22 2. RELIABILITY OF A COMPONENT UNDER CYCLIC LOAD
2.7 RELIABILITY OF A COMPONENT WITH AN
INFINITE LIFE
For a component with infinite life, the component endurance limit
S
e
will be used for the com-
ponent fatigue design. e limit state function of a component with an infinite life will be:
g
S
e
; K
f
;
eqa
D S
e
K
f
eqa
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure;
(2.26)
where S
e
is the component endurance limit at the critical section, which is defined by Equa-
tion (2.12). K
f
is the fatigue stress concentration factor at the critical section and is defined by
Equations (2.22)–(2.25).
eqa
is the component’s equivalent fully reversed cyclic stress ampli-
tude at the critical section and is determined by Equation (2.21).
e limit state function Equation (2.26) can be used to calculate the reliability of a com-
ponent with infinite life. e H-L, R-F, or Monte Carlo method, which were discussed in
Chapter 3 of Le [8], can be used to calculate its reliability. e concise description of proce-
dures of three methods is presented in Appendix A of this book. We will use three examples to
demonstrate how to calculate the reliability of a component with infinite life. e corresponding
MATLAB programs will be displayed in Appendix B for a reference.
Example 2.6
A machined constant circular bar with a diameter d D 1:250 ˙ 0:005
00
is subjected a cyclic axial
loading. e mean axial loading F
m
of the cyclic axial loading is a constant and equal to 12 (klb).
e loading amplitude F
a
of the cyclic loading follows a normal distribution with a mean
F
a
D
8:5 (klb) and a standard deviation
F
a
D 1:2 (klb). e ultimate material strength is 61.5 (ksi).
Its endurance limit S
0
e
follows a normal distribution with a mean
S
0
e
D 24:7 (ksi) and a standard
deviation
S
0
e
D 2:14 (ksi), which are based on fatigue tests under fully reversed bending stress.
is bar is designed to have an infinite life. (1) Establish the limit state function of this problem.
(2) Calculate the reliability of the bar under the cyclic axial loading.
Solution:
(1) Establish the limit state function of this problem.
We can use Equation (2.26) to establish the limit state function for this problem.
Per Equation (2.12), the component endurance limit will be:
S
e
D k
a
k
b
k
c
S
0
e
: (a)
2.7. RELIABILITY OF A COMPONENT WITH AN INFINITE LIFE 23
Per Equations (2.14), (2.15), and (2.16), the surface finish modification factor k
a
of the ma-
chined bar will be a normal distribution with the following distributed parameters:
k
a
D 2:7
.
S
ut
/
0:2653
D 2:7 61:5
0:2653
D 0:905 (b)
k
a
D
k
a
k
a
D 0:06 0:905 D 0:0543: (c)
Per Equation (2.17), the size modification factor k
b
will be treated as a constant and is equal to
1 due to the cyclic axial loading:
k
b
D 1: (d)
Per Equations (2.18), (2.19), and (2.20), the loading modification factor k
c
will be treated as a
normal distribution with the following mean and standard deviation:
k
c
D 0:774 (e)
k
c
D
k
c
k
c
D 0:774 0:163 D 0:1262: (f)
For this problem, the fatigue stress concentration factor K
f
will be equal to 1 due to a constant
circular cross-section.
e mean stress
m
and the stress amplitude
a
of the cyclic axial stress in this problem
can be calculated by the following equations:
m
D
F
m
d
2
=4
D
4F
m
d
2
D
4 12
d
2
D
15:279
d
2
.ksi/ (g)
a
D
F
a
d
2
=4
D
4F
a
d
2
D
1:273F
a
d
2
.ksi/: (h)
e diameter d of the bar can be treated as a normal distribution. Its mean
d
and standard
deviation
d
can be determined per Equation (1.1):
d
D 1:250;
d
D
0:005 .0:005/
8
D 0:00125: (i)
Since the cyclic axial stress is not a fully reversed cyclic stress, we need to use Equation (2.21)
to consider the effect of mean stress and converted it into a fully reversed cyclic stress with an
equivalent stress amplitude:
aeq
D
a
S
ut
S
ut
m
D
1:273F
a
d
2
0
B
@
61:5
61:5
15:279
d
2
1
C
A
D
78:290F
a
61:5d
2
15:279
: (j)
Now, by using all information from Equations (a)–(j), we can establish the limit state function
for this problem per Equation (2.26):
g
k
a
; k
c
; S
0
e
; d; F
a
D k
a
k
c
S
0
e
78:290F
a
61:5d
2
15:279
: (k)
24 2. RELIABILITY OF A COMPONENT UNDER CYCLIC LOAD
Table 2.3: e distribution parameters of random variables in Equation (k)
k
a
k
c
S
'
e
(ksi) d
(in) F
a
(klb)
μ
k
a
σ
k
a
μ
k
c
σ
k
c
μ
S
'
e
σ
S
'
e
μ
d
σ
d
μ
F
a
σ
F
a
0.905 0.0543 0.774 0.1262 24.7 2.14 1.250 0.0125 8.5 1.2
In this example, all random variables in the limit state function (k) are normal distributions.
eir distribution parameters are listed in Table 2.3.
(2) Use the H-L method to calculate the reliability of the bar.
e limit state function (k) contains five normally distributed random variable and is a
nonlinear function. We will follow the H-L method presented in Appendix A.1 and the program
flowchart in Figure A.1 to create a MATLAB program. is MATLAB program is displaced
in Appendix B as “B.1: e H-L method for Example 2.6.”
e iterative results are listed in Table 2.4. From the iterative results, the reliability index
ˇ and corresponding reliability R of the bar in this example are:
ˇ D 2:792757 R D ˆ
.
2:792757
/
D 0:9974:
Table 2.4: e iterative results of Example 2.6 by the H-L method
Iterative #
k
a
*
k
c
*
S
'
*
e
d
*
F
a
*
β
*
|∆β
*
|
1 0.905 0.774 24.7 1.25 17.85956 2.534002
2 0.865081 0.52188 22.42825 1.246358 10.37987 2.800277 0.266275
3 0.872351 0.481666 22.74402 1.247141 9.811213 2.793018 0.00726
4 0.874674 0.477329 22.8934 1.247324 9.816224 2.792766 0.000252
5 0.874928 0.476348 22.91546 1.24734 9.808635 2.792757 8.78E-06
Example 2.7
e critical section for a machined rotating shaft is on the shoulder section, as shown in Fig-
ure 2.4. e bending moment M (klb.in) on the shoulder section can be described by a lognor-
mal distribution with a log-mean
ln M
D 0:315 and a log-standard deviation
ln M
D 0:142.
e shaft material’s ultimate strength is 61.5 (ksi). Its endurance limit S
0
e
follows a normal dis-
tribution with a mean
S
0
e
D 24:7 (ksi) and a standard deviation
S
0
e
D 2:14 (ksi), which are
based on fatigue tests under fully reversed bending stress. is shaft is designed to have an in-
finite life. (1) Establish the limit state function of this shaft. (2) Calculate the reliability of the
shaft.
2.7. RELIABILITY OF A COMPONENT WITH AN INFINITE LIFE 25
1
16
R "
Ø 1.250±.005 Ø 1.500±.005
Figure 2.4: Schematic of a shoulder section of a shaft.
Solution:
(1) Establish the limit state function of the rotating shaft.
We can use Equation (2.26) to establish the limit state function for this problem.
Per Equation (
2.12), the component endurance limit will be:
S
e
D k
a
k
b
k
c
S
0
e
: (a)
e mean and standard deviation of the surface finish modification factor k
a
of the machined
shaft can be calculated per Equations (2.14), (2.15), and (2.16):
k
a
D 0:905I
k
a
D 0:0543: (b)
e size modification factor k
b
can be calculated per Equation (2.17):
k
b
D 0:8507: (c)
e loading modification factor k
c
will be 1 because the cyclic stress is cyclic bending stress:
k
c
D 1: (d)
e static stress concentration K
t
in this case, can be obtained from any design handbook or a
stress concentration website:
K
t
D 1:96: (e)
e mean and standard deviation of the fatigue stress concentration factor K
f
can be calculated
per Equations (2.22)–(2.25):
K
f
D 1:562I
K
f
D 0:1250: (f)
26 2. RELIABILITY OF A COMPONENT UNDER CYCLIC LOAD
For a rotating shaft, the bending moment will induce a fully reversed cyclic bending stress. Its
stress amplitude of the fully reversed cyclic bending stress will be:
a
D
M
d
2
I
D
M
d
2
64
d
4
D
32M
d
3
: (g)
e diameter d of the shaft can be treated as a normal distribution. Its mean
d
and standard
deviation
d
can be determined per Equation (1.1):
d
D 1:250;
d
D
0:005 .0:005/
8
D 0:00125: (h)
Now, by using all information from (a)–(h), we can establish the limit state function for this
problem per Equation (2.26):
g
M; k
a
; S
0
e
; K
f
; d
D k
a
k
b
S
0
e
K
f
32M
d
3
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(i)
In this example, M is a lognormal distribution, and the rest random variables in the limit state
function (i) are normal distributions. eir distribution parameters are listed in Table 2.5.
Table 2.5: e distribution parameters of random variables in Equation (i)
M (klb.in),
Lognormal
Distribution
k
a
, Normal
Distribution
S
'
e
(ksi), Normal
Distribution
K
f
, Normal
Distribution
d
(in), Normal
Distribution
μ
lnM
σ
lnM
μ
k
a
σ
k
a
μ
S
'
e
σ
S
'
e
μ
K
f
σ
K
f
μ
d
σ
d
0.315 0.142 0.905 0.0543 24.7 2.14 1.562 0.1250 1.250 0.00125
(2) Use the R-F method to calculate the reliability of the shaft.
e limit state function (i) contains four normally distributed random variable and one
lognormal distribution variable. We will use the R-F method to calculate the reliability of this
example. e procedure of the R-F method and the program flowchart are presented in Ap-
pendix A.2. e MATLAB program of the R-F method for this example is displayed in Ap-
pendix B as “B.2: e R-F method for Example 2.7.”
e iterative results are listed in Table 2.6. From the iterative results, the reliability index
ˇ and corresponding reliability R of the bar in this example are:
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