9 Reliability
Section 9.2: First Excursion Failure
9.1. Consider a car that is being hit by hail. The number of times that hail hits the car is modeled as
a Poisson process with the arrival rate of ν = 1 hail/s. Find the probability that the car will be hit
more than 100 times in 20 min. It is estimated that only one tenth of the hail is big enough to cause
permanent damage to the car. What is the probability that the car will have more than 5 such dents in
20 min.
Solution: The Poisson probability density is given by
p
N
(n, t) =
(νt)
n
e
−νt
n!
,
where p
N
(n, t) is the probability of exactly n arrivals in time t. The probability that the car will be hit more
than 100 times in 20 minutes (or 1200 seconds) is then
Pr (more than 100 in 20 minutes) =
∞
n=101
p
N
(n, 1200)
= 1 −
100
n=0
p
N
(n, 1200)
= 1 −
100
n=0
(1200)
n
e
−1200
n!
= 1 − 6. 799 8 × 10
−372
.
Since the number of hail that will cause permanent damage to the car is one tenth of the total number of
hail, the arrival rate for the big hail is ν = 1/10 hail/s. The probability that the car will have more than 5
such dents in 20 minutes is given by
Pr (more than 5 dents in 20 minutes) =
∞
n=6
p
N
(n, 1200)
= 1 −
5
n=0
p
N
(n, 1200)
= 1 −
5
n=0
(120)
n
e
−120
n!
= 1 − 1.658 5 × 10
−44
.
256
9.2. An electronic component is subjected to a random force that is a banded Gaussian process with a
spectrum given by
S
F F
(ω) =
S
0
ω
2
−ω
1
, ω
1
< |ω| < ω
2
.
When can we expect that the force will exceed its threshold of F
0
for the first time? The electronic
component is protected by a vibration isolator with damping and stiffness of c and k, respectively. The
electronic component will fail if the acceleration of the component exceeds F
0
/m. When can we expect
the component to fail if it is protected by the isolator?
Solution: The expected time to the first excursion is 1/ν
+
. The up-crossing rate for a Gaussian process
F (t) is obtained in Example 9.4 and given by
ν
+
=
σ
˙
F
2πσ
F
exp
−
1
2
F
0
σ
F
2
.
The σ
F
and σ
˙
F
are given by
σ
F
=
4
∞
−∞
S
F F
dω
=
S
0
σ
˙
F
=
4
∞
−∞
S
˙
F
˙
F
dω =
4
∞
−∞
ω
2
S
F F
dω
=
S
0
ω
2
2
+ ω
1
ω
2
+ ω
2
1
.
If the system is protected by a vibration isolator as seen in the figure below, the force transmitted is m
¨
Y ,
where Y is the acceleration of the component. That is, the component will fail if the transmitted force is
greater than F
0
m
¨
Y > F
0
or the acceleration is greater than F
0
/m. The new equations motion are given
by
m
¨
Y = c
˙
X −
˙
Y
+ k (X − Y )
0 = F − c
˙
X −
˙
Y
−k (X − Y ) .
Then, we have m
¨
Y = F, which is same as before. Therefore, there is no change in the expected time of first
excursion.
F
F
m
m
c
k
( )
a
( )
b
Y
Y
X
257
9.3. In the previous problem, assume that ω
2
−ω
1
is small so that the force can be considered a narrow-
band process. What is the expected time that the force will exceed F
0
for the first time with and without
the vibration isolator? How do the results change from the previous problem?
Solution: The expected time to the first excursion is 1/ν
+
. The crossing rate for the narrow band process
is given by (Section 9.2.4)
ν
+
= F
0
*
σ
2
˙
F
−ω
2
m
σ
2
F
√
2πσ
2
F
exp
−
1
2
F
2
0
σ
2
F
,
where
σ
F
=
4
∞
−∞
S
F F
dω
=
S
0
σ
˙
F
=
4
∞
−∞
S
˙
F
˙
F
dω =
4
∞
−∞
ω
2
S
F F
dω
=
S
0
ω
2
2
+ ω
1
ω
2
+ ω
2
1
ω
m
=
∞
0
ωS
o
F F
dω
∞
0
S
o
F F
dω
.
The expected time to the first excursion for the narrow band process is σ
F
σ
˙
F
/
F
0
√
2π
*
σ
2
˙
F
−ω
2
m
σ
2
F
times
that for the wide band process.
258
9.4. A delicate system is enclosed in a box that is also protected by a spring and damper system, as
shown in Figure 9.21. The delicate system will fail if its acceleration exceeds the threshold a
c
m/s
2
.
The box is subjected to a Gaussian random force F (t) with a white noise spectrum of intensity S
0
N
2
s.
What is the expected time of failure?
Figure 9.21: Two-degree-of-freedom system subjected to random force.
Solution: The expected time to the first excursion is the 1/ν
+
, and for a Gaussian process
ν
+
=
σ
˙
A
2πσ
A
exp
−
1
2
a
c
σ
A
2
,
where A is the acceleration
¨
Y . Then, the problem is reduced to obtaining σ
A
and σ
˙
A
. From free body diagram
shown next, the equations of motion are given by
M 0
0 m
¨
X
¨
Y
+
C + c −c
−c c
˙
X
˙
Y
+
K + k −k
−k c
X
Y
=
F
0
,
where X is the motion of the box with mass M and Y is the motion of the equipment mass m. Then, the
transfer function matrix is given by
[H (ω)] =
K + k −Mω
2
+ iω (C + c) −k −iωc
−k −iωc k −mω
2
+ iωc
−1
.
F
CX
KX
k(Y-X)
c(Y-X)
Free body diagram for Problem 9.4.
The response spectral can be found by
[S
XX
(ω)] = [H
∗
(ω)] [S
F F
(ω)] [H (ω)]
T
S
Y Y
= |H
21
|
2
S
0
H
21
=
k + iωc
(K + k −Mω
2
+ iω (C + c)) (k − mω
2
+ iωc) − (k + iωc)
2
259
The acceleration spectral density is S
AA
= ω
4
|H
21
|
2
S
0
and the spectral density for
˙
A is S
˙
A
˙
A
= ω
6
|H
21
|
2
S
0
.
The standard deviation are obtained by
σ
A
=
∞
−∞
S
AA
dω
σ
˙
A
=
∞
−∞
S
˙
A
˙
A
dω.
260
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