140 3. COMPUTATIONAL METHODS FOR THE RELIABILITY OF A COMPONENT
e iterative results are listed in Table 3.2. From the table, the reliability index ˇ and corre-
sponding reliability R of this component are:
ˇ D 2:9452 R D ˆ
.
ˇ
/
D ˆ
.
2:9452
/
D 0:9839:
Table 3.2: e iterative results for the limit state function (2) in Example 3.11
# S
y
*
Z
*
P
*
L
*
β
*
|∆ β
*
| < 0.0001
1 600000 0.0001 10 24 2.030685
2 496907.9 7.53E-05 12.47421 11.99154 2.885681 0.854996
3 443008.1 5.85E-05 12.5015 8.297361 2.953815 0.068134
4 448340.9 5.41E-05 12.1497 7.983943 2.947101 0.006714
5 457762.6 5.28E-05 12.0995 7.996524 2.945587 0.001514
6 462281 5.23E-05 12.08414 7.999504 2.94526 0.000327
7 464338.3 5.2E-05 12.07591 8.000162 2.945191 6.93E-05
When the H-L method is used, the reliability of the component in two different versions
of the limit state function of the same problem will be the same.
3.7 THE RACKWITZ AND FIESSLER (R-F) METHOD
When a limit state function of a component contains at least one non-normal distributed ran-
dom variables such as log-normal distribution or Weibull distribution, we need to use the R-F
(Rackwitz and Fiessler) method [3] to calculate the reliability of a component. e R-F method
is a modified H-L method. e R-F method still linearizes the limit state function at the design
point, which is the point on the surface of the limit state function. e key difference between
the R-F method and the H-L method is that any non-normally distributed random variable at
the design point will be first converted into an equivalent normally distributed random variable,
and then the H-L method is applied for calculating the reliability index.
In the R-F method, a non-normally distributed variable is converted into an equivalent
normal distribution variable at the design point. Let f
X
.
x
/
and F
X
.
x
/
represent the PDF and
the CDF of a non-normally distributed random variable X. Let
X
eq
and
X
eq
represent the
equivalent mean and the equivalent standard deviation of the equivalent normally distributed
variable at the design point X
. Two conditions for calculating the equivalent mean and the
equivalent standard deviation of the equivalent normal distribution at the design point are as
follows.
3.7. THE RACKWITZ AND FIESSLER (R-F) METHOD 141
(1) e PDF of a non-normal distribution variable at the design point will be equal to the
PDF of its equivalent normal distribution at the design point:
f
X
x
D
1
p
2
X
eq
e
X
X
eq
2
2
2
X
eq
D
1
X
eq
X
X
eq
X
eq
!
; (3.54)
where
.
/
is the PDF of the standard normally distributed random variable.
(2) e CDF of a non-normal distribution variable at the design point will be equal to the
CDF of its equivalent normal distribution at the design point:
F
X
X
D
Z
X
1
1
p
2
X
eq
e
x
X
eq
2
2
2
0X
eq
dx D ˆ
X
X
eq
X
eq
!
; (3.55)
where ˆ
.
/
is the CDF of the standard normally distributed random variable.
For the convenient calculation, we can introduce z
X
, which is the value of the standardized
equivalent normal distribution at the design point x
:
z
X
D
x
X
eq
X
eq
: (3.56)
We can rewrite the Equations (3.54) and (3.55) as:
f
X
x
D
1
X
eq
z
X
(3.57)
F
X
X
D ˆ
z
X
(3.58)
From Equation (3.58) we can get z
X
:
z
X
D ˆ
1
F
X
X
; (3.59)
where ˆ
1
.
/
is the inverse of the CDF of the standard normal distribution.
In Excel, the function for ˆ
1
.
/
is:
z
X
D ˆ
1
F
X
X
D NORM:S:INV
F
X
X
: (3.60)
In the MATLAB program, the command for ˆ
1
.
/
is:
z
X
D ˆ
1
F
X
X
D norminv
F
X
X
: (3.61)
142 3. COMPUTATIONAL METHODS FOR THE RELIABILITY OF A COMPONENT
Since F
X
.
X
/
of the non-normally distribution at the design point X
is known, z
X
can be
determined by Equations (3.60) or (3.61). en, from Equation (3.57), we can calculate the
equivalent standard deviation
X
eq
:
X
eq
D
1
f
X
.
X
/
z
X
: (3.62)
After the equivalent standard deviation
X
eq
is known, we can calculate the equivalent mean
X
eq
from Equation (3.56):
X
eq
D x
z
X
X
eq
: (3.63)
We can use Equations (3.59)–(3.63) to calculate the equivalent mean
X
eq
and the equivalent
standard deviation
X
eq
of the equivalent normal distribution of a non-normal distribution at
the design point X
:
Example 3.12
A random variable X follows a log-normal distribution with the distribution parameter
ln X
D
7:81 and the
ln X
D 0:192.
(a) Use Excel to determine the equivalent mean and the equivalent standard deviation of the
equivalent normal distribution at the design point X
D 2600.
(b) Use MATLAB to determine the equivalent mean and the equivalent standard deviation
of the equivalent normal distribution at the design point X
D 2600:
Solution:
(a) Use Excel to determine
X
eq
and
X
eq
at the design point
X
D
2600
.
e PDF is f
X
.
X
/
and the CDF is F
X
.
X
/
of the log-normal distribution at the design
point X
D 2600 with the distribution parameters
ln X
D 7:81 and the
ln X
D 0:192 are:
f
X
.
2600
/
D LOGNORM:DIST
.
2600; 7:81; 0:192 FALSE
/
D 7:68993 10
4
(a)
F
X
.
2600
/
D LOGNORM:DIST
.
2600; 7:81; 0:192; TRUE
/
D 0:609275; (b)
where LOGNORM:DIST
.
x;
ln X
;
ln X
; FALSEorTRUE
/
is the PDF with “FALSE” and the
CDF with “TRUE” of Excel function for a log-normal distribution.
From Equations (3.60) and (b), we have:
z
X
D ˆ
1
F
X
x
D NORM:S:INV
.
0:609275
/
D 0:277430; (c)
where NORM:S:INV
.
z
/
is the Excel function for the inverse of the standard normal distribution
with a probability z.
3.7. THE RACKWITZ AND FIESSLER (R-F) METHOD 143
From Equations (3.62), (a), and (c), we can have the equivalent standard deviation
X
eq
:
X
eq
D
1
f
X
.
x
/
z
X
D
NORM:S:DIST
.
0:277430
/
7:68993 10
4
D
0:383881
7:68993 10
4
D 499:2: (d)
From Equations (3.63), (c), and (d), we can have the equivalent mean
X
eq
:
X
eq
D X
z
X
X
eq
D 2600 0:277430 499:20 D 2461:5: (e)
(b) Use the MATLAB program to determine
X
eq
and
X
eq
at the design point X
D 2600.
From Equations (3.57), (3.60), and (3.61), we can compile a short program as listed here.
% Calculate an equivalent mean and standard deviation
% Input the lognormal distribution parameter
mln=7.81; % Log mean
sln=0.192; % Log standard deviation
xs=2600; % Design point x*
fx=lognpdf (xs, mln, sln); % PDF of log-normal distribution
Fx=logncdf (xs, mln, sln); % CDF of log-normal distribution
zs=norminv (Fx); % The inverse value of the normal
% distribution per Equation (3.61)
seq=normpdf (zs)/fx; % Equivalent standard deviation per
% Equation (3.64)
meq=xs-seq*zs; % Equivalent mean, Equation (3.65)
display ('Equivalent mean')
meq
display ('Equivalent standard deviation')
seq
MATLAB gives the following results:
Equivalent mean D 2461:5.
Equivalent standard deviation D 499:2.
e R-F method is an iterative process. e procedure is very similar to the H-L method.
However, at the beginning of each iterative process, any non-normally distributed random vari-
able will be converted into an equivalent normal distribution with the equivalent mean and the
equivalent standard deviation. Following is the general procedure for the R-F method.
Step 1: Calculate the mean for non-normal distributed random variables.
For a clear description of the R-F method, we can rearrange the limit state function, as
shown in Equation (3.64). In Equation (3.64), the first r random variables are non-normally
144 3. COMPUTATIONAL METHODS FOR THE RELIABILITY OF A COMPONENT
distributed random variables, and the rest .n r/ random variables are normally distributed
random variables:
g
.
X
1
; : : : ; X
r
; X
rC1
; : : : ; X
n
/
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(3.64)
e surface of this limit state function is
g
.
X
1
; : : : ; X
r
; X
rC1
; : : : ; X
n
/
D 0: (3.65)
For non-normally distributed random variable, we can calculate their means based on their type
of distributions, which has been discussed in Chapter 2. e means of some typical distribution
are listed here.
For a uniform distribution X with distribution parameters a and b, the mean is
X
D
a C b
2
: (3.66)
For a log-normal distribution X with the distribution parameter
ln x
and
ln x
, the mean is
X
D Exp
ln x
C
2
ln x
2
: (3.67)
For a two-parameter Weibull distribution X with the scale parameter and the shape parameter
ˇ, the mean is
X
D
1
ˇ
C 1
: (3.68)
For an exponential distribution X with the distribution parameter , the mean is
X
D
1
: (3.69)
Step 2: Pick an initial design point P
0
X
0
1
; X
0
2
; : : : ; X
0
n
.
e initial design point could be any point. e simple choice is to use the means of every
random variable as the design point. Since the design point must be on the surface of the limit
state function as specified by Equation (3.65), we can use the mean values for the first n 1
variables, and then determine the last one through Equation (3.71a). Since there is only one
unknown X
0
n
in Equation (3.71a), we can solve this unknown X
0
n
:
X
0
i
D
X
i
i D 1; 2; : : : ; n 1 (3.70)
g
X
0
1
; X
0
2
; : : : X
0
n1
; X
0
n
D 0: (3.71a)
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