# University of Florida/Eml4507/s13.team3.andersondavyr7
## Problem 7.1
### Given

The free body diagram to the right was given for the problem.

### Find

Solve 2-element frame system using same data for 2-bar truss system.
Solve 2-element truss system.

### Solution
#### Frame Problem

A basic form is shown below:
${\begin{bmatrix}a_{1}&0&0&-a_{1}&0&0\\0&12a_{2}&6La_{2}&0&-12a_{2}&6La_{2}\\0&6La_{2}&4L^{2}a_{2}&0&-6La_{2}&2L^{2}a_{2}\\-a_{1}&0&0&a_{1}&0&0&\\0&-12a_{2}&-6La_{2}&0&12a_{2}&-6La_{2}\\0&6La_{2}&4L^{2}a_{2}&0&-6La_{2}&4L^{2}a_{2}\\\end{bmatrix}}{\begin{bmatrix}u_{1}\\v_{1}\\t_{1}\\u_{2}\\v_{2}\\t_{2}\\\end{bmatrix}}={\begin{bmatrix}f_{x1}\\f_{y1}\\c_{c1}\\f_{x2}\\f_{y2}\\c_{c2}\\\end{bmatrix}}$
Equations derived from stiffness equations:
$a_{1}={\frac {E*A}{L}}$
$a_{2}={\frac {E*I}{L^{3}}}$
Globle stiffness equation is shown below:
${\begin{bmatrix}a_{1}^{(1)}&0&0&-a_{1}^{(1)}&0&0&0&0&0\\0&12a_{2}^{(1)}&6La_{2}^{(1)}&0&-12a_{2}^{(1)}&6La_{2}^{(1)}&0&0&0\\0&6La_{2}^{(1)}&4L^{2}a_{2}^{(1)}&0&-6La_{2}^{(1)}&2L^{2}a_{2}^{(1)}&0&0&0\\-a_{1}^{(1)}&0&0&a_{1}^{(1)}+a_{1}^{(2)}&0&0&-a_{1}^{(2)}&0&0\\0&-12^{(1)}a_{2}&-6La_{2}^{(1)}&0&12a_{2}^{(2)}-12a_{2}^{(1)}&6La_{2}^{(2)}-6La_{2}^{(1)}&0^{(2)}&-12a_{2}^{(2)}&6La_{2}^{(2)}\\0&6La_{2}^{(1)}&2L^{2}a_{2}^{(1)}&0&6La_{2}^{(2)}-6La_{2}^{(1)}&4L^{2}a_{2}^{(1)}+4L^{2}a_{2}^{(2)}&0^{(2)}&-6La_{2}^{(2)}&2L^{2}a_{2}^{(2)}\\0&0&0&-a_{1}^{(2)}&0&0&a_{1}^{(2)}&0&0\\0&0&0&0&-12a_{2}^{(2)}&-6La_{2}^{(2)}&0&-12a_{2}^{(2)}&-6La_{2}^{(2)}\\0&0&0&0&6La_{2}^{(2)}&2L^{2}a_{2}^{(2)}&0&-6La_{2}^{(2)}&4L^{2}a_{2}^{(2)}\\\end{bmatrix}}$
Q matrix is shown below:
$Q={\begin{bmatrix}u_{1}\\v_{1}\\t_{1}\\u_{2}\\v_{2}\\t_{2}\\u_{3}\\v_{3}\\t_{3}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\0\\u_{2}\\v_{2}\\t_{2}\\0\\0\\t_{3}\\\end{bmatrix}}$
Force matrix is show below:
$F={\begin{bmatrix}f_{x1}\\f_{y1}\\c_{c1}\\f_{x2}\\f_{y2}\\c_{c2}\\f_{x3}\\f_{y3}\\c_{c3}\\\end{bmatrix}}={\begin{bmatrix}f_{x1}\\f_{y1}\\c_{c1}\\0\\P\\0\\f_{x3}\\f_{y3}\\0\\\end{bmatrix}}$
Matrix for the system is shown below:
${\begin{bmatrix}a_{1}^{(1)}&0&0&-a_{1}^{(1)}&0&0&0&0&0\\0&12a_{2}^{(1)}&6La_{2}^{(1)}&0&-12a_{2}^{(1)}&6La_{2}^{(1)}&0&0&0\\0&6La_{2}^{(1)}&4L^{2}a_{2}^{(1)}&0&-6La_{2}^{(1)}&2L^{2}a_{2}^{(1)}&0&0&0\\-a_{1}^{(1)}&0&0&a_{1}^{(1)}+a_{1}^{(2)}&0&0&-a_{1}^{(2)}&0&0\\0&-12^{(1)}a_{2}&-6La_{2}^{(1)}&0&12a_{2}^{(2)}-12a_{2}^{(1)}&6La_{2}^{(2)}-6La_{2}^{(1)}&0^{(2)}&-12a_{2}^{(2)}&6La_{2}^{(2)}\\0&6La_{2}^{(1)}&2L^{2}a_{2}^{(1)}&0&6La_{2}^{(2)}-6La_{2}^{(1)}&4L^{2}a_{2}^{(1)}+4L^{2}a_{2}^{(2)}&0^{(2)}&-6La_{2}^{(2)}&2L^{2}a_{2}^{(2)}\\0&0&0&-a_{1}^{(2)}&0&0&a_{1}^{(2)}&0&0\\0&0&0&0&-12a_{2}^{(2)}&-6La_{2}^{(2)}&0&-12a_{2}^{(2)}&-6La_{2}^{(2)}\\0&0&0&0&6La_{2}^{(2)}&2L^{2}a_{2}^{(2)}&0&-6La_{2}^{(2)}&4L^{2}a_{2}^{(2)}\\\end{bmatrix}}{\begin{bmatrix}0\\0\\0\\u_{2}\\v_{2}\\t_{2}\\0\\0\\0\\\end{bmatrix}}={\begin{bmatrix}f_{x1}\\f_{y1}\\c_{c1}\\0\\P\\0\\f_{x3}\\f_{y3}\\0\\\end{bmatrix}}$
The reduced stiffness is
$k={\begin{bmatrix}5.75&0&0&0\\0&0.622&0.619&0.625\\0&0.619&-0.818&0.417\\0&0.625&0.417&0.833\\\end{bmatrix}}$
To solve for the displacements
${\begin{bmatrix}0\\6\\0\\0\\\end{bmatrix}}===={\begin{bmatrix}5.75&0&0&0\\0&0.622&0.619&0.625\\0&0.619&-0.818&0.417\\0&0.625&0.417&0.833\\\end{bmatrix}}{\begin{bmatrix}u_{2}\\v_{2}\\t_{2}\\t_{3}\\\end{bmatrix}}$
After matrix math.
${\begin{bmatrix}u_{2}\\v_{2}\\t_{2}\\t_{3}\\\end{bmatrix}}===={\begin{bmatrix}0\\24.5\\7.32\\-22.08\\\end{bmatrix}}$
After solving for the forces the result is
${\begin{bmatrix}fx_{1}\\fy_{1}\\C_{1}\\fx_{3}\\fy_{3}\\\end{bmatrix}}={\begin{bmatrix}0\\-594\\-4118\\0\\-6.1\\\end{bmatrix}}$
Below is a figure of the new deformed frame plotted with the undeformed frame.