186 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
Table 4.19: e iterative results for R
2
by the H-L method for Equation (c)
Iterative#
S
y
*
d
*
β
*
|∆β
*
|
1 34.5 0.3327409 2.7990799
2 25.796855 0.3847976 2.3650156 0.4340643
3 27.131775 0.3752119 2.3478514 0.0171642
4 27.186969 0.3748308 2.3478244 2.702E-05
Table 4.20: e iterative results for R
3
by the H-L method for Equation (d)
Iterative#
S
y
*
d
*
β
*
|∆β
*
|
1 34.5 0.3463277 1.8251188
2 28.823684 0.3788977 1.6262435 0.1988753
3 29.43552 0.3749392 1.6230807 0.0031628
4 29.445978 0.3748726 1.6230798 8.925E-07
4.6.2 RELIABILITY OF A ROD UNDER AXIAL LOADING FOR A
DEFORMATION ISSUE
When deformation of a component under axial loading is more than the specified deformation,
the component is treated as a failure. e following equation can calculate the deformation of a
bar under axial loading:
ı D
Z
L
0
F
a
.
x
/
dx
EA.x/
; (4.17)
where F
a
.
x
/
is the axial loading function along a rod. A.x/ is the rod’s cross-section area func-
tion along the rod. E is Young’s modulus of the rod material. L is the length of the rod under
consideration. ı is the axial deformation of the rod segment with a length L. e limit state
function of a rod under axial loading for a deformation issue will be:
g
.
E; A; L; F
a
/
D
Z
L
0
F
a
.
x
/
dx
EA
.
x
/
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.18)
where is the specified deformation, which is typically treated as a deterministic value. All
other variables in Equation (4.18) have the same meanings as those in Equation (4.17).
4.6. RELIABILITY OF A ROD UNDER AXIAL LOADING 187
When a rod can be simplified as several rod segments in each of which axial loading and
cross-section area are constant, the deformation of the rod can be expressed as:
ı D
X
F
ai
L
i
EA
i
; (4.19)
where F
ai
, L
i
, and A
i
are axial loading, length, and cross-section area of the i th segment. en,
the limit state function of a rod under axial loading for a deformation issue will be
g
.
E; A
i
; L; F
ai
/
D
X
F
ai
L
i
EA
i
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(4.20)
e limit state function of a rod for a deformation issue needs to be determined by the actual
problem. For example, there is an allowable envelope dimension L
e
, as shown in Figure 4.6. e
A
L
P
1
P
2
L
e
B
C
Figure 4.6: Schematic of a rod with an envelope dimension L
e
.
design specification could be that the actual physical entire length L of the bar is not allowed to
over the envelope dimension L
e
, or that a gap must be maintained to be at least between the
actual physical entire length L and the envelope dimension L
e
. In such a case, the limit state
function of a rod under axial loading will be
g
.
L
e
; E; A
i
; L
i
; F
ai
/
D .L
e
L/
D L
e
X
L
i
C
F
ai
L
i
EA
i
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure;
(4.21)
where L
e
is the envelope dimension, which could be a random variable or a deterministic value.
One example L
e
is a distance to another component. is the gap between the actual physical en-
tire length L and the envelope dimension L
e
and will be a deterministic value.
P
L
i
C
F
ai
L
i
EA
i
is the entire physical length L of the rod.
188 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
e limit state functions (4.18) or (4.20) or (4.21) can be used to calculate the reliability
of a rod under axial loading for a deformation issue.
When axial loading is expressed by a PMF, as shown in Equation (4.14), we can use
the total probability theorem Equation (2.24) in Chapter 2 to calculate the reliability of the
components. e reliability of a rod under axial loading for a deformation issue will be the same
as Equation (4.16) and is shown here again:
R D
n
X
iD1
.
p
i
R
i
/
: (4.16)
In this case, R
i
can be calculated by the limit state functions (4.18), (4.20), or (4.21) when the
axial loading F
a
is equal to F
ai
. p
i
is the PMF when the axial loading is equal to F
ai
.
We will use two examples to demonstrate how to calculate the reliability of a rod under
axial loading for a deformation issue. Example 4.10 will be the reliability of a rod under axial
loading, which is described by a PDF. Example 4.11 will be the reliability of a rod under axial
loading, which is described by a PMF.
Example 4.10
A key component: the stepped round-solid bar ABC is subjected to two axial loadings, as shown
in Figure 4.7. e bar ABC is fixed at the right end. ere is another component (does not
show) near the left end of the bar. e dimensions and loadings for this example are listed in
Table 4.21. e Young’s modulus of the bar material follows a normal distribution with a mean
E
D 2:76 10
7
(psi) and a standard deviation
E
D 6:89 10
5
(psi). e design specification
is that the gap between the deformation of the entire bar and the envelope dimension must be
at least 0:003
00
. Use the Monte Carlo method to calculate the reliability of the bar and its range
with a 95% confidence level.
Table 4.21: e dimensions and loading for the Example 4.10
d
1
(in) d
2
(in) L
1
(in) L
2
(in) L
e
P
1
(lb) P
2
(lb)
0.750 ± 0.002 0.500 ± 0.002 5.000 ± 0.005 3.000 ± 0.005 8.010 ± 0.005 1300 ± 120 500 ± 50
Solution:
1. Axial loading and geometric parameters in each segment.
e stepped bar ABC can be divided into two segments. e loading and corresponding
geometric parameters for each segment are listed in Table 4.22.
e deformation of the entire bar per Equation (4.19) will be:
ı D
.P
1
C P
2
/L
1
Ed
2
1
=4
C
P
2
L
2
Ed
2
2
=4
D
4.P
1
C P
2
/L
1
Ed
2
1
C
4P
2
L
2
Ed
2
2
: (a)
4.6. RELIABILITY OF A ROD UNDER AXIAL LOADING 189
A
L
1
L
2
d
2
P
1
P
2
L
e
B
C
d
2
Figure 4.7: A stepped bar with an envelope dimension.
Table 4.22: e loading and corresponding geometric parameters for each segment
Segment AB Segment BC
Length Cross-section area Axial loading Length Cross-section area Axial loading
L
1
πd
1
2
⁄4
P
1
+ P
2
L
2
πd
2
2
⁄4
P
2
2. e limit state function of the stepped bar for the deformation issue.
e limit state function of this example can be established per Equation (4.20):
g
.
E; L
e
; L
1
; L
2
; d
1
; d
2
; P
1
; P
2
/
D L
e
L
1
C L
2
C
4.P
1
C P
2
/L
1
Ed
2
1
C
4P
2
L
2
Ed
2
2
0:003
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(b)
e geometric dimension and the loading can be treated as normal distributions. For ge-
ometric dimensions L
e
; L
1
; L
2
; d
1
, and d
2
, we can use Equation (4.1) to calculate their
means and standard deviations. e axial P
1
and P
2
can also be treated as normal distri-
butions. For the loadings P
1
and P
2
, we can use Equation (4.2) to calculate their means
and standard deviations. All distribution parameters of random variables in the limit state
function Equation (b) are listed in Table 4.23.
3. e reliability of the stepped bar.
e limit state function in this example contains eight normally distributed random vari-
ables. We will use the Monte Carlo method to calculate the reliability of this example. e
Monte Carlo method has been discussed in Section 3.8. We can follow the Monte Carlo
method and the program flowchart in Figure 3.8 to create a MATLAB program. Since
the stepped bar is a key component, we will use the trial number N D 1,598,400 from
190 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
Table 4.23: e distribution parameters of Example 4.10
E (psi) L
e
(in)
L
1
(in)
L
2
(in)
μ
E
σ
E
μ
L
e
σ
L
e
μ
L
1
σ
L
1
μ
L
2
σ
L
2
2.76×10
7
6.89×10
5
8.010 0.00125 5.000 0.00125 3.000 0.00125
d
1
(in) d
2
(in) P
1
(lb) P
2
(lb)
μ
d
1
σ
d
1
μ
d
2
σ
d
2
μ
P
1
σ
P
1
μ
P
2
σ
P
2
0.750 0.0005 0.500 0.0005 1300 30.0 500 12.5
Table 3.2 in Section 3.8. e estimated reliability of this bar R by the MATLAB program
is:
R D 0:9971: (c)
e estimated probability of the bar failure F is
F D 1 R D 1 0:9971 D 0:0029: (d)
e relative error of the probability of the failure is
" D 0:0295: (e)
So, the range of the probability of the bar failure with a 95% confidence level will be:
F D 0:0029 ˙ 0:0029 0:0295 D 0:0029 ˙ 0:00009: (f)
erefore, the range of the reliability of the bar with a 95% confidence level will be:
R D 1 F D 0:9971 ˙ 0:00009: (g)
Example 4.11
A key component: a stepped plate as shown in Figure 4.8 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 0:20 F
a
D F
1
D 3150 (lb)
p
2
D 0:75 F
a
D F
2
D 3200 (lb)
p
3
D 0:05 F
a
D F
3
D 3250 (lb):
(a)
e Young’s modulus of the plate material follows a normal distribution with a mean
E
D
2:76 10
7
(psi) and a standard deviation
E
D 6:89 10
5
(psi). e design specification is that
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