4.7. RELIABILITY OF A COMPONENT UNDER DIRECT SHEARING 193
Table 4.25: e distribution parameters of the Example 4.11 for Equations (c), (d), and (e)
E (psi)
L
1
(in) L
2
(in) h
1
(in) h
2
(in) t (in)
μ
E
σ
E
μ
L
1
σ
L
1
μ
L
2
σ
L
2
μ
h
1
σ
h
1
μ
h
2
σ
h
2
μ
t
σ
t
2.76E7 6.89E5 10.000 7.5E-4 7.000 7.5E-4 1.000 7.5E-4 0.50 7.5E-4 0.375 7.5E-4
discussed in Section 3.8 and the program flowchart in Figure 3.8 to create a MATLAB
program. e estimated reliability of this component R is:
R D 0:9981:
4.7 RELIABILITY OF A COMPONENT UNDER DIRECT
SHEARING
Direct shearing refers that the failure of a component is due to direct shearing stress on the
shearing-off surface where the bending moment is small and can be negligible. Figure 4.9a
shows a bolted joint. e shearing-off section is shown in Figure 4.9b. In this case, the bending
moment on the shearing off section is small and can be negligible. is case can be called as a
single shear because there is one shearing off section per shearing-off item.
(a) A Bolted Joint
(b) The Shearing Off Section
P
P
P
V V
V V
Figure 4.9: Schematic of a bolted joint.
Another direct shearing example is a joint connection, as shown in Figure 4.10a. Shearing
off sections of the pin is shown in Figure 4.10b. In this case, the bending moment on the shearing
194 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
(a) A Pin Connection
(b) The Pin’s Shearing Off Section
P
P
P
V
V
Figure 4.10: Schematic of a pin connection.
off section is small and can be negligible. e pin has two shearing-off sections in this case. It
is also called as double shearing because there are two shearing-off sections per pin item.
For a component under direct shearing, the failure is due to shearing stress. When the
shear stress is more than the shearing strength such as the shearing yield strength for ductile
material and the ultimate shearing strength for brittle material, the component under direct
shearing is treated as a failure. e limit state function of a component made from a ductile
material under direct shearing is:
g
S
sy
; A; V
D S
sy
V
A
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.22)
where S
sy
is material shear yield strength. V is the shear force on a shearing-off section. A is the
area of a shearing-off section.
e limit state function of a component made from a brittle material under direct shearing
is:
g
.
S
su
; A; V
/
D S
su
V
A
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.23)
where S
su
is material shear ultimate strength. V and A have the same meaning as those in Equa-
tion (4.22).
4.7. RELIABILITY OF A COMPONENT UNDER DIRECT SHEARING 195
When the shear force on a direct shearing section is described by a PMF, we can use
the total probability theorem Equation (2.24) in Chapter 2 to calculate the reliability of the
components. In this case, the shearing force is expressed by a PMF as:
P
.
V
D
V
i
/
D p
i
i D 1; 2; : : : ; n; (4.24)
where p
i
is the PMF when the shearing force V is equal to V
i
. Since it is a PMF, the reliability
of the component under such the shearing force will be:
R D
n
X
iD1
.
p
i
R
i
/
; (4.25)
where R is the reliability of component under such shearing forces. R
i
is the reliability of a
component when the direct shearing force V is equal to V
i
. Even V
i
can be a constant value; we
still need to use Equations (4.22) or (4.23) to calculate R
i
because other variables in the limit
state function are still random variable.
We will use two examples of direct shearing to demonstrate how to calculate the reliability
of a component under direct shearing.
Example 4.12
A bar under a tensile loading P D 0:90 ˙ 0:0:08 (klb) is connected to the ground supporter
by a single shearing pin. e pin material is brittle. e ultimate shear strength S
su
(ksi) of
the pin follows a log-normal distribution with the distribution parameters
ln S
su
D 3:25 and
ln S
su
D 0:181. e pin has a diameter d D 0:25 ˙ 0:005
00
. Calculate the reliability of this pin.
Solution:
1. e limit state function of the pin.
e limit state function of this single shearing pin per Equation (4.22) will be:
g
.
S
su
; d; P
/
D S
su
P
d
2
=4
D S
su
4P
d
2
D
8
ˆ
<
ˆ
:
> 0 Safe
0
Limit state
< 0 Failure:
(a)
e loading can be simplified as a normal distribution per Equation (4.2). e pin diameter
will be treated as a normal distribution per Equation (4.1). e type of distributions and
corresponding distribution parameters of random variables in the limit state function (a)
are listed in Table 4.26.
2. e reliability of the single shearing pin.
ere are three random variables in the limit state function (a). One is a lognormal distribu-
tion, and two are normal distributions. We can follow the procedure of the R-F method in
196 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
Table 4.26: e distribution parameters of the limit state function (a) for Example 4.12
Ultimate shear strength S
su
(ksi)
Lognormal distribution
Diameter d (in)
Normal distribution
Loading P (klb)
Normal distribution
μ
lnS
su
σ
lnS
su
μ
d
σ
d
μ
P
σ
P
3.25 0.181 0.25 0.00125 0.90 0.02
Section 3.7 and the flowchart in Figure 3.7 to compile a MATLAB program. e iterative
results are listed in Table 4.27. From the table, the reliability index ˇ and corresponding
reliability R of the pin in this example are:
ˇ D 1:8683 R D ˆ
.
1:8683
/
D 0:9691:
Table 4.27: e iterative results of Example 4.12 by the R-F method
Iterative#
S
*
s
u
d
*
P
*
β
*
|∆β
*
|
1 26.21628 0.25 1.286889 1.562365
2 18.41154 0.249893 0.902998 1.868327 0.305963
3 18.44584 0.249872 0.904531 1.868335 7.81E-06
Example 4.13
e beam BCD is subjected to a concentrated load P D 2:01 ˙ :20 (klb), and supported by two
vertical bars AB and DE, as shown in Figure 4.11. e bar AB is connected to the supporter
through a double shearing pin, as shown in Figure 4.11. e pin material is ductile. Its diameter
is d D 3=16 ˙ 0:005
00
. e shear yield strength follows a normal distribution with a mean
S
sy
D
32:2 (ksi) and a standard deviation
S
sy
D 3:63 (ksi). Use the Monte Carlo method to calculate
the reliability of the double shearing pin and its range with a 95% confidence level.
Solution:
1. e shearing force on a shearing-off section.
e loading on the bar AB is F
AB
D PL
1
=L
2
. Since the pin A is a double shear pin, the
shear force on a shearing-off section will be F
AB
=2. So, the shearing force V will be:
V D
F
AB
2
D
PL
1
2L
2
: (a)
4.7. RELIABILITY OF A COMPONENT UNDER DIRECT SHEARING 197
Pin A
B
A E
P
C
L
1
= 24˝ ± 1/16˝
L
2
= 36˝ ± 1/16˝
D
Figure 4.11: A beam with a concentrated load P .
2. Limit state function of the double shear pin A.
For a pin made of ductile material, its limit state function per Equation (4.22) is:
g
S
sy
; d; L
1
; L
2
; P
D S
sy
2PL
1
L
2
d
2
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(b)
ere are five random variables in the limit state function (b). For geometric dimensions,
they can be treated as a normal distribution per Equation (4.1). For the loading, it can
be treated as a normal distribution per Equation (4.2). All these five random variables are
normal distributions. eir distribution parameters are listed in Table 4.28.
Table 4.28: e distribution parameters of the limit state function (b) for Example 4.13
S
sy
(ksi)
d (in) L
1
(in) L
2
(in) P (klb)
μ
S
sy
σ
S
sy
μ
d
σ
d
μ
L
1
σ
L
1
μ
L
2
σ
L
2
μ
P
σ
P
32.2 3.63 0.1875 0.00125 24 0.015625 36 0.015625 2.01 0.05
3. e reliability of the double shearing pin A.
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 limit
state function is not too complicated, we will use the trial number N D 1,598,400 from
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