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Vibrations Final Exam
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My Vibrations final exam in mathcad
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Content Preview
Ben Birrer
Problem 1
3600
m
lbf = 1.633 103 kg
g
Vibrations Final Exam
mV
ME 472
AS 10
where AS is the accelerometer
g
sensitivity.
Part A
According to the graph, the voltage amplitude is .002 Volts.
Therefore, the following applies for the acceleration amplitude.
.002 V
m
a =
1.961
AS
s2
To determine the frequency of the acceleration, I assumed exactly 3 periods
between 0 seconds and .025 seconds. Therefore, the follow applies.
-1
.025 s
f =
w
120 Hz
3
rad
w f
=
f
w 2
753.982
s
Part B
Becuase the frequency and mass are so high, it is recommended that to
isolate and reduce the vibrations, a vibration isolation pad be used. I will use
a Fabreeka M1393-P1 for my solution. http://www.fabreeka.com/documents/
file/products/fabreeka_transformer_mount_0610.pdf
Part C
A 3 in 3 in = 0.006 m2
F
450 lbf =
max
2.002 103 N
.06 in =
max
0.002 m
8 total pads must be used as
each pad can only handle
Fmax
N
450 lbf.
K
=
pad
1.313 106
max
m
2
8 Kpad
rad
w
=
n
80.217
m
s
Given the provided information, I must select an intelligent guess for the
damping ratio.
.2
kg
c 2 m w =
n
5.24 104
s
Part D
wf
r =
9.399
While this is overkill, it is acceptable.
wn
However, more economical studies may be
required as an r of over 9 may be expensive.
r > 3
1
TR =
0.011
Approximately 1.1% of the force is transmitted.
r2 - 1
R 1 - TR = 0.989
Reduction in 98.9% of initial force transmission.
m
a
TR a =
new
0.022
s2
Percent Difference = Reduction in Transmissibility
| a
- |
new
a
PD | |
100 = 98.855
|
a
|
Problem 2
m
m 100 kg
V
amp
.001
s
X
1 mm =
amp
0.001 m
Information obtained from Graph #1
T .1 s
1
f
=
f
10 Hz
T
1
w f
=
f
f 2
62.832
s
Information obtained from Graph #2
Just before .2 seconds the displacement is 0 and just after .4 seconds the
displacement is also 0. This change in time is over a distance of 2 Pi or 1
period.
T .42 s - .19 s = 0.23 s
1
f
=
n
4.348 Hz
T
rad
w f
=
n
n 2
27.318
s
wf
r =
2.3
wn
Now I must approximate a region of graph #2 to find the damping ratio. I will
use from time t=0 to t=.5. This reduction in amplitude is about .52.
XA e-(w
n
t )
Rearrange and solve for damping ratio.
-2 ln ((.52)) s-1
= 0.048
MathCAD has issues with units so I
wn
multiplied by inverse seconds to cancle
units properly.
r < 3
2
Therefore, approximately
2
1 + ((2 r))
TR
100 = 23.837
23.8 percent of the original
2
2
force is transmitted.
1 - r2 + ((2 r))
R 100 - TR = 76.163
Thus a reduction in 76.2%
of the transmitted force.
Problem 1
3600
m
lbf = 1.633 103 kg
g
Vibrations Final Exam
mV
ME 472
AS 10
where AS is the accelerometer
g
sensitivity.
Part A
According to the graph, the voltage amplitude is .002 Volts.
Therefore, the following applies for the acceleration amplitude.
.002 V
m
a =
1.961
AS
s2
To determine the frequency of the acceleration, I assumed exactly 3 periods
between 0 seconds and .025 seconds. Therefore, the follow applies.
-1
.025 s
f =
w
120 Hz
3
rad
w f
=
f
w 2
753.982
s
Part B
Becuase the frequency and mass are so high, it is recommended that to
isolate and reduce the vibrations, a vibration isolation pad be used. I will use
a Fabreeka M1393-P1 for my solution. http://www.fabreeka.com/documents/
file/products/fabreeka_transformer_mount_0610.pdf
Part C
A 3 in 3 in = 0.006 m2
F
450 lbf =
max
2.002 103 N
.06 in =
max
0.002 m
8 total pads must be used as
each pad can only handle
Fmax
N
450 lbf.
K
=
pad
1.313 106
max
m
2
8 Kpad
rad
w
=
n
80.217
m
s
Given the provided information, I must select an intelligent guess for the
damping ratio.
.2
kg
c 2 m w =
n
5.24 104
s
Part D
wf
r =
9.399
While this is overkill, it is acceptable.
wn
However, more economical studies may be
required as an r of over 9 may be expensive.
r > 3
1
TR =
0.011
Approximately 1.1% of the force is transmitted.
r2 - 1
R 1 - TR = 0.989
Reduction in 98.9% of initial force transmission.
m
a
TR a =
new
0.022
s2
Percent Difference = Reduction in Transmissibility
| a
- |
new
a
PD | |
100 = 98.855
|
a
|
Problem 2
m
m 100 kg
V
amp
.001
s
X
1 mm =
amp
0.001 m
Information obtained from Graph #1
T .1 s
1
f
=
f
10 Hz
T
1
w f
=
f
f 2
62.832
s
Information obtained from Graph #2
Just before .2 seconds the displacement is 0 and just after .4 seconds the
displacement is also 0. This change in time is over a distance of 2 Pi or 1
period.
T .42 s - .19 s = 0.23 s
1
f
=
n
4.348 Hz
T
rad
w f
=
n
n 2
27.318
s
wf
r =
2.3
wn
Now I must approximate a region of graph #2 to find the damping ratio. I will
use from time t=0 to t=.5. This reduction in amplitude is about .52.
XA e-(w
n
t )
Rearrange and solve for damping ratio.
-2 ln ((.52)) s-1
= 0.048
MathCAD has issues with units so I
wn
multiplied by inverse seconds to cancle
units properly.
r < 3
2
Therefore, approximately
2
1 + ((2 r))
TR
100 = 23.837
23.8 percent of the original
2
2
force is transmitted.
1 - r2 + ((2 r))
R 100 - TR = 76.163
Thus a reduction in 76.2%
of the transmitted force.
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