User:Eml4500.f08.ateam/hw 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)

A-Team Teampage

Homework Report #6

This page is intended for Homework Report #6 of the A-Team.

Due date: Friday, 21 November 08, 5 pm EST = 22:00 UTC.

The Equation of Motion
The equation of motion can be derived by prescribing initial conditions to any system. The initial conditions for the system at t = 0 can be defined below.

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

and

$$\frac{\partial u}{\partial t}(x,t=0)={\dot{u}}(x,t=0)=\overline{v}(x) $$ for a known velocity function.

The principle of virtual work (continuous) of dynamics of the elastic bar can be shown as a partial differential equation:

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

The Discrete Equation of Motion can then be represented as

$$-\mathbf{K}\mathbf{d} + \mathbf{F} = \mathbf{M}\mathbf{\ddot{d}}$$

or $$\mathbf{M}\mathbf{\ddot{d}} + \mathbf{K}\mathbf{d} = \mathbf{F}\ \ \ \ (2) $$

Where $$\mathbf{M}$$ is a mass matrix.

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial }{\partial x} \right] = -Kd$$

$$f\, =\, F$$

$$m\ddot{u} = M\ddot{d} $$

Note that equation (2) is the equation of motion for a system with multiple degrees of freedom. A system with a single degree of freedom will be shown in equation 3.

For a single degree of freedom



$$\mathbf{m}\mathbf{\ddot{d}} + \mathbf{k}\mathbf{d} = \mathbf{F}$$

All of these equations have been derived with the help of the following integral:

$$\int_{x=0}^{x=L} W(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right] + f - m\ddot{u}}\right)dx = 0 \ \ \ \ (3)$$

for all possible $$\mathbf {W(x)}$$, where $$\mathbf {W(x)}$$ is the weighting coefficient. As seen before, going from equation 1 to equation 3 is trivial but it is important to know how to get from equation 3 to equation 1. Equation 3 can be rewritten as:

$$\int_{x=0}^{x=L} w(x)g(x)dx = 0$$ for all $$\mathbf {w(x)}$$

Since equation 3 holds true for all values of $$\mathbf{w(x)}$$ by the principle of virtual work, we can choose $$ \mathbf {w(x) = g(x)}$$. This changes equation 3 to

$$\int_{x=0}^{x=L} g^2(x)dx = 0$$

Integrating by Parts
Integrating by parts is needed when trying to integrate two functions multiplied together. The formula for integrating by parts can be seen below. The symbols 'r' and 's' will represent the two different functions of x in this example.

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

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

Integrating both sides forms the equation:

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

This leads to the final equation of:

$$ \int r^'s = rs - \int rs^'$$

Now to continue with the Principle of Virtual Work, recall equation 3:

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

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

After integrating by parts, the equation yields:

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

Where the minus sign on the right side of the equation relates to the - Kd in the step between equations 1 and 2 above.

Replacing the x's with either L or 0, sets the right side equal to:

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

Now consider the model equation with boundary conditions:



At x = 0, select W(x) such that $$W\left(0\right) = 0$$ or so that it is kinematically admissible.

Motivation: Discrete PVW applied to equation below:

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

Where $$\left(\left[K \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1} - {F}_{6x1}\right)$$ becomes a 6x1 matrix and $$ \mathbf{F}^{T}=\left[F_{1}F_{2}F_{3}F_{4}F_{5}F_{6}\right] $$ where F1, F2, F5, and F6 are unknown reactions.

Note that W can be selected arbitrarily, but for simpliciity select W such that W1 = W2 = W5 = W6 = 0. This is so all equations involving unknown reactions can be eliminated, or rows 1,2,5 and 6 are eliminated. Equation 1 shown below can then be formed.

$$ \mathbf{\overline{K}_{2x2}\overline{d}_{2x1}=\overline{F}_{2x1}} $$

Where $$\mathbf\overline{d} = \begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}$$ and $$\mathbf\overline{F} = \begin{bmatrix}F_{3} \\ F_{4}\end{bmatrix}$$.

Note that $$ \mathbf{\overline{W}\left(\overline{K}\overline{d}-\overline{F}\right)}=0 $$ (2) is actually the step before equation 1.

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

This becomes the final equation:

$$ \int_{0}^{L}w(m\ddot{u})dx+\int_{0}^{L} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx

= W(L)F(t) + \int_{0}^{L}wfdx $$

These equations are for all $$ W\left(x\right)$$ such that $$ W\left(0\right) = 0 $$

Two cases of PVW


For the rest of this section, we will be focusing on element i in the above figures.

Assume the displacement for u(x) for $$x_i \le x \le x_{i+1}$$. (i.e., x $$\epsilon [x_i, x_{i+1}]$$)

The motivation for the linear interpolation of u(x) is shown below using a 2-bar truss.



The deformed shape is a straight line. This means there was an implicit assumption of the linear interpolation of the displacement between two nodes. Consider the case where there is only axial displacement, or zero transverse displacements. The question that needs to be solved is:

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

The answer to this question is defined by linear interpolation, or:

$$u(x) = N_i\left(x\right)d_i + N_{i+1}\left(x\right)d_{i+1}$$

where $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$ are linear functions of x.



Continuous PVW to Discrete PVW
Langrangian Interpolation

Motivation for form of $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$

1) $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$ are linear (straight lines), thus any linear combination of $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$ is also linear, and in particular the expression for $$u\left(x\right)$$.

$$N_i\left(x\right) = \alpha_i + \beta_i x \qquad \quad \quad \ \ (with \ \alpha_i\, +\, \beta_i\, being\ real\ numbers) $$

$$N_{i+1}\left(x\right) = \alpha_{i+1} + \beta_{i+1} x \qquad (with \ \alpha_{i+1}\ +\, \beta_{i+1}\,  being\ real\ numbers) $$

Linear combination of $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$:

$$N_i\left(x\right)d_i + N_{i+1}\left(x\right)d_{i+1} = (\alpha_i + \beta_i x)d_i + (\alpha_{i+1} + \beta_{i+1} x)d_{i+1}$$

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

Since the linear combination of this function is also a function of x (as shown in the last line), the initial assertion has been confirmed.

2) Recall the equation for $$ u(x)$$ (the interpolation of $$ u(x)$$) from the previous meeting:

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

$$N_{i+1}\left(x\right) = \frac {x - x_{i}}{x_{i+1} - x_{i}}$$

Using this relationship and plugging in $$x\, =\, x_i$$

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

$$N_{i+1}\left(x\right) = \frac {x_{i} - x_{i}}{x_{i+1} - x_{i}}\ = \ 0$$

Plugging these values into the equation for u(x):

$$u(x_i) = N_i\left(x_i\right)d_i + N_{i+1}\left(x_i\right)d_{i+1}$$

$$u(x_i) = (1)\, d_i + (0)\, d_{i+1}$$

FEM Via Principle of Virtual Work
The finite element method uses a weighting matrix W to solve for the solution. The first set of solutions is for a bar element i. The PVW solution followes a similar pattern with the displacement matrix being replaced by the weighting matrix W.

$$ U\,(x_{i+1}) = d_{i+1} $$

Apply the same interpretation for the W(x) matrix: For example,

$$ W(x) = N_i\, W_i + N_{i+1}\, W_{i+1} $$

Element Stiffness Matrix for element i:

$$ \int_{x_i}^{x_{i+1}}{} [N_iW_i + N_{i+1}W_{i+1}](EA)[N_id_i + N_{i+1}d_{i+1}]dx $$

The Definition of $$ N_i\, and\, N_{i+1} $$ can be seen below:

$$ N_i = \frac{dN_i(x)}{dx} $$       $$ N_{i+1} = \frac{dN_{i+1}(x)}{dx} $$

Note:

$$ u(x) = \N_i(x) \N_{i+1}(x)| \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} $$

The N(x) matrix is a 1x2 matrix and the d matrix is 2x1.

Taking the derivative yeilds the following:

$$ \frac{du(x)}{dx} = \N_i'(x) \N'_{i+1}(x)| \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} $$

The N'(x) matrix is a 1x2 matrix and is called the B(x) matrix.

Similarly for the W matrix from the Principle of Virtual Work (PVW)
$$ W(x) = N(x) \begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix} $$

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

Recall the Element DOFs:



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

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

From these relations, we can find the integral that contains both W and d.

$$ \int_{x_i}^{x_{i+1}}{} (BW^i)(EA)(Bd^i)dx $$

The B matrix relates d to W. (Principle of Virtual Work)

Our Goal now is to derive $$ W^i\, (k_id_i) $$ from the previous integral.

Re-write the integral with the EA in the front:

$$ \int_{x_i}^{x_{i+1}}{} (EA)(BW^i)(Bd^i)dx $$

Take the Transpose of the BW factor to eliminate the B matrix:

$$ \int_{x_i}^{x_{i+1}}{} (EA)(BW^i)^T(Bd^i)dx $$

Distribute the Transpose inside of the Parentheses:

$$ \int_{x_i}^{x_{i+1}}{} (EA)(B^TW^{iT})(Bd^i)dx $$

The W matrix is equal to its transpose so it can be removed:

$$ \int_{x_i}^{x_{i+1}}{} (EA)(B^TW^{i})(Bd^i)dx $$

What we are left with is ( With the W matrix pulled outside the integrand because it does not vary with x):

$$ W^{i} \int_{x_i}^{x_{i+1}}{}(B^T)(EA)(B)dx d^i$$

The B matrix is a function of x and is equal too:

$$ B(x) = \|\frac{-1}{L^i} \|\frac{-1}{L^i}| $$

Note: $$ L^{i}\, =\, x_{i+1}\, -\, x_i $$

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

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

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

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

 Error: See my comments in Team Bike (with annotation). Perhaps because your two teams were working together on this part, so you get the same error; it's OK to collaborate across the teams, but be careful of getting the same mistakes. Eml4500.f08 19:52, 26 November 2008 (UTC)

Determining Average Stiffness Matrix
Our goal is to use linear interpolation to determine $$N_i(\tilde{x})$$. If $$\tilde{x}= x-x_i$$, then

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

Now, the shape function can be written as follows:

$$ N_2^{(i)}(\tilde{x})=\frac{\tilde{x}}{L^{(i)}}= \begin{cases} & \text{0 at } \tilde{x}= 0\\ & \text{1 at } \tilde{x}= L \end{cases} $$

Our next goal is to compare the general $$\mathbf{k}^{(i)}$$ to the stiffness matrix obtained by using

$$\frac{1}{2}(A_1+A_2)$$ and $$\frac{1}{2}(E_1+E_2)$$, where $$E_1 \neq E_2$$. The resulting stiffness matrix is

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

This matrix is also known as the average stiffness matrix.

Now, we want to determine $$\mathbf{k}^{(i)}-\mathbf{k}_{avg}^{(i)}$$

$$ \mathbf{k}^{(i)}-\mathbf{k}_{avg}^{(i)}= \frac{(A_1+A_2)}{4L^{(i)}}(2E-E_1-E_2) $$

These results show that the average value of the stiffness matrix is not equal to the value of the stiffness matrix evaluated at the average value of x. This can be verified with the Mean Value Theorem.

Mean Value Theorem
Consider the 2D mass located in the xy plane. Its centroid is located at $$(\bar{x},\bar{y})$$.



The Mean Value Theorem (MVT) states

$$\int_{x=a}^{x=b}{f(x)dx}=f(\bar{x})[b-a]$$

for $$\bar{x}\epsilon[a,b]$$ or $$a \leq \bar{x} \leq b$$.

Similarly,

$$\int_{x=a}^{x=b}{f(x)g(x)dx}=f(\bar{x})g(\bar{x})[b-a]$$

for $$a \leq \bar{x} \leq b$$.

However,

$$f(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$

where $$\frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$ is the average value of f.

Similarly,

$$g(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{g(x)dx}$$

where $$\frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$ is the average value of g.

MATLAB: Eigenvalue Analysis of Rectangular Three-Bar and Four-Bar Truss Systems
This problem is a second attempt at plotting the eigenvectors for the truss problem described in Homework Report #4. The unstable three-bar truss system is depicted as the truss at the left of the of the following figure while the stable four-bar truss is shown at the right.



Unstable Three-Bar Truss System
The global stiffness matrix for the problem is assembled as follows.

Using the global stiffness matrix, the following MATLAB command can be used to determine the eigenvalues and vectors for the truss. These are displayed as follows with V corresponding to the eigenvectors and D corresponding to the eigenvalues.

Plotting Eigenvectors
By plotting the columns of the eigenvector matrix that correspond to the zero eigenvalues, a mode of the truss can be depicted. In this case, the zero eigenvalues fall in columns 1, 2, 3, 4, and 5 of the eigenvalue matrix, D. Therefore, the first five columns of the eigenvector matrix, V, are plotted similarly to plotting displacements. These values are simply added to the original nodal positions to give the mode shape.

The following code was appended to the original truss problem MATLAB file to plot the eigenvectors.

First Column Eigenvectors
This plot represents the eigenvectors corresponding to the first column of the eigenvector matrix, V.

Second Column Eigenvectors
This plot represents the eigenvectors corresponding to the second column of the eigenvector matrix, V.

Third Column Eigenvectors
This plot represents the eigenvectors corresponding to the third column of the eigenvector matrix, V.

Fourth Column Eigenvectors
This plot represents the eigenvectors corresponding to the fourth column of the eigenvector matrix, V.

Fifth Column Eigenvectors
This plot represents the eigenvectors corresponding to the fifth column of the eigenvector matrix, V.

Stable Four-Bar Truss System
The global stiffness matrix for the problem is assembled as follows.

Using the global stiffness matrix, the following MATLAB command can be used to determine the eigenvalues and vectors for the truss. These are displayed as follows with V corresponding to the eigenvectors and D corresponding to the eigenvalues.

Plotting Eigenvectors
By plotting the columns of the eigenvector matrix that correspond to the zero eigenvalues, a mode of the truss can be depicted. In this case, the zero eigenvalues fall in columns 1, 2, 3, and 4 of the eigenvalue matrix, D. Therefore, the first five columns of the eigenvector matrix, V, are plotted similarly to plotting displacements. These values are simply added to the original nodal positions to give the mode shape, just as in the previous truss system.

The following code was appended to the original truss problem MATLAB file to plot the eigenvectors.

First Column Eigenvectors
This plot represents the eigenvectors corresponding to the first column of the eigenvector matrix, V.

Second Column Eigenvectors
This plot represents the eigenvectors corresponding to the second column of the eigenvector matrix, V.

Third Column Eigenvectors
This plot represents the eigenvectors corresponding to the third column of the eigenvector matrix, V.

Fourth Column Eigenvectors
This plot represents the eigenvectors corresponding to the fourth column of the eigenvector matrix, V.

MATLAB: Electric Pylon Structure Analysis
The truss system in this problem is a electric-line pylon consisting of 91 elements. The MATLAB code for accurately assembling this truss system was obtained from Dr. Loc Vu-Quoc and can be found here or seen below.

The objective of this problem is to perform additional MATLAB programming to find specific properties of the truss system based on given specifications and an applied load P. P is applied vertically downward at the far right tip of the arm of the pylon. All elements are constructed of 300M steel and have identical cross-sectional areas. Properties for this material were given by Dr. Vu-Quoc and verified online here. They are displayed in the table below.

Running the provided code produces the following plot showing the structure of the electric pylon truss system.



Analysis Requirements

 * Scale the electric pylon so that its height is 60 m
 * Plot the undeformed shape (dotted line) and the deformed shape (solid line)
 * Compute the axial stress in each bar
 * Display the highest tensile stress and the highest compressive stress and the corresponding elements in which they occur
 * Indicate these elements with an arrow on the figure of undeformed and deformed shapes
 * Determine if the problem is statically determinant and provide justification through argument or MATLAB programming
 * Construct the lumped mass matrix of the electric pylon
 * Solve the generalized eigenvalue problem
 * Find the lowest three eigenpairs and, in 3 separate plots, plot the eigenvectors as deformed shapes (solid line) superposed onto the undedeformed shape (dotted line)
 * Find the 3 lowest vibrational periods of the electric pylon

MATLAB Analysis Code
The code below is the  file written to solve the problem as assigned. This program requires additional MATLAB functions to run. These functions are also detailed below.

Additional Functions
Similar to the five-bar truss system, the electric pylon truss system analysis MATLAB program requires additional functions to run. Some of hese functions are found at the course textbook website, Fundamental Finite Element Analysis and Applications. They include,  , and. Other essential functions include  and. These functions were independently written and cannot be found with the preceding functions. The details of these functions are outlined below.

This function generates the element stiffness matrix for each element. When the function is called it calculates the X and Y coordinate positions of the element ends from the coordinates passed in. It produces the element length from this information. Next it calculates the direction cosines of the element. Using the Young's modulus and cross-sectional area of the element, the element stiffness matrix, k, is assembled. The stiffness matrix is passed back to the.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

This function outputs the axial strain, axial stress, and axial force of each element.

This function returns the length of an element based on the element coordinates it receives. It was independently written and cannot be found with the preceding functions.

This function was written to output the reduced matrices, $$\bar{M}$$ and $$\bar{K}$$. It was independently written and cannot be found with the preceding functions.

Running Analysis Code
The code below is the complete results from running the MATLAB program. Each part is detailed in later sections of this homework report.

Plot of Undeformed and Deformed Truss System
The extended MATLAB code plots the deformation of the truss system as follows. To scale the electric pylon truss to be 60m tall, the nodal positions were multiplied by a factor of 8.0753701211306. It is important to note that the deformation has been magnified by a factor of 100 for viewing purposes.



Axial Stress of Elements
The axial stresses for each element are simply the second column of the results array. They are shown above as the result  when the MATLAB code is run. They are also shown here.

Highest Tensile Stress
As shown on the plot and results from running, the greatest tensile stress occurs in Element 81. The value of the stress is 9.0511x106 Pa.

Highest Compressive Stress
As shown on the plot and results from running, the greatest compressive stress occurs in Element 55. The value of the stress is 8.6957x106 Pa. It is important to note that the negative sign has been removed from this stress because it has been denoted as compressive stress. In the stress table above, negative stresses denote compressive stress. This aligns with common stress notation.

Lumped Mass Matrix
The lumped mass matrix was assembled as a diagonal matrix of the masses of both degrees of freedom for each node. To determine the masses, each elements length was multiplied by its cross-sectional area and its density. Once the global nodes of each end of the element were known, the half of the element's mass was assigned to both nodes. Continuing this for each element created the lumped mass matrix.

Eigenpair Analysis
The three smallest eigenvalues for the truss system were determined using the generalized eigenvalue problem shown below. In this relation $$\bar{K}$$ is the reduced stiffness matrix and $$\bar{M}$$ is the reduced lumped mass matrix.

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

By plotting the columns of the eigenvector matrix that correspond to the smallest eigenvalues, modes of the truss can be depicted. In this case, the smallest eigenvalues were 132.16, 2468.4 and 2940.3. Therefore, the columns of the eigenvector matrix, V, that correspond to these values are plotted similarly to plotting displacements. These values are simply added to the original nodal positions to give the mode shape.

The following code was appended to the original truss problem MATLAB file to plot the eigenvectors. It is important to note that the eigenvectors were multiplied by a factor of 10 for viewing purposes.

Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the smallest eigenvalue.

Second Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the second smallest eigenvalue.

Third Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the third smallest eigenvalue.

Lowest Vibrational Periods of the Electric Pylon
To find the three lowest vibration periods of the electric pylon, the eigenvalues are set equal to the angular frequencies as shown in the equation below.

$$\lambda = \omega^2$$

The angular frequencies are converted to vibrational frequencies as follows.

$$f = \frac{\omega}{2\cdot \pi}$$

From the vibrational frequencies, the vibrational periods can be determined using the following equation.

$$T = \frac{1}{f} = \frac{2\cdot \pi}{\omega}$$

Following these relations reveals the three lowest vibrational periods, shown in the table below.

Statically Determinant?
The electric pylon structure is not statically determinate. While two nodes of the system are fixed, these two nodes act as the termination point of four elements. Thus, there are a total of four unknown reaction forces in a system with only three equations with which to attempt to solve them. This cannot be solved using the principles of statics.

It is important to note that, while similar in number of fixed nodes to previous systems analyzed, this system differs in that the fixed nodes contain more than one element.

 Note: See my comments in Team Bike (with annotation). Eml4500.f08 19:52, 26 November 2008 (UTC)

Contributing Team Members
The following students contributed to this report:

Mark Barry Eml4500.f08.Ateam.barry 04:24, 20 November 2008 (UTC)

Daniel DiDomenico Eml4500.f08.ateam.didomenico 16:01, 17 November 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 18:17, 21 November 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 15:00, 19 November 2008 (UTC)

Matthew Philbin Eml4500.f08.Ateam.philbin 19:53, 20 November 2008 (UTC)

Ian Slevinski Eml.f08.Ateam.Slevinski 23:03, 20 November 2008 (UTC)