4.6. RELIABILITY OF A ROD UNDER AXIAL LOADING 183
tion 3.7. We can follow the procedure in Section 3.7 and the flowchart in Figure 3.7 to
compile a MATLAB program for this example. e iterative resulted are listed in Ta-
ble 4.16. From the iterative results, the reliability index ˇ and corresponding reliability R
of the plate in this example are:
ˇ D 1:4607 R D ˆ
.
1:4607
/
D 0:9280:
Table 4.16: e iterative results of Example 4.8 by the R-F method
Iterative#
S
u
*
K
t
*
d
*
h
*
t
*
P
*
β
*
|∆β
*
|
1
61.838053 2.36 0.75 2.5 0.25 11.463622 0.9259472
2
12.346059 2.3675912 0.7500011 2.499992 0.2499678 2.2810853 1.4606672 0.53472
3
12.17867 2.3665355 0.750001 2.4999931 0.2499722 2.251203 1.4607128 4.558E-05
Example 4.9
A circular rod as shown in Figure 4.5 is subjected to an axial loading F
a
which is defined by a
PMF:
P
.
F
a
D F
i
/
D
8
ˆ
<
ˆ
:
p
1
D 0:10 F
a
D F
1
D 2:75 klb
p
2
D 0:85 F
a
D F
2
D 3:00 klb
p
3
D 0:05 F
a
D F
3
D 3:25 klb:
(a)
e diameter of the rod d D 0:375 ˙0:005
00
. e rod material is ductile. e yield strength S
y
of the rod material follows a normal distribution with a mean
S
y
D 34:5 (ksi) and a standard
deviation
S
y
D 3:12 (ksi). Calculate the reliability of the rod.
A B
F
a
Figure 4.5: A rod under an axial loading.
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.
4.6. RELIABILITY OF A ROD UNDER AXIAL LOADING 185
Table 4.17: e distribution parameters of random variables in Equations (b), (c), and (d)
S
y
(ksi)
d (in)
μ
S
y
σ
S
y
μ
d
σ
d
34.5 3.12 0.375 0.00125
2. Reliability of the rod under axial loading F
a
.
e limit state functions (b), (c), and (d) in this example contains two normally distributed
random variables and is a nonlinear function. We can follow the H-L method discussed
in Section 3.6 and the program flowchart in Figure 3.6 to create a MATLAB program.
e iterative results for the limit state function (b) are listed in the following tables. e
reliability index ˇ and corresponding reliability R
1
is:
ˇ D 3:07292 R D ˆ
.
3:07292
/
D 0:99894:
Table 4.18: e iterative results for R
1
by the H-L method for Equation (b)
Iterative#
S
y
*
d
*
β
*
|∆β
*
|
1 34.5 0.3185751 3.9023328
2 22.370303 0.3956266 3.136871 0.7654618
3 24.722992 0.3763318 3.0733175 0.0635535
4 24.924507 0.3748074 3.0729205 0.0003971
5 24.92607 0.3747956 3.0729205 2.357E-08
e iterative results for R
2
of the rod when F
a
D F
2
D 3:0 (klb) are listed in Table 4.19.
From the iterative results, the reliability index ˇ and corresponding reliability R
2
is
ˇ D 2:3478 R D ˆ
.
2:3478
/
D 0:99056:
e iterative results for R
3
of the rod when F
a
D F
3
D 3:25 (klb) are listed in Table 4.20.
From the iterative results, the reliability index ˇ and corresponding reliability R
3
is:
ˇ D 1:6231 R D ˆ
.
1:6231
/
D 0:94771:
Per Equation (e) for this example, the reliability of the rod will be:
R D p
1
R
1
C p
2
R
2
C p
3
R
3
D 0:10 0:99894 C 0:85 0:99056 C 0:05 0:94771 D 0:9893:
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