University of Florida/Eml4507/s13.team3.GuzyR3

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Given

Spring-damper-body arrangement for R3.2 ^[1]

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

$M = [\begin{matrix} m_{1} & 0 \\ 0 & m_{2} \end{matrix}]$

$d = [\begin{matrix} d_{1} \\ d_{2} \end{matrix}]$

$C = [\begin{matrix} C_{1} + C_{2} & - C_{2} \\ - C_{2} & C_{2} + C_{3} \end{matrix}]$

$K = [\begin{matrix} (k_{1} + k_{2}) & - k_{2} \\ - k_{2} & (k_{2} + k_{3}) \end{matrix}]$

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)

File:Report31damper.png

Dampened:Spring-damper-body free body

File:Report311.png

Undamped:Spring-body free body

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:

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}

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:

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}

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:

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

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:

$M d^{″} + C d^{'} + k d = 0$

UNDAMPED

Analyzing the forces on mass one, we obtain:

m_{1} d^{'}'_{1} + k_{1} d_{1} - k_{2} d_{2} + k_{2} d_{1} = F_{1}

m_{1} d^{'}'_{1} - k_{2} d_{2} + (k_{1} + k_{2}) d_{1} = F_{1}

Analyzing the forces on mass two, we obtain:

m_{2} d^{'}'_{2} + k_{2} d_{2} - k_{2} d_{1} + k_{3} d_{2} = F_{2}

m_{2} d^{'}'_{2} + (k_{2} + k_{3}) d_{2} - k_{2} d_{1} = F_{2}

Plugging equations into matrix form we obtain:

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

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:

$M d^{″} + k d = 0$

Template:Reflist

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↑ https://docs.google.com/document/d/1lQLCUXKTYtMs4BqzLExS5jqLVcpAIP1FW4sd2sl1cqA/edit?pli=1

[1] ttps://docs.google.com/document/d/1lQLCUXKTYtMs4BqzLExS5jqLVcpAIP1FW4sd2sl1cqA/edit?pli=1

[1]

University of Florida/Eml4507/s13.team3.GuzyR3

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Given

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Solution

Free Body Diagrams (Question 1)

Solving the System (Question 2)

DAMPENED

UNDAMPED

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University of Florida/Eml4507/s13.team3.GuzyR3

Given

Find

Solution

Free Body Diagrams (Question 1)

Solving the System (Question 2)

DAMPENED

UNDAMPED

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