Structural Design
Essay by 24 • December 3, 2010 • 407 Words (2 Pages) • 1,491 Views
Objective
To learn to describe a structure in terms of natural dynamic characteristics such as frequency, damping, and mode shapes. The analyzed properties are applied to Structural Health Monitoring and Damage Detection, to vibration-based analysis.
Theoretical background
Briefing based on last experimentation, dynamic response of single degree of freedom system is described by the following equation.
в?'в-'F = my М?
-ky вЂ" cy Ðœ‡ = my Ðœ?
y is equal to; - C/2m Ð'± iв?Ñ™(k/m- (C/2m)^2 ) , and for the case of under-damped system, the value inside route is below 0.
Substituting П‰_D= в?Ñ™(k/m- гЂ-(C/2m)гЂ--^2 ) = в?Ñ™(k/m(1-C^2/4mk)) = в?Ñ™(k/m(1-C^2/гЂ-C^2гЂ--_cr ))
= в?Ñ™(k/m) в?Ñ™(гЂ-(1-(C/C_cr ))гЂ--^2 )= П‰Ð²?Ñ™(1-в„Ò^2, )where в„Ò = C/C_r
Replacing, y(t) = Ce^(-в„Òwt) cosвЃЎ(П‰_D t+ О±), and through estimating,
в„Ò = 1/2ПЂn lnвЃЎгЂ-y_i/y_(i+n) гЂ--
Another way of measuring shows in the Fig.1.
<Fig.1 Method to experimentally measure в„Ò>
Experimental set-ups and procedures
Pin steel plate on a table with shaking controller attached
Adjust to the starting point, by first trying shaking with the base only
Connect columns and the cover plate to the bottom plate
Set the frequency of the controller and let it run
Scan the graph showing the deflection of the bottom/top plate with respect to time
<Fig2. Picture of our experimentation setting>
Results & Discussion
First Experiment: Free Vibration of SDOF System.
<Fig.3 Graph of free vibration ([Units] x-axis; sec/ y-axis; mm)>
* Graph is not revised to start from the origin. Careful consideration was made while measuring the amplitude.
Finding П‰_d, the damped natural frequency
Shook 130 times in 50 seconds; f_d=130/50=2.6Hz П‰_d=2ПЂf_d=16.328 rad/s
Finding ОÑ*, the damping ratio
Based on the analyses of the Fig.3 x_1=25,гЂ- xгЂ--_130=3.36
ОÑ*= 1/(2ПЂ*130) lnвЃЎ(x_1/x_130 )=1/(2ПЂ*130) lnвЃЎ(25/3.36)=0.0025
Finding П‰_n, the natural frequency
П‰_n=П‰_d (в?Ñ™(1-ОÑ*^2 ))^(-1)=16.328(в?Ñ™(1-гЂ-0.0025гЂ--^2 ))^(-1)=16.328 rad/s
f_n=П‰_n/2ПЂ=2.60
Since ОÑ* is so small, П‰_n=П‰_d
Second
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