User:Eml4500.f08.bottle.vitello/HW 7

Homework Problem
Resuming the 2-bar truss system from lecture 34, we assign the elements the following properties: Element 1: $$ E^{(1)}_{1}=2, A^{(1)}_{1}=0.5 \! $$ $$ E^{(1)}_{2}=4, A^{(1)}_{2}=1.5 \! $$ Element 2: $$ E^{(2)}_{1}=3, A^{(2)}_{1}=1 \! $$ $$ E^{(2)}_{2}=7, A^{(2)}_{2}=3 \! $$

We can then compute the solution for the 2-bar truss system with tapered elements. The solution and plot of the deformed shapes can be found below.

Intro to Frame Elements
A frame element is composed of a truss (bar) element and a beam element. The truss element undergoes axial deformation while the beam element undergoes transverse deformation. A model problem can be found below. Note the rigid connection where the force $$ P \! $$ is applied. This implies that the angle between the two elements will remain constant after deformation.

Frame FBD’s
Due to the axial and transverse deformation, we now draw the free body diagrams for each element as seen below.

In general, we assume $$ d^{(e)}_{i} \! $$ correlates to $$ f^{(e)}_{i} \! $$ (This means that we generalize the forces, understanding that the rotational forces are the bending moments about that node), where $$ e = 1, 2 \! $$, $$ i = 1,2,3,4,5,6 \! $$

We can then draw the global dof as follows:

We are then able to construct 2 element stiffness matrices: $$ k^{(e)}_{6x6}, e = 1, 2 \! $$ $$ K_{9x9}=Ak^{(e)}_{6X6} \! $$

We can then display K as:

For ease of computation, we then chose a different coordinate system. The free body diagram for element 1 is

Where $$ \tilde{k}^{(e)}_{6x6}\tilde{d}^{(e)}_{6x6}=\tilde{f}^{(e)}_{6x6} \! $$ and $$\tilde{d}^{(e)}=\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \cdot \cdot \cdot \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} \! $$ and $$ \tilde{f}^{(e)}=\begin{Bmatrix} \tilde{f}_{1}^{(e)}\\ \cdot \cdot \cdot \\ \tilde{f}_{6}^{(e)}

\end{Bmatrix} \! $$

Note: Because of the moments about $$ \tilde{x} \! $$, $$ \tilde{f}^{(e)}_{3}=f^{(e)}_{3} \! $$ and $$ \tilde{f}^{(e)}_{6}=f^{(e)}_{6} \! $$

Constructing $$ \tilde{k} \! $$
We find

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

Dimensional Analysis
Note: For our dimensional analysis we will use the following symbol $$ [ ] \! $$ to denote "dimension of...".

$$ [\tilde{d}_{1}] = L = [\tilde{d}_{i}], i=1,2,4,5 \! $$ $$ [\tilde{d}_{3}]= 1 (dimensionless) = [\tilde{d}_{6}] \! $$

We use the following proof for $$ [\theta ] \! $$



$$ [\theta ] = \frac{[\bar{AB}]}{[R]}= \frac{L}{L} = 1 \! $$ $$ \sigma =E\epsilon \Rightarrow [\sigma ]=[E]\underbrace{[\epsilon ]}_{=1} \! $$ $$ [\epsilon ] = \frac{du}{dx}=\frac{L}{L}=1 \!$$ $$ [\sigma ]=[E]=\frac{F}{L^{2}} \! $$ $$ [A] = L^{2}, [I] = L^{4} \! $$ $$ [\frac{EA}{L}]=[\tilde{k}_{11}= \frac{\frac{F}{L^{2}}L^{2}}{L}= \frac{F}{L} \! $$ $$ [\tilde{k}_{11}\tilde{d_{1}}]=F \! $$ $$ [\tilde{k}_{23}\tilde{d}_{3}]=[\tilde{k}_{23}]1=\frac{[6][E][I]}{L^{2}}= \frac{1(\frac{F}{L^{2}})L^{4}}{L^{2}}=F \! $$

From Local to Global Coordinates
We now to the elemental FD relationship in global coordinates from the elemental FD relationship in local coordinates.

$$ k^{(e)}_{6x6}d^{(e)}_{6x6}=f^{(e)}_{6x6} \! $$ $$ k^{(e)}_{6x6}= \!$$$$\tilde{T}^{(e)} \!$$$$_{6x6}^{T} \!$$ $$\tilde{k}^{(e)}_{6x6}\tilde{T}^{(e)}_{6x6} \! $$ We can then construct the following matrices

$$\begin{Bmatrix} \tilde{d}_{1}\\ \tilde{d}_{2}\\ \tilde{d}_{3}\\ \tilde{d}_{4}\\ \tilde{d}_{5}\\ \tilde{d}_{6} \end{Bmatrix}_{6x1}= \underbrace{\begin{bmatrix} l^{(e)} & m^{(e)} & 0 &0 &0  &0 \\ -m^{(e)}&l^{(e)} &0  & 0 & 0 & 0\\ 0& 0 & 1 & 0 &0 &0 \\ 0 &0  &0  & l^{(e)} &m^{(e)}  & 0\\ 0 &0 &0  & -m^{(e)} & l^{(e)} &0 \\ 0& 0 &0 &0  & 0 & 1 \end{bmatrix}_{6x6}}_{\tilde{T}} \begin{Bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6} \end{Bmatrix}_{6x1} \! $$

where $$\begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)}& l^{(e)} \end{bmatrix} = R \! $$

The 2-element frame system is solved below. The same data used for the 2-bar truss system is used to solve the current system. The deformation of the two systems are then plotted on the same graph to give the reader a better understanding of a truss system vs a frame system.

PVW for Beams
$$ \int_{0}^{L}{W(x)[\frac{-\partial^{2} }{\partial x^{2}}((EI)\frac{\partial ^{2}v}{\partial x^{2}})+f_{t}-m\ddot{v}]dx}=0 |! $$ for all $$ W(x)\; equation 1 \! $$ We can use integration by parts on the first term:

$$ \alpha =\; \int_{0}^{L}{\underbrace{W(x)}_{s(x)}\frac{\partial ^{2}}{\partial x^{2}}((EI)\frac{\partial ^{2}v}{\partial x^{2}})dx} \! $$ where $$\underbrace{\frac{\partial }{\partial x}\underbrace{(\frac{\partial }{\partial x}(EI)\frac{\partial ^{2}v}{\partial x^{2}})}_{r(x)}}_{r'(x)} \! $$

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

By integrating the second part we achieve: $$ [\beta _{1}]^{L}_{0}-\underbrace{[\frac{dW}{dx}(EI)\frac{\partial ^{2}v}{\partial x^{2}}]^{L}_{0}}_{\beta _{2}}+\underbrace{\int_{0}^{L}{\frac{d^{2}W}{dx^{2}}((EI)\frac{\partial ^{2}v}{\partial x^{2}})dx}}_{\gamma } \! $$

Therefore, $$\alpha =\beta _{1}+\beta _{2}-\gamma \! $$ >br> and equation 1 can be written as $$ -\beta _{1}+\beta _{2}-\gamma +\int_{0}^{L}{wf_{t}dx}-\int_{0}^{L}{w\, m\, \ddot{v}dx}=0 \! $$

Shape Functions
Now we'll focus on the stiffness term $$ \gamma \! $$ to derive the beam stiffness matrix and to identify the shape functions.

Equation 2$$ = \; v(\tilde{x})=N_{2}(\tilde{x})\tilde{d}_{2}+ \! $$$$N_{3}(\tilde{x})\! $$$$\tilde{d}_{3}+ \! $$$$N_{5}(\tilde{x})\tilde{d}_{5}+N_{6}(\tilde{x})\tilde{d}_{6} \! $$ Recall equation 1 $$=\; u(\tilde{x})=\ N_{1}(\tilde{x})\tilde{d}_{1}+N_{4}(\tilde{x})\tilde{d}_{4} \! $$

The shape functions can be written as $$N_{2}(\tilde{x})=1-\frac{3\tilde{x}^{2}}{L^{2}}+\frac{2\tilde{x}^{3}}{L^{3}} \! $$ $$N_{3}(\tilde{x})=\tilde{x}-\frac{2\tilde{x}^{2}}{L}+\frac{\tilde{x}^{3}}{L^{2}} \! $$ $$N_{5}(\tilde{x})=\frac{3\tilde{x}^{2}}{L^{2}}-\frac{2\tilde{x}^{3}}{L^{3}} \! $$ $$N_{6}(\tilde{x})=-\frac{\tilde{x}^{2}}{L}+\frac{\tilde{x}^{3}}{L^{2}} \! $$ $$\tilde{d}^{(e)}_{6x1}=\tilde{T}^{(e)}_{6x6} d^{(e)}_{6x1} \! $$ Where $$d^{(e)}_{6x1} \! $$ is known after solving the FE system. Compute $$ u(\tilde{x}), v(\tilde{x})\! $$:  $$u(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+v(\tilde{x})\vec{\tilde{j}} \!$$ $$ = u_{x}(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+u_{y}(\tilde{x})\vec{\tilde{j}} \!$$ We can compute $$ u(\tilde{x}), v(\tilde{x})\! $$ by using equation 1 and 3. $$\begin{Bmatrix} u_{x}(\tilde{x})\\ u_{y}(\tilde{x}) \end{Bmatrix}= R^{T}\begin{Bmatrix} u(\tilde{x})\\ v(\tilde{x}) \end{Bmatrix} \! $$  $$\begin{Bmatrix} u(\tilde{x})\\ v(\tilde{x}) \end{Bmatrix}=\underbrace{\begin{bmatrix} N_{1} &0 &0  & N_{4} & 0 & 0\\ 0 & N_{2} &N_{3} & 0 & N_{5} & N_{6} \end{bmatrix}}_{\mathbb{N}(\tilde{x})}\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \tilde{d}_{3}^{(e)}\\ \tilde{d}_{4}^{(e)}\\ \tilde{d}_{5}^{(e)}\\ \tilde{d}_{6}^{(e)} \end{Bmatrix} \! $$  $$ \begin{Bmatrix} u_{x}(\tilde{x})\\ u_{v}(\tilde{x}) \end{Bmatrix}=R^{T}\mathbb{N}(\tilde{x})\tilde{T}^{(e)}d^{(e)} \! $$

In order to solve these equations, we use the dimensional analysis to solve the second equation assuming:

$$\begin{bmatrix} u \end{bmatrix}=L$$

From our previous presentation of solving linear functions in x, we assumed:

$$\begin{bmatrix} N_{1} \end{bmatrix}=\begin{bmatrix} N_{4} \end{bmatrix}=1$$

By plugging these values into the second equation for u:

$$\begin{bmatrix} N_{1} \end{bmatrix}\begin{bmatrix} \tilde{d}_{1} \end{bmatrix}+\begin{bmatrix} N_{4} \end{bmatrix}\begin{bmatrix} \tilde{d}_{4} \end{bmatrix}$$

we can then solve for the displacements and rotations for v:

$$\begin{bmatrix} v \end{bmatrix}=L$$

$$\begin{bmatrix} N_{2} \end{bmatrix}\begin{bmatrix} \tilde{d}_{3} \end{bmatrix}=L$$

$$\begin{bmatrix} N_{3} \end{bmatrix}\begin{bmatrix} \tilde{d}_{3} \end{bmatrix}=L$$

Derivation of beam shape functions
We recall the shapes for $$N_{2}, N_{3}, N_{5}, N_{6}$$ and the governing PDE for beams for the distorted frame load without inertia force, in which case, it is defined as a static case.

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

Considering constant EI:

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

We integrate four times to get four constants:

$$V(x)=C_{0}+C_{1}x^{1}+C_{2}x^{2}+C_{3}x^{3}$$

to obtain the values for $$N_{2}(x)$$.

$$V(0)=1$$

$$V(L)=0$$

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

We then use the above case to solve for $$C_{0},...,C_{3}$$:

$$V(0)=1=C_{0}$$ $$V(L)=1+C_{1}L+C_{2}L^{2}+C_{3}L^{3}=0$$ $$V'(0)=C_{1}=0$$ $$V'(L)=2C_{2}L+3C_{3}L^{2}=0$$ $$C_{3}=-\frac{2}{3}\frac{C_{2}}{L}$$ $$0=1+C_{2}L^{2}+\begin{pmatrix} -\frac{2}{3}\frac{C_{2}}{L} \end{pmatrix}L^{3} $$  $$0=1+C_{2}L^{2}\begin{bmatrix} 1-\frac{2}{3} \end{bmatrix} $$  which we then obtain the following results: $$C_{0}=1$$ $$C_{1}=0$$ $$C_{2}=-\frac{3}{L^{2}}$$ $$C_{3}=-\frac{2}{3}\frac{1}{L}\begin{pmatrix} -\frac{3}{L^{2}} \end{pmatrix}=\frac{2}{L^{3}} $$

We can then compare with the expression for $$N_{2}$$

For $$N_{3}$$: $$V(0)=V(L)=0$$ $$V'(0)=1$$ $$V'(L)=0$$ For $$N_{5}$$: $$V(0)=0$$ $$V(L)=1$$ $$V'(0)=V'(L)=0$$ For $$N_{6}$$ $$V(0)=V(L)=0$$ $$V'(0)=0$$ $$V'(L)=1$$

The plots for $$N_{5}, N_{6}$$ are shown above. With these points, we can then finally derive the coefficient for the element stiffness matrix. The coefficient for EA has been solved, what remains left is the coefficient for EI. This is solved for in the sample equation shown below:

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

$$\tilde{k}_{23}=\frac{6EI}{L^{2}}=\int_{0}^{L}{\frac{d^{2}N_{2}}{dx^{2}}(EI)\frac{d^{2}N_{3}}{dx^{2}}dx}$$

In general:

$$\tilde{k}_{ij}=\int_{0}^{L}{\frac{d^{2}N_{i}}{dx^{2}}(EI)\frac{d^{2}N_{i}}{dx^{2}}dx}$$

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

With these equations finally set up, we can then find the elastodynamics of trusses, frames, and of 2D and 3D frames. Starting with the model setup:

$$\mathbf{\bar{W}\cdot [\bar{M}\ddot{\bar{d}}+\bar{K}\bar{d}-\bar{F}]}=0$$ for all W's

$$\mathbf{\bar{M}\ddot{\bar{d}}+\bar{K}\bar{d}=\bar{F}(t)}$$ $$\mathbf{\bar{d}(0)=\bar{d}\circ }$$ $$\mathbf{\dot{\bar{d}}(0)=\bar{v}\circ }$$

next, complete the ordinary differential equations, using 2nd order in final and initial conditions governing the elastodynamics of discretion.

We first consider the unforced vibration problem:

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

assume $$\textbf{v}_{nx1}(t)=(\sin \omega t)\phi _{nx1}$$.

$$\ddot{\textbf{v}}=-\omega ^{2}\sin \omega t\phi -\omega ^{2}\sin \omega t\textbf{M}\phi +\sin \omega t\textbf{K}\phi =\textbf{0}$$

using cross-elimination gives us

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

which is the generalized evaluation problem. The general form is:

$$\mathbf{Ax}=\lambda \mathbf{Bx}$$

The above statement can be reduced to its mode form:

$$i\Rightarrow \textbf{v}_{i}(t)=(\sin \omega _{i}t)\mathbf{\phi }_{i}$$

with i = 1,...,n

Model superposition method
We first start off with the orthogonal proportion of eigenpairs:

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

Then, we use equations (1) and (2) listed above:

$$\mathbf{\bar{M}\phi _{j}}=\lambda _{j}\bar{\textbf{K}}\phi _{j}$$

$$\mathbf{\phi _{i}^{T}\bar{M}\phi _{j}}=\lambda _{j}\phi _{i}^{T}\mathbf{K\phi _{j}}$$

which comes out to:

$$\mathbf{\phi _{i}^{T}\bar{K}\phi _{j}}=\frac{1}{\lambda _{j}}\delta _{ij}$$

$$\bar{\textbf{d}}(t)\sum_{i=1}^{n}{Z_{i}(t)\phi _{i}}$$

Then, using these equations, we can then be applied to equation (1) listed above.

$$\bar{\textbf{M}}(\sum_{j}{Z_{j}\phi _{j}})+\bar{\textbf{K}}(\sum_{j}{Z_{i}\phi _{i}})=\textbf{F}$$

$$\sum_{j}{\ddot{Z}_{j}(\phi _{i}^{T}\bar{\textbf{M}}\phi _{j})}+\sum_{j}{Z_{i}(\phi _{i}^{T}\bar{\textbf{K}}\phi _{i})}$$

$$\ddot{Z}_{i}+\lambda _{i}Z_{i}=\mathbf{\phi _{i}^{T}F}$$

Lecture Problems
Revisitng the matrix below we must verify the dimensions for all terms multiplied by $$\tilde{d}$$. $$\tilde{k} =\! $$ $$\begin{bmatrix} \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} &0 &0 \\ \\ & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{-12EI}{L^{3}} &\frac{6EI}{L^{2}} \\ \\ & &  \frac{4EI}{L}&0  & \frac{-6EI}{L^{2}} & \frac{2EI}{L}\\ \\ & &  & \frac{EA}{L} &0  & 0\\ \\ & &  &  &\frac{12EI}{L^{3}}  &\frac{-6EI}{L^{2}} \\ \\ symmetric & &  &  &  & \frac{4EI}{L} \end{bmatrix} \! $$ Verify all $$\tilde{k}_{ij}\tilde{d}_{j}$$ for i = 1, ..., 6 and j = 1, ..., 6 We know from the lecture notes that the dimensions of the first term are

$$[\tilde{k}_{11}\tilde{d}_1] = \frac{[E][A]}{[L]}[\tilde{d}_{1}] = \frac{(\frac{F}{L^2})L^2}{L}(L) = F$$

Calculating the dimensions of the four other terms results in the dimensions

$$[\tilde{k}_{22}\tilde{d}_2] = \frac{[12][E][I]}{[L]^3}[\tilde{d}_{2}] = \frac{(\frac{F}{L^2})L^4}{L^3}(L) = F$$

$$[\tilde{k}_{23}\tilde{d}_2] = \frac{[6][E][I]}{[L]^2}[\tilde{d}_{2}] = \frac{(\frac{F}{L^2})L^4}{L^2}(L) = FL$$

$$[\tilde{k}_{33}\tilde{d}_3] = \frac{[4][E][I]}{[L]}[\tilde{d}_{3}] = \frac{(\frac{F}{L^2})L^4}{L}(1) = FL$$

$$[\tilde{k}_{36}\tilde{d}_6] = \frac{[2][E][I]}{[L]}[\tilde{d}_{6}] = \frac{(\frac{F}{L^2})L^4}{L}(1) = FL$$

After the dimensions of all terms have been determined, the results can be displayed in the dimension matrix below.

$$ \begin{bmatrix} F & 0 & 0 & -F & 0 & 0\\ 0 & F & FL & 0 & -F & FL\\ 0 & FL & FL & 0 & -FL & FL\\ -F & 0 & 0 & F & 0 & 0\\ 0 & -F & -FL & 0 & F & -FL\\ 0 & FL & FL & 0 & -FL & FL \end{bmatrix} $$ To solve for the global stiffness matrix k(e) in $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$ from $$\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{d}}^{(e)}=\tilde{\mathbf{f}}^{(e)}$$ we use the equation $$\mathbf{k}^{(e)}=\tilde{\mathbf{T}}^{(e)^{T}}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}$$ The stiffness matrix is first given as $$\tilde{k} = \begin{bmatrix} \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} &0 &0 \\ \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{-12EI}{L^{3}} &\frac{6EI}{L^{2}} \\ \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L}     & 0 & \frac{-6EI}{L^{2}}  & \frac{2EI}{L}\\ \\ \frac{-EA}{L} & 0     & 0                 & \frac{EA}{L}  &0        & 0\\ \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} \! $$ The Transfer matrix and its transpose are below. The R matrix used to assemble the Transfer matrix was discussed in Lecture 19. $$\tilde{T}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 & 0 & 0\\ -m^{(e)} & l^{e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & 0\\ 0 & 0 & 0 & -m^{(e)} & l^{e)} & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ $$\tilde{T}^{(e)^{T}} = \begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0 & 0 & 0\\ m^{(e)} & l^{e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & -m^{(e)} & 0\\ 0 & 0 & 0 & m^{(e)} & l^{e)} & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ Multiply the three matrices according to the equation listed above for the global stiffness matrix

We know from previously in the lecture notes that the units of displacement are length (L) and the units of rotation are simply 1. For the transverse displacement term N5 and the rotation of N6, dimensional analysis gives us the following results. $$ \left[N_{5} \right]\left[\tilde{d}_{5} \right] = (1)(L) = L $$ $$ \left[N_{6} \right]\left[\tilde{d}_{6} \right] = (L)(1) = L $$

Electric Pylon Analysis
Our team was asked to modify the pylon analysis code to use frame analysis. The properties of the pylon are as follows:

Material = 300 M Steel Height = 60 m Moment of Intertia = 1.3333e-8 Young's Modulus = 200 GPa Area = 4 cm2 Density = 7.8 g/cm3

MATLAB was implemented in solving for the maximum bending moment, shear stress, eigenvalues, and vibrational periods.

Code
The code implemented is as follows:

%********************************************************************* % Filename: pylon2.m % % PURPOSE: % FEA Frame Analysis % %  %  Team Bottle % EML 4500 %*********************************************************************  clear; close; dof = 2; % Node coordinates n_node = 46;            % number of nodes n_elem = 91;            % number of elements total_dof = 2 * n_node; % total dof of system position(:, 1) = [ 1.5; 0]; position(:, 2) = [ 4.53; 0]; position(:, 3) = [ 1.88; 1.18]; position(:, 4) = [ 3.03; 1.18]; position(:, 5) = [ 4.17; 1.18]; position(:, 6) = [ 2.2; 2.181]; position(:, 7) = [ 3.03; 2.181]; position(:, 8) = [ 3.85; 2.181]; position(:, 9) = [ 2.53; 3.181]; position(:, 10) = [ 3.53; 3.181]; position(:, 11) = [ 2.34; 3.681]; position(:, 12) = [ 3.03; 3.681]; position(:, 13) = [ 3.71; 3.681]; position(:, 14) = [ 2.16; 4.181]; position(:, 15) = [ 2.62; 4.181]; position(:, 16) = [ 3.44; 4.181]; position(:, 17) = [ 3.89; 4.181]; position(:, 18) = [ 2.36; 4.5]; position(:, 19) = [ 3.7; 4.5]; position(:, 20) = [ 1.8; 5.172]; position(:, 21) = [ 4.25; 5.172]; position(:, 22) = [ 0; 6]; position(:, 23) = [ 1.06; 6]; position(:, 24) = [ 1.5; 6]; position(:, 25) = [ 1.82; 6]; position(:, 26) = [ 2.22; 6]; position(:, 27) = [ 2.71; 6]; position(:, 28) = [ 3.34; 6]; position(:, 29) = [ 3.84; 6]; position(:, 30) = [ 4.23; 6]; position(:, 31) = [ 4.55; 6]; position(:, 32) = [ 5; 6]; position(:, 33) = [ 6.05; 6]; position(:, 34) = [ .8; 6.23]; position(:, 35) = [ 5.25; 6.23]; position(:, 36) = [ 1.24; 6.355]; position(:, 37) = [ 4.82; 6.355]; position(:, 38) = [ 1.5; 6.43]; position(:, 39) = [ 2; 6.43]; position(:, 40) = [ 2.48; 6.43]; position(:, 41) = [ 3.03; 6.43]; position(:, 42) = [ 3.58; 6.43]; position(:, 43) = [ 4.05; 6.43]; position(:, 44) = [ 4.55; 6.43]; position(:, 45) = [ 1.5; 7.43]; position(:, 46) = [ 4.55; 7.43]; scale = 8.0753701211306; for i = 1 : n_node; x(i) = scale * position(1,i); y(i) = scale * position(2,i); end node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 3; node_connect(1, 2) = 1;  % element 2 node_connect(2, 2) = 4; node_connect(1, 3) = 3;  % element 5 node_connect(2, 3) = 4; node_connect(1, 4) = 4;  % element 3 node_connect(2, 4) = 2; node_connect(1, 5) = 4;  % element 6 node_connect(2, 5) = 5; node_connect(1, 6) = 2;  % element 4 node_connect(2, 6) = 5; node_connect(1, 7) = 3;  % element 8 node_connect(2, 7) = 7; node_connect(1, 8) = 5;  % element 9 node_connect(2, 8) = 7; node_connect(1, 9) = 3;  % element 7 node_connect(2, 9) = 6; node_connect(1, 10) = 5;  % element 10 node_connect(2, 10) = 8; node_connect(1, 11) = 6;  % element 11 node_connect(2, 11) = 7; node_connect(1, 12) = 7;  % element 12 node_connect(2, 12) = 8; node_connect(1, 13) = 6;  % element 13 node_connect(2, 13) = 9; node_connect(1, 14) = 7;  % element 14 node_connect(2, 14) = 9; node_connect(1, 15) = 7;  % element 15 node_connect(2, 15) = 10; node_connect(1, 16) = 8;  % element 16 node_connect(2, 16) = 10; node_connect(1, 17) = 9;  % element 17 node_connect(2, 17) = 10; node_connect(1, 18) = 9;  % element 18 node_connect(2, 18) = 11; node_connect(1, 19) = 9;  % element 19 node_connect(2, 19) = 12; node_connect(1, 20) = 10;  % element 20 node_connect(2, 20) = 12; node_connect(1, 21) = 10;  % element 21 node_connect(2, 21) = 13; node_connect(1, 22) = 11;  % element 22 node_connect(2, 22) = 12; node_connect(1, 23) = 12;  % element 23 node_connect(2, 23) = 13; node_connect(1, 24) = 11;  % element 24 node_connect(2, 24) = 14; node_connect(1, 25) = 11;  % element 25 node_connect(2, 25) = 15; node_connect(1, 26) = 12;  % element 26 node_connect(2, 26) = 15; node_connect(1, 27) = 12;  % element 27 node_connect(2, 27) = 16; node_connect(1, 28) = 13;  % element 28 node_connect(2, 28) = 16; node_connect(1, 29) = 13;  % element 29 node_connect(2, 29) = 17; node_connect(1, 30) = 14;  % element 30 node_connect(2, 30) = 15; node_connect(1, 31) = 15;  % element 31 node_connect(2, 31) = 16; node_connect(1, 32) = 16;  % element 32 node_connect(2, 32) = 17; node_connect(1, 33) = 14;  % element 33 node_connect(2, 33) = 20; node_connect(1, 34) = 14;  % element 34 node_connect(2, 34) = 18; node_connect(1, 35) = 15;  % element 35 node_connect(2, 35) = 18; node_connect(1, 36) = 16;  % element 36 node_connect(2, 36) = 19; node_connect(1, 37) = 17;  % element 37 node_connect(2, 37) = 19; node_connect(1, 38) = 17;  % element 38 node_connect(2, 38) = 21; node_connect(1, 39) = 18;  % element 39 node_connect(2, 39) = 20; node_connect(1, 40) = 19;  % element 40 node_connect(2, 40) = 21; node_connect(1, 41) = 20;  % element 41 node_connect(2, 41) = 24; node_connect(1, 42) = 20;  % element 42 node_connect(2, 42) = 26; node_connect(1, 43) = 21;  % element 43 node_connect(2, 43) = 29; node_connect(1, 44) = 21;  % element 44 node_connect(2, 44) = 31; node_connect(1, 45) = 22;  % element 45 node_connect(2, 45) = 23; node_connect(1, 46) = 23;  % element 46 node_connect(2, 46) = 24; node_connect(1, 47) = 24;  % element 47 node_connect(2, 47) = 25; node_connect(1, 48) = 25;  % element 48 node_connect(2, 48) = 26; node_connect(1, 49) = 26;  % element 49 node_connect(2, 49) = 27; node_connect(1, 50) = 27;  % element 50 node_connect(2, 50) = 28; node_connect(1, 51) = 28;  % element 51 node_connect(2, 51) = 29; node_connect(1, 52) = 29;  % element 52 node_connect(2, 52) = 30; node_connect(1, 53) = 30;  % element 53 node_connect(2, 53) = 31; node_connect(1, 54) = 31;  % element 54 node_connect(2, 54) = 32; node_connect(1, 55) = 32;  % element 55 node_connect(2, 55) = 33; node_connect(1, 56) = 22;  % element 56 node_connect(2, 56) = 34; node_connect(1, 57) = 23;  % element 57 node_connect(2, 57) = 34; node_connect(1, 58) = 23;  % element 58 node_connect(2, 58) = 36; node_connect(1, 59) = 24;  % element 59 node_connect(2, 59) = 36; node_connect(1, 60) = 24;  % element 60 node_connect(2, 60) = 38; node_connect(1, 61) = 25;  % element 61 node_connect(2, 61) = 38; node_connect(1, 62) = 25;  % element 62 node_connect(2, 62) = 39; node_connect(1, 63) = 26;  % element 63 node_connect(2, 63) = 39; node_connect(1, 64) = 26;  % element 64 node_connect(2, 64) = 40; node_connect(1, 65) = 27;  % element 65 node_connect(2, 65) = 40; node_connect(1, 66) = 27;  % element 66 node_connect(2, 66) = 41; node_connect(1, 67) = 28;  % element 67 node_connect(2, 67) = 41; node_connect(1, 68) = 28;  % element 68 node_connect(2, 68) = 42; node_connect(1, 69) = 29;  % element 69 node_connect(2, 69) = 42; node_connect(1, 70) = 29;  % element 70 node_connect(2, 70) = 43; node_connect(1, 71) = 30;  % element 71 node_connect(2, 71) = 43; node_connect(1, 72) = 30;  % element 72 node_connect(2, 72) = 44; node_connect(1, 73) = 31;  % element 73 node_connect(2, 73) = 44; node_connect(1, 74) = 31;  % element 74 node_connect(2, 74) = 37; node_connect(1, 75) = 32;  % element 75 node_connect(2, 75) = 37; node_connect(1, 76) = 32;  % element 76 node_connect(2, 76) = 35; node_connect(1, 77) = 33;  % element 77 node_connect(2, 77) = 35; node_connect(1, 78) = 34;  % element 78 node_connect(2, 78) = 36; node_connect(1, 79) = 35;  % element 79 node_connect(2, 79) = 37; node_connect(1, 80) = 36;  % element 80 node_connect(2, 80) = 38; node_connect(1, 81) = 37;  % element 81 node_connect(2, 81) = 44; node_connect(1, 82) = 38;  % element 82 node_connect(2, 82) = 39; node_connect(1, 83) = 39;  % element 83 node_connect(2, 83) = 40; node_connect(1, 84) = 40;  % element 84 node_connect(2, 84) = 41; node_connect(1, 85) = 41;  % element 85 node_connect(2, 85) = 42; node_connect(1, 86) = 42;  % element 86 node_connect(2, 86) = 43; node_connect(1, 87) = 43;  % element 87 node_connect(2, 87) = 44; node_connect(1, 88) = 38;  % element 88 node_connect(2, 88) = 45; node_connect(1, 89) = 39;  % element 89 node_connect(2, 89) = 45; node_connect(1, 90) = 43;  % element 90 node_connect(2, 90) = 46; node_connect(1, 91) = 44;  % element 91 node_connect(2, 91) = 46;

figure(1) for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; plot(xx,yy,'--') hold on  end E = 200e9; A = 4e-4; P = -1000; dof = 2*n_node; K=zeros(dof); R = zeros(dof,1); R(66) = P;  debc = [1,2,3,4]; ebcVals = zeros(length(debc),1);

for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); location = [x(node_1),y(node_1); x(node_2),y(node_2)]; lm = [ node_1*2-1, node_1*2, node_2*2-1, node_2*2 ]; k = PlaneTrussElement(E,A,location); K(lm,lm) = K(lm,lm) + k;  end [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:n_elem node_1 = node_connect(1,i); node_2 = node_connect(2,i); location = [x(node_1),y(node_1); x(node_2),y(node_2)]; lm = [ node_1*2-1, node_1*2, node_2*2-1, node_2*2 ]; results = [results; PlaneTrussResults(E, A, ...      location, d(lm))]; end format short g  multiplier = 20; d = d*multiplier; for i = 1:n_node position_d(:,i) = [ position(1,i)+d(2*i-1), position(2,i)+d(2*i) ]; end for i = 1 : n_node; x(i) = scale * position_d(1,i); y(i) = scale * position_d(2,i); end

for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; plot(xx,yy,'-black') hold on  end dof = 3; % Node coordinates n_node = 46;              % number of nodes n_elem = 91;              % number of elements total_dof = dof * n_node; % total dof of system position(:, 1) = [ 1.5; 0]; position(:, 2) = [ 4.53; 0]; position(:, 3) = [ 1.88; 1.18]; position(:, 4) = [ 3.03; 1.18]; position(:, 5) = [ 4.17; 1.18]; position(:, 6) = [ 2.2; 2.181]; position(:, 7) = [ 3.03; 2.181]; position(:, 8) = [ 3.85; 2.181]; position(:, 9) = [ 2.53; 3.181]; position(:, 10) = [ 3.53; 3.181]; position(:, 11) = [ 2.34; 3.681]; position(:, 12) = [ 3.03; 3.681]; position(:, 13) = [ 3.71; 3.681]; position(:, 14) = [ 2.16; 4.181]; position(:, 15) = [ 2.62; 4.181]; position(:, 16) = [ 3.44; 4.181]; position(:, 17) = [ 3.89; 4.181]; position(:, 18) = [ 2.36; 4.5]; position(:, 19) = [ 3.7; 4.5]; position(:, 20) = [ 1.8; 5.172]; position(:, 21) = [ 4.25; 5.172]; position(:, 22) = [ 0; 6]; position(:, 23) = [ 1.06; 6]; position(:, 24) = [ 1.5; 6]; position(:, 25) = [ 1.82; 6]; position(:, 26) = [ 2.22; 6]; position(:, 27) = [ 2.71; 6]; position(:, 28) = [ 3.34; 6]; position(:, 29) = [ 3.84; 6]; position(:, 30) = [ 4.23; 6]; position(:, 31) = [ 4.55; 6]; position(:, 32) = [ 5; 6]; position(:, 33) = [ 6.05; 6]; position(:, 34) = [ .8; 6.23]; position(:, 35) = [ 5.25; 6.23]; position(:, 36) = [ 1.24; 6.355]; position(:, 37) = [ 4.82; 6.355]; position(:, 38) = [ 1.5; 6.43]; position(:, 39) = [ 2; 6.43]; position(:, 40) = [ 2.48; 6.43]; position(:, 41) = [ 3.03; 6.43]; position(:, 42) = [ 3.58; 6.43]; position(:, 43) = [ 4.05; 6.43]; position(:, 44) = [ 4.55; 6.43]; position(:, 45) = [ 1.5; 7.43]; position(:, 46) = [ 4.55; 7.43]; scale = 8.0753701211306; for i = 1 : n_node; x(i) = scale * position(1,i); y(i) = scale * position(2,i); end node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 3; node_connect(1, 2) = 1;  % element 2 node_connect(2, 2) = 4; node_connect(1, 3) = 3;  % element 5 node_connect(2, 3) = 4; node_connect(1, 4) = 4;  % element 3 node_connect(2, 4) = 2; node_connect(1, 5) = 4;  % element 6 node_connect(2, 5) = 5; node_connect(1, 6) = 2;  % element 4 node_connect(2, 6) = 5; node_connect(1, 7) = 3;  % element 8 node_connect(2, 7) = 7; node_connect(1, 8) = 5;  % element 9 node_connect(2, 8) = 7; node_connect(1, 9) = 3;  % element 7 node_connect(2, 9) = 6; node_connect(1, 10) = 5;  % element 10 node_connect(2, 10) = 8; node_connect(1, 11) = 6;  % element 11 node_connect(2, 11) = 7; node_connect(1, 12) = 7;  % element 12 node_connect(2, 12) = 8; node_connect(1, 13) = 6;  % element 13 node_connect(2, 13) = 9; node_connect(1, 14) = 7;  % element 14 node_connect(2, 14) = 9; node_connect(1, 15) = 7;  % element 15 node_connect(2, 15) = 10; node_connect(1, 16) = 8;  % element 16 node_connect(2, 16) = 10; node_connect(1, 17) = 9;  % element 17 node_connect(2, 17) = 10; node_connect(1, 18) = 9;  % element 18 node_connect(2, 18) = 11; node_connect(1, 19) = 9;  % element 19 node_connect(2, 19) = 12; node_connect(1, 20) = 10;  % element 20 node_connect(2, 20) = 12; node_connect(1, 21) = 10;  % element 21 node_connect(2, 21) = 13; node_connect(1, 22) = 11;  % element 22 node_connect(2, 22) = 12; node_connect(1, 23) = 12;  % element 23 node_connect(2, 23) = 13; node_connect(1, 24) = 11;  % element 24 node_connect(2, 24) = 14; node_connect(1, 25) = 11;  % element 25 node_connect(2, 25) = 15; node_connect(1, 26) = 12;  % element 26 node_connect(2, 26) = 15; node_connect(1, 27) = 12;  % element 27 node_connect(2, 27) = 16; node_connect(1, 28) = 13;  % element 28 node_connect(2, 28) = 16; node_connect(1, 29) = 13;  % element 29 node_connect(2, 29) = 17; node_connect(1, 30) = 14;  % element 30 node_connect(2, 30) = 15; node_connect(1, 31) = 15;  % element 31 node_connect(2, 31) = 16; node_connect(1, 32) = 16;  % element 32 node_connect(2, 32) = 17; node_connect(1, 33) = 14;  % element 33 node_connect(2, 33) = 20; node_connect(1, 34) = 14;  % element 34 node_connect(2, 34) = 18; node_connect(1, 35) = 15;  % element 35 node_connect(2, 35) = 18; node_connect(1, 36) = 16;  % element 36 node_connect(2, 36) = 19; node_connect(1, 37) = 17;  % element 37 node_connect(2, 37) = 19; node_connect(1, 38) = 17;  % element 38 node_connect(2, 38) = 21; node_connect(1, 39) = 18;  % element 39 node_connect(2, 39) = 20; node_connect(1, 40) = 19;  % element 40 node_connect(2, 40) = 21; node_connect(1, 41) = 20;  % element 41 node_connect(2, 41) = 24; node_connect(1, 42) = 20;  % element 42 node_connect(2, 42) = 26; node_connect(1, 43) = 21;  % element 43 node_connect(2, 43) = 29; node_connect(1, 44) = 21;  % element 44 node_connect(2, 44) = 31; node_connect(1, 45) = 22;  % element 45 node_connect(2, 45) = 23; node_connect(1, 46) = 23;  % element 46 node_connect(2, 46) = 24; node_connect(1, 47) = 24;  % element 47 node_connect(2, 47) = 25; node_connect(1, 48) = 25;  % element 48 node_connect(2, 48) = 26; node_connect(1, 49) = 26;  % element 49 node_connect(2, 49) = 27; node_connect(1, 50) = 27;  % element 50 node_connect(2, 50) = 28; node_connect(1, 51) = 28;  % element 51 node_connect(2, 51) = 29; node_connect(1, 52) = 29;  % element 52 node_connect(2, 52) = 30; node_connect(1, 53) = 30;  % element 53 node_connect(2, 53) = 31; node_connect(1, 54) = 31;  % element 54 node_connect(2, 54) = 32; node_connect(1, 55) = 32;  % element 55 node_connect(2, 55) = 33; node_connect(1, 56) = 22;  % element 56 node_connect(2, 56) = 34; node_connect(1, 57) = 23;  % element 57 node_connect(2, 57) = 34; node_connect(1, 58) = 23;  % element 58 node_connect(2, 58) = 36; node_connect(1, 59) = 24;  % element 59 node_connect(2, 59) = 36; node_connect(1, 60) = 24;  % element 60 node_connect(2, 60) = 38; node_connect(1, 61) = 25;  % element 61 node_connect(2, 61) = 38; node_connect(1, 62) = 25;  % element 62 node_connect(2, 62) = 39; node_connect(1, 63) = 26;  % element 63 node_connect(2, 63) = 39; node_connect(1, 64) = 26;  % element 64 node_connect(2, 64) = 40; node_connect(1, 65) = 27;  % element 65 node_connect(2, 65) = 40; node_connect(1, 66) = 27;  % element 66 node_connect(2, 66) = 41; node_connect(1, 67) = 28;  % element 67 node_connect(2, 67) = 41; node_connect(1, 68) = 28;  % element 68 node_connect(2, 68) = 42; node_connect(1, 69) = 29;  % element 69 node_connect(2, 69) = 42; node_connect(1, 70) = 29;  % element 70 node_connect(2, 70) = 43; node_connect(1, 71) = 30;  % element 71 node_connect(2, 71) = 43; node_connect(1, 72) = 30;  % element 72 node_connect(2, 72) = 44; node_connect(1, 73) = 31;  % element 73 node_connect(2, 73) = 44; node_connect(1, 74) = 31;  % element 74 node_connect(2, 74) = 37; node_connect(1, 75) = 32;  % element 75 node_connect(2, 75) = 37; node_connect(1, 76) = 32;  % element 76 node_connect(2, 76) = 35; node_connect(1, 77) = 33;  % element 77 node_connect(2, 77) = 35; node_connect(1, 78) = 34;  % element 78 node_connect(2, 78) = 36; node_connect(1, 79) = 35;  % element 79 node_connect(2, 79) = 37; node_connect(1, 80) = 36;  % element 80 node_connect(2, 80) = 38; node_connect(1, 81) = 37;  % element 81 node_connect(2, 81) = 44; node_connect(1, 82) = 38;  % element 82 node_connect(2, 82) = 39; node_connect(1, 83) = 39;  % element 83 node_connect(2, 83) = 40; node_connect(1, 84) = 40;  % element 84 node_connect(2, 84) = 41; node_connect(1, 85) = 41;  % element 85 node_connect(2, 85) = 42; node_connect(1, 86) = 42;  % element 86 node_connect(2, 86) = 43; node_connect(1, 87) = 43;  % element 87 node_connect(2, 87) = 44; node_connect(1, 88) = 38;  % element 88 node_connect(2, 88) = 45; node_connect(1, 89) = 39;  % element 89 node_connect(2, 89) = 45; node_connect(1, 90) = 43;  % element 90 node_connect(2, 90) = 46; node_connect(1, 91) = 44;  % element 91 node_connect(2, 91) = 46; % Undeformed plot for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; plot(xx,yy,':') hold on  end % Frame analysis E = 200e9; A = 4e-4; P = -1000; I = 1.3333e-8; K = zeros(dof*n_node); R = zeros(dof*n_node,1); R(66) = P;  debc = [1,2,3,4,5,6]; ebcVals = zeros(length(debc),1); % Assemble global matrix for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); coord = [x(node_1),y(node_1); x(node_2),y(node_2)]; lm = [ node_1*3-2, node_1*3-1, node_1*3, node_2*3-2, node_2*3-1, node_2*3 ]; [ke rq] = PlaneFrameElement(E,I,A,0,0,coord); K(lm, lm) = K(lm, lm) + ke; R(lm) = R(lm) + rq; end % Compute reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) fa = []; bma = []; Va = []; for i = 1:n_elem node_1 = node_connect(1,i); node_2 = node_connect(2,i); coord = [x(node_1),y(node_1); x(node_2),y(node_2)]; lm = [ node_1*3-2, node_1*3-1 node_1*3, node_2*3-2, node_2*3-1, node_2*3 ]; [f, bm, V] = PlaneFrameResults(E,I,A,0,0,location,d(lm)); fa = [fa; f]; bma = [bma; bm]; Va = [Va; V]; end % Highest bending moment moment = 0; for i = 1:n_elem*2 if abs(bma(i,3)) > moment moment = abs(bma(i,3)); momentnode = i;     end end moment if mod(momentnode,2) == 0 momentelement = momentnode/2; else momentelement = (momentnode+1)/2; end momentelement % Shear stress element shearstress = 0; for i = 1:n_elem*2 if abs(Va(i,3)) > shearstress shearstress = abs(Va(i,3)); shearstressnode = i;     end end shearstress if mod(momentnode,2) == 0 shearstresselement = shearstressnode/2; else shearstresselement = (shearstressnode+1)/2; end shearstresselement

multiplier = 20; d = d*multiplier; % Deformed positions for i = 1:n_node position_d(:,i) = [ position(1,i)+d(3*i-2), position(2,i)+d(3*i-1) ]; end for i = 1 : n_node; x(i) = scale * position_d(1,i); y(i) = scale * position_d(2,i); end for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; axis([-5 53 -5 70]) plot(xx,yy,'--') hold on  end title('Electric Pylon in Deformed and Undeformed State') xlabel('x') ylabel('y') rho = 7.8e3; M = zeros(dof); for i = 1 : n_node; x(i) = scale * position(1,i); y(i) = scale * position(2,i); end for i = 1 : n_elem node_1 = node_connect(1,i); node_2 = node_connect(2,i); location = [x(node_1),y(node_1); x(node_2),y(node_2)]; L(i) = PlaneTrussLength(location); m(i) = rho*L(i)*A; end i = 1; for j = 1 : n_node mass(j) = 0; end for j = 1 : n_node for k = 1 : n_elem if node_connect(1,k) == j              mass(j) = mass(j) + m(k)/2; elseif node_connect(2,k) == j                  local_elems(j,i) = k;                   mass(j) = mass(j) + m(k)/2; end end end for i = 1 : 46 M(3*i-2,3*i-2) = mass(i); M(3*i-1,3*i-1) = mass(i); M(3*i,3*i) = 0; end M;  Mf = ReduceMassMatrix(M, R, debc, ebcVals); Kf = ReduceStiffnessMatrix(K, R, debc, ebcVals); A = pinv(Mf)*Kf; [V2,D2] = eig(A); eig = [999999999999,9999999999999,9999999999999]; for i = 1 : 132 if (eig(1) > D2(i,i) && D2(i,i) ~= 0) eig(1) = D2(i,i); column(1) = i;          end end for i = 1 : 132 if (eig(2) > D2(i,i) && D2(i,i) > eig(1) && D2(i,i) ~= 0) eig(2) = D2(i,i); column(2) = i;      end end for i = 1 : 132 if (eig(3) > D2(i,i) && D2(i,i) > eig(2) && D2(i,i) ~= 0) eig(3) = D2(i,i); column(3) = i;      end end eig % Period Calculation for i = 1 : 3 omega(i) = sqrt(eig(i)); T(i) = 2*pi/omega(i); end T % Eigenvalue plots % First eigenvalue plot figure(2) for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; plot(xx,yy,'--') hold on  end

Z = zeros(6,132); V2 = [Z;V2]; for i = 1:n_node position_d(:,i) = [ scale*position(1,i) + multiplier*V2(3*i-2,column(1)), scale*position(2,i) + multiplier*V2(3*i-1,column(1))  ]; end for i = 1 : n_node dx(i) = position_d(1,i); dy(i) = position_d(2,i); end for i = 1 : n_elem node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [dx(node_1),dx(node_2)]; yy = [dy(node_1),dy(node_2)]; plot(xx,yy,'-black') hold on  end title('Effects of the Smallest Eigenvalue') xlabel('x') ylabel('y') % Second eigenvalue plot

figure(3)

for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; axis([-5 53 -5 70]) plot(xx,yy,'--') hold on  end for i = 1:n_node position_d(:,i) = [ scale*position(1,i) + multiplier*V2(3*i-2,column(2)), scale*position(2,i) + multiplier*V2(3*i-1,column(2)) ]; end for i = 1 : n_node dx(i) = position_d(1,i); dy(i) = position_d(2,i); end for i = 1 : n_elem node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [dx(node_1),dx(node_2)]; yy = [dy(node_1),dy(node_2)]; axis([-5 53 -5 70]) plot(xx,yy,'-black') hold on  end title('Effects of the Second Smallest Eigenvalue') xlabel('x') ylabel('y') % plot eigenvectors for 3rd smallest eigenvalue multiplier = 10; figure(4) for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; plot(xx,yy,'--') hold on  end for i = 1:n_node position_d(:,i) = [ scale*position(1,i) + multiplier*V2(3*i-2,column(3)), scale*position(2,i) + multiplier*V2(3*i-1,column(3)) ]; end for i = 1 : n_node dx(i) = position_d(1,i); dy(i) = position_d(2,i); end for i = 1 : n_elem node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [dx(node_1),dx(node_2)]; yy = [dy(node_1),dy(node_2)]; plot(xx,yy,'-black') hold on  end title('Effects of the Third Smallest Eigenvalue') xlabel('x') ylabel('y')

Results
The displacement results were:

d = 0           0            0            0  -0.00019023   0.00017465   -0.0001277 -1.4074e-006 -0.00016006  -0.0002967  -0.00012671   0.00032602  -0.00012798    -0.000128  -0.00012792  -0.00054327   0.00044932   0.00030901   0.00046381  -0.00061057     0.001017   0.00064583    0.0010325  -0.00019427    0.0010502  -0.00097027    0.0017167    0.0010338    0.0017162   0.00023259    0.0016833  -0.00064959     0.001683   -0.0013293    0.0023452   0.00064079    0.0021863   -0.0010297    0.0047122    0.0023902    0.0039576   -0.0024165    0.0063605    0.0057328    0.0063605    0.0037877    0.0063605    0.0029803    0.0063614     0.002384    0.0063657    0.0016463    0.0063711   0.00086194    0.0063182  -0.00024268    0.0062289   -0.0013182    0.0061363   -0.0023672    0.0060413   -0.0033231    0.0058839   -0.0051278    0.0055152    -0.012587    0.0067826    0.0042648     0.006996   -0.0063359     0.007012    0.0034574    0.0074006   -0.0043591    0.0071505    0.0029772    0.0071478    0.0020508    0.0071421    0.0012278     0.007182   0.00032026    0.0072816  -0.00073249    0.0074073   -0.0018633    0.0075643   -0.0033872    0.0090005    0.0029772     0.010455   -0.0033872

The maximum bending moment is:

moment = 472.49

This moment occurs at:

momentelement = 45

The maximum transverse shear stress is:

shearstress = 85.852

This stress occurs at:

shearstresselement =  45

Below is a plot of the undeformed (blue) and deformed (black) pylon structure:

Effects of Eigenvectors
The lowest eigenvalues and their corresponding oscillation periods are:

eig =  132.86       2471.2       2942.6

T =  0.54511      0.12639      0.11583

Plots of the effects of the three lowest eigenvalues are shown below. The blue dashed line represents the undeformed structure while the solid black line represents the deformed structure.



Statically Determinant
The final step in this problem was to determine whether the structure can be solved using statics. Since there are only two fixed points and four bars for which reactions must be computed, it can be determined that this problem is statically indeterminate.

Group Opinions on Mediawiki vs. E-Learning
Mediawiki provided an excellent opportunity for group collaboration. While this is also possible in E-learning, a lot of the interface is more difficult to adjust to. Also, it seems like the javascript programming in E-learning tends to make processes run a bit slower than Mediawiki, which is oftentimes frustrating. As is, I really enjoyed the opportunity to use Mediawiki, and the learning and experience gained from it will certainly be useful in future endeavors. Eml4500.f08.bottle.vitello 03:56, 22 November 2008 (UTC)

Really the two should not be compared, since they are completely different from one another. Group homework assignments should not be done on either platform for the type of homework reports that Vu-Quoc wants. Having said that,

Mediawiki Pros:
 * Attractive looking pages are relatively easy to create
 * Equations, pictures, etc. can be included

Mediawiki Cons:
 * Mediawiki code can be tedious and time consuming
 * There is no spell checker
 * Work can be vandalized
 * Work can be easily plagarized
 * Occasionally, an error will occur, thus all work is lost when trying to save a page
 * Work will not be saved if someone else edits the page while editing it

E-Learning Pros:
 * Team communication is easier
 * Work cannot be vandalized
 * Grade information is more easily accessible

E-Learning Cons: Eml4500.f08.bottle.barnes 15:13, 24 November 2008 (UTC)
 * Equations cannot be included

I agree with all of the points listed by Barnes. Please delete Brockmiller's statements. I agree, but that's not necessary to have in our homework report.

E-Learning and Mediawiki are two different systems for two different purposes.

E-Learning was created to submit homework assignments to professors and TAs. These homework assignments could be individual submissions or group projects. E-Learning provides a secure environment where no one else can access your work or your grades. It is much easier to receive grade info through E-Learning, where a student can login and click on the "My Grades" link rather than wait for an email from the team leader. E-Learning has an interface designed to receive file transfers, such as submitting a word document, an excel spreadsheet, a Matlab m-file, or a .JPG. Therefore, if it is necessary to turn in an equation, the student can simply write the equation in the Microsoft Word Equation Editor and then submit the word document containing the equation.

Mediawiki was designed to put information on display for the rest of the world to see. If the Professor desires our homework to be put on display, then by all means - Mediawiki has given us the opportunity to do that splendidly. Mediawiki definitely has its pros, but it is not necessarily "better" than E-Learning. They are simply different. They are completely different systems that accomplish different goals. They can certainly accomplish some of the same goals, such as getting information from Point A to Point B, but the "best" way to do this must be determined by the Professor based on what the Professor wants.

Every Professor has different expectations for homework submissions. Some require a few pieces of lined engineering paper. Some require only pencil scratch on loose-leaf. If a Professor simply wanted the answers to some homework problems, E-learning would be perfectly sufficient. Groups could collaborate on different parts and one person would be in charge of organizing the information and then submitting it. However, if the Professor wants his or her students to display their work for everyone to see and take up students' time by making it as attractive as possible, then Mediawiki is the clear choice. Mediawiki is also more effective for collaborations between UF students and non-UF students.

Eml4500.f08.bottle.loschak 06:03, 30 November 2008 (UTC)

I have learned a whole lot about sharing information worldwide across the internet using wikiversity. I believe this should be taught to every engineering and science major student who are preparing to share their discoveries and knowledge to the world.

Eml4500.f08.bottle.hipps 1:45, 7 December 2008 (UTC)

Recommended Software that Improves Productivity
For writing Wiki articles, Team Bottle primarily wrote all code and text directly into the intrinsic Wiki editing space. There never seemed to be issues doing so, as the formulation of articles and their constituents seemed direct enough. Hence, there was never an overt need to use alternative programs, though the necessitation for doing so may arise with more complicated article structures and content. Occasionally, some of the editing was done beforehand in a Word or Notepad editor and simply copy and pasted into the Wiki editing section, however such occurrences were rare.

For writing latex equations, the simplest method seemed to be directly using the built-in Wiki commands such as. While this may not have been the most efficient method for coding equations, it proved to be effective enough, especially as the semester progressed and familiarity and comfort increased. Some members of the group used the equation editor provided by codecogs.com to create equations and easily paste them into Wikipedia between tags.

For drawing figures, most image files were generated using Microsoft Paint or Paint.NET. Direct approaches to this included taking screenshots of MATLAB images, pasting them into Paint, then exporting the files as a .JPEG or .BMP file. Uploading these images into MediaWiki became a mindless process, as the recursive image requisite per homework made it such that everyone in Team Bottle could generate Wiki image files seemingly automatically. On occasion, Microsoft Powerpoint was used to generate arrows and vectors, but again, this simply involved producing a screenshot, pasting into Paint, copying the edited image into Powerpoint, editing the image in Powerpoint, copying the new image and finally pasting into Paint, where it was exported to an uploadable format for MediaWiki. As such, no additional programs beyond Microsoft Office were necessary.

Thus, while the processes described herein for homework generation may not have been the most efficient, they proved simple enough for user interface, as well as being effective as methods for producing acceptable results. If productivity were to increase, it would lay in the streamlining of generation via the distribution of workload in an assembly line system. But for the scale of the homeworks and the time allotted to satisfy their completion deadlines, ideal efficiency could be considered trivial. Indeed, it could be argued that the time investment necessary to simulate models of time efficiency based upon an experimental comparison using Monte Carlo parameters over an iterative data set to produce the ideal methodology sequence would itself be a waste of time that could have been invested in simply following the Occam's Razor type algorithm internalized by the group organically.

Hence, the proposals made by Professor Vu-Quoc about increasing productivity through alternative programs and methods, given the scale of the data set under consideration, may not necessarily be true for every member of the group.

Two-Bar Truss
This is a continuation of the two-bar truss system found throughout the semester and in previous homeworks.

Here, it is defined that

L(1) = 4.0

A(1) = 1.0

E(1) = 3.0

θ(1) = pi/3

and

L(2) = 2.0

A(2) = 2.0

E(2) = 5.0

θ(2) = pi/4

From Homework 3, the two-bar truss system was originally analyzed as

The image below represents the two-bar truss system plotted in Matlab. The dotted line corresponds to the two-bar truss system before deformation, and the

solid line includes deformation. The arrow pointing from Global Node 2 to Global Node 6 (which is the same as Global Node 2 for the deformed case) is the

deformation vector.

Here, new elements are defined such that

A(1)(1) = 0.5

A(1)(2) = 1.5

E(1) (1) = 2.0

E(1)(2) = 4.0

and

A(2)(1) = 1.0

A(2)(2) = 3.0

E(2)(1) = 3.0

E(2)(2) = 7.0

This system could be analyzed using the following MATLAB code:

Where the code Planesol.m is defined as

The results of running the above code yields:

To plot the solution of this system, the following code was developed:



Individual Contributions
Eml4500.f08.bottle.vitello 03:44, 22 November 2008 (UTC) Eml4500.f08.bottle.brockmiller 14:10, 22 November 2008 (UTC) Eml4500.f08.bottle.barnes 20:02, 24 November 2008 (UTC) Eml4500.f08.bottle.butler 22:20, 6 December 2008 (UTC) Eml4500.f08.bottle.hipps 1:41, 7 December 2008 (UTC) Eml4500.f08.bottle.loschak 15:02, 9 December 2008 (UTC) Eml4500.f08.bottle.ranto 21:00, 9 December 2008 (UTC)