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System Modeling Coursework
P.R. VENKATESWARANFaculty, Instrumentation and Control Engineering,
Manipal Institute of Technology, ManipalKarnataka 576 104 INDIAPh: 0820 2925154, 2925152
Fax: 0820 2571071Email: [email protected], [email protected]
Web address: http://www.esnips.com/web/SystemModelingClassNotes
Class 19-20: Free (undamped and damped) vibrations
July – December 2008 prv/System Modeling Coursework/MIT-Manipal 2
WARNING!
•
I claim no originality in all these notes. These are the compilation from various sources for the purpose of delivering lectures. I humbly acknowledge the wonderful help provided by the original sources in this compilation.
•
For best results, it is always suggested you read the source material.
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Contents
•
Modeling of free vibrations–
Free vibrations
–
Forced vibrations–
Characteristics of response
•
Numerical on the derivations
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Introduction
•
In theory, the single degree-of-freedom spring-mass system, once set into motion, would continue to move up and down for ever.
•
In practice all systems are damped, which means that energy is dissipated, and the amplitude of the motion gradually gets smaller and smaller until it stops altogether.
•
Damping can be introduced from various sources, and is hard to model accurately.
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System with damping
•
One model, known as viscous damping, is that of a force that is proportional to the velocity of the mass, and which opposes its motion.
•
This is represented by a dashpot, which is given the symbol shown in Fig.
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Equations of motion
•
Free body diagram
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Solutions to this equation
•
The solution to this equation is given by
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Output graphs
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Most important case
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Output graph
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Summary
•
If ζ
> 1, the system is heavily damped, and no vibration occurs.
•
If ζ
= 1, the system is critically damped, and the damping is only just sufficient to prevent vibration.
•
If ζ
< 1, as it is in the majority of cases, there is insufficient damping in the system to prevent vibration and the motion is oscillatory.
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Numerical
•
Let us return to the system in Figure, and, as before, assume that the weight is given an initial displacement, X0
, and then released from rest. Let X0
= 100 mm. Therefore, the initial conditions are that at t = 0, x = 100 mm, and x
= 0 . Find the solution for various values of ζ.
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Case (1): Over damped system with ζ
= 2
•
The displacement is given by•
Differentiating with respect to time
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Output for over damped system with ζ
= 2
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Case (2): Critically damped system, ζ
= 1
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Case (3): Under damped system with ζ
= 0.05
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Case (3): Under damped system with ζ
= 0.05
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Summary
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Summary
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Numerical -
1
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Solution to Numerical 1
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Numerical -
2
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Numerical -
2
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Numerical 3
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Solution to Numerical 3
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And, before we break…
•
Reality is often a matter of personal perception as much as objective fact.
Thanks for listening…