182 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.51: e iterative results of Example 3.21 by the modified R-F method
Iterative #
K
0
*
k
a
*
k
c
*
F
a
*
1
F
a
*
2
K
f
*
d
*
|∆d
*
|
1
1.43E+18 0.8588 0.774 22.15 12.45 1.6624 1.126794
2
1.28E+18 0.851578 0.725936 24.34357 12.74353 1.687255 1.232736 0.105941
3
1.31E+18 0.854803 0.745879 24.73924 1.190743 1.675838 1.204671
0.028065
4
1.31E+18 0.854718 0.745943 24.73765 1.185953 1.67627 1.204815
0.000144
5
1.31E+18 0.854722 0.745971 24.73794 1.185973 1.676253 1.20479
2.58E-05
According to the result obtained from the program, the mean of the diameter with the
reliability 0.99 is
d
D 1:205
00
: (g)
erefore, the diameter of the bar with the required reliability 0.99 under the specified loading
will be
d D 1:205 ˙ 0:005
00
:
3.4.4 PIN UNDER CYCLIC DIRECT SHEARING LOADING SPECTRUM
e limit state function of a pin under any cyclic shear loading spectrum can be established per
Equations (2.87) or (2.88), which have been discussed in Section 2.9.7. After the limit state
function of a component under cyclic direct shearing 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 pin with the required reliability under cyclic shear loading spectrum.
Example 3.22
A machined single-shear pin is under a cyclic shearing loading spectrum, which is a zero-to-
maximum direct shear loading. e maximum shear loading V
max
of this cyclic shearing loading
can be treated as a constant V
max
D 25:75 (klb). e number of cycles n
L
of this cyclic shear-
ing loading is also treated as a constant n
L
D 500;000 (cycles). 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 pin with a reliability 0.95 when its dimension tolerance is ˙0:005
00
.
Solution:
(1) Preliminary design for determining the size modification factor k
b
.
3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 183
In this example, the size modification factor k
b
is a dimension-dependent parameter. Per
the discussion in Section 3.2.1, it will be
k
b
D 0:87: (a)
(2) e cyclic stress and the component fatigue damage index.
e mean shear stress
m
and the shear stress amplitude
a
of the pin due to this zero-to-
maximum cyclic shearing loading are:
m
D
V
m
A
D
V
max
=2
d
2
=4
D
2V
max
d
2
(b)
a
D
V
a
A
D
V
max
=2
d
2
=4
D
2V
max
d
2
: (c)
Since this is non-zero-mean cyclic shear stress, the equivalent stress amplitude of a fully reversed
cyclic shear stress is:
aeq
D
a
S
u
.
S
u
m
/
D
2V
max
S
u
.
d
2
S
u
2V
max
/
: (d)
e component fatigue damage index of this pin under model #1 cyclic shear stress per Equa-
tion (2.84) is:
D D n
L
2V
max
S
u
.
d
2
S
u
2V
max
/
8:21
: (e)
(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
: (f )
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 pin. e mean and 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 pin under model #1 cyclic direct shearing loading spectrum
per Equation (2.87) is:
g
.
K
0
; k
a
; k
c
; d
/
D
.
k
a
k
b
k
c
/
m
K
0
n
L
2V
max
S
u
.
d
2
S
u
2V
max
/
8:21
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(g)
184 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.52: e distribution parameters of random variables in Equation (g)
K
0
(lognormal) k
a
k
c
d (in)
μ
lnK
0
σ
lnK
0
μ
k
a
σ
k
a
μ
k
c
σ
k
c
μ
d
σ
d
41.738 0.357 0.8588 0.05153 0.774 0.1262
μ
d
0.00125
ere are four random variables in the limit state function (g). K
0
is a log-normal distribution.
e rests are normal distributions. 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 param-
eters in the limit state function (g) are listed in Table 3.52.
(4) Use the modified R-F method to determine the diameter d .
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 3.3 to compile a MATLAB program. e iterative
results are listed in Table 3.53.
Table 3.53: e iterative results of Example 3.22 by the modified R-F method
Iterative #
k
0
*
k
a
*
k
c
*
d
*
|∆d
*
|
1 1.43E+18 0.8588 0.774 1.03942
2 1.3E+18 0.843918 0.674958 1.111562 0.072142
3 1.31E+18 0.845409 0.673574 1.115417 0.003855
4 1.31E+18 0.845455 0.673535 1.115592 0.000175
5 1.31E+18 0.845456 0.673534 1.1156 8.01E-06
According to the result obtained from the program, the mean of the diameter with a
reliability 0.99 is
d
D 1:116
00
: (h)
erefore, the diameter of the pin with the required reliability 0.99 under the specified
loading will be
d D 1:116 ˙ 0:005
00
:
Example 3.23
A machined double-shear pin is subjected to model #3 cyclic shear loading spectrum as listed in
Table 3.54. e ultimate material strength S
u
is 75 (ksi). e three distribution parameters of
3.4. DIMENSION OF A COMPONENT UNDER CYCLIC LOADING SPECTRUM 185
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 pin with a reliability 0.99 when its dimension
tolerance is ˙0:005
00
.
Table 3.54: Model #3 cyclic shear loading for Example 3.23
Number of Cycles n
L
Mean of the Cyclic
Shear Loading V
m
(klb)
Amplitude of Cyclic Shear Loading
V
a
(klb) (normal distribution)
μ
V
a
σ
V
a
600,000 3.422 4.815 0.6
Solution:
(1) Preliminary design for determining the size modification factor k
b
.
In this example, the size modification factor k
b
is a dimension-dependent parameter. Per
the discussion in Section 3.2.1,
k
b
D 0:87: (a)
(2) e cyclic shear stress and the component fatigue damage index.
e mean shear stress
m
and the shear stress amplitude
a
of the double-shear pin in this
example are:
m
D
V
m
=2
A
D
V
m
=2
d
2
=4
D
2V
m
d
2
(b)
a
D
V
a
=2
A
D
V
a
=2
d
2
=4
D
2V
a
d
2
: (c)
Since this is non-zero-mean cyclic shear stress, the equivalent stress amplitude of a fully reversed
cyclic shear stress is:
aeq
D
a
S
u
.S
u
m
/
D
2V
a
S
u
.
d
2
S
u
2V
m
/
: (d)
e component fatigue damage index of this pin under model #3 cyclic shear stress per Equa-
tion (2.84) is:
D D n
L
2V
a
S
u
.
d
2
S
u
2V
m
/
8:21
: (e)
(3) e limit state function.
186 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
e component fatigue strength index K can be calculated per Equation (2.79):
K D
.
k
a
k
b
k
c
/
m
K
0
: (f)
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 pin. 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 double-shear pin under model #1 cyclic shearing loading
spectrum per Equation (2.87) is:
g
.
K
0
; k
a
; k
c
; V
a
; d
/
D
.
k
a
k
b
k
c
/
m
K
0
n
L
2V
a
S
ut
.
d
2
S
ut
2V
m
/
8:21
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(g)
ere are five random variables in the limit state function (g). K
0
is a lognormal distribution. e
rests are normal distributions. 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 (g) are listed in Table 3.55.
Table 3.55: e distribution parameters of random variables in Equation (g)
K (lognormal) k
a
k
c
V
a
(klb) d (in)
μ
lnK
σ
lnK
μ
k
a
σ
k
a
μ
k
c
σ
k
c
μ
V
a
σ
V
a
μ
d
σ
d
41.738 0.357 0.8588 0.05153 0.774 0.1262 4.815 0.6
μ
d
0.00125
(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 3.56.
According to the result obtained from the program, the mean of the diameter with a
reliability 0.99 is
d
D 0:550
00
: (h)
erefore, the diameter of the pin with the required reliability 0.99 under the specified loading
will be
d D 0:550 ˙ 0:005
00
:
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