equilibrium balance of node 2.



The equilibrium of global node 2 can be verified using trigonometry. P = 7 P1(1) = 5.1243 P2(2) = 6.2760

$$sum\ F_y = 7 - 5.1243sin(30) - 6.2760sin(45) = 0$$ $$sum\ F_x = 5.1243cos(30) - 6.2760cos(45) = 0 $$

Homework 3
Derivation of Element FD wrt Global Coordinate System

The element force displacement (FD) relation can be determined with respect to the global coordinate system. The following equation shows the general force displacement relation for an element member.

$$\mathbf{k}^{(e)}_{4x4}\,\mathbf{d}^{(e)}_{4x1} = \mathbf{f}^{(e)}_{4x1}$$

The local free body diagram for the element member is pictured in the figure below.



The local axial force displacement diagram for the element member is pictured in the figure below.



The following force displacement equation can is constructed from above axial force displacement diagram.

$$\mathbf{k}^{(e)}\begin{bmatrix}1 & -1\\-1 & 1\end{bmatrix}\begin{Bmatrix}q_1^{(e)}\\q_2^{(e)}\end{Bmatrix}=\begin{Bmatrix}p_1^{(e)}\\ p_2^{(e)}\end{Bmatrix}$$
 * $$ \mathbf{\hat{\; k}}^{(e)}_{2x2}\,\;\;\;\;\;\;\;\; \mathbf{q}^{(e)}_{2x1} \;\; = \;\;\;\; \mathbf{p}^{(e)}_{2x1}$$

Where:
 * qi(e) = axial displacement of element e at local node i.
 * pi(e) = axial force of element e at local node i.

Want to find the relation between $$\begin{cases} \mathbf{q}^{(e)}_{2x1} & \text{ and}\;\;\;\;\;   \mathbf{d}^{(e)}_{4x1}  \\ \mathbf{p}^{(e)}_{2x1} & \text{ and}\;\;\;\;\;  \mathbf{f}^{(e)}_{4x1} \end{cases}$$

The above relation can be expressed in the form below.

$$\mathbf{q}^{(e)}_{2x1} = \mathbf{T}^{(e)}_{2x4}\, \mathbf{d}^{(e)}_{4x1}$$

The figure below depicts the displacement vector of local node 1 as well as a portion of the element member e.



In the above figure, the displacements $$d^{(e)}_1$$ and $$d^{(e)}_2$$ in the $$\vec{i}\,$$ and $$\,\vec{j}$$ directions are shown as components of $$\mathbf{\hat{\; d}}\,$$1(e). The equation of the $$\vec{i}\,$$ and $$\,\vec{j}\,$$ components of $$\mathbf{\hat{\; d}}\,$$1(e) can be seen in the following equation.

$$\bar{d}^{(e)}_1=d^{(e)}_1\vec{i}+d^{(e)}_2\vec{j}$$

q1(e) = axial displacement of node 1 is the orthagonal projection of the displacement vector $$\bar{d}^{(e)}_1$$ of node 1 on the $$\tilde{x}$$ axis of element e. This is represented in the below equation.

$$q^{(e)}_1=\vec{d}^{(e)}_1\cdot \vec{\tilde{i}}$$

The displacement vector of node 1 can be broken down into components. $$q^{(e)}_1= (d^{(e)}_1\vec{i}+d^{(e)}_2\vec{j}) \cdot \vec{\tilde{i}}$$

Which is equivalent to. $$q^{(e)}_1= d^{(e)}_1(\vec{i}\cdot \vec{\tilde{i}})+d^{(e)}_2(\vec{j} \cdot \vec{\tilde{i}})$$
 * $$ \cos \theta _e \;\;\;\;\; \sin \theta _e$$
 * $$ l^{(e)} \;\;\;\;\;\;\;\;\;\; m^{(e)}$$

q1(e) is a 1x1 scalar which can be calculated with the displacements of node 1 and the director cosines as shown in the following equations.

$$q_1^{(e)}=l^{(e)}d_1^{(e)}+m^{(e)}d_2^{(e)}$$

$$q_1^{(e)}=\begin{bmatrix}l^{(e)} & m^{(e)} \end{bmatrix}\begin{Bmatrix}d_1^{(e)}\\ d_2^{(e)}\end{Bmatrix}$$ $$1x1 \;\;\;\;\;\;\;\;\; 1x2 \;\;\;\;\;\;\;\;\; 2x1$$

The matrices that result from combining the displacement equations for nodes 1 and 2 are listed below. $$\begin{Bmatrix}q_1^{(e)}\\q_2^{(e)}\end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\end{bmatrix}\begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\end{Bmatrix}$$ $$\mathbf{q}^{(e)}=\mathbf{T}^{(e)}\mathbf{d}^{(e)}$$

Homework 4
(10/17/2008)

Equation 1 from above can be rewritten as follows: $$f_3^{(1)}+f_1^{(2)}=0\; \; \;(1a)$$

Equation 2 from above can be rewritten as follows: $$f_4^{(1)}+f_2^{(2)}=0\; \; \;(2a)$$

For Equation 1a, the forces can be written in terms of element stiffness(k) and degrees of freedom(d).$$\mathbf{F=kd}$$: $$\left[k_{31}^{(1)}d_1^{(1)} +k_{32}^{(1)}d_2^{(1)}+k_{33}^{(1)}d_3^{(1)}+k_{34}^{(1)}d_4^{(1)}\right]$$ $$+\left[k_{11}^{(2)}d_1^{(2)} +k_{12}^{(2)}d_2^{(2)}+k_{13}^{(2)}d_3^{(2)}+k_{14}^{(2)}d_4^{(2)}\right]\; \; \;(1b)$$ $$=0$$

Local Degrees of Freedom to Global Degrees of Freedom Relation: $$d_1^{(1)}=d_1$$ $$d_2^{(1)}=d_2$$ $$d_3^{(1)}=d_2^{(2)}=d_3$$ $$d_4^{(1)}=d_1^{(2)}=d_4$$ $$d_3^{(2)}=d_5$$ $$d_4^{(2)}=d_6$$

The local degrees of freedom(dof) of Equation 1b can be replaced with the global degrees of freedom using the above relations.

$$\left[k_{31}^{(1)}d_1 +k_{32}^{(1)}d_2+k_{33}^{(1)}d_3+k_{34}^{(1)}d_4\right]$$ $$+\left[k_{11}^{(2)}d_3 +k_{12}^{(2)}d_4+k_{13}^{(2)}d_5+k_{14}^{(2)}d_6\right]$$ $$=0$$

Assembly of Global Stiffness Matrix from Element Stiffness Matrices

The global stiffness matrix $$\mathbf{k}$$ can be assembled from the element stiffness matrices $$\mathbf{k^{(e)}}, e=1,...,n_{el}$$ where nel is the number of elements:

$$\mathbf{k}_{nxn}=A\mathbf{k}_{n_{el}xn_{el}}^{(e)}$$

New Nomenclature: n: total number of global degrees of freedom before elimination of boundary conditions ned: number of element degrees of freedom
 * (ned << n)

A: assembly operation
 * Principal of Virtual Work (PVW)
 * Used to eliminate the rows for corresponding boundary conditions to obtain $$\mathbf{\bar{k}}_{2x2}$$

Deriving Finite Element Method for Partial Differential Equations

The force displacement (FD) relation for a bar or spring is $$kd=F$$. This implies that $$kd-f=0\; \; \;(3)$$ Which is equivalent to $$w(kd-F)=0\; \; \;(4)$$ for all values of w.

Proof: $$(3)\Rightarrow (4)$$ trivial $$(4)\Rightarrow (3)$$ not trivial

Since Equation 4 is valid for all w, select w=1, then Equation 4 becomes: $$1\left(kd-F \right)=0\Rightarrow (3)$$

Homework 5
$$\hat{\textbf{w}}_{2x1}$$ = virtual axial displacement corresponding to $$\textbf{q}^{(e)}$$. $$\textbf{w}_{4x1}$$ = virtual displacement in global coordinate system, corresponding to $$\textbf{d}^{(e)}$$.

Substituting Equations (3) and (4) into Equation (2) yields the following Equation: $$\left(\textbf{T}^{(e)}\textbf{W} \right)\cdot \left[\mathbf{\hat{k}}^{(e)}\left(\mathbf{T}^{(e)}\mathbf{\bar{d}}^{(e)}-\mathbf{p}^{(e)} \right) \right]=0$$ for all $$\textbf{w}_{4x1} \; \; \; \; \; \; \;(5)$$

In order to solve Equation (5), a few matrix operations must be recalled:

$$\left(\textbf{AB} \right)^T=\textbf{B}^T\textbf{A}^T \; \; \; \; \; \; \; (6)$$

$$\textbf{a}_{nx1}\cdot\mathbf{b}_{nx1} = \textbf{a}^t_{1xn}\mathbf{b}_{nx1} \; \; \; \; \; \; \; (6)$$

To solve Equation (5), Equation (6) and (7) must be applied:

The dot product must be replaced with the transpose as shown below from Equation (7). $$\left(\textbf{T}^{(e)}\textbf{W} \right)^T\left[\mathbf{\hat{k}}^{(e)}\left(\mathbf{T}^{(e)}\mathbf{\bar{d}}^{(e)}-\mathbf{p}^{(e)} \right) \right]=0$$ for all $$\textbf{w}_{4x1}$$

The matrix transpose is replaced with the product of individual transposes from Equation (6). $$\textbf{W}^{T}\textbf{T}^{(e)T}\left[\mathbf{\hat{k}}^{(e)}\left(\mathbf{T}^{(e)}\mathbf{\bar{d}}^{(e)}-\mathbf{p}^{(e)} \right) \right]=0$$ for all $$\textbf{w}_{4x1}$$

After distributing the transpose matrix T the following equation is the result. $$\textbf{W}^{T}\cdot\left[\left(\textbf{T}^{(e)T}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}\right)\mathbf{\bar{d}}^{(e)}-\left(\textbf{T}^{(e)T}\mathbf{p}^{(e)} \right) \right]=0$$ for all $$\textbf{w}_{4x1}$$

$$\mathbf{k}^{(e)}$$ replaced $$\textbf{T}^{(e)T}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$, and $$\mathbf{f}^{(e)}$$ replaced $$\textbf{T}^{(e)T}\mathbf{p}^{(e)}$$.

The resulting equation is: $$\mathbf{w}\cdot\left[\mathbf{k}^{(e)}\mathbf{d}^{(e)}-\mathbf{f}^{(e)}\right]=0$$ for all $$\mathbf{w}$$

After dividing both sides by w, and moving f(e) to the other side yielded: $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

The previous methods have all been for discrete, non-continuous cases (matrices), and the new analytic methods will be for continuous cases (PDEs).

Motivational Model Problem: An elastic bar with varying A(x) and E(x) is subject to a varying axial load (distributed), concentrated load, and inertia force (dynamic). Both the axial and concentrated loads are time dependent. The figure below depicts the problem.

Homework 6
Elastic Bar Interpolation Example

It is assumed that the displacement u(x) is for $$x_i\leq x\leq x_{i+1}$$.

The following is an example of linear interpolation of u(x) for a 2-bar truss.



The deformed shape is a straight line, which means there was an assumption of linear interpolation of displacement between the two nodes. Below is a case where there is only axial displacement, meaning zero transverse displacement.

u(x) is expressed in terms of di=u(xi) and di+1=u(xi+1) as a linear function in x.

$$u(x)=N_i(x)d_i+N_{i+1}(x)d_{i+1}$$ where $$N_i(x) and N_{i+1}(x)$$ are linear functions in x.



Homework 7
Motivation: The deformed shape of the truss element id the interpolation of the transverse displacement V(s), where $$s=\tilde{x}$$.

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

Integration by parts of the first term: $$\alpha \equiv \int_{0}^{L}{w(\tilde{x})}\frac{\partial^2}{\partial x^2}\left\{(EI)\frac{\partial^2v}{\partial x^2} \right\}dx$$

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

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

The equation α can be rewritten as follows: $$\alpha =\left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2} \right\} \right]^L_0-\int_{0}^{L}\frac{dw}{dx}\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2} \right\}dx$$

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

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

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

and where, $$\gamma =\int_{0}^{L}{\frac{d^2w}{dx^2}(EI)\frac{\partial ^2v}{\partial x^2}}dx$$

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

The stiffness term γ is now the focus to derive the beam stiffness matrix and to identify the beam shape functions.



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

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



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

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

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

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

Recommended Software to Improve Productivity When Using Mediawiki
WikED To increase efficiency and productivity in the writing and editing of Mediawiki articles, WikED was used throughout the semester. The WikED software saved time because it inserted Mediawiki code at the click of a button. Without WikED, all Mediawiki format code would have been typed by hand, requiring much excess typing. An example of the typing savings is formatting a simple superscript. To insert a superscript, the code $$undefined$$ must be typed around the characters to be made into superscripts. With WikED, the characters only had to be highlighted and the superscript button clicked, which input the $$undefined$$ automatically. Wik Ed allowed for easy insertion of pictures, changing fonts, as well as formatting all text and symbols. WikED increased efficiency and productivity in the writing and editing of Mediawiki articles.

LaTeX Editor A free online LaTeX editor powered by Codecogs was used to generate the equations made in the homework reports throughout the semester. The website allowed for easy and efficient writing of equations through the use of buttons that generated LaTeX code. Without the LaTeX editor, all of the equation codes would have been manually typed, which would increase the amount of time to write the equaitons. The LaTeX editor is also useful, because it continually updates the typed equaiton, so all formatting changes can be seen instantly. The LaTeX editor also reduces the amount of code that must be remembered, because its buttons show equation formats, and it automatically inserts appropriate code at the click of a button. To insert LaTeX code into Mediawiki, the LaTeX button must be clicked to insert the $$$$ code, and then the code generated by Codecogs can be copied and pasted between the math command. The following hyperlink is a link to the LaTeX editor Codecogs LaTeX Editor.

Drawing Figures Figures for the homework reports were generated in Microsoft PowerPoint. The figures were drawn using the drawing tools of the program, and then when the figures were finished the image was viewed full screen in a sideshow. The print screen command was used to copy the image, which was then pasted into Microsoft paint and saved as a .jpg file. To post the images on Mediawiki, the image button was clicked and the file was uploaded with the appropriate licensing and description.