Vibration Solved
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Transcript of Vibration Solved
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Mechanical Vibrations
ME444
Dr Khaled Zied
Lecture 5:
Base excitation
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Consi er t e ase excite system
of the figure. The goal of the analysis
response x(t) (typically acceleration or
displacement of the mass) given the base
motion y(t).
rom t e ree o y agram, app cat on
of Newtons second law leads directly to
yckykx xc xm &&&& +=++
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Assume t at t e ase motion is armonic,
t i
oeY t y
=)( ,t ie X t x
~)( =
~ .
function are derived in the same manner as for the force excited system. The transfer function is:
icmk H
+=
)()( 2
222
2
222
22 )2(1)()(
r ck H
X o
+=
+==o
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2
In non imensiona orm
222 )2()1( r r Y oo
+=
system is shown in the figure.
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Assuming t at t e ase motion is armonic,
t i
oeY t y
=)( And assuming the response is also harmonic,
t i ~
Following the procedure as described above, the transfer function is:
icmk
m H
+=
)()( 2
222
2
)(
m
H Y Z o
==
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2
In non imension ess orm,
222 )2()1( r r Y oo
+=
shown in the figure:
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Again note t at t e trans er unction or re ative ve ocity an acce eration
responses can be derived by multiplying m 2
icmk +=
)( 2
by i and
^2,
respectively.
The gain functions for velocity and acceleration responses can be obtained
222
2
)()()(
cmk
m H
Y
Z
o
o
+==
222
2
)2()1( r r
r
Y
Z
o
o
+=
by and ^2, respectively.
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Gain functions for force excited systems
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Gain functions for base excited systems
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Because r is t e ratio o t e orcing requency to t e natura requency, it follows that r will also be the ratio of the natural period to the forcing period. Thus, 4
To compute k, first note that the natural frequency is
.8
===
nr
.25.01
Hz f n
n ==
g W
k f n
/2
1
=
We compute k=27667 N/m
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Fina y su stituting into
1 X o =
Xo=5.87 m.
)2()1( r r o +
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For resonance please download the Millenium
brid e and Tacoma Brid e videos in the website of the course.
