3.3. RELIABILITY OF A COMPONENT WITH TWO RANDOM VARIABLES 119
Since the limit state function g
.
S; Q
/
is a normally distributed random variable, the reliability
of the component can be calculated by the following equation if the Excel function is used:
R D P
Œ
g
.
S; Q
/
0
D 1 P
Œ
g
.
S; Q
/
< 0
D 1 NORM:DIST
0;
g
;
g
; true
: (3.9)
e reliability of the component can be calculated by the following equation if the MATLAB
command is used:
R D P
Œ
g
.
S; Q
/
0
D 1 P
Œ
g
.
S; Q
/
< 0
D 1 normcdf
0;
g
;
g
;
: (3.10)
If the standard normal distribution table is used, the reliability of the component can be calcu-
lated by the following equation:
R D P
Œ
g
.
S; Q
/
0
D 1 P
Œ
g
.
S; Q
/
< 0
D 1 ˆ
g
g
:
D ˆ
g
g
D ˆ
0
B
@
S
Q
q
2
S
C
2
Q
1
C
A
D ˆ
.
ˇ
/
: (3.11)
e term
g
g
or
S
Q
q
2
S
C
2
Q
will be frequently shown in a lot of formula or iterative process.
We can use ˇ to represent it, that is:
ˇ D
g
g
D
S
Q
q
2
S
C
2
Q
: (3.12)
ˇ is called the reliability index, which can be directly used to calculate the reliability, as shown
in Equation (3.11).
Example 3.5
e failure of a component will be a static failure of brittle material. According to the design
information, the component material ultimate strength S
u
follows a normal distribution with a
mean
S
u
D 62:5 (ksi) and a standard deviation
S
u
D 5:8 (ksi). e maximum normal stress
max
at the critical section follows a normal distribution with a mean
max
D 30:3 (ksi) and a
standard deviation
max
D 15:4 (ksi). Calculate the reliability of this component.
Solution:
In this example with a component of a brittle material, the component strength index is the
ultimate material strength S
u
and the component stress index is the maximum normal stress