3.3. RELIABILITY OF A COMPONENT WITH TWO RANDOM VARIABLES 115
e limit state function g
.
S
e
; h; b; M
/
is the function of four parameters S
e
, h, b; and M .
3.3 RELIABILITY OF A COMPONENT WITH TWO
RANDOM VARIABLES
When the limit state function of a component consists of only two mutually independent ran-
dom variables, we can have an explicit formula for determining the reliability of a component.
In this section, we will derive an explicit formula for a general case. en, four special cases will
be discussed, which are: both uniform distributions, both normal distributions, both log-normal
distributions, and both exponential distributions.
3.3.1 INTERFERENCE METHOD
Let f
S
.
s
/
and F
S
.
s
/
represent the PDF and CDF of component strength index S. Let f
Q
.
q
/
and F
Q
.
q
/
represent the PDF and CDF of component stress index Q. In the following, we
assume that S and Q are statistically independent.
e interference method is to use the following Equations (3.5) or (3.6) to calculate the
reliability of a component when there are only two statistically independent random variables in
a limit state function.
Now, we will derive these two equations. Based on Equation (3.2), the reliability of a
component is
R D P
Œ
g
.
S; Q
/
0
D
ZZ
SQ0
f
S
.s/f
Q
.q/dsdq D
ZZ
SQ
f
S
.s/f
Q
.q/dsdq: (3.4)
If we run the integration concerning the random variable S first, as shown in Figure 3.2,
the reliability of the component will become:
R D P
Œ
g
.
S; Q
/
0
D
ZZ
SQ
f
S
.s/f
Q
.q/dsdq D
Z
C1
1
f
Q
.q/
Z
C1
q
f
S
.s/ds
dq:
D
Z
C1
1
f
Q
.q/
1
Z
q
1
f
S
.s/ds
dq D
Z
C1
1
f
Q
.q/
Œ
1 F
S
.q/
dq: (3.5)
116 3. COMPUTATIONAL METHODS FOR THE RELIABILITY OF A COMPONENT
f
Q
(q)
q s > q
s , q
f
S
(s)
f
S
(s),f
Q
(q)
Figure 3.2: e schematic for integrating s first.
If we run the integration concerning the random variable Q first, as shown in Figure 3.3, the
reliability of the component will become:
R D P
Œ
g
.
S; Q
/
0
D
ZZ
SQ
f
S
.
s
/
f
Q
.
q
/
dsdq D
Z
C1
1
f
S
.
s
/
Z
s
1
f
Q
.
q
/
dq
ds
D
Z
C1
1
f
S
.
s
/
F
Q
.
s
/
ds : (3.6)
f
Q
(q)
sq < s
s , q
f
S
(s)
f
S
(s),f
Q
(q)
Figure 3.3: e schematic for integrating q first.
3.3. RELIABILITY OF A COMPONENT WITH TWO RANDOM VARIABLES 117
Equations (3.5) and (3.6) are the explicit formula for calculating the reliability of a com-
ponent when the limit state function of a component contains only two random variables.
Example 3.4
For a bar under axial loading, the maximum normal stress follows a uniform distribution. Its
PDF is:
f
.
/
D
8
<
:
1
40
15 .ksi/ 55 .ksi/
0 otherwise:
(a)
e material yield strength also follows a uniform distribution with the following PDF:
f
S
y
s
y
D
8
<
:
1
30
45 .ksi/ s
y
75 .ksi/
0 otherwise:
(b)
Determine the reliability of this bar under the specified axial loading.
Solution:
For this problem, the component strength index will be the material yield strength, and the
component stress index will be the maximum normal stress induced by the axial loading. e
limit state function will be:
g
.
S; Q
/
D S Q D S
y
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(c)
en, we could use Equations (3.5) or (3.6) to calculate the reliability. In the following, we will
use Equation (3.5) to calculate the reliability of the bar.
To use Equation (3.6) to run the calculation, we need to get the CDF of the yield strength.
Based on the PDF specified in Equation (b), the CDF of the yield strength will be:
F
S
y
S
y
D
8
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
:
0 < 45 .ksi/
S
y
45
30
45 .ksi/ 75 .ksi/
1 > 75 .ksi/:
(d)
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