CE 573: Structural Dynamics
HW#2 Solution
Problem #1
Given
:
A viscously damped structure is set into free vibration with an initial velocity. The resulting
damped oscillations are shown above.
Required
:
(a) Determine the logarithmic decrement, the damping factor
ξ
, and the undamped natural
period of vibration for this structure.
(b) Suppose that the displacement measurements have an accuracy of
in. (You may
assume that the time measurements have negligible error.) Determine the range of possible values
for the damping factor. Discuss the influence that the measurement errors have on
0.02
±
.
Solution
:
(a) I should point out that, since all of the results will depend upon the accuracy with which
you read the displacement values off of the graph, it is expected that there will be some variations in your
answers. As long as your results are reasonably close, they will be accepted.
Let’s start by estimating the logarithmic decrement
δ
. I chose to work with the positive peaks, starting
with peak #1 (at
seconds):
0.08
t
≈
1
0.60
u
≈
inches.
Next, I looked at peak #4 (at
seconds):
0.60
t
≈
4
0.18
u
≈
inches.
Thus, the value of
can be evaluated as:
11
0
.
6
0
ln
ln
0.40
30
.
1
8
i
ik
u
ku
+
⎛⎞
===
⎜⎟ ⎜⎟
⎝⎠
Answer
Based on this, the damping factor
can be estimated according to the formulas we derived in class:
()
22
2
2
0.40
0.064
4
0.40
4
δπ
π
==
=
+
+
Answer
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentFinally, we can estimate
T
as follows. The length of time between peak #4 and peak #1 is approximately
. There are three cycles of damped oscillation during this
time, so the damped natural period can be estimated as
14
0.60
0.08 seconds
0.52 seconds
t
→
∆≈ −
=
0.173
3
d
t
T
→
∆
≈=
seconds.
But
is related to the natural period of vibration
T
via
d
T
()
(
)
2
2
0.173 seconds
1
0.064
0.172 seconds.
1
d
T
TT
ξ
=⇒
=
− =
−
Answer
(b) Since the range in
relates directly to the range in
δ
, I will discuss the range in
first. Notice
that, since
depends on a ratio of peak amplitudes, the largest possible value of
comes when the
amplitude ratio is as large as possible, while the smallest possible value of
occurs when the amplitude
ratio is as small as possible. The measurement errors give us a way to estimate these largest and smallest
ratios. (I will use the same peaks as in part (a) to estimate the extreme ratios.)
•
For the largest amplitude ratio, we want to assume that the measured peak #1 is underestimated
while the measured peak #4 is overestimated. This gives us:
11
44
max
max
max
22
max
0.60
0.02
0.62 in.
0.18
0.02
0.16 in.
10
.
6
2
ln
0.4515
0.072.
30
.
1
6
4
real
measured
real
measured
uu
u
u
δξ
δπ
=+
∆
=
+
=
=−
∆
=
−
=
⎛⎞
⇒≈
=
⇒=
=
⎜⎟
⎝⎠
+
•
For the smallest amplitude ratio, we want to assume that the measured peak #1 is overestimated
while the measured peak #4 is underestimated. This gives us:
min
min
min
min
0.60
0.02
0.58 in.
0.18
0.02
0.20 in.
.
5
8
ln
0.3549
0.056.
.
2
0
4
real
measured
real
measured
u
u
=
−
∆
=−=
∆
=
+
=
=
=
+
Thus, based on the measurements used above and the measurement error, we can estimate the range of
damping ratios as 0.056
0.072
≤≤
Answer
. Notice that, as a percentage of our assumed “correct”
damping ratio
0.064
=
, the range of damping ratios goes from 12.5% below to 10.8% above the
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '05
 Whalen
 Ceq, amplitude ratio, total resisting force

Click to edit the document details