92 2. RELIABILITY OF A COMPONENT UNDER CYCLIC LOAD
From the iterative results, the reliability index ˇ and corresponding reliability R of the shaft in
this example are:
ˇ D 2:278915 R D ˆ
.
2:278915
/
D 0:9887:
Table 2.48: e distribution parameters of random variables in Equation (e)
K (lognormal) T
a
(klb.in) d (in)
μ
lnK
σ
lnK
μ
T
a
σ
T
a
μ
d
σ
d
37.308 0.518 8.9 0.85 1.500 0.00125
Table 2.49: e iterative results of Example 2.27 by the R-F method
Iterative #
K
*
T
a
*
d
*
β
*
|∆β
*
|
1 1.82E+16 1.5 11.57999 2.220042
2 2.13E+15 1.499878 8.913915 2.158924 0.061117
3 5.11E+15 1.499899 9.915022 2.265797 0.106873
4 7.37E+15 1.499887 10.36785 2.278682 0.012884
5
7.76E+15 1.499883 10.43223 2.278909 0.000228
6
7.75E+15 1.499882 10.43026 2.278915 5.31E-06
2.9.9 RELIABILITY OF A BEAM UNDER CYCLIC BENDING LOADING
Per Equation (2.87) or Equation (2.88), we can establish the limit state function of a beam under
any type of cyclic bending loading spectrum and then calculate its reliability. In this section, we
will use two examples to demonstrate how to calculate the reliability of a beam under cyclic
bending loading spectrum.
Example 2.28
e critical section of a beam with a rectangular cross-section is subjected to model #5 cyclic
bending loading spectrum listed in Table 2.50. e height h and the width b of the critical
cross-section are h D 2:500 ˙ 0:010
00
and b D 4:00 ˙ 0:010
00
. e ultimate material strength S
u
of the beam is 61.5 (ksi). ree parameters of the component fatigue strength index K on the
critical section for the cyclic bending loading are m D 6:38,
ln K
D 32:476, and
ln K
D 0:279.
For the component fatigue strength index K, the stress unit is ksi. Calculate the reliability of
the beam.
2.9. THE PROBABILISTIC FATIGUE DAMAGE THEORY (THE K-D MODEL) 93
Table 2.50: Model #5 cyclic bending loading spectrum for Example 2.28
Stress
Level
# i
Cyclic Bending Moment (klb.in)
Number of Cycles N
Li
(normal distribution)
M
mi
M
ai
μ
N
Li
σ
N
Li
1 80.2 61.55 250,000 3,000
2 80.2 108.17 5,000 450
Solution:
(1) e cyclic bending stress and the component fatigue damage index.
For the stress level #1, the mean bending stress
m1
, the bending stress amplitude
a1
,
and its corresponding equivalent bending stress amplitude
eq1
are
m1
D
M
m1
h= 2
I
D
M
m1
h= 2
bh
3
=12
D
6M
m1
bh
2
(a)
a1
D
M
a1
h=2
I
D
M
a1
h= 2
bh
3
=12
D
6M
a1
bh
2
(b)
eq1
D
a1
S
u
S
u
m1
D
6M
a1
S
u
bh
2
S
u
6M
m1
: (c)
For the stress level #2, by repeating (a), (b), and (c), we have
eq2
D
a2
S
u
S
u
m2
D
6M
a1
S
u
bh
2
S
u
6M
m2
: (d)
e component fatigue damage index of this beam under model #5 cyclic bending stress per
Equation (2.85) is:
D D N
L1
6M
a1
S
u
bh
2
S
u
6M
m1
6:38
C N
L2
6M
a2
S
u
bh
2
S
u
6M
m2
6:38
: (e)
(2) e limit state function of the beam.
e limit state function of the beam under model #5 cyclic loading spectrum per Equa-
tion (2.87) is:
g
.
K; N
L1
; N
L2
; b; h
/
D K N
L1
6M
a1
S
u
bh
2
S
u
6M
m1
6:38
N
L2
6M
a2
S
u
bh
2
S
u
6M
m2
6:38
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(f)
94 2. RELIABILITY OF A COMPONENT UNDER CYCLIC LOAD
ere are five random variables in the limit state function (f). e dimensions h and b can
be treated as normal distributions. eir means and standard deviations can be calculated per
Equation (1.1). e distribution parameters in the limit state function (f) are listed in Table 2.51.
(3) e reliability of the beam.
We will follow the Monte Carlo method and the program flowchart in Appendix A.3 to
create a MATLAB program. Since the limit state function is not too complicated, we will use
the trial number N D 1;598;400. e reliability R of this beam by the Monte Carlo method is
R D
1;514;162
1;598;400
D 0:9473:
Table 2.51: e distribution parameters of random variables in Equation (e)
K (lognormal) N
L1
(normal) N
L2
(normal) b (in) h (in)
μ
lnK
σ
lnK
μ
N
L1
σ
N
L1
μ
N
L2
σ
N
L2
μ
b
σ
b
μ
h
σ
h
32.476 0.279 250,000 3,000 5,000 450 4.000 0.00125 2.500 0.00125
Example 2.29
e critical section of a beam with a circular cross-section is subjected to model #2 cyclic bending
spectrum listed in Table 2.52. e diameter d of the critical cross-section is d D 2:500 ˙0:010
00
.
e ultimate material strength S
u
of the beam is 61.5 (ksi). ree parameters of the component
fatigue strength index K on the critical section for the cyclic shear loading are m D 6:38,
ln K
D
32:476, and
ln K
D 0:279. For the component fatigue strength index K, the stress unit is ksi.
Calculate the reliability of the beam.
Table 2.52: e model #2 cyclic bending loading spectrum for Example 2.29
Number of
Cycles
Mean Bending
Moment M
m
(klb.in)
Normally Distributed Bending Moment
Amplitude M
a
(klb.in)
μ
M
a
σ
M
a
550,000 28.2 16.8 3.19
Solution:
(1) e cyclic bending stress and the component fatigue damage index.
2.9. THE PROBABILISTIC FATIGUE DAMAGE THEORY (THE K-D MODEL) 95
e mean bending stress
m
, the bending stress amplitude
a
, and its corresponding equiv-
alent bending stress amplitude
eq
of the beam due to model #2 cyclic bending loading are:
m
D
M
m
d=2
I
D
M
m
d=2
d
4
=64
D
32M
m
d
3
(a)
a
D
M
a
d=2
I
D
M
a
d=2
d
4
=64
D
32M
a
d
3
(b)
eq
D
a
S
u
S
u
m
D
32M
a
S
u
d
3
S
u
32M
m
: (c)
e component fatigue damage index of this beam under model #2 cyclic bending stress per
Equation (2.84) is:
D D n
L
32M
a
S
u
d
3
S
u
32M
m
6:38
: (d)
(2) e limit state function of the beam.
e limit state function of the beam under model #2 cyclic bending loading spectrum per
Equation (2.87) is:
g
.
K; M
a
; d
/
D K n
L
32M
a
S
u
d
3
S
u
32M
m
6:38
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(e)
ere are three random variables in the limit state function (e). e dimension d can be treated
as a normal distribution. Its mean and standard deviation can be calculated per Equation (1.1).
e distribution parameters in the limit state function (e) are listed in Table 2.53.
(3) e reliability of the beam.
e limit state function (e) contains two normal distributions and one lognormal distri-
bution. We will use the R-F method to calculate its reliability. We can follow the procedure and
the flowchart in Appendix A.2 to create a MATLAB program. e iterative results are listed in
Table 2.54. From the iterative results, the reliability index ˇ and corresponding reliability R of
the beam in this example are:
ˇ D 1:624707 R D ˆ
.
1:624707
/
D 0:9479:
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