# University of Florida/Eml4507/s13.team4ever.R7
## Problem R7.1a: Verify the dim of the ${\underline {k}}_{6x6}^{(e)}{\underline {d}}_{6x1}^{(e)}$ matrices (fead.f08.mtgs.[37-41] pg. 2)

On our honor, we did this assignment on our own.

### Given: The desired dimensions for k and d with index of 6

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

### Find:
#### 1.Use an index of 6 to verify the dimensions of k and d

Obtain:
${\underline {k}}_{6x6}^{(e)}{\underline {d}}_{6x1}^{(e)}={\underline {f}}_{6x1}^{(e)}$
With supporting: ${\underline {k}}_{6x6}^{(e)}={\underline {\tilde {T}}}_{6x6}^{(e)T}{\underline {\tilde {k}}}_{6x6}^{(e)}{\underline {\tilde {T}}}_{6x6}^{(e)}$
From:
${\underline {\tilde {k}}}_{6x6}^{(e)}{\underline {\tilde {d}}}_{6x1}^{(e)}={\underline {\tilde {f}}}_{6x1}^{(e)}$

### Solution: Verify the dimensions of k and d

The constructed 6x6 matrix shown below represents ${\underline {\tilde {T}}}$
${\begin{Bmatrix}{\tilde {d}}_{1}\\{\tilde {d}}_{2}\\{\tilde {d}}_{3}\\{\tilde {d}}_{4}\\{\tilde {d}}_{5}\\{\tilde {d}}_{6}\end{Bmatrix}}={\begin{Bmatrix}R_{11}&R_{12}&0&0&0&0\\R_{21}&R_{22}&0&0&0&0\\0&0&1&0&0&0\\0&0&0&R_{11}&R_{12}&0\\0&0&0&R_{21}&R_{22}&0\\0&0&0&0&0&1\end{Bmatrix}}{\begin{Bmatrix}d_{1}\\d_{2}\\d_{3}\\d_{4}\\d_{5}\\d_{6}\end{Bmatrix}}$
The following matrix represents the element stiffness matrix in local coordinates.
${\underline {\tilde {k}}}_{6x6}^{(e)}={\begin{bmatrix}{\frac {EA}{L}}&0&0&{\frac {-EA}{L}}&0&0\\0&{\frac {12EI}{L^{3}}}&{\frac {6EI}{L^{2}}}&0&{\frac {12EI}{L^{3}}}&{\frac {6EI}{L^{2}}}\\0&{\frac {6EI}{L^{2}}}&{\frac {4EI}{L}}&0&{\frac {-6EI}{L^{2}}}&{\frac {2EI}{L}}\\{\frac {-EA}{L}}&0&0&{\frac {EA}{L}}&0&0\\0&{\frac {-12EI}{L^{3}}}&{\frac {-6EI}{L^{2}}}&0&{\frac {12EI}{L^{3}}}&{\frac {-6EI}{L^{2}}}\\0&{\frac {6EI}{L^{2}}}&{\frac {2EI}{L}}&0&{\frac {-6EI}{L^{2}}}&{\frac {4EI}{L}}\end{bmatrix}}$
Using the above two matricies and the transpose of T, you can construct the following equation:
${\underline {k}}_{6x6}^{(e)}={\underline {\tilde {T}}}_{6x6}^{(e)T}{\underline {\tilde {k}}}_{6x6}^{(e)}{\underline {\tilde {T}}}_{6x6}^{(e)}$
Multiplying ${\underline {\tilde {k}}}_{6x6}^{(e)}$ by ${\underline {\tilde {d}}}_{6x1}^{(e)}$ will yield the required ${\underline {\tilde {f}}}_{6x1}^{(e)}$

## Problem R7.1b: Solve 2 element frame system (fead.f08.mtgs.[37-41] pg. 3)

On our honor, we did this assignment on our own.

### Given: Information on the two element truss system

Assume a square cross section. Also use same data from fea.f08.mtgs.p5-4.
Element length: (1) =4
Element length: (2) =2
Young's modulus: (1) =3
Young's modulus: (2) =5
Cross section area: (1) =1
Cross section area: (2) =2
Inclination angle: (1) = 30 deg
Inclination angle: (2) = -45 deg

### Solution:
#### 1. Plot Undeformed Shape

Undeformed Shape.
View of element with single force member view.

#### 2. Plot Deformed Shape 2-bar Truss

Plotting the deformed shape 2-bar truss for the given problem from p5-2 of the fead08 lecture notes. The deformation can be seen the truss using the forces given from p5-2.

#### 3. Plot Deformed Shape 2-bar Frame

Plotting the deformed shape 2-bar frame for the given problem from p5-2 of the fead08 lecture notes.
The constraints are preserved during the analysis in this mode to keep the frame.

#### ref
### Problem R7.2
#### Description

We are to resolve problem 5.7. We are to solve for motion of the truss using modal superposition using the three lowest eigenvalues.

#### Solution

The following code was used for problem 5.7 to obtain the K and M matrices and to obtain the eigenvalues and the eigenvectors:
From this, we obtain M and K matrices which are:
K =
M =
This will also output the lowest eigenpairs:
$\gamma _{1},\phi _{1}$
$\gamma _{2},\phi _{2}$
$\gamma _{3},\phi _{3}$
Then, with these, we will find what the modal equations are according to the equation:
$z^{''}+\gamma _{i}z=\phi _{i}^{T}F(t)$
This will yield 3 unique differential equations. We solve for the complete solution to these differential equation using the boundary conditions from:
$z_{i}(0)={\bar {\phi }}_{i}^{T}Md(0)$ and
$z_{i}^{'}(0)={\bar {\phi }}_{i}^{T}Md^{'}(0)$
With these solutions for z, we can then find and plot the actual displacements from modal superposition using:
$d(t)=\Sigma _{j=1}^{3}z_{j}\phi _{j}$

## Problem R7.2
### Description

We are to resolve problem 5.7. We are to solve for motion of the truss using modal superposition using the three lowest eigenvalues.

### Solution

The following code was used for problem 5.7 to obtain the K and M matrices and to obtain the eigenvalues and the eigenvectors:
From this, we obtain M and K matrices which are:
K =
M =
This will also output the lowest eigenpairs:
$\gamma _{1},\phi _{1}$
$\gamma _{2},\phi _{2}$
$\gamma _{3},\phi _{3}$
Then, with these, we will find what the modal equations are according to the equation:
$z^{''}+\gamma _{i}z=\phi _{i}^{T}F(t)$
This will yield 3 unique differential equations. We solve for the complete solution to these differential equation using the boundary conditions from:
$z_{i}(0)={\bar {\phi }}_{i}^{T}Md(0)$ and
$z_{i}^{'}(0)={\bar {\phi }}_{i}^{T}Md^{'}(0)$
With these solutions for z, we can then find and plot the actual displacements from modal superposition using:
$d(t)=\Sigma _{j=1}^{3}z_{j}\phi _{j}$

## Table of Assignments R7

Problem Assignments R5 :
| Problem # | Solved&Typed by | Reviewed by |
| 1a | Vernon Babich, Chad Colocar | All |
| 1b | Vernon Babich, Tyler Wulterkens | All |
| 2 | David Bonner, Chad Colocar | All |