University of Florida/Eml4507/s13.team3.GuzyR3

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.

Free Body Diagrams (Question 1)
The free body diagrams for a dampened system and undamped system can be seen to the right.

DAMPENED
Analyzing the forces on mass one, we obtain:
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$$\displaystyle m_{1}d''_{1}+ k_{1}d_{1} +C_{1}d'_{1} - k_{2}d_{2}+ k_{2}d_{1}-C_{2}d'_{2}+C_{2}d'_{1} = F_{1}$$
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$$\displaystyle m_{1}d''_{1} + C_{1}d'_{1} + C_{2}(d'_{1}- d'_{2}) +k_{1}d_{1}+ k_{2}(d_{1}- d_{2})= F_{1}$$
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Analyzing the forces on mass two, we obtain:
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$$\displaystyle m_{2}d''_{2} + C_{2}(d'_{2}- d'_{1}) + k_{2}(d_{2}- d_{1}) +C_{3}d'_{2} +k_{3}d_{2} = F_{2}$$
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$$\displaystyle m_{2}d''_{2} + C_{2}(d'_{2}- d'_{1}) +C_{3}d'_{2}+ k_{2}(d_{2}- d_{1}) +k_{3}d_{2} = F_{2}$$
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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:
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$$\displaystyle m_{1}d''_{1}+ k_{1}d_{1} - k_{2}d_{2}+ k_{2}d_{1} = F_{1}$$
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$$\displaystyle m_{1}d''_{1} - k_{2}d_{2} +(k_{1}+k_{2})d_{1} = F_{1}$$
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Analyzing the forces on mass two, we obtain:


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$$\displaystyle m_{2}d''_{2}+ k_{2}d_{2} - k_{2}d_{1}+ k_{3}d_{2} = F_{2}$$
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$$\displaystyle m_{2}d''_{2}+ (k_{2}+k_{3})d_{2} - k_{2}d_{1}= F_{2}$$
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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:

$$ Md''+kd=0 $$