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User:Eml4500.f08.bike.mcdonald/homework report 7

Introduction to Frame Elements
A frame element is defined as an element with a truss or bar element and a beam element. This gives the frame element both axial and transverse deformations.

The frame shown above has a rigid connection between its two elements. This means that the angle between the two elements will stay constant after deformation.

Elemental Breakdown of the Frame Element
Next the frame will be broken down into its two elements. The free body diagrams for these elements are shown below.

In the free body diagram above, a new degree of freedom is introduced for each node. This new degree of freedom is the moment around the node and adds two degrees of freedom. This brings the total to six degrees of freedom for element 1.

As with element 1, element 2 has the two new degrees of freedom.

Global Components of the Frame Element
In general, the degrees of freedom shown above for each element would be the general forces previously used. However, the new rotational degrees of freedom, $$d_3^{(e)}$$ and $$d_6^{(e)}$$, are now the bending moments, $$f_3^{(e)}$$ and $$f_6^{(e)}$$ instead of forces.

The global diagram contains nine degrees of freedom now. All three nodes now contain bending moments just like the elements do.

The elemental stiffness matrices $$\textbf{k}^{(e)}_{6x6}$$ where e = 1,2 are used to make the global stiffness matrix $$\textbf{K}_{9x9}$$. For the current frame element, the global matrix would look like the picture shown below.

In order to build the local stiffness matrix for each element that go in the assembly above, the following diagram shows the local element degrees of freedom of element 1.

Using the diagram above, the local stiffness matrix $$ \mathbf{K}^{(e)} $$ in the $$ \tilde{X}$$ and $$ \tilde{Y}$$ directions is shown below.

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

Verification of Dimensions and Element Force Displacement Relations
When doing a dimensional analysis on the global degrees of freedom of the system, the following conclusions are found.

$$\left[\tilde{d}_{i} \right]=L$$

These dimensions are true when $$i=1,2,4,5.$$

$$\left[\tilde{d}_{3} \right]=1$$

This is because $$ \tilde{d}_{3} $$ is a rotational degree of freedom, and it does not have a dimensional parameter. This is shown below using the definition of an arc length AB.

$$AB=R \times \theta $$

The arc length AB is shown below.

Then, using the definition shown above, the following dimensional analysis shows that the rotational degrees of freedom are dimensionless. $$\left[\theta \right]=\frac{\left[AB \right]}{\left[R \right]}=\frac{L}{L}=1$$

Now, a dimensional analysis is done on other important parameters.

$$ \left[\epsilon\right] = \frac{[du]}{[dx]} = \frac{L}{L} = 1 $$

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

$$ \left[A\right] = L^2, [I] = L^4 $$

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

$$ \left[\tilde{k}_{11}\tilde{d_1}\right] = [\tilde{k}_{11}\tilde{d_1}] = F $$

$$ \left[\tilde{k}_{23}\tilde{d_3}\right] = [\tilde{k}_{23}][\tilde{d_3}] $$ where $$ \tilde{d_3} = 1 $$

$$ = \frac{\left[6\right][E][I]}{L^2} $$

$$ = \frac{\left(1\right)\left(\frac{F}{L^2}\right)(L^4)}{L^2} = F $$

The element force displacement relation in global coordinates from the element force displacement relation in local coordinates yields:

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

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

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

The derivation of $$ \tilde{K}^{(e)} $$ from PVW results in the equation shown below, focusing on the bending effects.

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

Deformed Shape of Truss Element and Interpolation of Transverse Displacement.
Motivation: The deformed shape of the truss element and 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: $$u(\tilde{x})=N_1(\tilde{x})\tilde{d}_1+N_4(\tilde{x})\tilde{d}_4$$

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

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

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

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

Determination of the Displacement Vector
The interpolations of the applied internal forces $$ N_{5}\left(\tilde{x}\right) $$ and $$ N_{6}\left(\tilde{x}\right) $$ are shown below.

Recall the equation of the displacement matrix in the tilde coordinate system shown below, where $$ \underline{d}_{6x1}^{(e)} $$ is known after solving the finite element system.

$$ \underline{\tilde{d}}_{6x1}^{(e)} = \underline{\tilde{T}}_{6x6}^{(e)}\underline{d}_{6x1}^{(e)}$$

Now we will look at the computation of $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$.

$$ \underline{u}\left(\tilde{x}\right) = u(\tilde{x})\tilde{i} + v(\tilde{x})\tilde{j} = u_{x}(\tilde{x})\tilde{i} + u_{y}(\tilde{x})\tilde{j} $$

Computing $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$ required using the equations shown below.

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

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

The computation of $$ u_{x}\left(\tilde{x}\right) $$ and $$ u_{y}\left(\tilde{x}\right) $$ from $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$ is shown below.

$$\begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \underline{R}^{T}\begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix}$$

$$ \begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix} = \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix}\begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} $$ where $$ \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix} $$ is represented by $$ \underline{N}\left(\tilde{x}\right) $$ and $$\begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix}$$ is denoted $$ \underline{\tilde{d}}^{(e)} $$

$$\begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \underline{R}^{T}\underline{N}\left(\tilde{x}\right)\underline{\tilde{T}}^{(e)}\underline{d}^{(e)} $$

Dimensional Analysis with the shape functions
Now, a dimensional analysis will be done to the axial as well as the transverse displacements. Axial Displacements

$$ \left[u \right] =L $$ $$ \left[N_1 \right] =[N_4]=1 $$ $$ [\tilde{d}^{(e)}_1]= [\tilde{d}^{(e)}_4]= 1$$ Therefore, $$ [N_1][\tilde{d}^{(e)}_1]= [N_4][\tilde{d}^{(e)}_4]= L $$

Transverse Displacements

$$ \left[v \right] =L $$ The shape functions for the transverse displacements are shown below. $$\left[N_2 \right]=\left[N_3 \right]=\left[N_5 \right]=\left[N_6 \right]=1$$ The displacements corresponding rotation are shown below, which are dimensionless.

$$ [\tilde{d}^{(e)}_3]= [\tilde{d}^{(e)}_6]= 1$$

The rest of the displacements are in the $$ \tilde{Y}$$ direction, and those are shown below.

$$ [\tilde{d}^{(e)}_2]= [\tilde{d}^{(e)}_5]= 1$$

Therefore,

$$ [N_2][\tilde{d}^{(e)}_2]= [N_4][\tilde{d}^{(e)}_5]= L $$

$$ [N_1][\tilde{d}^{(e)}_3]= [N_4][\tilde{d}^{(e)}_6]= 1 $$

Derivation of Beam Shape Functions
First, the governing PDE for beams is analyzed. This PDE is considered without a distributed transverse load and also without an inertia force.

$$ m \ddot{v} $$(static case) is shown below.

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

Also, let's consider a constant EI, and the final equation is shown below.

$$\frac{\partial^{4} }{\partial x^{4}}=0$$

Solving for this fourth degree derivative, the following equation is obtained.

$$ v(x)= C_0 + C_{1}x+ C_{3}x^{2}+C_{3}x^{3}$$

Now, to get $$ N_2(x) $$, and since it is in the axial direction, $$\tilde{x} = x$$.

This results in the following values using the boundary conditions.

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

Use above boundary conditions to solve for $$c_0$$ through $$c_3$$

$$V(0) = 1 = C_0$$

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

$$V'(x) = C_1 + 2C_2x + 3C_3x^2$$

$$V'(0) = C_1 = 0$$

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

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

(1) $$\Rightarrow 0 = 1 + C_2L^2 + (-\frac{2}{3}\frac{C_2}{L}) - L^3$$

Where the second, third, and fourth terms combine to be:

$$C_2 = -\frac{2}{L^2}$$

This is similar to the result for $$N_2$$ found previously.

For $$N_3$$

Boundary Condition

$$V(0) = V(L) = 0$$

$$\tilde{d}_3$$ rotational

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

For $$N_5$$:

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

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

For $$N_2$$:

$$V(0) = V(L) = 0$$

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

$$\tilde{d}_6$$ rotational

The plots for $$N_5 and N_6$$ where previously provided.

Now we must derive the coefficients in $$\mathbf{\tilde{k}}$$ ( the element stiffness matrix)

The coefficients with EA are already complete, however the ones with EI still need to be computed.

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

Vibrational Analysis using FEA
$$\hat{k}_{23} = \frac{6EI}{L^2} = \int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}dx}$$

In general,

$$\hat{k}_{ij} = \int_{0}^{L}{\frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}dx}$$

With i,j = 2,3,5,6

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

Analyzing the Modal problem from lecture 31 using discrete Principle of Virtual Work is as follows:

$$ \mathbf{\bar{w}} = [\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{K}}\mathbf{\bar{d}} - \mathbf{\bar{F}}] = 0$$

In the above equation, the boundary conditions are already applied and the K matrix represents the reduced stiffness matrix.

This equation is true for all $$\mathbf{\bar{w}}$$.

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

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

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

The above equations will be referred to as (1) for the rest of this discussion.

These are the complete ordinary differential equations (ODEs) which are second order in time with initial conditions governing the elastodynamics of the discretized continuous problem with multiple degrees of freedom.

Now, solving the method for solving equation (1) is as follows:

1) Consider the unforced vibrations problem:

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

This equation is equal to zero becuase it is unforced. This will be referred to as equation (2) from now on.

Assume:

$$\mathbf{v}(t)_{nx1} = (sinwt)\mathbf{\phi _{nx1}}$$

Where the phi matrix is not time dependent. This solution gives the oscillating motions that have been seen in the FEA animations that have been shown.

Thus:

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

This means equation (2) becomes:

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

Therefore,

$$\mathbf{\bar{k}}\mathbf{\phi } = \omega ^2\mathbf{\bar{M}}\mathbf{\phi }$$

Which is the generalized eigenvalue problem, as it is of the form:

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

Where lambda is an eigenvalue.

A standard eigenvalue problem is given when the B matrix is equal to the identity matrix and the equation becomes:

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

This means that

$$\lambda =\omega ^2$$ is an eigenvalue.

Also the eigenpairs can be represented as:

$$(\lambda _i,\mathbf{\phi _i})$$

with i = 1 through n

Now the 'ith' mode for the animation can be represented as:

$$\mathbf{v}_i(t) = (sinw_it)\mathbf{\phi _i}$$

for i = 1 through n

Now step 2 in the solution is as follows:

2) Modal superposition method:

By using the orthogonal properties of the eigenpairs we can equate:

$$\mathbf{\phi _i}^T_{1xn}\mathbf{\bar{M}}_{nxn}\mathbf{\phi }_{nx1} = \delta _{ij} = \begin{cases} & \text{1 if } i=j \\ & \text{0 if } i\neq j \end{cases}$$

This delta 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) and two gives:

$$\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j \cdot \mathbf{\bar{k}}\mathbf{\phi }_j$$

$$ \mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j\phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j$$

where the left hand side is equal to the Kronecker delta, therefore:

$$\phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j = \frac{1}{\lambda _j}\delta _{ij}$$

$$\mathbf{\bar{d}}_{nx1}(t) = \sum{\zeta _i(t)\mathbf{\phi }_{inx1}}$$

This means equation (1) can be written as:

$$\mathbf{\bar{M}}(\sum_{j}^{}{\ddot{\zeta }_j\mathbf{\phi }_j}) + \mathbf{\bar{K}}(\sum_{j}^{}{\zeta _j\mathbf{\phi }_j}) = \mathbf{F}$$

In the above equation, the first term in parenthesis is equal to the second derivative of the d matrix, while the second term in parenthesis is equal to the d matrix itself. We can also write:

$$\sum_{j}^{}{\ddot{\zeta }_j(\mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j} + \sum_{j}^{}{\zeta _j(\mathbf{\phi }_i^T\mathbf{\bar{K}}\mathbf{\phi }_j} = \mathbf{\phi }_i^T\mathbf{F}$$

The first term in parenthesis in this equation is equal to the Kronecker delta, while the second term in parenthesis is equal to an eigenvalue times the Kronecker delta. This means that this equation can be written as an ordinary differential equation as follows:

$$\ddot{\zeta } + \lambda _i\zeta _i = \mathbf{\phi }_i^T\mathbf{F}$$

with i = 1 through n

Pros and Cons of Mediawiki vs. E-Learning for Collaboration
MediaWiki has a lot of advantages over ELS. These advantages improve the usefulness of the online portion of the class and create an additional resource for life.

The first and most obvious advantage of MediaWiki is its accessibility. Unlike ELS, MediaWiki is available for everyone to view since it is not protected or passworded like ELS. This creates the opportunity to supplement your own lecture notes with the rest of the class so the material can be better understood. If this trend was to catch on throughout universities, it would create an advantage to all students worldwide since there would be a vast database of knowledge, comparable to multiple textbooks. This database would be subjected to scrutiny from many people and thus the information would be constantly updated and checked for accuracy. New methods or improvements on old methods could easily be added, unlike a textbook which would need reprinting. Also, ELS course sections are subject to removal from the professor after the semester is ended and access to ELS may be revoked after graduation. MediaWiki will not remove the content added by us, creating a resource for future projects.

Another advantage of MediaWiki over ELS is the ability to share thoughts and work together as a group. Although ELS has a discussion board, it does not have a section for group work. When creating an entry on MediaWiki, the team can edit and critique each other’s work without having to share documents. This makes it very easy for a team to collaborate and put together their best work. The MediaWiki page can be signed and therefore submitted by all members, contrasting ELS which requires one person to submit all the information if the project was done by a group.

MediaWiki and ELS both share the capabilities to include audio, video, and figures. The difference is how this media can be connected to lectures or assignments. The ELS system requires each file to be uploaded and then can be accessed individually. Video and audio cannot be included in a submission unless it is already embedded in the file, such as audio with a PowerPoint presentation. MediaWiki allows for all of these to be embedded directly in the page after being uploaded. This approach makes adding or changing the media in an assignment easy and user-friendly. MediaWiki accommodates equations using LaTeX while ELS requires the use of an equation editor with a word processor such as Microsoft Word. The equations on MediaWiki can be edited on the fly while a Word document would have to be resubmitted on ELS to accept changes to an already submitted document.

Both MediaWiki and ELS have problems with access to their websites. The difference is the frequency and speed of recovery. ELS gets bogged down with users during high system loads since it is hosted by UF, which makes access almost impossible at times. Easy access to the system does not return until the load decreases. This could be a long time depending on timeframe, such as during exam preparation. MediaWiki has a lot more servers to host these pages and never gets bogged down. The only issue is when editing a page sometimes the server is not available for the update. By simply going back on the browser and resubmitting the changes, the error is usually eliminated. This makes the recovery time almost instant.

The advantage that ELS has over MediaWiki is the ability for privacy. Grades for each person can be shown on ELS allowing a student to verify everything in the computer is correct. Currently there is no system to verify this with our class using MediaWiki. Privacy would also confirm that an individual’s work is original. As already proven in our class, plagiarism can arise as an issue during projects. With cheating an ever-present problem in the academic world, there is no protection using MediaWiki.

Both systems allow for practice using computer skills which will be helpful in the future. Knowledge of how MediaWiki works allows the user to edit other pages on Wikipedia or Wikiversity and continue to contribute to the system.

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.

Original 2-Bar Truss System
An image showing the setup of the original 2-bar truss system is shown in the figure below.

The properties of the original 2-bar truss system are shown in the table below.

Plot of Undeformed and Deformed Shapes of the Original 2-Bar Truss System
A plot of the undeformed and deformed shapes of the original 2-bar truss system solved using the Finite Element Method is shown in the figure below.

2-Bar Truss System with General Stiffness Matrix compared with Average Stiffness Matrix
The original 2-bar truss system analyzed previously was modified by introducing bar elements with variable cross-sectional area (A) and modulus of elasticity (E). An image showing the setup of the 2-bar truss system is shown in the figure below.

The properties of the modified 2-bar truss system with variable cross-sectional area and modulus of elasticity values are shown in the table below.

In order to determine the deformed shape of the modified 2-bar truss system, both the general stiffness matrix and the average stiffness matrix can be used.

The general stiffness matrix for each bar element is defined as shown below.

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

The average stiffness matrix for each bar element is defined as shown below.

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

Plot of Undeformed and Deformed Shapes of 2-Bar Truss System with General and Average Stiffness Matrices
A plot of the undeformed and deformed shapes of the 2-bar truss system solved using the general and average stiffness matrices is shown in the figure below. The dotted blue line represents the undeformed shape of the truss system, while the solid red line represents the deformed shape of the truss system using the general stiffness matrix, and the solid green line represents the deformed shape of the truss system using the average stiffness matrix.

Introduction
The following image shows the 2-Element frame system to be analyzed using MATLAB. This system has the same geometric characteristics as the 2-Bar truss system analyzed in Homework Report 3. The only difference between these two structures is Element 1 in the frame/truss system is assumed to be a frame element with a square cross-section, while Element 2 is assumed to be a truss element.

The properties of the system are shown in the table below. Because the cross-sections of both elements were assumed to be square, the moments of inertia, $$I$$, were calculated using the following equation, where $$s_{1}$$ and $$s_{2}$$ correspond to side 1 and side 2 of the square, respectively, and $$s_{1}=s_{2}$$.

$$ I=\frac{s_{1}s_{2}^{3}}{12} $$

The value of the applied force $$P$$ at global node 2 is equal to 7.

MATLAB Code Developed to Analyze the 2-Element Frame/Truss System
The following MATLAB code was written by Andrew McDonald in order to solve for the displacements of the 2-Element frame/truss system and plot the undeformed and deformed shapes of the frame system to compare with the deformed shape of the 2-bar truss system.

Plot of Undeformed Shape of 2-Element Frame/Truss System Compared with Deformed Shape for Truss and Frame Systems
A plot comparing the undeformed shape of the 2-Element frame/truss system with its deformed shape and the deformed shape of the 2-bar truss system with the same geometric characteristics is shown below. In this plot, the blue dotted line represents the undeformed shape of the 2-Element frame/truss system, the red solid line represents the deformed shape of the 2-Element frame/truss system, and the green solid line represents the deformed shape of the 2-bar truss system.

It is important to note that the deformed shape of the 2-bar truss system was multiplied by a scaling factor of 0.25 in order to produce a more aesthetically pleasing plot.

Because Element 1 in the frame/truss system is a frame element, the deformation in this element is not simply a straight line between the deformed nodal positions. The above plot shows the curved deformation of Element 1, whereas the deformation for Element 1 of the 2-bar truss system is just a straight line between the deformed nodal positions.

MATLAB Analysis of Electric Pylon Structure Model as Frame System
The setup of the Electric Pylon Frame Model is shown in the figure below. This system will be analyzed using the Finite Element Method in MATLAB.

The properties of the Electric Pylon Frame Model are shown in the table below.

Analysis Requirements

 * 1) Perform Finite Element Analysis using MATLAB on the Electric Pylon Truss Structure
 * 2) Plot undeformed and deformed shape of frame model and compare with deformed shape of truss model
 * 3) Identify the frame element with the highest nodal bending moment
 * 4) Identify the frame element with the highest transverse shear force
 * 5) Determine if the frame model is statically determinate?
 * 6) Compare reactions obtained from truss model with those obtained from frame model
 * 7) Construct the lumped mass matrix for frame model and solve the generalized eigenvalue problem
 * 8) Find the lowest three eigenpairs of the frame system
 * 9) Plot the lowest three eigenvectors of the frame system and plot them as deformed shapes of the Electric Pylon Structure
 * 10) Find the lowest three vibrational periods of the frame system and compare with the lowest three vibrational periods of the truss system

Command Window Printout Produced by Running ElectricPylonAnalysisFrame.m
d =

0        0         0         0   -0.0002    0.0002   -0.0001   -0.0000   -0.0002   -0.0003   -0.0001    0.0003   -0.0001   -0.0001   -0.0001   -0.0005    0.0004    0.0003    0.0005   -0.0006    0.0010    0.0006    0.0010   -0.0002    0.0011   -0.0010    0.0017    0.0010    0.0017    0.0002    0.0017   -0.0006    0.0017   -0.0013    0.0023    0.0006    0.0022   -0.0010    0.0047    0.0024    0.0040   -0.0024    0.0064    0.0057    0.0064    0.0038    0.0064    0.0030    0.0064    0.0024    0.0064    0.0016    0.0064    0.0009    0.0063   -0.0002    0.0062   -0.0013    0.0061   -0.0024    0.0060   -0.0033    0.0059   -0.0051    0.0055   -0.0126    0.0068    0.0043    0.0070   -0.0063    0.0070    0.0035    0.0074   -0.0044    0.0072    0.0030    0.0071    0.0021    0.0071    0.0012    0.0072    0.0003    0.0073   -0.0007    0.0074   -0.0019    0.0076   -0.0034    0.0090    0.0030    0.0105   -0.0034

reactions =

1.0e+003 *

0.1496  -0.5017   -0.1496    1.5017

results =

1.0781e-005 2.1562e+006       862.48 -6.5356e-006 -1.3071e+006     -522.85 6.7339e-006 1.3468e+006       538.71 6.4556e-006 1.2911e+006       516.45 -3.5152e-006 -7.0304e+005     -281.22 -2.3798e-005 -4.7596e+006     -1903.8 -1.2325e-005 -2.465e+006      -986.01 7.1182e-006 1.4236e+006       569.46 1.9269e-005 3.8538e+006       1541.5 -2.8828e-005 -5.7655e+006     -2306.2 -1.8941e-007      -37882      -15.153 8.778e-009      1755.6      0.70224 1.9328e-005 3.8655e+006       1546.2 1.4698e-005 2.9395e+006       1175.8 -1.8494e-005 -3.6988e+006     -1479.5 -2.883e-005 -5.7661e+006     -2306.4 1.7948e-006 3.5896e+005       143.58 2.621e-005  5.242e+006       2096.8 9.8986e-006 1.9797e+006       791.89 -1.8874e-005 -3.7749e+006       -1510 -3.258e-005 -6.5159e+006     -2606.4 2.7892e-006 5.5784e+005       223.14 3.2194e-006 6.4388e+005       257.55 2.9828e-005 5.9656e+006       2386.3 -4.0852e-006 -8.1704e+005     -326.82 -1.9808e-005 -3.9616e+006     -1584.6 1.16e-005   2.32e+006       928.02 -4.0653e-006 -8.1307e+005     -325.23 -2.8778e-005 -5.7556e+006     -2302.2 -1.4237e-007      -28474       -11.39 -4.9677e-006 -9.9353e+005     -397.41 -7.6072e-008      -15214      -6.0857 2.9608e-005 5.9216e+006       2368.6 2.7907e-007       55813       22.325 -2.4358e-005 -4.8717e+006     -1948.7 6.9573e-006 1.3915e+006       556.59 -1.5116e-008     -3023.3      -1.2093 -2.8794e-005 -5.7588e+006     -2303.5 -2.427e-005 -4.854e+006      -1941.6 6.9522e-006 1.3904e+006       556.17 -9.4154e-007 -1.8831e+005     -75.323 1.129e-005  2.258e+006       903.22 -3.16e-006  -6.32e+005       -252.8 -2.0051e-005 -4.0102e+006     -1604.1 4.0531e-019 8.1063e-008  3.2425e-011 4.8822e-019 9.7644e-008  3.9058e-011 3.2073e-007       64147       25.659 1.3501e-006 2.7001e+005       108.01 1.3575e-006 2.7149e+005        108.6 -1.0389e-005 -2.0778e+006     -831.13 -2.2136e-005 -4.4271e+006     -1770.9 -2.9388e-005 -5.8777e+006     -2351.1 -3.6774e-005 -7.3549e+006     -2941.9 -4.331e-005 -8.662e+006      -3464.8 -4.3478e-005 -8.6957e+006     -3478.3 0           0            0            0            0            0            0            0            0            0            0            0 -8.8522e-007 -1.7704e+005      -70.818 1.1034e-006 2.2069e+005       88.276 -9.5965e-007 -1.9193e+005     -76.772 9.9436e-007 1.9887e+005       79.549 1.0732e-005 2.1464e+006       858.56 -1.0415e-005 -2.083e+006      -833.19 1.1448e-005 2.2895e+006       915.81 -1.1321e-005 -2.2643e+006     -905.71 1.0517e-005 2.1035e+006       841.38 -1.0732e-005 -2.1464e+006     -858.56 7.0689e-006 1.4138e+006       565.51 -6.8859e-006 -1.3772e+006     -550.87 7.9177e-006 1.5835e+006       633.42 -1.8464e-005 -3.6928e+006     -1477.1 -4.8696e-007      -97393      -38.957 -1.1845e-007      -23690      -9.4762 1.5604e-007       31208       12.483 4.5239e-005 9.0479e+006       3619.2 -7.0445e-019 -1.4089e-007 -5.6356e-011 4.5158e-005 9.0317e+006       3612.7 -1.1908e-018 -2.3815e-007 -9.5262e-011 4.5256e-005 9.0511e+006       3620.5 -6.5877e-007 -1.3175e+005     -52.702 -1.4822e-006 -2.9645e+005     -118.58 8.9829e-006 1.7966e+006       718.63 2.2438e-005 4.4876e+006         1795 3.3117e-005 6.6233e+006       2649.3 3.8878e-005 7.7755e+006       3110.2 1.6111e-019 3.2222e-008  1.2889e-011 -9.6069e-020 -1.9214e-008 -7.6855e-012 -1.441e-019 -2.8821e-008 -1.1528e-011 1.0741e-019 2.1482e-008  8.5927e-012

d =

0           0            0            0            0            0  -0.00019023   0.00017464  9.1757e-006 -0.00012769 -1.4086e-006 -2.4036e-005 -0.00016005  -0.0002967  9.1254e-006 -0.0001267  0.00032601 -4.1467e-005 -0.00012797   -0.000128 -6.1763e-005 -0.00012791 -0.00054327 -3.4931e-005 0.00044932    0.000309  -0.00011737   0.00046381  -0.00061057  -0.00011826     0.001017   0.00064582  -0.00016353    0.0010325  -0.00019427  -0.00013889    0.0010502  -0.00097027  -0.00015648    0.0017167    0.0010338  -0.00024125    0.0017162    0.0002326  -0.00017832    0.0016833  -0.00064959  -0.00016107     0.001683   -0.0013293  -0.00019661    0.0023452   0.00064079  -0.00028233    0.0021862   -0.0010297  -0.00022023     0.004712    0.0023901   -0.0003608    0.0039576   -0.0024164  -0.00033295    0.0063604    0.0057326  -0.00022721    0.0063604    0.0037875  -0.00022759    0.0063604    0.0029802  -0.00022392    0.0063612    0.0023839  -0.00023194    0.0063655    0.0016463  -0.00021304    0.0063709   0.00086187  -0.00020968    0.0063181   -0.0002427  -0.00024756    0.0062287   -0.0013181  -0.00031262    0.0061362   -0.0023672  -0.00036099    0.0060411    -0.003323  -0.00041992    0.0058837   -0.0051276  -0.00057906    0.0055151    -0.012586   -0.0010709    0.0067824    0.0042646  -0.00022701    0.0069958   -0.0063358  -0.00065704    0.0070118    0.0034572  -0.00022802    0.0074004    -0.004359  -0.00048349    0.0071503    0.0029771    -0.000229    0.0071476    0.0020507  -0.00022519    0.0071419    0.0012277  -0.00021034    0.0071818   0.00032022  -0.00022274    0.0072814  -0.00073249  -0.00027283    0.0074071   -0.0018632  -0.00034933    0.0075641   -0.0033871  -0.00041223    0.0090003    0.0029771  -0.00023027     0.010455   -0.0033871  -0.00035946

reactions =

149.65     -501.65    -0.045833      -149.65       1501.7    -0.047459

fa =

12.113           0      -1520.2       15.182       9.5289      -1520.2       12.113            0      -654.71       24.468       9.5289      -654.71       15.182       9.5289        538.7       24.468       9.5289        538.7       24.468       9.5289       662.83       36.581            0       662.83       24.468       9.5289      -281.21       33.674       9.5289      -281.21       36.581            0      -1285.2       33.674       9.5289      -1285.2       15.182       9.5289       404.52       24.468       17.612       404.52       33.674       9.5289       209.49       24.468       17.612       209.49       15.182       9.5289       598.85       17.766       17.612       598.85       33.674       9.5289       302.98        31.09       17.612       302.98       17.766       17.612      -15.156       24.468       17.612      -15.156       24.468       17.612      0.70819        31.09       17.612      0.70819       17.766       17.612         5419       20.431       25.688         5419       24.468       17.612       5115.2       20.431       25.688       5115.2       24.468       17.612       5243.7       28.506       25.688       5243.7        31.09       17.612       5583.1       28.506       25.688       5583.1       20.431       25.688       143.57       28.506       25.688       143.57       20.431       25.688        10514       18.896       29.725        10514       20.431       25.688       8170.8       24.468       29.725       8170.8       28.506       25.688       7967.7       24.468       29.725       7967.7       28.506       25.688        10932        29.96       29.725        10932       18.896       29.725        223.1       24.468       29.725        223.1       24.468       29.725       257.54        29.96       29.725       257.54       18.896       29.725        13045       17.443       33.763        13045       18.896       29.725        12088       21.157       33.763        12088       24.468       29.725        10475       21.157       33.763        10475       24.468       29.725       9970.7       27.779       33.763       9970.7        29.96       29.725        11038       27.779       33.763        11038        29.96       29.725        11797       31.413       33.763        11797       17.443       33.763      -11.502       21.157       33.763      -11.502       21.157       33.763      -397.41       27.779       33.763      -397.41       27.779       33.763       -6.051       31.413       33.763       -6.051       17.443       33.763        28144       14.536       41.766        28144       17.443       33.763        16535       19.058       36.339        16535       21.157       33.763        15140       19.058       36.339        15140       27.779       33.763        12107       29.879       36.339        12107       31.413       33.763        13426       29.879       36.339        13426       31.413       33.763        21371        34.32       41.766        21371       19.058       36.339        26805       14.536       41.766        26805       29.879       36.339        20207        34.32       41.766        20207       14.536       41.766        18542       12.113       48.452        18542       14.536       41.766        17643       17.927       48.452        17643        34.32       41.766        24351       31.009       48.452        24351        34.32       41.766        23438       36.743       48.452        23438            0       48.452 -4.3247e-005 8.5599      48.452 -4.3247e-005 8.5599      48.452    0.0012651       12.113       48.452    0.0012651       12.113       48.452       25.644       14.697       48.452       25.644       14.697       48.452       108.06       17.927       48.452       108.06       17.927       48.452       108.68       21.884       48.452       108.68       21.884       48.452      -831.05       26.972       48.452      -831.05       26.972       48.452      -1770.8       31.009       48.452      -1770.8       31.009       48.452        -2351       34.159       48.452        -2351       34.159       48.452      -2941.8       36.743       48.452      -2941.8       36.743       48.452      -3464.7       40.377       48.452      -3464.7       40.377       48.452      -3477.9       48.856       48.452      -3477.9            0       48.452       5022.9       6.4603        50.31       5022.9       8.5599       48.452        12045       6.4603        50.31        12045       8.5599       48.452        16214       10.013       51.319        16214       12.113       48.452        14666       10.013       51.319        14666       12.113       48.452        18199       12.113       51.925        18199       14.697       48.452        14585       12.113       51.925        14585       14.697       48.452        16713       16.151       51.925        16713       17.927       48.452        16041       16.151       51.925        16041       17.927       48.452        15305       20.027       51.925        15305       21.884       48.452        15662       20.027       51.925        15662       21.884       48.452        14986       24.468       51.925        14986       26.972       48.452        16141       24.468       51.925        16141       26.972       48.452        19380        28.91       51.925        19380       31.009       48.452        20754        28.91       51.925        20754       31.009       48.452        24395       32.705       51.925        24395       34.159       48.452        27010       32.705       51.925        27010       34.159       48.452        26392       36.743       51.925        26392       36.743       48.452        35087       36.743       51.925        35087       36.743       48.452        30192       38.923       51.319        30192       40.377       48.452        37749       38.923       51.319        37749       40.377       48.452        32431       42.396        50.31        32431       48.856       48.452        17622       42.396        50.31        17622       6.4603        50.31       4967.8       10.013       51.319       4967.8       42.396        50.31       8950.5       38.923       51.319       8950.5       10.013       51.319       5070.9       12.113       51.925       5070.9       38.923       51.319         5787       36.743       51.925         5787       12.113       51.925      -52.735       16.151       51.925      -52.735       16.151       51.925      -118.65       20.027       51.925      -118.65       20.027       51.925       718.56       24.468       51.925       718.56       24.468       51.925         1795        28.91       51.925         1795        28.91       51.925       2649.3       32.705       51.925       2649.3       32.705       51.925       3110.1       36.743       51.925       3110.1       12.113       51.925        18327       12.113           60        18327       16.151       51.925        16416       12.113           60        16416       32.705       51.925        27005       36.743           60        27005       36.743       51.925        28638       36.743           60        28638

bma =

12.113           0     0.022994       15.182       9.5289    -0.018105       12.113            0    0.0081229       24.468       9.5289    -0.016338       15.182       9.5289    -0.029397       24.468       9.5289     0.010324       24.468       9.5289      0.01673       36.581            0   -0.0084122       24.468       9.5289    -0.033186       33.674       9.5289     0.052397       36.581            0    -0.052714       33.674       9.5289     0.057599       15.182       9.5289    -0.013139       24.468       17.612    -0.017589       33.674       9.5289     0.036925       24.468       17.612    -0.067784       15.182       9.5289     0.048154       17.766       17.612     -0.07998       33.674       9.5289    -0.044293        31.09       17.612     0.016607       17.766       17.612    -0.046562       24.468       17.612     0.030412       24.468       17.612    -0.023904        31.09       17.612     0.045514       17.766       17.612      0.12186       20.431       25.688     -0.16946       24.468       17.612      0.22807       20.431       25.688     -0.26091       24.468       17.612     0.048105       28.506       25.688    -0.081478        31.09       17.612      0.10335       28.506       25.688     -0.15577       20.431       25.688    0.0075086       28.506       25.688   -0.0080992       20.431       25.688      0.78058       18.896       29.725     -0.83757       20.431       25.688        0.102       24.468       29.725     -0.12211       28.506       25.688      0.55491       24.468       29.725     -0.57418       28.506       25.688      0.17591        29.96       29.725     -0.22341       18.896       29.725     0.013049       24.468       29.725     0.010529       24.468       29.725     0.010025        29.96       29.725    -0.027106       18.896       29.725       1.0433       17.443       33.763      -1.1399       18.896       29.725       0.2737       21.157       33.763     -0.29075       24.468       29.725      0.71635       21.157       33.763     -0.75663       24.468       29.725      0.18104       27.779       33.763     -0.20369        29.96       29.725      0.79459       27.779       33.763     -0.79992        29.96       29.725      0.32132       31.413       33.763     -0.37118       17.443       33.763       0.0198       21.157       33.763     0.070548       21.157       33.763     0.095069       27.779       33.763     -0.08117       27.779       33.763    -0.062272       31.413       33.763     0.010112       17.443       33.763      0.82756       14.536       41.766     -0.90244       17.443       33.763      0.66144       19.058       36.339      -0.7335       21.157       33.763       1.6168       19.058       36.339      -1.7837       27.779       33.763      0.31976       29.879       36.339     -0.41471       31.413       33.763       1.6244       29.879       36.339      -1.6665       31.413       33.763      0.21493        34.32       41.766     -0.30033       19.058       36.339       1.2596       14.536       41.766      -1.3188       29.879       36.339      0.13701        34.32       41.766     -0.22274       14.536       41.766      0.89572       12.113       48.452     -0.79307       14.536       41.766      0.45313       17.927       48.452     -0.34803        34.32       41.766       1.0151       31.009       48.452      -1.0006        34.32       41.766      0.52748       36.743       48.452     -0.59271            0       48.452   0.00020201       8.5599       48.452  -0.00043773       8.5599       48.452    -0.003897       12.113       48.452    0.0094171       12.113       48.452    -0.025727       14.697       48.452    0.0091642       14.697       48.452    -0.013444       17.927       48.452     0.044649       17.927       48.452     0.055357       21.884       48.452    -0.050826       21.884       48.452     0.016314       26.972       48.452    -0.056021       26.972       48.452     0.011473       31.009       48.452    -0.097414       31.009       48.452    -0.022087       34.159       48.452     -0.05983       34.159       48.452     0.066556       36.743       48.452     -0.18818       36.743       48.452     -0.10405       40.377       48.452      -0.1295       40.377       48.452      -0.2579       48.856       48.452    -0.051456            0       48.452     0.020854       6.4603        50.31    -0.020695       8.5599       48.452       2.2693       6.4603        50.31      -2.2682       8.5599       48.452      0.62209       10.013       51.319      -0.6228       12.113       48.452       1.6189       10.013       51.319      -1.6251       12.113       48.452       1.0354       12.113       51.925      -1.0433       14.697       48.452       1.3603       12.113       51.925      -1.3567       14.697       48.452      0.60013       16.151       51.925     -0.59057       17.927       48.452       1.3159       16.151       51.925      -1.3325       17.927       48.452      0.42978       20.027       51.925     -0.42623       21.884       48.452       1.2303       20.027       51.925      -1.2312       21.884       48.452      0.32859       24.468       51.925     -0.34468       26.972       48.452       1.3859       24.468       51.925      -1.3549       26.972       48.452      0.53436        28.91       51.925     -0.56825       31.009       48.452       1.7494        28.91       51.925      -1.6971       31.009       48.452      0.76098       32.705       51.925     -0.81165       34.159       48.452       2.0868       32.705       51.925      -2.0703       34.159       48.452       0.5265       36.743       51.925     -0.58963       36.743       48.452        1.838       36.743       51.925      -1.8262       36.743       48.452      0.68173       38.923       51.319     -0.77585       40.377       48.452       3.9141       38.923       51.319      -3.7556       40.377       48.452      0.96025       42.396        50.31      -1.1119       48.856       48.452       4.4339       42.396        50.31      -4.1055       6.4603        50.31     0.037973       10.013       51.319    -0.039427       42.396        50.31       5.0697       38.923       51.319      -4.8137       10.013       51.319     0.063035       12.113       51.925    -0.065423       38.923       51.319       6.2872       36.743       51.925      -6.1193       12.113       51.925   -0.0066913       16.151       51.925     0.011718       16.151       51.925     0.032676       20.027       51.925    -0.012235       20.027       51.925     0.036539       24.468       51.925    -0.051432       24.468       51.925    0.0087058        28.91       51.925    -0.068847        28.91       51.925    0.0017025       32.705       51.925     -0.10921       32.705       51.925    -0.028188       36.743       51.925    -0.054888       12.113       51.925      0.45455       12.113           60     -0.45539       16.151       51.925      0.58389       12.113           60     -0.58688       32.705       51.925      0.32594       36.743           60     -0.33192       36.743       51.925      0.78189       36.743           60     -0.74704

Va =

12.113           0   -0.0041054       15.182       9.5289   -0.0041054       12.113            0   -0.0015677       24.468       9.5289   -0.0015677       15.182       9.5289    0.0042772       24.468       9.5289    0.0042772       24.468       9.5289   -0.0016313       36.581            0   -0.0016313       24.468       9.5289    0.0092965       33.674       9.5289    0.0092965       36.581            0     0.011073       33.674       9.5289     0.011073       15.182       9.5289  -0.00036142       24.468       17.612  -0.00036142       33.674       9.5289   -0.0085469       24.468       17.612   -0.0085469       15.182       9.5289    -0.015099       17.766       17.612    -0.015099       33.674       9.5289    0.0071762        31.09       17.612    0.0071762       17.766       17.612     0.011484       24.468       17.612     0.011484       24.468       17.612     0.010483        31.09       17.612     0.010483       17.766       17.612    -0.034257       20.431       25.688    -0.034257       24.468       17.612     -0.05416       20.431       25.688     -0.05416       24.468       17.612    -0.014353       28.506       25.688    -0.014353        31.09       17.612    -0.030561       28.506       25.688    -0.030561       20.431       25.688   -0.0019328       28.506       25.688   -0.0019328       20.431       25.688     -0.37463       18.896       29.725     -0.37463       20.431       25.688    -0.039248       24.468       29.725    -0.039248       28.506       25.688     -0.19773       24.468       29.725     -0.19773       28.506       25.688    -0.093051        29.96       29.725    -0.093051       18.896       29.725  -0.00045215       24.468       29.725  -0.00045215       24.468       29.725   -0.0067618        29.96       29.725   -0.0067618       18.896       29.725     -0.50876       17.443       33.763     -0.50876       18.896       29.725     -0.12197       21.157       33.763     -0.12197       24.468       29.725     -0.28209       21.157       33.763     -0.28209       24.468       29.725    -0.073681       27.779       33.763    -0.073681        29.96       29.725     -0.34748       27.779       33.763     -0.34748        29.96       29.725     -0.16137       31.413       33.763     -0.16137       17.443       33.763     0.013661       21.157       33.763     0.013661       21.157       33.763    -0.026615       27.779       33.763    -0.026615       27.779       33.763     0.019919       31.413       33.763     0.019919       17.443       33.763     -0.20319       14.536       41.766     -0.20319       17.443       33.763     -0.45879       19.058       36.339     -0.45879       21.157       33.763      -1.0232       19.058       36.339      -1.0232       27.779       33.763     -0.22101       29.879       36.339     -0.22101       31.413       33.763      -1.0976       29.879       36.339      -1.0976       31.413       33.763    -0.060517        34.32       41.766    -0.060517       19.058       36.339     -0.36502       14.536       41.766     -0.36502       29.879       36.339    -0.051301        34.32       41.766    -0.051301       14.536       41.766     -0.23747       12.113       48.452     -0.23747       14.536       41.766     -0.10686       17.927       48.452     -0.10686        34.32       41.766     -0.27015       31.009       48.452     -0.27015        34.32       41.766     -0.15751       36.743       48.452     -0.15751            0       48.452 -7.4737e-005 8.5599      48.452 -7.4737e-005 8.5599      48.452    0.0037471       12.113       48.452    0.0037471       12.113       48.452     0.013502       14.697       48.452     0.013502       14.697       48.452     0.017985       17.927       48.452     0.017985       17.927       48.452    -0.026835       21.884       48.452    -0.026835       21.884       48.452    -0.014218       26.972       48.452    -0.014218       26.972       48.452    -0.026968       31.009       48.452    -0.026968       31.009       48.452    -0.011984       34.159       48.452    -0.011984       34.159       48.452    -0.098576       36.743       48.452    -0.098576       36.743       48.452   -0.0070022       40.377       48.452   -0.0070022       40.377       48.452     0.024347       48.856       48.452     0.024347            0       48.452    -0.006181       6.4603        50.31    -0.006181       8.5599       48.452      -1.6187       6.4603        50.31      -1.6187       8.5599       48.452     -0.38731       10.013       51.319     -0.38731       12.113       48.452     -0.91293       10.013       51.319     -0.91293       12.113       48.452     -0.59863       12.113       51.925     -0.59863       14.697       48.452     -0.62771       12.113       51.925     -0.62771       14.697       48.452     -0.31631       16.151       51.925     -0.31631       17.927       48.452     -0.67899       16.151       51.925     -0.67899       17.927       48.452     -0.21095       20.027       51.925     -0.21095       21.884       48.452     -0.62506       20.027       51.925     -0.62506       21.884       48.452     -0.15555       24.468       51.925     -0.15555       26.972       48.452     -0.64027       24.468       51.925     -0.64027       26.972       48.452     -0.27727        28.91       51.925     -0.27727       31.009       48.452     -0.84934        28.91       51.925     -0.84934       31.009       48.452     -0.40695       32.705       51.925     -0.40695       34.159       48.452      -1.1043       32.705       51.925      -1.1043       34.159       48.452     -0.25786       36.743       51.925     -0.25786       36.743       48.452      -1.0552       36.743       51.925      -1.0552       36.743       48.452     -0.40469       38.923       51.319     -0.40469       40.377       48.452      -2.3862       38.923       51.319      -2.3862       40.377       48.452     -0.75534       42.396        50.31     -0.75534       48.856       48.452      -1.2704       42.396        50.31      -1.2704       6.4603        50.31    -0.020954       10.013       51.319    -0.020954       42.396        50.31      -2.7331       38.923       51.319      -2.7331       10.013       51.319    -0.058786       12.113       51.925    -0.058786       38.923       51.319      -5.4826       36.743       51.925      -5.4826       12.113       51.925    0.0045592       16.151       51.925    0.0045592       16.151       51.925    -0.011586       20.027       51.925    -0.011586       20.027       51.925    -0.019807       24.468       51.925    -0.019807       24.468       51.925    -0.017461        28.91       51.925    -0.017461        28.91       51.925    -0.029222       32.705       51.925    -0.029222       32.705       51.925   -0.0066126       36.743       51.925   -0.0066126       12.113       51.925     -0.11268       12.113           60     -0.11268       16.151       51.925     -0.12967       12.113           60     -0.12967       32.705       51.925    -0.072864       36.743           60    -0.072864       36.743       51.925     -0.18933       36.743           60     -0.18933

max_bending_moment =

6.2872

max_bending_moment_element =

81

max_transverse_shear_stress =

5.4826

max_transverse_shear_stress_element =

81

lambda =

132.86      2471.2       2942.6

column =

74   81    80

T =

0.54511     0.12639      0.11583

Identify Frame Element with Highest Nodal Bending Moment
As shown in the above command window printout, the frame element with the highest nodal bending moment is element 81. The nodal bending moment in this frame element was calculated to be 6.2872 N-m.

Identify Frame Element with Highest Transverse Shear Force
As shown in the above command window printout, the frame element with the highest transverse shear force was element 81. The transverse shear force in this element was calculated to be 5.4826 Pa.

MATLAB Plot of Undeformed Electric Pylon Structure Compared with Deformed Shape of Truss and Frame Models
A plot of the undeformed and deformed shapes of the electric pylon structure for the truss and frame models is shown in the figure below. The dotted blue line represents the undeformed shape of the electric pylon structure, the solid red line represents the deformed shape of the electric pylon frame structure, and the solid green line represents the deformed shape of the electric pylon truss structure.

The difference between the deformations in the frame and truss models can not be discerned in the above plot. A zoomed in plot of the above figure at Node 33 is shown below, showing the small difference in deformations between the two models. Node 33 is the node where the vertical downward force was applied to the system.

Is the Frame Model Statically Determinate?
No, the electric pylon frame structure is not statically determinate. While this system is similar to the 2-bar truss system analyzed in previous Homework Reports in that there are only two nodes that have reactions, it is dissimilar to that problem due to the fact that each of these nodes has two frame members meeting at them. Because of this, each of the nodes has two unknown forces along the direction of each of those members. This means that the system has four unknown reaction forces, with only two equations to attempt to solve them with. These two equations are developed by summing the forces in the positive X direction and summing the forces in the positive Y direction.

The reactions of the frame system were calculated by the MATLAB code described previously and are compared with the reactions of the electric pylon truss structure in the table below.

Compare Frame Model Reactions with Truss Model Reactions
The reactions of the electric pylon frame and truss model have been tabulated and are compared below. These values were shown in the command window printout above.

From the above table, it can be seen that the reactions for the electric pylon truss and frame structures are nearly identical. The major difference between the two systems is the presence of the moment reactions in the frame structure.

Find the Lowest 3 Eigenpairs
The eigenvalues of the electric pylon structure were determined from the generalized eigenvalue problem shown below, involving the reduced global stiffness matrix and the reduced lumped mass matrix.

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

Lowest Eigenvalue
The lowest eigenvalue of the electric pylon frame structure was determined to be 132.86. This value was shown above in the MATLAB command window printout as well.

2nd Lowest Eigenvalue
The 2nd lowest eigenvalue of the electric pylon frame structure was determined to be 2471.2. This value was shown above in the MATLAB command window printout as well.

3rd Lowest Eigenvalue
The 3rd lowest eigenvalue of the electric pylon frame structure was determined to be 2942.6. This value was shown above in the MATLAB command window printout as well.

Plot of Lowest Eigenvector as Deformed Shape of Electric Pylon Frame Structure
A plot of the undeformed and deformed shapes of the electric pylon frame structure using the lowest eigenvector as the deformations of the system is shown in the figure below. The dotted blue line represents the undeformed shape of the electric pylon frame structure and the solid red line represents the deformed shape of the structure.

Plot of 2nd Lowest Eigenvector as Deformed Shape of Electric Pylon Frame Structure
A plot of the undeformed and deformed shapes of the electric pylon frame structure using the 2nd lowest eigenvector as the deformations of the system is shown in the figure below. The dotted blue line represents the undeformed shape of the electric pylon frame structure and the solid red line represents the deformed shape of the structure.

Plot of 3rd Lowest Eigenvector as Deformed Shape of Electric Pylon Frame Structure
A plot of the undeformed and deformed shapes of the electric pylon frame structure using the 3rd lowest eigenvector as the deformations of the system is shown in the figure below. The dotted blue line represents the undeformed shape of the electric pylon frame structure and the solid red line represents the deformed shape of the structure.

Find the Lowest 3 Vibrational Periods of Frame Model
The three lowest vibrational periods of the electric pylon frame structure were determined from the lowest eigenvalues determined above as follows.

$$ \lambda =\omega ^2 $$

Taking the square root of both sides of this equation gives the following equation, which allows for the circular frequency of vibration of the system to be determined.

$$ \omega= \sqrt{\lambda } $$

Substituting the above value for the circular frequency of vibration into the following relation gave the vibrational period of the electric pylon structure for each of the lowest eigenvalues.

$$ T=\frac{2\pi }{\omega } $$

Lowest Vibrational Period
The lowest vibrational period of the electric pylon frame structure was determined to be 0.11583 s. This value was also shown in the MATLAB command window printout above. Because the mode with the longest period is actually called the first fundamental vibration mode, this vibrational period is defined as the "First Mode with the Third Longest Period".

2nd Lowest Vibrational Period
The 2nd lowest vibrational period of the electric pylon frame structure was determined to be 0.12639 s. This value was also shown in the MATLAB command window printout above. Because the mode with the longest period is actually called the first fundamental vibration mode, this vibrational period is defined as the "First Mode with the Second Longest Period".

3rd Lowest Vibrational Period
The 3rd lowest vibrational period of the electric pylon frame structure was determined to be 0.54511 s. This value was also shown in the MATLAB command window printout above. Because the mode with the longest period is actually called the first fundamental vibration mode, this vibrational period is defined as the "First Mode with the Longest Period".

Compare with Lowest 3 Vibrational Periods of Truss Model
(INSERT TABLE COMPARING LOWEST 3 VIBRATIONAL PERIODS OF TRUSS AND FRAME MODEL)

Animation of Lowest Eigenmode of Electric Pylon Truss Model
The animation of the deformation of the Electric Pylon Truss Model at the lowest eigenmode is shown below. The deformations were multiplied by a factor of 50 in order to clearly show the difference between the undeformed and deformed shape.

Animation of 2nd Lowest Eigenmode of Electric Pylon Truss Model
The animation of the deformation of the Electric Pylon Truss Model at the 2nd lowest eigenmode is shown below. The deformations were multiplied by a factor of 20 in order to clearly show the difference between the undeformed and deformed shape.

Animation of 3rd Lowest Eigenmode of Electric Pylon Truss Model
The animation of the deformation of the Electric Pylon Truss Model at the 3rd lowest eigenmode is shown below. The deformations were multiplied by a factor of 20 in order to clearly show the difference between the undeformed and deformed shape.

Contributing Team Members
Andrew McDonald - Eml4500.f08.bike.mcdonald 23:27, 4 December 2008 (UTC) Garrett Pataky - Eml4500.f08.bike.pataky 17:59, 5 December 2008 (UTC) Sam Bernal - Eml4500.f08.bike.bernal 15:48, 9 December 2008 (UTC) Bobby Sweeting - Eml4500.f08.bike.sweeting 23:52, 4 December 2008 (UTC) Shawn Gravois - Eml4500.f08.bike.gravois 03:29, 26 November 2008 (UTC) Eric Viale - Eml4500.f08.bike.viale 01:06, 9 December 2008 (UTC)


text:
User:Arbabian~enwikiversity

Design Studio Based Course (DSBC), a new interface for e-learning

E-learning or electronic learning is a time and location free environment. In the past, e-learning was limited to non-studio based courses and a little attention has been given to such an interface in e-learning. I explored the DSBC related to revolutionize e-learning method. Following introducing of a primitive and prototype model, an interactive interface was introduced at SIAT/SFU. In the first step, it showed that four tools of; Multimedia, Interactive Intelligent Images, Database Library, and a Video Conferencing System in a customized parallel interface, are essential. The interface would facilitate a reasonable and progressive learning performance would possess the most important impact on the e-learning environment. Enhancing the Design Studio Based Course system will improve the users' design knowledge, ability, outcomes and quality in adoption of such an e-learning interface. A positive attitude in the system is introduced in which the effectiveness of the interface leads to its greater adoption by the students and increases the probability of successful results comparing with the conventional operation of the studio courses.

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

title: User:Arbabian~enwikiversity


text:
User:Eml4500.f08.Ateam.Slevinski/Homework1

EML4500 - Week 1 Summary

Overview: EML4500 is an upper level engineering course in the field of Finite Element Analysis and Design. The course, taught by Dr. L. Vu-Quoc at the University of Florida is an overview of the Finite Element Method and its applications. Throughout the course of this semester we will be introduced to the different aspects of the FEM and be applying these to solve different engineering problems. The curriculum will be presented to us through lectures from the professor and homework assignments that will assist in our understanding of the material. To ease in the understanding of the material we will work in small groups that will collaborate on weekly homework reports to be submitted using Wikipedia.

Wiki Homework Reports: As stated previously, we will be using Wikipedia (MediaWiki, WikiVersity,...) to submit our weekly HW reports. Working in small teams, we will be collaborating on the current HW questions and lecture notes to form a weekly HW report. These reports will consist of two parts: the HW problem statement/solution and a summary of that week's lecture notes. For the problem statement and solution we will work with our team to come up with a solution to the problem to the best of our ability and form this into the report. After proof-reading and, if needed, editing each of our team members contribution to the report, our team leader will submit a final copy using our Wiki page. The second half of our weekly reports will consist of a summary of the prevous week's lecture notes. For this section, we will take the lecture notes, rewrite them in our own words and also add any comments or suggestions to clarify and topics that we might not fully understand. In addition to the notes we will also redraw any figures or examples that would assist in our learning of the material.

Pros/Cons of Using Wikipedia: Using Wikipedia to work together and submit our HW reports has its pros and cons. Working in our teams will help our understanding of the material, but also presents some difficulties. This is where using Wiki will help. With this, we will be able to collaborate no matter where we are. As Dr. Vu-Quoc stated, "Even if a group member is in Australia, they will still be able to collaborate with the team." This will be beneficial further in the semester when our schedules keep the entire team from being able to meet in person. As with most good things there is the bad part. For our situation, that problem is the "vandalizing" of Wikipedia pages. Wikipedia is a free online encyclopedia that can be edited by anyone on the internet. This is an issue for us when it comes to submitting our weekly reports. Anyone with malicious intent could go to our page and edit our work to provide incorrect results or even just vandalize it with inappropriate content as seen by an example in class. To counteract this issue, we will be submitting the archived version of our report. Once a page on Wikipedia has been archived, that version can not be changed. The following versions might have been edited, but the archived copy will remain the same.


text:
User:Eml4500.f08.Ateam.Slevinski/Homework2

Class Notes - 9/10

Element 1: From the FBD of element 1, we see that the angle from the x-axis is 30°.

θ(1) = 30ο l(1) = cosθ(1) = cos30ο = $$\frac{\sqrt3}{2}$$ m(1) = sinθ(1) = sin30° = $$\frac{1}{2}$$

For element 1, the stiffness is given as: $$K^{(1)} = \frac{E^{(1)}A^{(1)}}{L^{(1)}}$$ = $$\frac{3}{4}$$

And the matrix of k(1) is given as:

$$ \underline K^{(1)}= K*\begin{bmatrix} k^{(1)}(l^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(l^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} \\ k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(m^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(m^{(1)})^2 \\ -k^{(1)}(l^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(l^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} \\ -k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(m^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(m^{(1)})^2 \end{bmatrix} $$

In this matrix, only three values need to be calulated. The rest only differ by being positive or negative. As you can see this matrix is symetric. Therefore,

 kij(1) = kji(1)

And in general:

 kij(e) = kji(e) and  k(e)T = k(e)

Element 2:

From the FBD of element 2, we see that the angle from the x-axis is -45°.

θ(2) = -45°

l(2) = cosθ(2) = cos(-45°) = $$\frac{\sqrt2}{2}$$

m(2) = sinθ(2) = sin(-45°) = $$\frac{-\sqrt2}{2}$$

For element 2, the stiffness is given as: $$K^{(2)} = \frac{E^{(2)}A^{(2)}}{L^{(2)}}$$ = 2.5

And the matrix of k(2) is given as:

$$ \underline K^{(2)}= K*\begin{bmatrix} k^{(2)}(l^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(l^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} \\ k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(m^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(m^{(2)})^2 \\ -k^{(2)}(l^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(l^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} \\ -k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(m^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(m^{(2)})^2 \end{bmatrix} $$


text:
User:Eml4500.f08.Ateam.Slevinski/Homework3

HW#3: 9/26/08 Lecture Notes (Meeting 14)
From the relationship betweeen qi(e) and di(e):

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$  where  q(e) = T(e)×d(e)

we can use the same argument to solve for the axial displacements, qi(e), and the axial forces, pi(e).

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} T^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$         where  P(e) = T(e)×f(e)

We now can recall the force-displacement relationship:

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

The goal here is to have the original force-displacement relationship: $$ k^{(e)}d^{(e)} = f^{(e)} $$. To accomplish this we would need to move matrix $$ T^{(e)} $$ from the right hand side to the left hand side. This could be achieved by multiplying the equation by the inverse $$ T^{-1} $$ of the matrix $$ T^{(e)} $$. As seen above though, the matrix $$ T^{(e)} $$ is a rectangular matrix and therefore cannot be inverted.

To overcome this problem, we will use the transpose of $$ T^{(e)} $$ and replace $$ k^{(e)} $$ with $$ [T^{(e)T}k^{(e)}T^{(e)}] $$. The force-displacement relationship now becomes:

$$ [T^{(e)T}k^{(e)}T^{(e)}]d^{(e)} = f^{(e)} $$

We can recall from previous lectures that $$ \theta^{(1)} = 30^{o} $$, $$ \theta^{(2)} = 45^{o} $$ and $$ l^{(e)} = cos(\theta^{(e)}) $$, $$ m^{(e)} = sin(\theta^{(e)}) $$. From these values and equations we can determine the matrix $$ T^{(e)} $$.

$$ \begin{Bmatrix} T^{(1)} \end{Bmatrix} = \left[ \begin{matrix} l^{(1)} & m^{(1)} & 0 & 0 \\ 0 & 0 & l^{(1)} & m^{(1)} \end{matrix} \right] = \left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$

And the transpose of $$ T^{(1)} $$ is:

$$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$

From lecture 12 we know that, $$ k^{(1)} = \left[ \begin{matrix} 1 & -1 \\ -1 & 1 \end{matrix} \right] $$

So therefore,

$$ [T^{(1)T}k^{(1)}T^{(1)}] $$ = $$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$'$$ \left[ \begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right] $$'$$\left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$ = $$\left[ \begin{matrix} 0.7499 & 0.433 & -0.7499 & -0.433 \\ 0.433 & 0.25 & -0.433 & -0.25 \\ -0.7499 & -0.433 & 0.7499 & 0.433 \\ -0.433 & -0.25 & 0.433 & 0.25 \end{matrix} \right] $$

And if we recall from previous lectures, we will see that these are the coefficients of the 4x4 matrix $$ \underline k^{(1)} $$.

Calculating the eigenvalues of this matrix leads to values of:

$$ \lambda = \left[ \begin{matrix} 0 & 2 & 0 & 0 \end{matrix} \right] $$

The three zero eigenvalues in this matrix correspond to the three possible rigid body motions. Two of these being translational and the other rotational.

HW #3
From the relationship betweeen qi(e) and di(e):

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$  where  q(e) = T(e)×d(e)

we can use the same argument to solve for the axial displacements, qi(e), and the axial forces, pi(e).

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} T^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$         where  P(e) = T(e)×f(e)

We now can recall the force-displacement relationship:

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

The goal here is to have the original force-displacement relationship: $$ k^{(e)}d^{(e)} = f^{(e)} $$. To accomplish this we would need to move matrix $$ T^{(e)} $$ from the right hand side to the left hand side. This could be achieved by multiplying the equation by the inverse $$ T^{-1} $$ of the matrix $$ T^{(e)} $$. As seen above though, the matrix $$ T^{(e)} $$ is a rectangular matrix and therefore cannot be inverted.

To overcome this problem, we will use the transpose of $$ T^{(e)} $$ and replace $$ k^{(e)} $$ with $$ [T^{(e)T}k^{(e)}T^{(e)}] $$. The force-displacement relationship now becomes:

$$ [T^{(e)T}k^{(e)}T^{(e)}]d^{(e)} = f^{(e)} $$

We can recall from previous lectures that $$ \theta^{(1)} = 30^{o} $$, $$ \theta^{(2)} = 45^{o} $$ and $$ l^{(e)} = cos(\theta^{(e)}) $$, $$ m^{(e)} = sin(\theta^{(e)}) $$. From these values and equations we can determine the matrix $$ T^{(e)} $$.

$$ \begin{Bmatrix} T^{(1)} \end{Bmatrix} = \left[ \begin{matrix} l^{(1)} & m^{(1)} & 0 & 0 \\ 0 & 0 & l^{(1)} & m^{(1)} \end{matrix} \right] = \left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$

And the transpose of $$ T^{(1)} $$ is:

$$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$

From lecture 12 we know that, $$ k^{(1)} = \left[ \begin{matrix} 1 & -1 \\ -1 & 1 \end{matrix} \right] $$

So therefore,

$$ [T^{(1)T}k^{(1)}T^{(1)}] $$ = $$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$'$$ \left[ \begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right] $$'$$\left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$ = $$\left[ \begin{matrix} 0.7499 & 0.433 & -0.7499 & -0.433 \\ 0.433 & 0.25 & -0.433 & -0.25 \\ -0.7499 & -0.433 & 0.7499 & 0.433 \\ -0.433 & -0.25 & 0.433 & 0.25 \end{matrix} \right] $$

And if we recall from previous lectures, we will see that these are the coefficients of the 4x4 matrix $$ k^{(1)} $$.

Given the 2x2 matrix $$ k^{(1)} = \left[ \begin{matrix} 1 & -1 \\ -1 & 1 \end{matrix} \right]$$ we can calculate the eigenvalues of this matrix as follows:

 det$$ \left[ \begin{matrix} (1-\lambda) & -1 \\ -1 & (1-\lambda) \end{matrix} \right]$$ = $$ (1-\lambda)^{2} - 1 = 0 $$.

Therefore, the eigenvalues (or roots) of this equation are:

$$ \lambda_{1} = 0 $$ and $$ \lambda_{2} = 2 $$


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User:Eml4500.f08.Ateam.Slevinski/Homework4

HW 4 - 10/06/08

From the diagram of the two bar truss above, we can see the relationships between the global and local nodes of elements 1 and 2. From these relationships, we can derive the connectivity array. In Matlab, this is denoted as conn(e,j). In this array, the rows represent the elements (e) and the columns represent the local node numbers (j). The values inside the array are the global node numbers. Using the two bar truss diagram above, the connectivity array becomes:

conn = $$ \left[ \begin{array}{cc} {1} & {2} \\ {2}& {3} \end{array} \right]$$

For example, in this case the local node of element 1 corresponds to the global node 1. Therefore, the value of in the first row and column is 1.

Another array that we will need to derive is the location matrix master array or "LMM". This array, will represent the relationships between the local degrees of freedom (DOF's) and the global DOF's numbers. As in the connectivity array, the rows in the LMM represent the elements (e), but the columns represent the local DOF's (j). In Matlab this command is lmm(e,j).

The above diagrams show the elements of the two bar truss and their local DOF's. From these diagrams, we will populate our location matrix master array. The array becomes:

lmm = $$ \left[ \begin{array}{cccc} {1} & {2} & {3} & {4} \\ {3} & {4} & {5} & {6} \end{array} \right]$$

HW 4 - 10/13/08
When the eigenvalues of the 6x6 matrix $$ \underline{K}$$ are calculated, there are 4 zero values. Three of these values correspond to the three rigid body motions and the other corresponds to the mechanism.

Eigenvalue problem: For this problem we will use the following equation: $$ \underline{K}\underline{v}=\lambda\underline{v} $$

For this matrix, we will let {$$ \underline{u}_{1}, \underline{u}_{2}, \underline{u}_{3}, \underline{u}_{4} $$} be the pure eigenvectors corresponding to the four zero eigenvalues. The equation above now becomes:

$$ \underline{K}\underline{u}_{i}=0\cdot\underline{u}_{i}=\underline{0} $$

Here $$ \underline{K}$$ is a 6x6 matrix, $$ \underline{u}$$ is a 6x1 matrix, and i=1,...,4.

We can now compute a linear combination on the values of $$ \underline{u}_{i}, i=1,...,4$$ with a summation:

$$ \sum_{i=1}^4 \alpha_{i} \underline{u}_{i} =: \underline{W} $$.

In this summation, $$ \alpha_{i}$$ is a 1x1 matrix, $$ \underline{u}_{i}$$ is a 6x1 matrix, $$ \underline{W}$$ is a 6x1, and $$ \alpha_{i}$$ are real numbers. The $$ =$$ sign here means "equal by definition" and $$ \underline{W}$$ is a zero eigenvector corresponding to a zero eigenvalue. Our original equation for the eigenvalue problem now becomes:

$$ \underline{K}\underline{W}=\underline{K}(\sum_{i=1}^4\alpha_{i} \underline{u}_{i})=\sum_{i=1}^4 \alpha_{i}(\underline{K}\underline{u}_{i})=\underline{0}=0\cdot\underline{W} $$

Using the equation $$ \underline{K}\underline{v}=\lambda\underline{v} $$ we will now evaluate a three and four bar truss.

Given the values in the diagram, we can compute the stiffness matrix $$ \underline{K}$$.

HW: Plot the eigenvectors corresponding to the zero eigenvalues of the two-bar truss system.

For the two bar truss system, our matrix $$ \underline{K}$$ is defined as:

With this matrix, we can now plot the eigenvalues and eigenvectors using the Matlab code [V,D]=eig(K). This will display the eigenvectors (V) and eigenvalues (D):

HW2: Plot the eigenvectors corresponding to the zero eigenvalues of the three-bar truss system above. The matrix $$ \underline{K}$$ has been determined to be:

Using the same command from earlier ([V,D]=eig(A)) we can determine the eigenvalues and eigenvectors of this matrix:


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User:Eml4500.f08.Ateam.Slevinski/Homework5

HW 5 - Derivation of FD Relation in Global Coordinates Using the PVW
Derive the transformation matrix based on the lecture for the 2-D truss elements:

$$a_1^{(e)} = d_1^{(e)} \cdot \vec\tilde{i} = ( d_1 \tilde{i} + d_2 \tilde{j} + d_3 \tilde{k} ) \cdot \vec\tilde{i} = d_1 \cdot ( \tilde{i} \cdot \vec\tilde{i} ) + d_2 \cdot ( \tilde{j} \cdot \vec\tilde{i} ) + d_3 \cdot ( \tilde{k} \cdot \vec\tilde{i} )$$

where $$( \tilde{i} \cdot \vec\tilde{i} ) = l^{(e)}, ( \tilde{j} \cdot \vec\tilde{i} ) = m^{(e)}, and ( \tilde{k} \cdot \vec\tilde{i} ) = n^{(e)}$$

$$a_1^{(e)} = l^{(e)}d_1 + m^{(e)}d_2 + n^{(e)}d_3$$

$$a_1^{(e)} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}\begin{vmatrix} d_1\\ d_2\\ d_3 \end{vmatrix}$$

$$a_2^{(e)} = d_2^{(e)} \cdot \vec\tilde{i} = ( d_4 \tilde{i} + d_5 \tilde{j} + d_6 \tilde{k} ) \cdot \vec\tilde{i} = d_4 \cdot ( \tilde{i} \cdot \vec\tilde{i} ) + d_5 \cdot ( \tilde{j} \cdot \vec\tilde{i} ) + d_6 \cdot ( \tilde{k} \cdot \vec\tilde{i} )$$

where $$( \tilde{i} \cdot \vec\tilde{i} ) = l^{(e)}, ( \tilde{j} \cdot \vec\tilde{i} ) = m^{(e)}, and ( \tilde{k} \cdot \vec\tilde{i} ) = n^{(e)}$$

$$a_2^{(e)} = l^{(e)}d_4 + m^{(e)}d_5 + n^{(e)}d_6$$

$$a_2^{(e)} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}\begin{vmatrix} d_4\\ d_5\\ d_6 \end{vmatrix}$$

$$\begin{vmatrix} a_1^{(e)}\\ q_2^{(e)} \end{vmatrix} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix} \begin{vmatrix} d_1\\d_2 \\d_3 \\ d_4 \\ d_5 \\ d_6

\end{vmatrix}$$

Note: The transformation matrix is in terms of the three director cosines.

Recall the relation between axial DOF's and element DOF's in the global coordinate system:

$$\underline{q} = \underline{T} \underline{d}$$

and also the relation between axial forces and element forces in the global coordinate system:

$$\underline{P} = \underline{T} \underline{f}$$

Using the principle of virtual work (PVW) we can derive the force-displacement relationship in the global coordinate system.

$$\underline{k} \underline{d} = \underline{F}$$

therefore,

$$\underline{k} \underline{d} - \underline{F} = 0$$

With the PVW, this becomes:

$$\underline{W} \cdot ( \underline{k} \underline{d} - \underline{F} ) = 0$$

where $$\underline{W}$$ is the weighting matrix. For this system we have 6 choices for $$\underline{W}$$:

Choice 1:

$$ W = \begin{vmatrix}1 & 0 & 0 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{1,j}d_j - F_1 = 0}$$

$$\sum_{j=1}^{6}{k_{1,j}d_j = F_1}$$

Choice 2:

$$ W = \begin{vmatrix}0 & 1 & 0 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{2,j}d_j - F_2 = 0}$$

$$\sum_{j=1}^{6}{k_{2,j}d_j = F_2}$$

Choice 3:

$$ W = \begin{vmatrix}0 & 0 & 1 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{3,j}d_j - F_3 = 0}$$

$$\sum_{j=1}^{6}{k_{3,j}d_j = F_3}$$

Choice 4:

$$ W = \begin{vmatrix}0 & 0 & 0 & 1 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{4,j}d_j - F_4 = 0}$$

$$\sum_{j=1}^{6}{k_{4,j}d_j = F_4}$$

Choice 5:

$$ W = \begin{vmatrix}0 & 0 & 0 & 0 & 1 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{5,j}d_j - F_5 = 0}$$

$$\sum_{j=1}^{6}{k_{5,j}d_j = F_5}$$

Choice 6:

$$ W = \begin{vmatrix}0 & 0 & 0 & 0 & 0 & 1\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{6,j}d_j - F_6 = 0}$$

$$\sum_{j=1}^{6}{k_{6,j}d_j = F_6}$$

$$\begin{vmatrix} k_{1,1} & k_{1,2} & k_{1,3} & k_{1,4} & k_{1,5} & k_{1,6}\\ k_{2,1} & k_{2,2} & k_{2,3} & k_{2,4} & k_{2,5} & k_{2,6}\\ k_{3,1} & k_{3,2} & k_{3,3} & k_{3,4} & k_{3,5} & k_{3,6}\\ k_{4,1} & k_{4,2} & k_{4,3} & k_{4,4} & k_{4,5} & k_{4,6}\\ k_{5,1} & k_{5,2} & k_{5,3} & k_{5,4} & k_{5,5} & k_{5,6}\\ k_{6,1} & k_{6,2} & k_{6,3} & k_{6,4} & k_{6,5} & k_{6,6}\\ \end{vmatrix} \begin{vmatrix} d_1\\d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6

\end{vmatrix} = \begin{vmatrix} F_1\\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6

\end{vmatrix}$$

HW 5 - Derivation of Elemental Stiffness Matrix in Global Coordinates Using PVW
The following is an expression of the element stiffness matrix $$\underline{K}_{6x6}^{(e)}$$ in terms of the transformation matrix $$\underline{T}^{(e)}$$ and the element stiffness matrix $$\underline\hat{k}^{(e)}$$ from the axial force displacement relationship. We will show that by using the priciple of virtual work, we can derive the equation for the element stiffness matrix in global coordinates.

$$\underline{K}^{(e)} = \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)}$$

Recall the force displacement relationship with axial DOF's $$ q^{(e)}$$

$$\underline\hat{k}_{2x2}^{(e)} \underline{q}_{2x1}^{(e)} = \underline{p}_{2x1}^{(e)}$$

Rearranging this equation yields:

$$ \underline\hat{k}^{(e)} \underline{q}^{(e)} - \underline{p}^{(e)} = \underline{0}_{2x1}$$    (1)

We can now apply the principle of virtual work (PVW) to get:

$$\underline\hat{W} \cdot ( \underline\hat{k}^{(e)} \underline{q}^{(e)} - \underline{p}^{(e)} ) = 0^{'}_{1x1}$$   (2)  for all $$\underline\hat{W}_{2x1}$$

We now can recall the equation:

$$\underline{q}_{2x1}^{(e)} = \underline{T}_{2x6}^{(e)} \underline{d}_{6x1}^{(e)}$$    (3)

Similarly,

$$\hat{W}_{2x1} = \underline{T}^{(e)}_{2x6} \underline{W}_{6x1}$$ (4)

Now, substitute equations (3) & (4) into equation (2) which yields:

$$( \underline{T}^{(e)} \underline{W} ) \cdot [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - \underline{p}^{(e)} ] = 0$$  for all $$ \underline{W}_{6x1}$$   (5)

Recall that:

$$ ( \underline{A}_{pxq} \underline{B}_{qxr} )^{T} = \underline{B}^{T} \underline{A}^{T}$$  (6)

And also that:

$$\underline{a}_{nx1} \cdot \underline{b}_{nx1} = \underline{a}^{T}_{1xn} \underline{b}_{nx1}$$   (7)

Now we can apply equations (6) and (7) into (5) to get:

$$ ( \underline{T}^{(e)} \underline{W} )^{T} [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - ( \underline{p}^{(e)} ] = 0 $$ for all $$\underline{W}_{6x1}$$

In this equation, $$\underline\hat{W}_{2x1}$$ represents the virtual axial displacement and from eq. (4) and the matrix $$\underline{W}_{6x1}$$ represents the virtual displacement in the global coordinate system corresponding to $$\underline{d}^{(e)}$$

The previous equation now becomes:

$$\underline{W}^{T} \underline{T}^{(e)T} [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - \underline{p}^{(e)} ] = 0$$

$$\underline{W} \cdot [ ( \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)} ) \underline{d}^{(e)} - ( \underline{T}^{(e)T} \underline{p}^{(e)}) ] = 0$$ for all $$\underline{W}_{6x1}$$

In the equation above, the element stiffness matrix $$\underline{K}_{6x6}^{(e)}$$ is represented by $$( \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)} )$$and the force matrix $$\underline{f}_{6x1}^{(e)}$$ is represented by $$\underline{T}^{(e)T} \underline{p}^{(e)}$$

The equation now becomes:

$$\underline{W} \cdot [ \underline{K}^{(e)} \underline{d}^{(e)} - \underline{f}^{(e)} ] = 0$$ for all $$\underline{W}$$

We have the original expression for the elemental stiffness matrix in terms of the axial force displacement relationship.

$$\underline{K}^{(e)} \underline{d}^{(e)} = \underline{f}^{(e)}$$

HW 5 - Element DOF's and Element Forces in Global xyz-Coordinate System for 6x1 Matrices

Using the previous diagram, we can write the following relationships.

$$\underline{k}^{(e)} \underline{q}^{(e)} = \underline{p}^{(e)}$$

$$\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix} \begin{bmatrix}q_{1}^{(e)}\\ q_{2}^{(e)}\end{bmatrix} = \begin{bmatrix}p_{1}^{(e)}\\ p_{2}^{(e)}\end{bmatrix}$$

From the diagram above, we can write the element DOF's and forces for a 3-D system.

$$\underline{f}^{(e)} = \begin{vmatrix}F_1\\F_2 \\F_3 \\ F_4\\ F_5\\ F_6\end{vmatrix}_{6x1}$$ and $$\underline{d}^{(e)} = \begin{vmatrix} d_1\\d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{vmatrix}_{6x1}$$

We can now use the transformation matrix $$\underline{T}$$ to transform the 6 element DOF's in the global xyz-coordinate system to the 2 axial DOF's $$p$$ and $$q$$.

$$\underline{q}^{(e)} = \underline{T} \underline{d}^{(e)}$$

This equation expands to:

$$\begin{vmatrix}q_{1}^{(e)}\\q_{2}^{(e)} \end{vmatrix}_{2x1} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}_{2x6} \begin{vmatrix}d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{vmatrix}_{6x1}$$

Similarly, this can be applied to the equation:

$$\underline{p} = \underline{T} \underline{f}$$

From either of the above equations, to solve for the element forces $$\underline{f}$$ or displacements $$\underline{d}$$ we can use the transpose of the transformation matrix $$\underline{T}^{T}$$.

$$\underline{f} = \underline{T}^{T} \underline{p}$$

$$\underline{d} = \underline{T}^{T} \underline{q}$$


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User:Eml4500.f08.Ateam.Slevinski/Homework6

HW 6 - Meeting 33 - 11/12/08
Continuing the finite element method via the principle of virtual work:

$$u(x_{i+1}) = d_{i+1}$$

Using the same method for the interpolation of $$u(x)$$ we can apply this to $$ W(x)$$:

$$ w(x) = N_{i}(x)W_{i} + N_{i+1}(x)W_{i+1}$$

Element Stiffness Matrix for Element i:

$$\beta = \int_{x_{i}}^{x_{i 1}}{[N'_{i}W_{i}  N'_{i+1}W_{i+1}](EA)[N'_{i}d_i   N_{i+1}d_{i+1}]dx}$$   (1)

In the above equation, $$[N'_{i}W_{i}  N'_{i+1}W_{i+1}]$$ represents $$W'(x)$$, and $$u'(x) = [N'_{i}d_i   N_{i+1}d_{i+1}]$$

Also, $$N'_i$$ is the derivative of $$N_i(x)$$ which is represented as: $$N'_i = \frac{dN_i(x)}{dx}$$ The same applies for $$N'_{i+1}$$

Note: From the previous equation, $$u(x) = \begin{bmatrix} N_i(x) & N_{i+1}(x) \end{bmatrix}_{1x2} \begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix}_{2x1}$$ where $$\begin{bmatrix} N_i(x) & N_{i+1}(x) \end{bmatrix}_{1x2}$$ is the 1x2 matrix $$\underline{N}(x)$$

Now, the derivative of this is: $$\frac{du(x)}{dx} = \begin{bmatrix} N'_i(x) & N'_{i+1}(x)\end{bmatrix}_{1x2} \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}_{2x1}$$ and in this equation, $$\begin{bmatrix} N'_i(x) & N'_{i+1}(x)\end{bmatrix}_{1x2}$$ is the 1x2 matrix $$\underline{B}(x)$$

Now, with the matrices $$\underline{N}(x)$$ and $$\underline{B}(x)$$ we can develop the following relationships:

$$W(x) = \underline{N}(x) \begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix}$$

$$\frac{dW(x)}{dx} = \underline{B}(x) \begin{Bmatrix} W_i \\ W_{i 1} \end{Bmatrix}$$

Now we need to recall the element degrees of freedom:

From the figure above, we have: $$\begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix} = \begin{Bmatrix} d_1^{(i)} \\ d_{2}^{(i)} \end{Bmatrix} = \underline{d}^{(i)}$$ and also the relationship for the weighting coefficient $$\begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix} = \begin{Bmatrix} W_1^{(i)} \\ W_{2}^{(i)} \end{Bmatrix} = \underline{W}^{(i)}$$

We can now rewrite equation (1) using the previous relationships. Equation (1) becomes:

$$\beta = \int_{x_i}^{x_{i+1}}{(\underline{B}\underline{W}^{(i)})(EA)(\underline{Bd}^{(i)})dx}$$ (2)

$$ = \underline{W}^{(i)} \cdot (\underline{k}^{(i)} \underline{d}^{(i)})$$

Now, rearranging equation (2) with the dot product we have:

$$\beta = \int_{x_i}^{x_{i+1}}{(EA)(\underline{BW}^{(i)})\cdot(\underline{Bd}^{(i)})dx}$$ (3)

From equation (3) we can simplify even further using the transpose of weighting matrix:

$$(\underline{BW}^{(i)})^T (\underline{Bd}^{(i)}) = $$

$$ \underline{W}^{(i)T} \underline{B}^T = $$

$$\underline{W}^{(i)} \cdot \underline{B}^T$$

Once again, we can rewrite equation (3) to form:

$$\beta = \underline{W}^{(i)} \cdot (\int {\underline{B}^T(EA)\underline{B}}dx)\underline{d}^{(i)}$$ (4)

In equation (4), the integral represents the 2x2 element stiffness matrix:

$$\underline{k}^{(i)} = \int_{x_i}^{x_{i+1}}{\underline{B}^T(x)(EA)\underline{B}(x)}dx$$

Transform of variable from $$x$$ to $$ \tilde{x}$$:

$$ \tilde{x} = x-x_i$$

$$ d\tilde{x} = dx$$

$$ \underline{k}^{(i)} = \int_{\tilde{x} = 0}^{\tilde{x} = L^{(i)}}{\underline{B}^T(\tilde{x})(EA)(\tilde{x})\underline{B}(\tilde{x})}d\tilde{x}$$


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User:Eml4500.f08.Ateam.Slevinski/Homework7

Lecture Notes - 11/17/08
Element 1: $$E_1^{(1)} = 2$$  $$ E_2^{(1)} = 4$$ $$A_1^{(1)} = 0.5$$ $$ A_2^{(1)} = 1.5$$

Element 2: $$E_1^{(1)} = 2$$  $$ E_2^{(1)} = 4$$ $$A_1^{(1)} = 0.5$$ $$ A_2^{(1)} = 1.5$$

Model Frame With Two Elements:

In the prevoius 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 $$\underline{K}_{9x9} = A\underline{k}_{6x6}^{(e)}$$

Lecture Notes - 11/21/08
$$[\varepsilon ] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$

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

$$[A] = L^2$$

$$[I] = L^4$$

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

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

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

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

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


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text:
User:Jestermeister/Seed/elevate

Windows Vista introduced a new security feature: User Account Control (UAC). On one look, it seems to mimic the elevation feature that Linux has. However, Vista does not provide a command-line equivalent tool to UAC. The best that one can do is to do a "run as administrator" on command prompt and start running everything from there.

Elevate is a bunch of scripts created to address this issue. With it, running applications/scripts with elevated rights are now much easier. One can do it even from "Run..." in the START menu.

If you copy and paste all the files into the Windows/System32 folder, you will be able to execute elevate from anywhere. An example of its usage will be:

elevate cmd

This command launches the command prompt in elevated privileges.

title: User:Jestermeister/Seed/elevate


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text:
User:Jestermeister/Seed/psinfo

PsInfo is a tool provided by Sysinternals. It is used to show information of the local system. A useful way to use this tool is to collect the Hotfixes and Installed Software from the user and store them in a text file.

psinfo -s -h > outputfile.txt

title: User:Jestermeister/Seed/psinfo


text:
Talk:Egyptology/Resources

Just so there's a link to the work you did before:. The Jade Knight (d'viser) 10:40, 25 November 2008 (UTC)
 * OK. Fine. very thoughtful of you. I may use it in a new context later on perhaps with something related to Ancient Egypt. Will see. Happy Thanksgiving 2008! Dilos1 07:30, 26 November 2008 (UTC)
 * Happy Thanksgiving to you too! The Jade Knight (d'viser) 10:14, 26 November 2008 (UTC)

title: Talk:Egyptology/Resources


text:
User:Jestermeister/Seed/devcon

DEVice CONsole
Devcon in short will be: Device management on the command-line.

It is a Microsoft tool offered free of charge to the public. This site here provides some scripts making use of Devcon. Some of the scripts that are provided include Backing up of Drivers, Listing all installed printers and printer ports, List all Drivers and Check if a HP Deskjet Printer is installed.

The following is one of the scripts provided from the website. It can be run as a batch file (.bat). The objective of this script will be to list out all printers and printer ports in the system.

@ECHO OFF
 * Check Windows version

IF "%OS%"=="Windows_NT" (SETLOCAL) ELSE (GOTO Syntax) VER | FIND "Windows NT" >NUL && GOTO Syntax

 * Check command line arguments

IF NOT "%~1"=="" IF /I NOT "%~1"=="/A" GOTO Syntax

 * Check if DEVCON.EXE is available and if not, offer to download it

SET DevconAvailable= SET Download= DEVCON.EXE /? >NUL 2>&1 IF ERRORLEVEL 1 (	SET DevconAvailable=No	ECHO This batch file requires Microsoft's DEVCON untility.	SET /P Download=Do you want to download it now? [y/N] )

 * Start download if requested

IF /I "%Download%"=="Y" (	START "DevCon" "http://support.microsoft.com/default.aspx?scid=kb;EN-US;Q311272"	ECHO.	ECHO Install the downloaded file and make sure DEVCON.EXE is in the PATH.	ECHO Then try again. )

 * Abort if DEVCON.EXE is not available yet

IF "%DevconAvailable%"=="No" GOTO End

 * List header

ECHO Printer name:	Port:

 * List all printers

FOR /F "tokens=1* delims=: " %%A IN ('DEVCON.EXE FindAll ^=Printer ^| FIND.EXE ":"') DO CALL :ShowPrn "%%~A" "%%~B"

ENDLOCAL GOTO:EOF

 * ShowPrn

SETLOCAL SET HwID="%~1" SET HwID=%HwID:&= % SET HwID=%HwID:"= % FOR %%X IN (%HwID%) DO SET Port=%%X FOR /F "tokens=1 delims=:." %%X IN ('ECHO.%Port%') DO SET Port=%%X SET Name=%~2 ECHO %Name%	%Port% ENDLOCAL GOTO:EOF

 * Syntax

ECHO. ECHO ShowPRNT.bat, Version 1.00 for Windows 2000 / XP ECHO Use DEVCON to list installed printers and their ports in tab delimited format ECHO. ECHO Usage: SHOWPRNT ECHO. ECHO Notes: [1] This batch file requires Microsoft's DEVCON.EXE, available at ECHO             http://support.microsoft.com/default.aspx?scid=kb;EN-US;Q311272 ECHO            You will be prompted for download if it isn't found. ECHO        [2] Both active and inactive printers are shown. ECHO        [3] Contrary to its VBScript counterpart, this batch file only lists ECHO            "physical" printers. ECHO        [3] This batch file's VBScript counterpart also lists driver names, ECHO            server and share names. ECHO. ECHO Written by Rob van der Woude ECHO http://www.robvanderwoude.com

title: User:Jestermeister/Seed/devcon


text:
Talk:Instructional design/Psychomotor behaviors/Introduction/Answer Key

This page needs to be moved to be specific to whatever it is being used for: it would be ideal as a subpage. The Jade Knight (d'viser) 12:04, 25 November 2008 (UTC)
 * ✅ —Internoob (Disc.·Cont.·Wikt.) 00:38, 26 January 2010 (UTC)


text:
MediaWiki talk:Privacypage

Huh? The Jade Knight (d'viser) 13:07, 25 November 2008 (UTC)

title: MediaWiki talk:Privacypage


text: User:CWardUSC

title: User:CWardUSC