3.3. DIMENSION OF A COMPONENT UNDER STATIC LOADING 157
erefore, the diameter d of the double-shear pins at the points B and C with the required
reliability 0.99 under the specified loading is
d D 0:243
00
˙ 0:005
00
: (f)
3.3.4 SHAFT UNDER STATIC TORSION LOADING
e limit state function of a component and its reliability calculation under torsion for strength
issue and deformation issue have been discussed in detail in Section 4.8 of Volume 1 [1]. After
the limit state function of a component under static torsion loading is established, we can run
the dimension design with the required reliability. Now we will use examples to show how to
conduct component dimension design.
Example 3.11
e solid shaft is subjected to a torque T which follows a uniform distribution between 8.5
(klb.in) and 12.50 (klb.in). e shear yield strength of the shaft material follows a normal dis-
tribution with a mean
S
sy
D 32:2 (ksi) and the standard deviation
S
sy
D 3:63 (ksi). Use the
modified R-F method to design the diameter of the shaft with the required reliability 0.99 when
shaft diameter tolerance is ˙0:005.
Solution:
In this example, there is no dimension-dependent parameter.
(1) Limit state function.
e shear stress induced by the torque T is
D
Td=2
J
D
Td=2
d
4
=32
D
16T
d
3
: (a)
e limit state function of the shaft for this problem is
g
T; S
sy
; d
D S
sy
16T
d
3
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(b)
In the limit state function (b), there are three random variables. T follows a uniform distribu-
tion. e standard deviation of the shaft diameter
d
can be calculated per Equation (1.1). eir
distribution parameters are listed in Table 3.25.
(2) Use the modified R-F method to determine the dimension.
158 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.25: Distribution parameters for the limit state function (d)
S
ys
(klb)
Normal Distribution
T (klb)
Uniform Distribution
d (in)
Normal Distribution
μ
S
ys
μ
S
ys
a b μ
d
σ
d
32.2 3.63 8.5 12.5
μ
d
0.00125
We can follow the modified R-F method procedure discussed in Section 3.2.4 and the
flowchart shown in Figure 3.3 to compile a MATLAB program. e iterative results are listed
in Table 3.26.
Table 3.26: e iterative results of Example 3.11 by the modified R-F method
Iterative #
T
*
S
y
*
d
*
|∆d
*
|
1 10.5 32.2 1.184226
2 11.57505 30.38601 1.247211 0.062985
3 11.35539 29.91543 1.245735 0.001476
4 11.38899 30.03348 1.245327 0.000408
5 11.38338 30.01924 1.245319 7.52E-06
According to the result obtained from the program, the mean of the diameter with a
reliability 0.99 is
d
D 1:246
00
: (d)
erefore, the diameter of the shaft with the required reliability 0.99 under the specified loading
will be
d D 1:246 ˙ 0:005
00
:
Example 3.12
e solid shaft with a constant cross-section and a length L D 23:75
00
˙ 0:032
00
is subjected to a
torque T . Per design specification, the torque follows a normal distribution with a mean
T
D
4:5 (klb.in) and a standard deviation
T
D 0:39 (klb.in). e shear yield strength of the shaft
material follows a normal distribution with a mean
S
sy
D 32:2 (ksi) and a standard deviation
S
sy
D 3:63 (ksi). e shear Young’s modulus follows a normal distribution with the mean
G
D
1:117 10
4
(ksi) and the standard deviation
G
D 2:793 10
2
(ksi). e allowable angle of twist
of the shaft is 4
ı
. Use the modified Monte Carlo method to design the diameter of the shaft
with the required reliability 0.99 when shaft diameter tolerance is ˙0:005.
3.3. DIMENSION OF A COMPONENT UNDER STATIC LOADING 159
Solution:
In this example, there is no dimension-dependent parameter.
(1) e limit state functions.
e shear stress induced by the torque T is
D
Td=2
J
D
Td=2
d
4
=32
D
16T
d
3
: (a)
e limit state function of the shaft for the strength issue is
g
S
sy
; T; d
D S
sy
16T
d
3
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(b)
e angle of twist of the shaft due to the torque is
D
TL
GJ
D
TL
Gd
4
=32
D
32TL
Gd
4
: (c)
e allowable angle of twist 4
ı
is equal to 0.069813 radians. e limit state function of the shaft
for the deformation issue is
g
.
G; T; L; d
/
D 0:069813
32TL
Gd
4
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(d)
In these two limit state functions, there are five normally distributed random variables. e mean
and the standard deviation of normally distributed L can be determined per Equation (1.1).
e standard deviation of normally distributed d can be determined per Equation (1.1). eir
distribution parameters are listed in Table 3.27.
Table 3.27: Distribution parameters for Example 3.12
S
y
(ksi) G (ksi) T (klb.in) L (in) d (in)
μ
S
y
σ
S
y
μ
G
σ
G
μ
T
σ
T
μ
L
σ
L
μ
d
σ
d
34.5 3.12
1.117×10
4
2.793×10
2
4.5 0.39 23.75 0.008
μ
d
0.00125
(2) Use the Monte Carlo method to determine the mean
d
of the diameter d .
160 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.28: e iterative results of Example 3.12 for the strength issue
Iterative #
μ
d
*
R
*
∆R
*
1 0.945022 0.969788 -0.02021
2 0.946022 0.971405 -0.01859
… … … …
20 0.964022 0.990068 6.77E-05
21 0.965022 0.99066 0.00066
Following the modified Monte Carlo method procedure discussed in Section 3.2.5 and
the flowchart shown in Figure 3.5, we can make MATLAB programs for two limits state func-
tions (c) and (d).
e iterative results for the limit state function (b) are listed in Table 3.28.
According to the result for the strength issue obtained from the program, the mean of the
diameter with a reliability 0.99 is
d
D 0:966
00
: (e)
e iterative results for the limit state function (d) are listed in Table 3.29.
Table 3.29: e iterative results of Example 3.12 for the deformation issue
Iterative #
μ
d
*
R
*
∆R
*
1 1.104389 0.765295 -0.22471
2 1.105389 0.778078 -0.21192
… … … …
37 1.140389 0.989331 -0.00067
38 1.141389 0.990534 0.000534
According to the result for the deformation issue obtained from the program, the mean
of the diameter with a reliability 0.99 is
d
D 1:142
00
: (f)
erefore, the diameter of the shaft with the required reliability 0.99 under the specified loading
and the deformation requirement will be the larger one of Equations (e) and (f). It is
d D 1:142 ˙ 0:005
00
:
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