3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 187
Table 3.56: e iterative results of Example 3.23 by the modified Monte Carlo method
Iterative #
μ
d
*
R
*
∆R
*
1 0.462279 0.806384 -0.18362
2 0.463279 0.811958 -0.17804
… … … …
87 0.548279 0.989843 -0.00016
88 0.549279 0.990191 0.000191
3.4.5 SHAFT UNDER CYCLIC TORSION LOADING SPECTRUM
e limit state function of a shaft under any type of cyclic torsion loading spectrum can be
established per Equations (2.87) or (2.88), which have been discussed in Section 2.9.8. After
the limit state function of a component under cyclic torsion 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 shaft with the required reliability under cyclic torsion fatigue spectrum.
Example 3.24
e critical section of a machined shaft with a shoulder is at the shoulder section, as shown in
Figure 3.12. It is subjected to model #4 cyclic torsion loading spectrum as listed in Table 3.57.
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 shaft with a reliability 0.99 when its dimension
tolerance is ˙0:005
00
.
1
32
R "
Ø 1.750"
d
Figure 3.12: Schematic of the segment of a shaft with a shoulder.
Solution:
(1) Preliminary design for determining dimension-dependent parameters.
188 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.57: e model #4 cyclic fatigue spectrum for Example 3.24
Loading Level #
Number of
Cycles n
Li
Mean T
mi
of the Cyclic
Torque (klb.in)
Amplitude T
ai
of the
Cyclic Torque (klb.in)
1 500,000 2.25 5.13
2 6,000 2.25 6.42
For the component under cyclic torsion loading, there are three dimension-dependent
parameters k
b
, K
t
, and K
f
in this fatigue problem. According to the schematic of the shaft with
a shoulder, we can assume that it has a sharp-fillet. e preliminary static stress concentration
factor per Table 3.2 will be
K
ts
D 2:2: (a)
e fatigue stress concentration factor K
fs
can be calculated and updated per Equations (2.22)–
(2.25).
e preliminary size modification factor k
b
per Equation (3.3) will be
k
b
D 0:87: (b)
(2) e cyclic shear stress and the component fatigue damage index.
e mean shear stress
mi
, the shear stress amplitude
ai
and its corresponding equivalent
shear stress amplitude
eqi
of the shaft due to model #4 cyclic torque loading are as follows.
For the loading level #1
m1
D K
fs
T
m1
d=2
J
D K
fs
16T
m1
d
3
: (c)
a1
D K
fs
T
a1
d=2
J
D K
fs
16T
a1
d
3
(d)
eq1
D K
fs
a1
S
u
S
u
K
fs
m1
D K
fs
16
a1
S
u
d
3
S
u
16K
fs
T
m1
: (e)
For the loading level #2, repeating (c), (d), and (e), we have
eq2
D K
fs
a2
S
u
S
u
K
fs
m2
D K
fs
16
a2
S
u
d
3
S
u
16K
fs
T
m2
: (f)
e component fatigue damage index of this shaft under model #4 cyclic shear stress per Equa-
tion (2.85) is:
D D n
L1
K
fs
16
a1
S
u
d
3
S
u
16K
fs
T
m1
8:21
C n
L2
K
fs
16
a2
S
u
d
3
S
u
16K
fs
T
m2
8:21
: (g)
3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 189
(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
: (h)
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 shaft. 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 shaft for this example per Equation (2.88) is
g
K
0
; k
a
; k
c
; K
fs
; d
D
.
k
a
k
b
k
c
/
m
K
0
n
L1
K
fs
16
a1
S
u
d
3
S
u
16K
fs
T
m1
8:21
n
L2
K
fs
16
a2
S
u
d
3
S
u
16K
fs
T
m2
8:21
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(i)
ere are five random variables in the limit state function (i). e K
0
is a lognormal distribution.
e dimension d can be treated as a normal distribution, and its mean and standard deviation
can be calculated per Equation (1.1). e distribution parameters in the limit state function (i)
are listed in Table 3.58.
Table 3.58: e distribution parameters of random variables in Equation (i)
K (lognormal) k
a
k
c
K
fs
d (in)
μ
lnK
σ
lnK
μ
k
a
σ
k
a
μ
k
c
σ
k
c
μ
K
fs
σ
K
fs
μ
d
σ
d
41.738 0.357 0.8588 0.05153 0.774 0.1262 1.6552 0.1324
1.25
0.00125
(4) Use the modified R-F method to determine the diameter d .
In the limit state function (i), there are one lognormal distribution and four normal distri-
bution. Now, 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 3.4 to compile a MATLAB program. e iterative
progress is listed in Table 3.59.
According to the result obtained from the program, the mean of the diameter with a
reliability 0.99 is
d
D 1:363
00
: (j)
190 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.59: e iterative results of Example 3.24 by the modified R-F method
Iterative #
k
0
*
k
a
*
k
c
*
K
f
*
s
d
*
|∆d
*
|
1 1.43E+18 0.8588 0.774 1.6552 1.365609
2 1.33E+18 0.854412 0.744798 1.660519 1.365754 0.000145
3 1.33E+18 0.85575 0.753018 1.529347 1.362163
0.003591
4 1.34E+18 0.857546 0.76545 1.526983 1.362098
6.49E-05
erefore, the diameter of the shaft with the required reliability 0.99 under the specified loading
will be
d D 1:363 ˙ 0:005
00
:
Example 3.25
A machined constant circular shaft is subjected to model #5 cyclic torsion loading spectrum as
listed in Table 3.60. e ultimate material strength S
u
is 75 (ksi). e three distribution param-
eters of material fatigue strength index K
0
for the standard specimen under fully-reversed bend-
ing 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 shaft with a reliability 0.99 when
its dimension tolerance is ˙0:005
00
.
Table 3.60: e model #5 cyclic fatigue spectrum for Example 3.25
Loading
Level#
Torque Mean T
m
(klb.in)
Torque Amplitude T
a
(klb.in)
Number of Cycles n
L
μ
n
Li
σ
n
Li
1 3.5 4.2 300,000 5,000
2 3.5 7.5 4,000 200
Solution:
(1) Preliminary design for determining the size modification factor k
b
.
In this example, the size modification factor k
b
is the only one dimension-dependent
parameter. Per the discussion in Section 3.2.1, it will be:
k
b
D 0:87: (a)
(2) e cyclic shear stress and the component fatigue damage index.
3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 191
e mean shear stress
m
, the shear stress amplitude
a
and its corresponding equivalent
shear stress amplitude
eq
of the shaft due to the model #5 cyclic torque loading are as follows.
For the loading level #1,
m1
D
T
m1
d=2
J
D
T
m1
d=2
d
4
=32
D
16T
m1
d
3
(b)
a1
D
T
a1
d=2
J
D
T
a1
d=2
d
4
=32
D
16T
a1
d
3
(c)
eq1
D
a1
S
u
S
u
m1
D
16T
a1
S
u
d
3
S
u
16T
m1
: (d)
For the loading level #2, repeating (b), (c), and (d), we have:
eq2
D
a2
S
u
S
u
m2
D
16T
2a
S
u
d
3
S
u
16T
m2
: (e)
e component fatigue damage index of this shaft under model #5 cyclic shear stress per Equa-
tion (2.85) is:
D D n
L1
16
a1
S
u
d
3
S
u
16T
m1
8:21
C n
L2
16T
2a
S
u
d
3
S
u
16T
m2
8:21
: (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 shaft. 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 shaft for this example per Equation (2.87) is:
g
.
K
0
; k
a
; k
c
; n
L1
; n
L2
; d
/
D
.
k
a
k
b
k
c
/
m
K
0
n
L1
16
a1
S
us
d
3
S
us
16T
m1
8:21
n
L2
16
a2
S
us
d
3
S
us
16T
m2
8:21
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(h)
ere are six random variables in the limit state function (h). K
0
is a lognormal distribution. e
rests are normal distributions. e dimension d can be treated as a normal distribution, and its
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