vibration measurement present
DESCRIPTION
for vibrationTRANSCRIPT
Vibration Measurement
Presented byMr.Kiratikorn Kaewpankan553040033-2
Mechanical Engineering KhonKaen University
ObjectiveTo determine the damped natural
frequency and damping ratio of vibration system by testing.
To determine the damped and undamped natural frequency of vibration system by theory.
Compare frequency from testing and theory.
How to Experiment1. Turn on vibration measurement and setting.
2. Install the sensor at the end of the beam.
3. Use rubber hammer to tap on the beam and wait until the scope shown result in graph.
4. Record the graph and use Vipview to change the graph to the number that use to plot.
Theory
Mass moment of inertia
π½π=13πππ
2π½ s=ππ (π ΒΏΒΏ2+h2)ΒΏ
β ππ= π½ οΏ½ΜοΏ½π½ οΏ½ΜοΏ½=βπ1 π¦1π1πβπ2 π¦2 π2π+π1 οΏ½ΜοΏ½+π2 οΏ½ΜοΏ½
πΉ=π π¦π½ οΏ½ΜοΏ½+π1 π1
2π+π2 π22π=0
ΟΟ
From
Change
We can find
=full volume of beam
=inner volume of beam
β π½ π2+(πΒΏΒΏ1 π12+π2 π2
2)=0ΒΏ
ππ=β (πΒΏΒΏ1 π12+π2 π2
2)π½
ΒΏ
(Ο
Now we want to find but this vibration system canβt be find pure β ππ=0
πΉ 1π1 cosπ+πΉ2 π2πππ π=ππ€π€ cosπ
πΉ=ππ¦π1 π¦1π1πππ π+π2 π¦2 π2πππ π=ππ€π€ cosπ
π1 π¦1π1+π2 π¦2π2=ππ€π€
π¦ π₯
ππ₯= π¦π€
Similar triangles
π1π1π¦π€π1+π2π2
π¦π€π2=ππ€π€
π1π12+π2π2
2=ππ€π€π€π¦
Using = 10 kg = 98.1 N
π1π12+π2π2
2=2275 . 470π .π
ResultsSection 1. Iβm use the average of 5graph to fine natural frequency and damping ratio
When n= cycle of vibration V0 = maximum velocity=
t0, tn=time for vibration
we use for five graph
π π π=27 .5229+27 .5229+27 .1003+27 .1257+27 .2537
5
π ππ=27 .3111π»π§
ππ=2π π ππππ=2π (27 .3111π»π§ )=171 .6007πππ /π
= 0.1145Logarithmic decrement
=πΏ
β(2π)2+πΏ2Damping Ratio
0 .1145
β(2π)2+0 .11452=0 .0182=
Find Damping Ratio ()
Section 2. From Theory
ππ=β (πΒΏΒΏ1 π12+π2 π2
2)π½
ΒΏ
π1π12+π2π2
2=2275.470π .π
= = = )() = =
We can change Undamped into Damped
πππ‘=ππβ1β2From testing
= 27.9246 Hz
πΈππππ=| π ππ‘β π π
π ππ‘|Γ100%ΒΏ|27.9246Hzβ27.3111Hz27.9246Hz |Γ100%=2.1969%
Error between and and
Bonus >> No weight. How about frequency ?
π½=[ 13 0 .23377614 (0 .598)2]New J ΒΏ0 .027866 ππ .π2
Change Undamped into Damped
πππ‘=ππβ1β2
πππ‘= (287 .203 rad /s )Γβ1β0 .01822
πππ‘=287 .155 rad /s
= 45.702 Hz
SummaryFrom testingDamped natural frequency () = 27.3111 Hz Damping ratio =0.0182
From TheoryUndamped Natural frequency () =27.9292 Hzdamped Natural frequency () =27.9249 Hz
Bonus no weightUndamped Natural frequency () = 45.470 Hzdamped Natural frequency () = 45.702 Hz
Error of Damped natural frequency between testing and theory =2.19 %