192 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.61: e distribution parameters of random variables in Equation (h)
K (lognormal) k
a
k
c
n
L1
n
L2
d (in)
μ
lnK
σ
lnK
μ
k
a
σ
k
a
μ
k
c
σ
k
c
μ
n
L1
σ
n
L1
μ
n
L2
σ
n
L2
μ
d
σ
d
41.738 0.357 0.8588 0.05153 0.774 0.1262 3E5 5000
4000 200
μ
d
1.25E-3
mean and standard deviation can be calculated per Equation (1.1). e distribution parameters
in the limit state function (h) are listed in Table 3.61.
(4) Use the modified Monte Carlo method to determine the diameter d .
We will use the modified Monte Carlo method to conduct this component dimension
design, which has been discussed in Section 3.2.5. We can follow the procedure discussed in
Section 3.2.5 and the program flowchart shown in Figure 3.5 to compile a MATLAB program.
e iterative results are listed in Table 3.62.
Table 3.62: e iterative results of Example 3.25 by the modified Monte Carlo method
Iterative #
μ
d
*
R
*
∆R
*
1 1.160466 0.849113 -0.14089
2 1.161466 0.852414 -0.13759
… … … …
111 1.270466 0.990013 1.28E-05
112 1.271466 0.990263 0.000263
According to the result obtained from the program, the mean of the diameter with a
reliability 0.99 is
d
D 1:272
00
: (i)
erefore, the diameter of the shaft with the required reliability 0.99 under the specified loading
will be
d D 1:272 ˙ 0:005
00
:
3.4.6 BEAM UNDER CYCLIC BENDING LOADING SPECTRUM
e limit state function of a beam under any type of cyclic bending loading spectrum can be
established per Equations (2.87) or (2.88), which have been discussed in Section 2.9.9. After
the limit state function of a component under cyclic bending loading is established, we can run
3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 193
the dimension design with the required reliability. In this section, we will show how to determine
the dimension of a beam with the required reliability under cyclic bending loading spectrum.
Example 3.26
e critical section of a machined circular beam is at the shoulder section, as shown in Figure 3.13
and is subjected to model #3 cyclic bending spectrum listed in Table 3.63. e ultimate material
strength S
u
is 61.5 (ksi). e three distribution parameters of material fatigue strength index K
0
for the standard specimen under fully-reversed bending stress are m D 6:38;
ln K
0
D 32:476,
and
ln K
0
D 0:279. For the material fatigue strength index K
0
, the stress unit is ksi. Determine
the diameter of the beam with a reliability 0.99 when its dimension tolerance is ˙0:005
00
.
1
16
R "
Ø 2.000"
d
Figure 3.13: Schematic of the segment of a circular beam with a shoulder.
Table 3.63: e model #3 cyclic bending loading spectrum for Example 3.26
Number of Cycles n
L
Bending Moment Mean
M
m
(klb.in)
Bending Moment Amplitude M
a
(klb.in)
μ
M
a
σ
M
a
250,000 0.46 1.08 0.19
Solution:
(1) Preliminary design for determining the dimension-dependent parameters k
b
, K
t
, and K
f
.
Per Section 3.2.1, the preliminary static stress concentration factor per Table 3.2 will be
K
t
D 2:7: (a)
e preliminary size modification factor k
b
per Equation (3.3) will be
k
b
D 0:87: (b)
e fatigue stress concentration factor K
f
can be calculated and updated per Equations (2.22)–
(2.25) when K
t
and the fillet radius are known.
194 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
(2) e cyclic bending stress and the component fatigue damage index.
e mean bending stress
m
, the bending stress amplitude
a
and its corresponding equiv-
alent bending stress amplitude
eq
of the beam due to the model #3 cyclic bending loading listed
in the above table are
m
D K
f
M
m
d=2
I
D K
f
M
m
d=2
d
4
=64
D K
f
32M
m
d
3
(c)
a
D K
f
M
a
d=2
I
D K
f
M
a
d=2
d
4
=64
D K
f
32M
a
d
3
(d)
eq
D K
f
a
S
u
S
u
m
D
32M
a
S
u
K
f
d
3
S
u
32K
f
M
m
: (e)
e component fatigue damage index of this shaft under model #3 cyclic bending stress per
Equation (2.84) is:
D D n
L
32M
a
S
u
K
f
d
3
S
u
32K
f
M
m
m
: (f)
(3) e limit state function.
e component fatigue strength index K can be calculated per Equation (2.79):
K D
.
k
a
k
b
k
c
/
m
K
0
: (g)
e surface finish modification factor k
a
follows a normal distribution. Its mean and standard
deviation can be determined per Equations (2.14), (2.15), and (2.16). k
b
is treated as a determin-
istic value and can be calculated per Equation (2.17). Its value will be updated in each iterative
step by using the newly available diameter of the beam. Since this is cyclic bending stress the
load modification factor k
c
will be 1.
e limit state function of the beam in this example per Equation (2.87) is:
g
K
0
; k
a
; K
f
; M
a
; d
D
.
k
a
k
b
/
m
K
0
n
L
32M
a
S
u
K
f
d
3
S
u
32K
f
M
m
m
(3.37)
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(h)
ere are five random variables in the limit state function (h). K
0
is a lognormal distribution. e
rests are normal distributions. e dimensions d can be treated as normal distributions, and its
mean and standard deviation can be calculated per Equation (1.1). e distribution parameters
in the limit state function (h) are listed in Table 3.64. K
f
and k
b
will be updated in each iterative
step by newly available diameter
d
.
3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 195
Table 3.64: e distribution parameters of random variables in Equation (e)
K
0
(lognormal) k
a
K
f
M
a
d (in)
μ
lnK
0
σ
lnK
0
μ
k
a
σ
k
a
μ
K
f
σ
K
f
μ
M
a
σ
M
a
μ
d
σ
d
32.476 0.279 0.9053 0.05432 2.0337 0.1627 1.08 0.19
μ
d
0.00125
(4) Use the modified R-F method to determine the diameter d .
We will use the modified R-F to conduct this component dimension design, which has
been discussed in Section 3.2.4. We can follow the procedure discussed in Section 3.2.4 and the
program flowchart shown in Figure 3.3 to compile a MATLAB program. e iterative results
are listed in Table 3.65.
Table 3.65: e iterative results of Example 3.26 by the modified R-F method
Iterative #
K
0
*
k
a
*
K
f
*
M
a
*
d
*
|∆d
*
|
1 1.32E+14 0.9053 2.0337 1.08 1.111808
2 1.21E+14 0.892354 2.091971 1.212769 1.171788 0.059979
3 1.2E+14 0.888834 2.110718 1.687193 1.307916
0.136129
4 1.18E+14 0.885946 2.118994 1.68845 1.316975
0.009059
5 1.18E+14 0.884797 2.116866 1.688034 1.317533
0.000558
6 1.18E+14 0.884717 2.116761 1.687974 1.317565
3.2E-05
According to the result obtained from the program, the mean of the diameter with a
reliability 0.99 is
d
D 1:318
00
: (i)
erefore, the diameter of the beam with the required reliability 0.99 under the specified loading
will be
d D 1:318 ˙ 0:005
00
:
Example 3.27
e rectangular beam is subjected to model #4 cyclic bending spectrum listed in Table 3.66.
e ultimate material strength S
u
is 61.5 (ksi). e three distribution parameters of material
fatigue strength index K
0
for the standard specimen under fully-reversed bending stress are
m D 6:38;
ln K
0
D 32:476, and
ln K
0
D 0:279. For the material fatigue strength index K
0
, the
stress unit is ksi. e width of the beam is b D 2:000 ˙ 0:010
00
. Determine the height h of the
beam with a reliability 0.99 when its dimension tolerance is ˙0:010
00
.
196 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.66: Model #4 cyclic bending loading spectrum for Example 3.27
Stress Level # Number of Cycles
Cyclic Bending Moment (klb.in)
M
mi
M
ai
1 550,000 38.5 28.92
2 5,000 38.5 42.16
Solution:
(1) Preliminary design for determining the size modification factor k
b
.
In this example, there is no stress concentration area. e size modification factor k
b
is
the only one dimension-dependent parameter. Per the discussion in Section 3.2.1,
k
b
D 0:87: (a)
(2) e cyclic bending stress and the component fatigue damage index.
For the cyclic bending loading level #1, the mean bending stress
m1
, the bending stress
amplitude
a1
and its corresponding equivalent bending stress amplitude
eq1
of the beam are
m1
D
M
m1
h= 2
I
D
M
m1
h= 2
bh
3
=12
D
6M
m1
bh
2
(b)
a1
D
M
a1
h=2
I
D
M
a1
h=2
bh
3
=12
D
6M
a1
bh
2
(c)
eq1
D
a1
S
u
S
u
m1
D
6M
a1
S
u
bh
2
S
u
6M
m1
: (d)
For the cyclic bending loading level #2, repeating (b), (c), and (d), we have
eq2
D
6M
a2
S
u
bh
2
S
u
6M
m2
: (e)
e component fatigue damage index of this beam under model #4 cyclic bending moment per
Equation (2.85) is:
D D n
L1
6M
a1
S
u
bh
2
S
u
6M
m1
m
C n
L2
6M
a2
S
u
bh
2
S
u
6M
m2
m
: (f )
(3) e limit state function.
e component fatigue strength index K can be calculated per Equation (2.79).
K D
.
k
a
k
b
k
c
/
m
K
0
: (g)
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