3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 177
erefore, the diameter of the shaft with the required reliability 0.99 under the specified cycle
bending loading is
d D 1:091 ˙ 0:005
00
:
3.4.3 ROD UNDER CYCLIC AXIAL LOADING SPECTRUM
e limit state function of a rod under any type of cyclic axial loading spectrum can be expressed
per Equations (2.87) or (2.88), which have been discussed in Section 2.9.6. After the limit state
function of a component under cyclic axial loading is established, we can run the dimension
design with the required reliability. In this section, we will show how to determine the dimension
of a rod with the required reliability under cyclic axial loading spectrum.
Example 3.20
A machined constant circular bar is subjected to model #2 cyclic axial loading spectrum as listed
in Table 3.46. e ultimate material strength S
u
is 75 (ksi). e three distribution parameters of
material fatigue strength index K
0
for the standard specimen under fully-reversed bending stress
are m D 8:21;
ln K
0
D 41:738, and
ln K
0
D 0:357. For the material fatigue strength index K
0
,
the stress unit is ksi. Determine the diameter of the bar with a reliability 0.95 when its dimension
tolerance is ˙0:005
00
.
Table 3.46: e model #2 cyclic axial loading spectrum for Example 3.20
Mean of the Cyclic
Axial Loading F
m
(klb)
Amplitude of Cyclic
Axial Loading F
a
(klb)
Number of Cycles n
L
(normal
distribution)
μ
n
L
σ
n
L
16.78 10.39 1.13 × 10
5
4.52 × 10
3
Solution:
For this example, there is no stress concentration area. Because the loading is an axial loading,
the size modification factor k
b
D 1. erefore, there are no dimension-dependant parameters.
(1) e cyclic axial stress and the component fatigue damage index.
e mean stress
m
and the stress amplitude
a
of the bar due to the cyclic axial loading
are:
m
D
F
m
d
2
=4
D
4F
m
d
2
;
a
D
F
a
d
2
=4
D
4F
a
d
2
: (a)
Since the cyclic stress is a no-zero mean cyclic stress, we need to convert it into a fully reversed
cyclic stress per Equation (2.83). e equivalent stress amplitude of this converted fully reversed
178 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
cyclic stress is:
aeq
D
a
S
u
.
S
u
m
/
D
4F
a
S
u
.
S
u
d
2
4F
m
/
: (b)
e component fatigue damage index D of the bar under model #2 cyclic fatigue spectrum per
Equation (2.84) is
D D n
L
K
f
a
8:21
D n
L
4F
a
S
u
.
S
u
d
2
4F
m
/
8:21
: (c)
(2) 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
: (d)
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 1 for cyclic axial
loading. e mean and the standard deviation of the load modification factor k
c
can be calculated
per Equations (2.18), (2.19), and (2.20).
e limit state function of the bar per Equation (2.87) is
g
.
K
0
; k
a
; k
c
; n
L
; d
/
D
.
k
a
k
c
/
m
K
0
n
L
4F
a
S
u
.
S
u
d
2
4F
m
/
8:21
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(e)
In this limit state function, we have five random variables.
K
0
is a log-normal distribution and
the rests are normal distributions. e diameter d follows a normal distribution. Its mean and
standard deviation can be determined per Equation (1.1). e distribution parameters of these
five random variables are listed in Table 3.47.
Table 3.47: e distribution parameters of random variables in Equation (e)
K
0
(lognormal) k
a
k
c
n
L
d (in)
μ
lnK
0
σ
lnK
0
μ
k
a
σ
k
a
μ
k
c
σ
k
c
μ
n
L
σ
n
L
μ
d
σ
d
41.738 0.357 0.8588 0.05153 0.774 0.1262 1.13×10
5
4.52×10
3
μ
d
0.00125
(3) Use the modified Monte Carlo method to determine the diameter d .
In the limit state function (e), there are one lognormal distribution and four normal dis-
tribution. We can 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
3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 179
Table 3.48: e iterative results of Example 3.20 by the modified Monte Carlo method
Iterative #
μ
d
*
R
*
∆R
*
1 0.921746 0.705623 -0.28438
2 0.922746 0.71141 -0.27859
… … … …
142 1.062746 0.990046 4.63E-05
143 1.063746 0.990293 0.000293
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 3.48.
According to the result obtained from the program, the mean of the diameter with a
reliability 0.99 is
d
D 1:064
00
: (f)
erefore, the diameter of the bar with the required reliability 0.99 under the specified loading
will be
d D 1:064 ˙ 0:005
00
:
Example 3.21
A machined stepped circular bar, as shown in Figure 3.11 is subjected to model #6 cyclic axial
loading spectrum listed in Table 3.49. e ultimate material strength S
u
is 75 (ksi). e three
distribution parameters of material fatigue strength index K
0
for the standard specimen under
fully-reversed bending stress are m D 8:21;
ln K
0
D 41:738, and
ln K
0
D 0:357. For the mate-
rial fatigue strength index K
0
, the stress unit is ksi. Determine the diameter of the bar with the
reliability 0.99 when its dimension tolerance is ˙0:005
00
.
1
8
R "
Ø 2.000"
d
Figure 3.11: Schematic of the segment of a stepped bar.
180 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.49: Model #6 cyclic axial loading spectrum for Example 3.21
Cyclic Number Level
Number of Cycles n
Li
(constant)
Fully Reversed Axial Loading Amplitude
F
ai
(klb) (normal distribution)
μ
F
ai
σ
F
ai
1 5,000 22.15 3.25
2 200,000 12.45
1.5
Solution:
(1) Preliminary design for determining K
t
and K
f
.
For the component under cyclic axial loading, the size modification factor k
b
will be 1.
However, there are two dimension-dependent parameters K
t
and K
f
in this fatigue problem.
According to the schematic of the stepped shaft, we can assume that it has a well-round fillet.
Per Table 3.2, we have the preliminary static stress concentration factor
K
t
D 1:9: (a)
K
f
follows a normal distribution. Its mean and standard deviation can be calculated per Equa-
tions (2.22)–(2.26).
(2) e cyclic axial stress and the component fatigue damage index.
In the cyclic number level #1 with n
L1
D 5000 (cycles), the fully reversed axial stress am-
plitude
a1
is
a1
D K
f
F
a1
d
2
=4
D K
f
4F
a1
d
2
; (b)
where F
a1
in klb is the fully reversed axial loading amplitude at the cyclic number level #1.
In the cyclic number level #2 with n
L2
D 200;000 (cycles), the fully reversed axial stress
amplitude
a2
is
a2
D K
f
F
a2
d
2
=4
D K
f
4F
a2
d
2
; (c)
where F
a2
in klb is the fully reversed axial loading amplitude at the cyclic number level #2.
e component fatigue damage index D of the bar under this model #6 cyclic axial loading
per Equation (2.85) is:
D D n
L1
K
f
4F
a1
d
2
m
C n
L2
K
f
4F
a2
d
2
m
D
4K
f
d
2
m
n
L1
.
F
a1
/
m
C n
L2
.
F
a2
/
m
: (d)
(3) e limit state function.
3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 181
e component fatigue strength index K can be calculated per Equation (2.79):
K D
.
k
a
k
b
k
c
/
m
K
0
: (e)
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 1 for cyclic axial
loading. e mean and the standard deviation of the load modification factor k
c
can be calculated
per Equations (2.18), (2.19), and (2.20).
e limit state function of the stepped bar in this example per Equation (2.88) is
g
K
0
; k
a
; k
c
; F
a1
; F
a2
; K
f
; d
D
.
k
a
k
c
/
m
K
0
4K
f
d
2
m
n
L1
.
F
a1
/
m
C n
L2
.
F
a2
/
m
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(f)
e diameter d will be treated as a normal distribution. Its mean and standard deviation can be
determined per Equation (1.1). ere are seven random variables in the limit state function (f).
K
0
is a log-normal distribution. e rests are normal distributions. eir distribution parameters
in Equation (f) are listed in Table 3.50. In this table, the mean and standard deviation of K
f
will be updated in each iterative step by using a new dimension of the bar.
Table 3.50: e distribution parameters of random variables in Equation (d)
K
0
(lognormal) k
a
k
c
F
a1
(klb)
μ
lnK
0
σ
lnK
0
μ
k
a
σ
k
a
μ
k
c
σ
k
c
μ
F
a1
σ
F
a1
41.738 0.357 0.8588 0.05153 0.774 0.1262 22.15 3.25
F
a2
(klb) K
f
d (in)
μ
F
a2
σ
F
a2
μ
K
f
σ
K
f
μ
d
σ
d
12.45 1.5 1.6624 0.1330
μ
d
0.00125
(4) Use the modified R-F method to determine the diameter d .
In the limit state function (f), there are one lognormal distribution and six normal dis-
tribution. 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. (Note: In this
MATLAB program, the subroutine for calculating the stress concentration factor is required.)
e iterative results are listed in Table 3.51.
..................Content has been hidden....................
You can't read the all page of ebook, please click here login for view all page.