102 2. RELIABILITY OF A COMPONENT UNDER CYCLIC LOAD
tion 2.9 such as Examples 2.24, 2.28, and 2.30 use the Monte Carlo method to calculate the
reliability of component under a cyclic loading spectrum.
2.9.12 THE COMPARISON OF RESULTS BY THE K-D MODEL WITH THE
RESULTS BY THE P-S-N CURVES
Both the P-S-N curves and the K-D model are probabilistic fatigue theory and can be used to
describe material fatigue test data, that is, material fatigue strength. Both can be used to calculate
the reliability of a component under cyclic loading spectrums.
Table 2.11 in Section 2.8.1 list fatigue test results of 195 fatigue tests on the 10 gauge
6060-T6 sheet-type flat test specimen under 5 different cyclic axial stress levels. e P-S-N
curves on these set of fatigue data are shown in Table 2.13 and are redisplayed here for conve-
nience. e K-D model on these set of fatigue test data is listed in Table 2.35 and redisplayed
here for conveniences. Now, the following three Examples 2.32–2.34 will be used to compare
the results from both fatigue theories.
Table 2.13: e P-N distribution at different fatigue strength levels
Axial Stress
Level #
σ
a
(ksi) σ
m
(ksi)
Equivalent Stress
σ
a˗eqi
(ksi)
Lognormal Distributed Fatigue Life
μ
lnN
σ
lnN
1 20.833 20.833 35.126 11.4736 0.238106
2 22.083 22.083 38.832 11.0410 0.227320
3 22.5 22.5 40.139 10.9551 0.242377
4 22.917 22.917 41.485 10.7831 0.225735
5 23.333 23.333 42.871 10.716 0.241955
Table 2.35: K
0
with three distribution parameters
Material: 6061-T6
10-Gauge Sheet
Sample Size:
195
Test Conditions: Cyclic Axial Stress, Milled
Machined Sheet-type Flat Specimen
Slope of the Traditional S-N
Curve: m
Log Normally Distributed K
0
Log Mean: μ
lnK
0
Log Standard Deviation: σ
lnK
0
3.8812 25.3014 0.245451
Example 2.32
e aluminum 6061-T6 10 Gauge sheet-type flat fatigue specimen, as shown in Figure 2.8 is
subjected to cyclic axial stress with a stress amplitude 20.833 (ksi) and mean stress 20.833 (kis).
e number of cycles of this cyclic axial loading is 60,000 (cycles). e ultimate tensile strength
2.9. THE PROBABILISTIC FATIGUE DAMAGE THEORY (THE K-D MODEL) 103
of this material is S
u
D 51:2 (ksi). e P-S-N curve of this material is listed in Table 2.13, and
the K-D model is listed in Table 2.35. Use the P-S-N curve and the K-D model to calculate its
reliability.
Solution:
(1) e P-S-N curve approach.
e P-S-N curve in Table 2.13 shows that the number of cycles to failures N under this
cyclic stress follows a lognormal distribution with a mean
ln
D 11:4736 and the standard de-
viation
ln N
D 0:238106. Since the number of cycles at the specified cyclic stress is a constant,
the reliability of this specimen under the specified cyclic stress will be:
R D P
.
N n
/
D P
.
ln.N / ln.n/
/
D ˆ
ln N
ln.n/
ln N
D ˆ
11:4736 ln.60000/
0:238106
D ˆ.1:98021/ D 0:97616;
where ˆ
.
/
is the CDF of a standard normal distribution.
(2) e K-D probabilistic model.
e cyclic axial stress is with a stress mean
m
D 20:833 (ksi) and the stress amplitude
a
D 20:833 (ksi). It is not a fully reversed cyclic stress. It will be converted into an equivalent
stress amplitude of a fully reversed cyclic stress per Equation (2.21):
eq
D
a
S
u
.
S
u
m
/
D
20:83333 51:2
.51:2 20:83333/
D 35:12623 .ksi/:
e K-D model of this material under cyclic axial loading is shown in Table 2.35. e
fatigue damage index D due to the specified loading per Equation (2.84) will be:
D D n
m
eq
D 60000 .35:12623/
3:8812
D 5:98489 10
10
:
Since the D in this example is a constant, the reliability of the specimen under the specified
cyclic loading will be:
R D P
.
K
0
D
/
D P
.
ln
.
K
0
/
ln.D/
/
D ˆ
ln K
0
ln.D/
ln K
0
D ˆ
25:3014 24:8151
0:238106
D ˆ
.
1:981209
/
D 0:97622:
e results from the P-S-N curve and the K-D model are almost identical. e relative difference
is 0.006%.
104 2. RELIABILITY OF A COMPONENT UNDER CYCLIC LOAD
Example 2.33
e aluminum 6061-T6 10 Gauge sheet-type fatigue specimen, as shown in Figure 2.8 is sub-
jected to cyclic axial stress with a stress amplitude 20.833 (ksi), and mean stress 20.833 (ksi).
e number of cycles n
L
of this cyclic axial stress follows a lognormal distribution with a mean
ln n
L
D 10:5 and a standard deviation
ln n
L
D 0:35. e ultimate tensile strength of this ma-
terial is S
u
D 51:2 (ksi). e P-S-N curve of this material is listed in Table 2.13, and the K-D
model is listed in Table 2.35. Use the P-S-N curve and the K-D probabilistic model to calculate
its reliability.
Solution:
(1) e P-S-N curve approach.
Since both the fatigue life N and the number of cycles n
L
at the cyclic axial stress level are
log-normal distributions, the reliability of this specimen under the specified cyclic axial stress
will be:
R D P .N n/ D P .ln.N / ln.n// D P .ln.N / ln.n/ 0/
D ˆ
0
B
@
ln N
ln N
q
.
ln N
/
2
C
.
ln n
/
2
1
C
A
D ˆ
11:4736 10:5
p
.0:238016/
2
C .0:35/
2
!
D ˆ.2:29995/ D 0:98927:
(2) e K-D probabilistic model.
e equivalent stress amplitude of a completely reversed cyclic stress per Equation (2.21)
is
eq
D
a
S
u
.
S
u
m
/
D
20:83333 51:2
.
51:2 20:83333
/
D 35:126 .ksi/: (a)
e fatigue damage index D due to the specified cyclic axial stress per Equation (2.84) will be:
D D n
L
m
eq
D n
L
.
35:126
/
3:8812
: (b)
is equation can be expressed as:
ln
.
D
/
D ln
.
n
L
/
C m ln
eq
D ln
.
n
L
/
C 13:812989: (c)
Since n
L
follows a lognormal distribution, D will also follow a lognormal distribution according
to the above equation. e mean and standard deviation of the lognormal distributed D will be:
ln D
D
ln n
L
C 13:812989 D 10:5 C 13:812989 D 24:312989 (d)
ln D
D
ln n
L
D 0:35: (e)
2.9. THE PROBABILISTIC FATIGUE DAMAGE THEORY (THE K-D MODEL) 105
Since both the specimen fatigue strength index K
0
and the specimen fatigue damage index D
are log-normal distributions, the reliability of this specimen under the specified cyclic axial stress
will be:
R D P
.
K
0
D
/
D P
.
ln K
0
ln D
/
D P
.
ln K
0
ln D 0
/
D ˆ
0
B
@
ln K
0
ln D
q
ln K
0
2
C
.
ln D
/
2
1
C
A
D ˆ
0
B
@
.25:3014 24:312989/
q
.
0:238106
/
2
C
.
0:35
/
2
1
C
A
D ˆ
.
2:312136
/
D 0:98962:
e results from the P-S-N curve and the K-D model are almost identical. e relative difference
is 0.035%.
Example 2.34
e aluminum 6061-T6 10 Gauge sheet-type fatigue specimen, as shown in Figure 2.8 is sub-
jected to two levels of cyclic axial stress listed in Table 2.60. e ultimate tensile strength of
this material is S
u
D 51:2 (ksi). e P-S-N curve of this material is listed in Table 2.13, and
the K-D model is listed in Table 2.35. Use the P-S-N curve and the K-D model to calculate its
reliability.
Table 2.60: Two levels of axial cyclic stress for Example 2.34
Level #
σ
ai
(ksi) σ
mi
(ksi)
Number of Cycles n
Li
1 20.833 20.833 30,000
2 22.5 22.5 15,000
Solution:
(1) e P-S-N curve approach.
is cyclic loading is model #4 cyclic loading spectrum. With the concept of equivalent
damage approach discussed in Section 2.8.6, the fatigue damage due to a cyclic stress level #1 can
be transferred to stress level #2 with an equivalent cyclic number per Equations (2.45), (2.47),
and (2.48):
ˇ
1
D
ln N1
ln
.
n
L1
/
ln N1
D
11:4736 ln.30;000/
0:238106
D 4:89130: (a)
106 2. RELIABILITY OF A COMPONENT UNDER CYCLIC LOAD
With the same fatigue damage, that is, the same reliability index ˇ, the equivalent number of
cycles n
12eq
of the stress level #1 into the stress level #2 is:
n
eq12
D exp
.
ln N 2
ˇ
1
ln N 2
/
D 17;493:1: (b)
Now, at the stress level 2, the total number of cycles n
2eq
including transferred equivalent
number of cycles will be:
n
2eq
D n
L2
C n
eq12
D 15;000 C 17;493:1 D 32;493:1: (c)
erefore, the reliability of the specimen under two stress levels per Equation (2.49) will be:
R D P
N
2
n
2eq
D P
ln
.
N
2
/
ln
n
2eq
D ˆ
ln N
2
ln
.
n
2total
/
ln N
2
D ˆ
10:9551 ln.32;493:1/
0:242377
D ˆ.2:33651/ D 0:99027:
(2) e K-D model.
e cyclic axial stresses in this example are non-zero-mean cyclic stresses. We need to
convert them into a fully reversed cyclic stress per Equation (2.21).
For the stress level #1 with
a1
D 20:833 (ksi) and
m1
D 20:833 (ksi), the fully reversed
equivalent stress amplitude of
eq1
is
eq1
D
a1
S
u
.
S
u
m1
/
D
20:83333 51:2
.51:2 20:83333/
D 35:12623 .ksi/: (d)
For the stress level #2 with
a2
D 22:5 (ksi) and
m2
D 22:5 (ksi), the fully reversed equivalent
stress amplitude of
eq2
is
eq2
D
a2
S
u
.
S
u
m2
/
D
22:5 51:2
.
51:2 22:5
/
D 40:139 .ksi/: (e)
e fatigue damage index D due to the specified cyclic axial stresses per Equation (2.85)
will be:
D D n
L1
m
eq1
C n
L2
m
eq2
D 30;000
.
35:126
/
3:8812
C 15;000
.
40:139
/
3:8812
D 5:50358 10
10
: (f)
In this example, the fatigue damage index D is deterministic. e reliability of the specimen
under the specified cyclic axial stress by the definition of reliability will be:
R D P
.
K
0
D
/
D P
.
ln
.
K
0
/
ln
.
D
//
D ˆ
ln K
0
ln.D/
ln K
0
D ˆ
25:3014 24:73125
0:238106
D ˆ
.
2:322869
/
D 0:98991:
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