184 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
Solution:
1. e limit state function.
For this rod made of a ductile material, the limit state function per Equation (4.12) is:
g
S
y
; d; F
a
D S
y
F
a
d
2
=4
D S
y
4F
a
d
2
D
8
ˆ
<
ˆ
:
> 0
Safe
0 Limit state
< 0 Failure:
(a)
In this example, F
a
is a discrete random variable and described by a PMF. e limit state
function of the rod can be expressed as the following three different limit state functions.
When F
a
D F
1
D 2:75 (klb), the limit state function of the rod is:
g
S
y
; d
D S
y
4F
1
d
2
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(b)
When F
a
D F
2
D 3:00 (klb), the limit state function of the rod is:
g
S
y
; d
D S
y
4F
2
d
2
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(c)
When F
a
D F
3
D 3:25 (klb), the limit state function of the rod is:
g
S
y
; d
D S
y
4F
3
d
2
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(d)
Per Equation (4.15), the reliability of the rod in this example will be:
R D p
1
R
1
C p
2
R
2
C p
3
R
3
; (e)
where p
1
, p
2
, and p
3
are the PMF for the axial loading F
a
when it is equal to F
1
D 2:75
(klb), F
2
D 3:00 (klb) and F
3
D 3:25 (klb), respectively. R
1
is the reliability of the rod
when F
a
D F
1
D 2:75 (klb), which is determined by the limit state function (b). R
2
is
the reliability of the rod when F
a
D F
2
D 3:00 (klb), which is determined by the limit
state function (c). R
3
is the reliability of the rod when F
a
D F
3
D 3:25 (klb), which is
determined by the limit state function (d).
In this example, the limit state functions (b), (c), and (d) contain only two random vari-
ables. e mean and the standard deviation of rod diameter can be calculated by Equa-
tion (4.1). e distribution parameters are listed in Table 4.17.