University of Florida/Eml4507/s13.team3.andersondavyr7

Problem 7.1
On my honor, I have neither given nor received unauthorized aid in doing this assignment.

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.

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^2a_{2}&0&-6La_{2}&2L^2a_{2}\\

-a_{1}&0&0&a_{1}&0&0&\\

0&-12a_{2}&-6La_{2}&0&12a_{2}&-6La_{2}\\

0&6La_{2}&4L^2a_{2}&0&-6La_{2}&4L^2a_{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^{(1)}_{2}&6La^{(1)}_{2}&0&-12a^{(1)}_{2}&6La^{(1)}_{2}&0&0&0\\

0&6La^{(1)}_{2}&4L^2a^{(1)}_{2}&0&-6La^{(1)}_{2}&2L^2a^{(1)}_{2}&0&0&0\\

-a^{(1)}_{1}&0&0&a^{(1)}_{1}+a^{(2)}_{1}&0&0&-a^{(2)}_{1}&0&0\\

0&-12^{(1)}a_{2}&-6La^{(1)}_{2}&0&12a^{(2)}_{2}-12a^{(1)}_{2}&6La^{(2)}_{2}-6La^{(1)}_{2}&0^{(2)}&-12a^{(2)}_{2}&6La^{(2)}_{2}\\

0&6La^{(1)}_{2}&2L^2a^{(1)}_{2}&0&6La^{(2)}_{2}-6La^{(1)}_{2}&4L^2a^{(1)}_{2}+4L^2a^{(2)}_{2}&0^{(2)}&-6La^{(2)}_{2}&2L^2a^{(2)}_{2}\\

0&0&0&-a^{(2)}_{1}&0&0&a^{(2)}_{1}&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^2a^{(2)}_{2}&0&-6La^{(2)}_{2}&4L^2a^{(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^{(1)}_{2}&6La^{(1)}_{2}&0&-12a^{(1)}_{2}&6La^{(1)}_{2}&0&0&0\\

0&6La^{(1)}_{2}&4L^2a^{(1)}_{2}&0&-6La^{(1)}_{2}&2L^2a^{(1)}_{2}&0&0&0\\

-a^{(1)}_{1}&0&0&a^{(1)}_{1}+a^{(2)}_{1}&0&0&-a^{(2)}_{1}&0&0\\

0&-12^{(1)}a_{2}&-6La^{(1)}_{2}&0&12a^{(2)}_{2}-12a^{(1)}_{2}&6La^{(2)}_{2}-6La^{(1)}_{2}&0^{(2)}&-12a^{(2)}_{2}&6La^{(2)}_{2}\\

0&6La^{(1)}_{2}&2L^2a^{(1)}_{2}&0&6La^{(2)}_{2}-6La^{(1)}_{2}&4L^2a^{(1)}_{2}+4L^2a^{(2)}_{2}&0^{(2)}&-6La^{(2)}_{2}&2L^2a^{(2)}_{2}\\

0&0&0&-a^{(2)}_{1}&0&0&a^{(2)}_{1}&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^2a^{(2)}_{2}&0&-6La^{(2)}_{2}&4L^2a^{(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.