# University of Florida/Eml4507/s13.team3.GuzyR3

Problem 2: Dampened and Undamped System

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

## Given

Spring-damper-body arrangement as shown. Two separate forces applied to masses.

$M={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}$ $d={\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}$ $C={\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}$ $K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}$ ## Find

1.Draw the FBDs of all components in the spring-mass-damper system on p.53-13, with the known disp dofs being at the left and right supports:

$d_{3}=0$ $d_{4}=0$ 2. Derive (1) p.53-12, and (1)-(3) p. 53-13.

## Solution

### Free Body Diagrams (Question 1)

The free body diagrams for a dampened system and undamped system can be seen to the right.

### Solving the System (Question 2)

#### DAMPENED

Analyzing the forces on mass one, we obtain:

 $\displaystyle m_{1}d''_{1}+k_{1}d_{1}+C_{1}d'_{1}-k_{2}d_{2}+k_{2}d_{1}-C_{2}d'_{2}+C_{2}d'_{1}=F_{1}$ $\displaystyle m_{1}d''_{1}+C_{1}d'_{1}+C_{2}(d'_{1}-d'_{2})+k_{1}d_{1}+k_{2}(d_{1}-d_{2})=F_{1}$ Analyzing the forces on mass two, we obtain:

 $\displaystyle m_{2}d''_{2}+C_{2}(d'_{2}-d'_{1})+k_{2}(d_{2}-d_{1})+C_{3}d'_{2}+k_{3}d_{2}=F_{2}$ $\displaystyle m_{2}d''_{2}+C_{2}(d'_{2}-d'_{1})+C_{3}d'_{2}+k_{2}(d_{2}-d_{1})+k_{3}d_{2}=F_{2}$ Plugging equations into matrix form we obtain:

${\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}{\begin{bmatrix}d''_{1}\\d''_{2}\\\end{bmatrix}}+{\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}{\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}{\begin{bmatrix}d'_{1}\\d'_{2}\\\end{bmatrix}}+{\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}$ \

Notice how this form matches each matrix given in the problem statement. Plugging in the original yields and proves the following equation of motion for an undamped system:

$Md''+Cd'+kd=0$ #### UNDAMPED

Analyzing the forces on mass one, we obtain:

 $\displaystyle m_{1}d''_{1}+k_{1}d_{1}-k_{2}d_{2}+k_{2}d_{1}=F_{1}$ $\displaystyle m_{1}d''_{1}-k_{2}d_{2}+(k_{1}+k_{2})d_{1}=F_{1}$ Analyzing the forces on mass two, we obtain:

 $\displaystyle m_{2}d''_{2}+k_{2}d_{2}-k_{2}d_{1}+k_{3}d_{2}=F_{2}$ $\displaystyle m_{2}d''_{2}+(k_{2}+k_{3})d_{2}-k_{2}d_{1}=F_{2}$ Plugging equations into matrix form we obtain:

${\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}{\begin{bmatrix}d''_{1}\\d''_{2}\\\end{bmatrix}}+{\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}{\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}$ \

Notice how this form matches each matrix given in the problem statement. Plugging in the original yields and proves the following equation of motion for an undamped system:

$Md''+kd=0$ 