# University of Florida/Eml4507/s13.team4ever.R6
### Problem R.6.1
#### Honor Pledge:

On my honor I have neither given nor received unauthorized aid in doing this problem.

#### Problem Description:

We are tasked with completing Pb.53.9 on p.53.19b by transforming the generalized eigenvalue problem into a standard eigenvalue problem.

#### Given

$m_{1}=3,m_{2}=2$ $k_{1}=10,k_{2}=20,k_{3}=15$

#### Solution

Solving the FBD of both $m_{1}$ and $m_{2}$ gives us the equations of motion. $m={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}={\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}$ $k={\begin{bmatrix}k_{1}-k_{2}&k_{2}\\k_{2}&-(k_{2}+k_{3})\\\end{bmatrix}}={\begin{bmatrix}-10&20\\20&-35\\\end{bmatrix}}$
Equation of Motion in Matrix Form.
${\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}{\begin{bmatrix}{\ddot {x_{1}}}\\{\ddot {x_{2}}}\\\end{bmatrix}}+{\begin{bmatrix}-10&20\\20&-35\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}$
Now transforming the generalized eigenvalue problem into the standard eigenvalue problem.
$kx={\lambda }mx$..............................................(1)
The $m,m^{-1/2},$ and $m^{1/2}$ matrix will be needed so they are solved and provided below.
$m={\begin{bmatrix}3&0\\0&2\\\end{bmatrix}},m^{-1/2}={\begin{bmatrix}1/{\sqrt {3}}&0\\0&1/{\sqrt {2}}\\\end{bmatrix}},m^{1/2}={\begin{bmatrix}{\sqrt {3}}&0\\0&{\sqrt {2}}\\\end{bmatrix}}$
Derivation of the new value $x^{*}$.First premultiply by $m^{-1/2}$
$m^{-1/2}kx={\lambda }m^{-1/2}mx={\lambda }m^{1/2}x$
Let, $x^{*}=m^{1/2}x$
Then, $x=m^{-1/2}x^{*}$
Plugging Into Eq. 1 above.
$(m^{-1/2}km^{1/2})x^{*}=k^{*}x^{*}={\lambda }x^{*}=0$
Where, $k^{*}=m^{-1/2}km^{1/2}={\begin{bmatrix}3.333&8.165\\8.165&-17.5\\\end{bmatrix}}$
Now, $(k^{*}-{\lambda }I)x^{*}=0$............................................(2)
First, Solving for the eigenvalues by setting the determinant of $(k^{*}-{\lambda }I)$equal to zero.
$det(k^{*}-{\lambda }I)=det({\begin{bmatrix}-3.333-{\lambda }&8.165\\8.165&-17.5-{\lambda }\\\end{bmatrix}})=0$
$(-3.33-{\lambda })(-17.5-{\lambda })=0$
${\lambda }^{2}+20.83{\lambda }-8.392=0$
Solving for ${\lambda }_{1}$ and ${\lambda }_{2}$ using the quadratic.
${\lambda }_{1}=0.395$ and ${\lambda }_{2}=-21.225$
Plugging these values in one at a time into Eq.2 above will yield two respective eigenvectors.
Solving for $x_{1}^{*}$,
${\begin{bmatrix}3.333-0.935&8.165\\8.165&-17.50.395\\\end{bmatrix}}{\begin{bmatrix}x_{11}^{*}\\x_{12}^{*}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}$
$-3.725x_{11}^{*}+8.165x_{12}^{*}=0$ $8.165x_{11}^{*}-17.895x_{12}^{*}=0$
So, $x_{1}^{*}={\begin{bmatrix}2.19\\1\\\end{bmatrix}}$
Now solving for $x_{2}^{*}$,
${\begin{bmatrix}3.333+21.225&8.165\\8.165&-17.5+21.225\\\end{bmatrix}}{\begin{bmatrix}x_{21}^{*}\\x_{22}^{*}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}$
$17.895x_{21}^{*}+8.165x_{22}^{*}=0$ $8.165x_{21}^{*}+3.725x_{22}^{*}=0$
So, $x_{2}^{*}={\begin{bmatrix}-0.456\\1\\\end{bmatrix}}$

#### Final Solution

Using the transformation $x=M^{-1/2}x^{*}$, it is possible to obtain the eigenvectors $x_{1}$ and $x_{2}$.
$x_{1}=M^{-1/2}x_{1}^{*}={\begin{bmatrix}1/{\sqrt {3}}&0\\0&1/{\sqrt {2}}\\\end{bmatrix}}{\begin{bmatrix}2.19\\1\\\end{bmatrix}}={\begin{bmatrix}1.264\\0.707\\\end{bmatrix}}$
$x_{2}=M^{-1/2}x_{2}^{*}={\begin{bmatrix}1/{\sqrt {3}}&0\\0&1/{\sqrt {2}}\\\end{bmatrix}}{\begin{bmatrix}-0.456\\1\\\end{bmatrix}}={\begin{bmatrix}-0.263\\0.707\\\end{bmatrix}}$
So,
$x_{1}={\begin{bmatrix}1.264\\0.707\\\end{bmatrix}}$ $x_{2}={\begin{bmatrix}-0.263\\0.707\\\end{bmatrix}}$

### Problem R6.2
#### Description

We are to resolve problem 4.4/5.4 but by using the method of transforming the generalized eigenvalue problem into a standard eigenvalue problem and then proceeding in the calculations.

#### Solution

This problem is accomplished by using the following simple transformation:
$K^{*}=M^{-1/2}*K*M^{-1/2}$
The problem is carried out exactly the same as from report 5 until right before the eigenvectors are found. Instead, the eigenvalues and eigenvectors are found from the $K^{*}$ matrix. Once this is done, the true eigenvalues for our system are found according to:
$x=M^{-1/2}*x^{*}$
The code to solve this problem are presented below.
References for Matlab code: Team 3 Report 4 Team 7 Report 4 Team 4 Report 5
The resulting displacements were:

### R6.3 Problem Computer reactions and compare
#### Problem Statement

Redo R3.1. Compute the reactions, and compare the reactions in the case of non-zero prescribed disp dofs with the reactions in the case of zero prescribed disp dofs.

#### Solution

Given zero displacement DOFs, the following MATLAB program solves for the unknown global displacement DOF's, unknown for components (reactions), and member forces in the system.
Global Displacements
Unknown disp at node 1(m)
Element forces (N)
Element stresses (Pa)
Verifying with Calfem and plotting
OUTPUT
Original values of non-zero displacement degrees of freedom: (R3.1) Global displacement DOF's (m)
$u_{1}=-0.005$
$v_{1}=-0.0103$ $u_{2}=0.02$ $v_{2}=-0.01$ $u_{3}=-0.03$ $v_{3}=0.05$ $u_{4}=-0.0257$ $v_{4}=0.0764$
Element Reaction Forces (N)
$P^{1}=-854900$ $P^{2}=-418180$ $P^{3}=740079$
Element Stresses (GPa)
$\sigma ^{1}=-8.5490$ $\sigma ^{2}=-4.1818$ $\sigma ^{3}=7.4008$

## Problem 6.4 Deformation in 3D Space

On my honor, I have neither given nor received unauthorized aid in doing this assignment.

### Problem Statement

There is a 3D truss system with applied forces. We need to optimize each beam to meet a factor of safety of 1.5.

### Conclusion

The final area of each member was able to be reduced to its minimum of 0.1 inches and stay above a factor of safety of 1.5. Final stresses are in psi:
* 13.94
* -8487.15
* -8487.15
* 8611.63
* 8611.63
* -12871.54
* 13045.99
* -12871.54
* 13045.99
* 63.42
* 63.42
* 788.95
* -778.48
* -5171.02
* 5174.49
* -5171.02
* 5174.49
* -7787.74
* -7787.74
* 7792.39
* 7792.39
* 14680.46
* -14679.76
* -14679.76
* 14680.46

### CALFEM Verification

The Matlab calculations were confirmed using the CALFEM toolbox in Matlab.
The following presents the constants used and options for different designs
This Edof matrix uses column 1 as the as the element number. The next three columns are the DOFs for the first node and the next three columns are for the second node since this is a three dimensional problem.
Coord contains the x, y, and z coordinates for each node.
The bar3e is used to make the stiffness matrix for a 3D system.
The bar3s is used to find the member forces in a 3D system.
One can see that the 3 denotes the dimension of the system being observed. The stresses can be found by dividing the forces by the areas.

## Problem 6.5 3d Mode Shapes

On my honor, I have neither given nor received unauthorized aid in doing this assignment.

### Problem Statement

Determine the first 3 mode shapes of the 3D truss system from problem 6.4

### Solution

The set up of the problem was the same as problem 6.4. The eigenvalues can be seen below.

#### Gif's of mode shapes
##### Mode 1
###### Top View
##### Mode 2
###### Top View
##### Mode 3
###### Top View
## Table of Assignments R6

Problem Assignments R5 :
| Problem # | Solved&Typed by | Reviewed by |
| 1 | Bryan Tobin, Chad Colocar | All |
| 2 | David Bonner, Vernon Babich | All |
| 3 | Chad Colocar, Bryan Tobin | All |
| 4 | Vernon Babich, Tyler Wulterkens | All |
| 5 | Tyler Wulterkens, David Bonner | All |