3.2. DIMENSION DESIGN WITH REQUIRED RELIABILITY 145
Table 3.14: e iterative results of Example 3.6 by the modified R-F method
Iterative #
K
0
*
k
a
*
k
c
*
V
a
*
d
*
|∆d
*
|
1 1.43E+18 0.8588 0.583 8.72 0.687351
2 1.21E+18 0.841437 0.533467 8.802078 0.728019 0.040669
3 1.23E+18 0.841917 0.53143 8.776322 0.715777 0.012242
4 1.23E+18 0.84222 0.532131 8.795148 0.715286 0.000491
5 1.23E+18 0.842223 0.53219 8.796141 0.715262 2.38E-05
3.2.5 DIMENSION DESIGN BY THE MODIFIED MONTE CARLO
METHOD
When a limit state function of a component is established, the Monte Carlo method can be
utilized to calculate the reliability of a component under specified loading, which is discussed in
Section 3.8 of Volume 1 [1] and is also displayed in Appendix A.3 of this book. e Monte Carlo
method can be modified to determine the component dimension with the required reliability
under specified loading. e idea is to pick an initial value for the dimension and then use the
Monte Carlo method to calculate the reliability of the component by using a new dimension
with a small incremental in the dimension such as 0.001
00
until the calculated reliability is slightly
over the required reliability. en, this dimension will be the dimension of the component with
the required reliability under specified loading.
e general procedure by using the modified Monte Carlo method to conduct the dimen-
sion design is explained and listed here.
Step 1: Preliminary design for determining K
t
for static design or K
f
and k
b
for fatigue design.
Per Section 3.2.1, we need to determine the dimension-dependent parameters K
t
for static
design or K
f
and k
b
for fatigue design if necessary.
Step 2: Establish the limit state function.
Per the question under consideration, we establish its limit state function. We can use the general
limit state function (3.1) and redisplay it here:
g
.
X
1
; : : : ; X
n
; d
/
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(3.1)
146 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
where X
i
.i D 1; 2; : : : ; n/ is a random variable related to component strength or loading, which
could be any type of distributions. d is a normal distribution dimension with a mean
d
and a
standard deviation
d
.
Step 3: Determine the initial value for the mean
0
d
of the dimension d .
We can choose any value as the initial value
0
d
for the mean
d
of the dimension d . To save
computing time, we could use the following approach to determine the initial value
0
d
.
For a strength-related question, we can assume that all random variables except the ma-
terial strength variable will be deterministic and are replaced by their means. For a deflection
related question, we can assume that all random variables except the material Young’s modulus E
or shear Young’s modulus G will be deterministic and are replaced by their means. Let us assume
that X
1
in the limit state function (3.1) is component strength variable such as yield strength,
ultimate strength, fatigue strength, Young’s modulus E, or shear Young’s modulus G. Its CDF
is F
X
1
.x
1
/. We can use the required reliability R to calculate the point x
0
1
for the component
strength to satisfy this
R D P
X
1
> x
0
1
D 1 P
X
1
< x
0
1
D 1 F
X
1
x
0
1
: (3.31)
Rearrange the Equation (3.31), we have
x
0
1
D F
1
X
1
.
1 R
/
; (3.32)
where R is the required reliability for the dimension design. F
1
X
1
.
/
is the inverse CDF of com-
ponent strength X
1
.
0
d
can be determined by following equations:
g
x
0
1
;
X
2
; : : : ;
X
n
;
0
d
D 0; (3.33)
where
X
i
.i D 2; 3; : : : ; n/ is the mean of the random variable X
i
. In Equation (3.33),
0
d
is
the only unknown variable and can be solved per actual limit state function.
Step 4: Update the
d
and the dimension-dependent parameters.
e increment in the dimension
d
will be 0.001
00
. So, the recurrence equation is
d
D
0
d
C 0:001
00
: (3.34)
After this
d
is known, the geometric dimensions for the stress concentration areas are all
known. If necessary, we need to update the dimension-dependent parameters. We can update
the static stress concentration factor K
t
. en use Equations (2.22), (2.23), and (2.24) to update
the fatigue stress concentration factor K
f
and use Equation (2.17) to update the size modifica-
tion factor k
b
.
Step 5: Use the Monte Carlo method to calculate the reliability R
1
of the component.
3.2. DIMENSION DESIGN WITH REQUIRED RELIABILITY 147
Use
d
as the mean of the normally distributed dimension d . en we can use the Monte Carlo
method per the limit state function (3.1) to calculate the reliability R
1
. Since the limit state
function of a component is typically not too complicated, the trial number for the Monte Carlo
method can be N D 15,998,400.
Step 6: Check the convergence condition.
e convergence condition can be
R D R
1
R > 0:0001: (3.35)
If the convergence condition (3.35) is not satisfied, we will update the
0
d
by the following
equation and go back to Step 4.
0
d
D
d
: (3.36)
If the convergence condition (3.35) is satisfied, the
d
will be the mean of the dimension d . e
program flowchart by using the modified Monte Carlo method to determine the dimension with
the required reliability under specified loading is shown in Figure 3.5.
Example 3.7
A circular rod is subjected to an axial loading F which follows a uniform distribution between
7.00 (klb) and 9.00 (klb). e rod is made of a ductile material. e yield strength S
y
of the rod’s
material follows a normal distribution with a mean
S
y
D 34:5 (ksi) and a standard deviation
S
y
D 3:12 (ksi). Determine the diameter of the rod with a reliability 0.99 when the dimension
tolerance is ˙0:005.
Solution:
In this example, there is no dimension-dependent parameter.
(1) e limit state function.
e normal stress of the rod caused by the axial loading F is
D
F
A
D
F
d
2
=4
D
4F
d
2
: (a)
e limit state function of the rod is
g
S
y
; T; d
D S
y
4F
d
2
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(b)
In the limit state function, there are three random variables. S
y
and d follow normal distribu-
tions. e axial loading F follows a uniform distribution. e standard deviation of normally
distributed d can be determined per Equation (1.1). eir distribution parameters in the limit
state function are listed in Table 3.15.
148 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
K
t
, K
f
, and k
b
for fatigue design
Figure 3.5: e flowchart of the Monte Carlo method for dimension design.
3.2. DIMENSION DESIGN WITH REQUIRED RELIABILITY 149
Table 3.15: Distribution parameters for Example 3.7
S
y
(ksi)
Normal Distribution
F (lb.in)
Uniform Distribution
d (in)
Normal Distribution
μ
S
y
σ
S
y
a b μ
d
σ
d
34.5 3.12 7.0 9.0
μ
d
0.00125
(2) Use the modified Monte Carlo method to determine the mean
d
of the diameter d .
Following the modified Monte Carlo method procedure discussed above and the
flowchart shown in Figure 3.5, we can compile a MATLAB program. e program is listed
in Appendix B.6 as “M-Monte Carlo program-Example 3.7.” e iterative results are listed in
Table 3.16.
Table 3.16: e iterative results of Example 3.7 by the modified Monte Carlo method
Iterative #
μ
d
*
R
*
∆R
*
1 0.61248 0.977524 -0.01248
2 0.61348 0.978904 -0.0111
… … … …
13 0.62448 0.989836 -0.00016
14 0.62548 0.990514 0.000514
According to the result obtained from the program, the mean of the diameter with the
reliability 0.99 is
d
D 0:626
00
: (c)
erefore, the diameter of the rod with the required reliability 0.99 under the specified loading
will be
d D 0:626 ˙ 0:005
00
:
Example 3.8
A circular bar with a length L D 17:000 ˙ 0:010
00
is subjected to an axial loading F which follows
a normal distribution with a mean
F
D 8:92 (klb) and a standard deviation
F
D 0:675 (klb).
e Young’s modulus of the bar material follows a normal distribution with the mean
E
D
2:76 10
4
(ksi) and the standard deviation
E
D 6:89 10
2
(ksi). e design specification is
that allowable deformation of the entire bar is less than 0:014
00
. (1) Determine the diameter of the
bar with a reliability 0.99 when the dimension tolerance is ˙0:005. (2) Calculate the reliability of
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