User:Eml4500.f08.bike.mcdonald/homework report 6

 See my comments below. Please don't remove these comment boxes; just add your comment to these comment boxes in case you fix the problems. Eml4500.f08 19:52, 26 November 2008 (UTC)

Initial Conditions and the Continuous Principle of Virtual Work
Figure 1 shows the loaded elastic bar system in question.



The initial conditions for the system in Figure 1 at t = 0 are

$$ u\left(x,t=0\right)=\overline{u}(x) $$ for the known displacement function

and

$$ \frac{\partial u}{\partial t}(x,t=0)={\ddot{u}}(x,t=0)=\overline{V}(x) $$ for the known velocity function.

The partial differential equation for the continuous principle of virtual work(PVW) of dynamics for the elastic bar in Figure 1 can be seen below in Equation 1.

$$ PDE: \frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f=m{\ddot{u}} $$ ⇨ $$ Equation \left(1\right)$$

The discrete equation of motion can then be shown by Equation 2, where $$ \underline{M} $$ is the mass matrix. Equation 2 represents a multiple defree of freedom system(MDOF). Note that a single degree of freedom system(SDOF) has a very similar equation of motion, shown be Equation 3.

MDOF System ⇨ $$ \underline{M}\underline{\ddot{d}}+\underline{k}\underline{d}=\underline{F} $$ ⇨ $$ Equation \left(2\right)$$

SDOF System ⇨ $$ m\ddot{d}+kd=F $$ ⇨ $$ Equation \left(3\right)$$

The integral shown in Equation 4 was used to derive Equation 2 from Equation 1. The term $$ w(x) $$ represents the weighting function.

$$ \int_{x=0}^{x=L} w(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)dx=0 $$ for all possible $$ w(x) $$ ⇨ $$ Equation \left(4\right)$$

To go from Equation 1 to Equation 4 is a trivial task, but going from Equation 4 to Equation 1 is not trivial. This process is shown below.

For simplicity, let $$ \left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)=g(x) $$

Equation 4 then becomes $$ \int_{x=0}^{x=L} w(x)g(x)dx=0 $$ for all $$ w\left(x\right) $$

Since Equation 4 holds for all $$ w\left(x\right) $$, select $$ w\left(x\right)=g\left(x\right) $$,

then Equation 4 becomes $$ \int_{x=0}^{x=L}g^2dx=0 $$ ⇨ $$ g\left(x\right)=0 $$

Integration by Parts and the Discrete Principle of Virtual Work
Recall the steps for and integration by parts:

$$ \left(rs\right)^{'}=r^{'}s+rs^{'} $$

$$ r^{'}=\frac{dr}{dx} $$ and $$ s^{'}=\frac{ds}{dx} $$

$$ \int \left(rs\right)^{'}=\int r^{'}s + \int rs^{'} $$

Which yields a final result of $$ \int r^{'}s=rs-\int rs^{'} $$

Now Recall the Cont. Principle of Virtual Work(PVW) shown below.

$$ \int_{x=0}^{x=L} w(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)dx=0 $$ for all possible $$ w(x) $$

In the first term, let $$ s\left(x\right)=w(x) $$ and $$ r\left(x\right)=(EA)\frac{\partial u}{\partial x} $$

Solving using integration by parts yields

$$ \int_{x=0}^{x=L} w\left(x\right)\frac{\partial}{\partial x}\left[\left(EA\right)\frac{\partial u}{\partial x}\right]dx = \left[w(EA)\frac{\partial u}{\partial x}\right]_{x=0}^{x=L} - \int_{x=0}^{x=L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx $$

$$ = w\left(L\right)(EA)(L)\frac{\partial u}{\partial x}(L,t)-w(0)(EA)(0)\frac{\partial u}{\partial x}(0,t)-\int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx $$

where $$ \left(EA\right)(L)\frac{\partial u}{\partial x}(L,t) = N(L,t)=F $$

and $$ (EA)(0)\frac{\partial u}{\partial x}(0,t)=N(0,t) $$

Now lets consider the model problem with 2 boundary conditions shown below.

At $$ x=0 $$ select $$ w(x) $$ such that $$ w(0)=0 $$. This means that the system is kinematically admissible.

Motivation: discrete PVW applied to the equation shown below.

$$ \underline{W}_{6x1}\left(\left[ \underline K \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1}-\underline{F}_{6x1}\right)=0_{1x1} $$ for all $$ \underline{W} $$

where $$ \underline{F}^{T}=\left[F_{1}F_{2}F_{3}F_{4}F_{5}F_{6}\right] $$

$$ F_{3}, F_{4} $$ are known reactions and $$ F_{1}, F_{2}, F_{5}, F_{6} $$ are unknown reactions.

Since $$ \underline{w} $$ can be selected arbitrarily, select $$ \underline{w} $$ such that $$ w_{1}=w_{2}=w_{5}=w_{6}=0 $$ so to eliminate equations involving unknown reactions. Simply put, rows 1, 2, 5, and 6 are eliminated. The result is shown below in Equation 1.

$$ \underline{W}\left(\underline{K}\underline{d}-\underline{F}\right)=0 $$ for all $$ \underline{W} $$

$$ \underline{K}_{2x2}\underline{d}_{2x1}=\underline{F}_{2x1} $$ ⇨ $$ Equation \left(1\right) $$

Back to the principle of virtual work, the unknown reaction is

$$ N(0,t)=(EA)(0)\frac{\partial u}{\partial x}(0,t) $$

The equation then becomes

$$ W\left(L\right)F(t)-\int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx+\int w(x)\left[f-m\ddot{u}\right]dx=0 $$ for all $$ w\left(x\right) $$ such that $$ w\left(0\right)=0 $$

The final equation can be written as

$$ \int_{0}^{L}w\left(m\ddot{u}\right)dx+\int_{0}^{L} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx=W(L)F(t)+\int_{0}^{L}wfdx $$ for all $$ w\left(x\right) $$ such that $$ w\left(0\right)=0 $$

Motivation for the Linear Interpolation of u(x)
Elastic Bar Interpolation Example

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

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



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

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

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



 Good essay. Eml4500.f08 20:08, 26 November 2008 (UTC)

Now, motivation is shown for the form of $$ N_{i}(x)+N_{i+1}(x)$$.

$$ N_{i}(x)$$ and $$N_{i+1}(x)$$ are considered to be linear, meaning they are represented as straight lines. Therefore, any linear combination of $$ N_{i}(x)$$ and $$N_{i+1}(x)$$ is also linear. The expressions below show $$ N_{i}(x)$$ and $$N_{i+1}(x)$$ written in linear form.

$$ N_{i}(x)=\alpha_{i}+\beta_{i}x$$

$$ N_{i+1}(x)=\alpha_{i+1}+\beta_{i+1}x$$

Where $$\alpha_{i}$$ and $$\beta_{i}$$ both represent real numbers.

Next, the linear combination of $$ N_{i}(x)$$ and $$N_{i+1}(x)$$ can be shown as follows.

$$ N_{i}d_{i}+N_{i+1}d_{i+1}=(\alpha_{i}+\beta_{i}x)d_{i}+(\alpha_{i+1}+\beta_{i+1}x)d_{i+1}$$

The above equation can be rearranged and written as shown below.

$$ U_{x}=(\alpha_{i}d_{i}+\alpha_{i+1}d_{i+1})+(\beta_{i}d_{i}+\beta_{i+1}d_{i+1})x$$

This final equation shows that the linear combination is clearly a linear function in x.

A second motivation for $$ N_{i}(x)+N_{i+1}(x)$$ is shown below.

Recall the linear interpolation for u(x).

$$N_i(x)=\frac{x-x_{i+1}}{x_{i}-x_{i+1}}$$ $$N_{i+1}(x)=\frac{x-x_i}{x_{i+1}-x_i}$$

Now, plugging in $$ x_{i}$$ for x, it can be rewritten as follows:

$$N_i(x_{i})=\frac{x_{i}-x_{i+1}}{x_{i}-x_{i+1}}=1$$

$$N_{i+1}(x_{i})=\frac{x_{i}-x_i}{x_{i+1}-x_i}=0$$

Rewriting $$ u(x_{i})$$,

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

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

This shows that $$ u(x_{i}) $$ is equal to the weighting coefficient $$d_{i}$$

Elemental Stiffness Matrix for Element i
The elemental stiffness matrix for element i is found using the integral below.

$$\int_{x_i}^{x_{i+1}}[N'_i(x)W_i + N'_{i+1}W_{i+1}](EA)[N'_id_i + N'_{i+1}d_{i+1}]dx$$

where $$ N'_i = \frac{dN_i(x)}{dx}$$ and $$ N'_{i+1} = \frac{dN_{i+1}(x)}{dx}$$

Note:

$$u(x) = \underbrace{\begin{bmatrix}N'_i(x) & N'_{i+1}(x)\end{bmatrix}}_{\textbf{N}(x)_{1x2}}\begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix}_{2x1}$$

$$\frac{du(x)}{dx} = \underbrace{\begin{bmatrix}N_i(x) & N_{i+1}(x)\end{bmatrix}}_{\textbf{B}(x)_{1x2}}\begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix}_{2x1}$$

Similarly, the $$W(x)$$ matrix can be described as:

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

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

Recall the elemental degrees of freedom used with this element.



$$\begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix} = \begin{Bmatrix} d_1^{(i)} \\ d_2^{(i)} \end{Bmatrix} = \textbf{d}^{(i)} $$

$$\begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix} = \begin{Bmatrix} W_1^{(i)} \\ W_2^{(i)} \end{Bmatrix} = \textbf{W}^{(i)} $$

Using these matrices, the following equation can be used to find $$\beta$$

$$\beta = \int_{x^i}^{x^{i+1}}(\textbf{B}\textbf{W}^{(i)})_{1x1}(EA)_{1x1}(\textbf{B}\textbf{d}^{(i)})_{1x1}dx = \textbf{W}^i \cdot (\textbf{k}^{(i)} \textbf{d}^{(i)}) $$

Since each of these matrices becomes a scalar, they can be reorganized and multiplied in any order. The substitutions shown can also be made.

$$\beta = \int_{x^i}^{x^{i+1}}(EA)(\textbf{B}\textbf{W}^{(i)})_{1x1} \cdot (\textbf{B}\textbf{d}^{(i)})_{1x1}dx = (\textbf{BW}^{(i)})^T(\textbf{Bd}^{(i)}) = \textbf{W}^{(i)T}\textbf{B}^T(\textbf{Bd}^{(i)}) = \textbf{W}^{(i)}\cdot\textbf{B}(\textbf{Bd}^{(i)})$$

Ultimately, the $$\beta$$ matrix can be finalized to the equation below.

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

Using B and the transpose of B, the global stiffness matrix K can be determined like the stiffness matrices were determined previously with the transfer matrix and its transpose.

$$\textbf{k}^{(i)}_{2x2} = \int_{x_i}^{x_{i+1}} \textbf{B}^T(x)_{2x1}(EA)_{1x1} \textbf{B}(x)_{1x2}dx $$

Considering EA as a constant, the K then becomes the equation shown below.

$$\textbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Transfer of Variable Coordinates from $$x$$ to $$\tilde{x}$$
$$\tilde{x} := x - x_i $$

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

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

When EA and the length of the element are considered constant, the K matrix becomes the equation shown below.

$$\textbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

An element with varying areas and moduli of elasticity can now be solved. To prove this, consider the element shown below.



The area and modulus of elasticity equations are shown below.

$$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

$$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

The elemental stiffness matrix can then be solved as a function of $$\begin{Bmatrix} A_1, & A_2, & E_1, & E_2, & L^{(i)}\end{Bmatrix}$$

$$k_{2x2}(\tilde{x})^{(i)} = \frac{\begin{bmatrix} N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2\end{bmatrix}\begin{bmatrix}N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2\end{bmatrix}}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

 Error: The integral was missing; the above expression should be integrated with respect to the variable $$\displaystyle \tilde x$$ over the length of the element, since otherwise the stiffness matrix would be a function of $$\displaystyle \tilde x$$ as shown, and that would not make sense. This error is the same as in Team ATeam(Sean) (with annotations). Eml4500.f08 20:08, 26 November 2008 (UTC)

For the sixth homework, the class was asked to evaluate the function:

$$N_i(\tilde{x})=$$

Which equals:

$$N_i(\tilde{x})=\frac{\tilde{x}-\tilde{x}_{i+1}}{\tilde{x}_{i}-\tilde{x}_{i+1}}$$

Where:

$$\tilde{x} = x - x_{i}$$

Using the linear interpolation method described in an earlier lecture and defining $$\tilde{x}$$ as the length from the beginning of the interpolation to the value that is to be found.

A description of the shape function $$N_{2}^{(i)}$$ using the $$\tilde{x}$$ notation is as follows:

$$N_{2}^{(i)}(\tilde{x}) = \frac{\tilde{x}}{L^{(i)}} = \begin{cases} & \text{0 if } \tilde{x}= 0\\ & \text{1 if } \tilde{x}= L \end{cases}$$

Where the boundary conditions are provided, at $$\tilde{x} = 0$$ and $$\tilde{x} = L$$ respectively.

The class was also asked to complete the example provided in the book on page 159 while setting $$E_{1} = E_{2} = E$$ and to let $$A(\tilde{x})$$ be linear as it was described in the previous lecture. $$\mathbf{k}^{(i)}$$ is found from the problem given in the previous lecture and compared to the expression given in the book, which is:

$$\frac{E(A_{1}+A_{2})}{L^{(i)}2}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} = \mathbf{k}^{(i)}$$

As seen from the previous lecture, this result for the stiffness matrix is identically the same.



Next, the general $$\mathbf{k}^{(i)}$$ matrix from the previous lecture is to be compared with the matrix obtained by using $$\frac{(A_{1}+A_{2})}{2}$$ and $$\frac{(E_{1}+E_{2})}{2}$$ while noting that in this case $$E_{1}\neq E_{2}$$. This produces the stiffness matrix:

$$\frac{(E_{1}+E_{2})(A_{1}+A_{2})}{4L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} = \mathbf{k}_{ave}^{(i)}$$

The average stiffness matrix is then subtracted from the general stiffness matrix to find that:

$$\frac{(2E - E_{1}+E_{2})(A_{1}+A_{2})}{4L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} = \mathbf{k}^{(i)} - \mathbf{k}_{ave}^{(i)}$$

In order to interpret this, it is useful to review the Mean Value Theorem from calculus.

Redo of Eigenvector Plots of Rectangular Truss Systems
In Homework Report 4, Professor Vu Quoc assigned each team in EML 4500 the task of plotting the eigenvectors of a rectangular 3-bar truss system. No teams properly completed this assignment, thus Professor Vu Quoc asked for each team to redo the task. The redo of this assignment is shown below for the basic 3-bar rectangular truss system as well as the rectangular truss system with a cross-member added for structural rigidity.

3-Bar Rectangular Truss System
The setup for the 3-bar rectangular truss system presented by Professor Loc Vu Quoc is shown below.

The global stiffness matrix of the above rectangular system was determined in Homework Report 4 and is shown below.

K =

0    0     0     0     0     0     0     0     0     6     0    -6     0     0     0     0     0     0     6     0    -6     0     0     0     0    -6     0     6     0     0     0     0     0     0    -6     0     6     0     0     0     0     0     0     0     0     6     0    -6     0     0     0     0     0     0     0     0     0     0     0     0     0    -6     0     6

The eigenvectors (V) and eigenvalues (D) of the above global stiffness matrix were determined using MATLAB and are shown below.

V =

1.0000        0         0         0         0         0         0         0         0   -0.7071         0         0         0         0         0   -0.7071         0         0   -0.7071         0         0   -0.7071         0         0         0   -0.7071         0         0         0         0         0    0.7071         0         0   -0.7071         0         0    0.7071         0         0         0         0         0         0   -0.7071         0   -0.7071         0         0         0         0    1.0000         0         0         0         0         0         0         0         0   -0.7071         0    0.7071         0

D =

0    0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0    12     0     0     0     0     0     0     0     0    12     0     0     0     0     0     0     0     0    12

 Note: As explained in class, you should apply the boundary conditions to obtain the reduced stiffness matrix (dimension 4x4), then in that case you would only have one zero eigenpair corresponding to one mechanism, instead of 5 zero eigenpairs, which are 5 mechanisms. If you had applied the boundary conditions (blocking 4 dofs), only the 3rd eigenpair (swaying motion) would be the only zero eigenpair. See below. Eml4500.f08 20:08, 26 November 2008 (UTC)

Plot of the Eigenvectors
The plots of the eigenvectors corresponding to the zero eigenvectors of the 3-bar rectangular truss system are shown below. The solid red lines correspond to the deformed shape of the 3-bar rectangular truss system and the dotted blue lines correspond to the undeformed shape of the 3-bar rectangular truss system.

MATLAB Code Used to Produce Eigenvector Plots
The following MATLAB code was used to produce the above plots.

Addition of Cross-Member to 3-Bar Rectangular Truss System


The global stiffness matrix of the above modified rectangular truss system was also determined in Homework Report 4 and is shown below.

K =

2.1213   2.1213         0         0   -2.1213   -2.1213         0         0    2.1213    8.1213         0   -6.0000   -2.1213   -2.1213         0         0         0         0    6.0000         0   -6.0000         0         0         0         0   -6.0000         0    6.0000         0         0         0         0   -2.1213   -2.1213   -6.0000         0    8.1213    2.1213         0         0   -2.1213   -2.1213         0         0    2.1213    8.1213         0   -6.0000         0         0         0         0         0         0         0         0         0         0         0         0         0   -6.0000         0    6.0000

The eigenvectors (V) and eigenvalues (D) of the above global stiffness matrix were determined using MATLAB and are shown below.

V =

0.2372   0.1828   -0.6190   -0.3565    0.5948   -0.0000    0.0000    0.2150    0.2512   -0.5293    0.1627    0.1742    0.1586    0.2656    0.5127    0.4914    0.5834    0.0231    0.0449   -0.2390   -0.4362   -0.5767   -0.0263    0.2764    0.2512   -0.5293    0.1627    0.1742    0.4362   -0.2656   -0.5127   -0.2764    0.5834    0.0231    0.0449   -0.2390   -0.1586    0.5767    0.0263   -0.4914   -0.0950   -0.3695   -0.5012    0.0566   -0.1586   -0.3112    0.4863   -0.4914    0.3447    0.3636   -0.2394    0.8317    0.0000    0.0000    0.0000    0.0000   -0.0950   -0.3695   -0.5012    0.0566   -0.4362    0.3112   -0.4863    0.2764

D =

-0.0000        0         0         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0         0         0    0.0000         0         0         0         0         0         0         0         0    3.8183         0         0         0         0         0         0         0         0   12.0000         0         0         0         0         0         0         0         0   12.0000         0         0         0         0         0         0         0         0   16.6670

 Note: These 4 zero eigenpairs are linear combinations of 4 pure eigenmodes (3 rigid body rotations and 1 mechanism). Similar to the above remark, if you had applied the boundary conditions to obtain the reduced stiffness matrix (dimension 4x4), then in that case you would only have one zero eigenpair corresponding to one mechanism. Eml4500.f08 20:08, 26 November 2008 (UTC)

MATLAB Code Used to Produce Eigenvector Plots
The following MATLAB code was used to produce the above plots.

Analysis of 91-Element Electric Pylon Structure
The setup of the 91-element electric pylon structure is shown in the figure below. This system is to be analyzed using the Finite Element Method.



Each of the 91-elements that make up the electric pylon structure were made of 300 M steel with Young's Modulus of 200 GPa and had a cross-sectional area of 4 cm2.

The following topics will be covered in this analysis:


 * 1) Develop a MATLAB code to solve the electric pylon structure
 * 2) Plot the undeformed and deformed shape
 * 3) Compute the axial stress in each element of the electric pylon structure
 * 4) Determine which elements have the highest tensile and compressive stress in the structure
 * 5) Insert arrows on the undeformed and deformed plot to show the highest tensile and compressive stress elements
 * 6) Determine if the problem is statically determinate
 * 7) Construct the lumped mass matrix of the structure by assuming that the mass of each element is divided by two and lumped at each of the two nodes of the element
 * 8) Solve the generalized eigenvalue problem
 * 9) Find the lowest three eigenpairs
 * 10) Plot the eigenvectors corresponding to the lowest three eigenvalues as deformed shapes superposed onto the undeformed plot of the electric pylon structure
 * 11) Find the lowest three vibrational periods of the electric pylon

MATLAB Code to Solve the Electric Pylon Problem Using the Finite Element Method
The MATLAB code developed by Team Bike to solve the electric pylon structure using the Finite Element Method is shown below.

Additional Functions Required
In order to properly run the above MATLAB code, the following additional functions must be saved in the working directory of MATLAB.

MATLAB Command Window Printout for Electric Pylon Structure and Results
Running the MATLAB code developed by Team Bike, shown above, produced the following results in the MATLAB command window.

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

reactions =

149.64     -501.65      -149.64       1501.7

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 7.093e-019 1.4186e-007  5.6744e-011 2.4411e-019 4.8822e-008  1.9529e-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 1.2903e-019 2.5807e-008  1.0323e-011 -1.5471e-019 -3.0942e-008 -1.2377e-011 2.6985e-019  5.397e-008  2.1588e-011 -6.1024e-019 -1.2205e-007 -4.8819e-011 -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 -4.6964e-019 -9.3927e-008 -3.7571e-011 4.5158e-005 9.0317e+006       3612.7 -1.5877e-018 -3.1754e-007 -1.2702e-010 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 -1.441e-019 -2.8821e-008 -1.1528e-011 -9.6069e-020 -1.9214e-008 -7.6855e-012 0           0            0

Bar_Stress =

2.1562e+006 -1.3071e+006 1.3468e+006 1.2911e+006 -7.0304e+005 -4.7596e+006 -2.465e+006 1.4236e+006 3.8538e+006 -5.7655e+006 -37882      1755.6  3.8655e+006 2.9395e+006 -3.6988e+006 -5.7661e+006 3.5896e+005 5.242e+006 1.9797e+006 -3.7749e+006 -6.5159e+006 5.5784e+005 6.4388e+005 5.9656e+006 -8.1704e+005 -3.9616e+006 2.32e+006 -8.1307e+005 -5.7556e+006 -28474 -9.9353e+005 -15214 5.9216e+006 55813 -4.8717e+006 1.3915e+006 -3023.3 -5.7588e+006 -4.854e+006 1.3904e+006 -1.8831e+005 2.258e+006 -6.32e+005 -4.0102e+006 1.4186e-007 4.8822e-008 64147 2.7001e+005 2.7149e+005 -2.0778e+006 -4.4271e+006 -5.8777e+006 -7.3549e+006 -8.662e+006 -8.6957e+006 2.5807e-008 -3.0942e-008 5.397e-008 -1.2205e-007 -1.7704e+005 2.2069e+005 -1.9193e+005 1.9887e+005 2.1464e+006 -2.083e+006 2.2895e+006 -2.2643e+006 2.1035e+006 -2.1464e+006 1.4138e+006 -1.3772e+006 1.5835e+006 -3.6928e+006 -97393      -23690        31208  9.0479e+006 -9.3927e-008 9.0317e+006 -3.1754e-007 9.0511e+006 -1.3175e+005 -2.9645e+005 1.7966e+006 4.4876e+006 6.6233e+006 7.7755e+006 3.2222e-008 -2.8821e-008 -1.9214e-008 0

Comp_Stress =

-8.6957e+006

Comp_Stress_Element_Number =

55

Tens_Stress =

9.0511e+006

Tens_Stress_Element_Number =

81

lambda =

132.16      2468.4       2940.3

column =

79   85    86

T =

0.54654     0.12647      0.11587

Axial Stress in Each Bar
The axial stress in each bar are shown in the above MATLAB command window printout as the variable. These values are replicated in the table below for ease of viewing.

Highest Tensile Stress Element in the Electric Pylon Structure
As shown in the MATLAB command window printout above, the highest tensile stress in the electric pylon structure exists in element number 81, and has a value of 9.0511 x 106 Pa.

Highest Compressive Stress Element in the Electric Pylon Structure
As shown in the MATLAB command window printout above, the highest compressive stress in the electric pylon structure exists in element number 55, and has a value of -8.6957 x 106 Pa.

Plot of the Undeformed and Deformed Shapes of Electric-Pylon Structure
The figure shown below is a plot of the undeformed and deformed shapes of the 91-element electric pylon structure. This plot was produced using the MATLAB code shown below. This MATLAB code solved the 91-element electric pylon structure by use of the Finite Element Method. The solid red lines in the plot below represent the deformed shape of the 91-element electric pylon structure, whereas the dotted blue lines represent the undeformed shape of the 91-element electric pylon structure. In order to more clearly show the deformation experienced by the electric pylon structure, the deformations were multiplied by a magnification factor of 120.



The arrows in the above plot show the bar elements of the electric pylon structure with the highest compressive and highest tensile stress.

Is The Electric Pylon Structure Statically Determinate?
No, the electric pylon 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 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.

 Note: Then you should compute the reactions at the base of the pylon using the finite element method; let's do this computation in HW7. Eml4500.f08 19:52, 26 November 2008 (UTC)

Lowest Eigenvalues of the Electric Pylon Structure
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 structure was determined to be 132.16. This value was shown above in the MATLAB command window printout as well.

2nd Lowest Eigenvalue
The 2nd lowest eigenvalue of the electric pylon structure was determined to be 2468.4. This value was shown above in the MATLAB command window printout as well.

3rd Lowest Eigenvalue
The 3rd lowest eigenvalue of the electric pylon structure was determined to be 2940.3. This value was shown above in the MATLAB command window printout as well.

Lowest Eigenpair
The following plot of the lowest eigenpair as the deformed shape of the electric pylon structure is shown below. The deformed structure was multiplied by a magnification factor of 50 to allow for visualization of the deformation.



2nd Lowest Eigenpair
The following plot of the 2nd lowest eigenpair as the deformed shape of the electric pylon structure is shown below. The deformed structure was multiplied by a magnification factor of 20 to allow for visualization of the deformation.



3rd Lowest Eigenpair
The following plot of the 3rd lowest eigenpair as the deformed shape of the electric pylon structure is shown below. The deformed structure was multiplied by a magnification factor of 30 to allow for visualization of the deformation.



Three Lowest Vibrational Periods of the Electric Pylon
The three lowest vibrational periods of the electric pylon 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 structure was determined to be 0.11587 s. This value was also shown in the MATLAB command window printout above.

2nd Lowest Vibrational Period
The 2nd lowest vibrational period of the electric pylon structure was determined to be 0.12647 s. This value was also shown in the MATLAB command window printout above.

3rd Lowest Vibrational Period
The 3rd lowest vibrational period of the electric pylon structure was determined to be 0.54654 s. This value was also shown in the MATLAB command window printout above.

 Note: The mode with longest period is called the first fundamental vibration mode, then the second mode is the one with the second longest period, etc. So your "3rd lowest vibrational mode" is actually the first mode with the longest period. Eml4500.f08 20:08, 26 November 2008 (UTC)

Contributing Team Members
Andrew McDonald - Eml4500.f08.bike.mcdonald 18:36, 19 November 2008 (UTC) Garrett Pataky - Eml4500.f08.bike.pataky 06:17, 19 November 2008 (UTC) Sam Bernal - Eml4500.f08.bike.bernal 01:36, 21 November 2008 (UTC) Bobby Sweeting - Eml4500.f08.bike.sweeting 13:01, 19 November 2008 (UTC) Shawn Gravois - Eml4500.f08.bike.gravois 22:16, 19 November 2008 (UTC) Eric Viale - Eml4500.f08.bike.viale 20:10, 20 November 2008 (UTC)