192 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
In this example, the axial loading F is a discrete random variable and described by a PMF.
e limit state function of the stepped plate can be expressed as the following three dif-
ferent limit state functions.
When F D F
1
D 3150 (lb), the limit state function of the stepped plate is:
g
.
E; L
1
; L
2
; h
1
; h
2
; t
/
D 0:008
3150
.
L
1
h
2
C L
2
h
1
/
Eh
1
h
2
t
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(c)
When F
a
D F
2
D 3200 (lb), the limit state function of the stepped plate is:
g
.
E; L
1
; L
2
; h
1
; h
2
; t
/
D 0:008
3200
.
L
1
h
2
C L
2
h
1
/
Eh
1
h
2
t
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(d)
When F
a
D F
3
D 3250 (lb), the limit state function of the stepped plate is:
g
.
E; L
1
; L
2
; h
1
; h
2
; t
/
D 0:008
3250
.
L
1
h
2
C L
2
h
1
/
Eh
1
h
2
t
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(e)
Per Equation (4.16), the reliability of the plate in this example will be:
R D p
1
R
1
C p
2
R
2
C p
3
R
3
; (f )
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
3150 (lb), F
2
D 3200 (lb), and F
3
D 3250 (lb), respectively. R
1
is the reliability of the
stepped plate determined by the limit state function (c). R
2
is the reliability of the stepped
plate determined by the limit state function (d). R
3
is the reliability of the stepped plate
determined by the limit state function (e).
e geometric dimensions and the loading can be treated as normal distributions. For
geometric dimensions L
1
; L
2
; h
1
; h
2
, and t , we can use Equation (4.1) to calculate their
means and standard deviations. All distribution parameters of random variables in the
limit state function Equations (c), (d), and (e) are displayed in Table 4.25.
3. e reliability of the stepped plate.
We will use the Monte Carlo simulation method to calculate the reliability R
1
, R
2
,
and R
3
of this example. en, we can use Equation (f) to calculate the reliability of the
stepped plate. Since the stepped bar is a key component, we will use the trial number
N D 1,598,400 from Table 3.2 in Section 3.8. We can follow the Monte Carlo method