164 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.33: e iterative results for the strength issue in this problem
Iterative #
S
y
*
T
*
L
*
d
*
|∆d
*
|
1 34.5 4.5 24 1.814967
2 27.56354 4.58536 24.00025 1.968272 0.153306
3 27.44461 4.568076 24.0002 1.96863 0.000358
4 27.44438 4.568041 24.0002 1.968631 2.14E-07
e iterative results for the strength in this problem are listed in Table 3.33.
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 1:969
00
: (e)
e iterative results for the deformation issue in this problem are listed in Table 3.34.
Table 3.34: e iterative results for the strength issue in this problem
Iterative #
E
*
T
*
L
*
d
*
|∆d
*
|
1 27600 4.5 24 3.171039
2 26,530.6 4.715882 24.0019 3.263924 0.092885
3 26,480.34 4.707326 24.00191 3.264013 8.92E-05
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 3:265
00
: (f)
e diameter of the beam with the required reliability 0.99 under the specified loading and
deformation requirement will be the larger one of Equations (e) and (f). erefore, the diameter
of the beam is
d D 3:265 ˙ 0:005
00
:
3.3.6 COMPONENT UNDER STATIC COMBINED LOADING
e limit state function of a component and its reliability calculation under combined loading
for strength issue have been discussed in detail in Section 4.10 of Volume 1 [1]. After the limit
state function of a component under static combined loading is established, we can run the
3.3. DIMENSION OF A COMPONENT UNDER STATIC LOADING 165
dimension design with the required reliability. Now we will use examples to show how to conduct
component dimension design under combined loading.
Example 3.15
On the critical section of a circular shaft, the resultant internal torsion and an internal resultant
bending moment of a shaft are T D 2:5 ˙ 0:18 (klb.in) and M D 4:6 ˙ 0:34 (klb.in), as shown
in Figure 3.7. e yield strength S
y
of the shaft’s material follows a normal distribution with the
mean
S
y
D 34:5 (ksi) and the standard deviation
S
y
D 3:12 (ksi). Use the distortion energy
theory with the modified Monte Carlo method to design the diameter d of the shaft with the
required reliability 0.99 when the diameter d has a tolerance ˙0:005.
(a) Schematic of Bending
and Torque
(b) Stress Element
at Point A
(c) Stress Element
at Point B
z
B
A
x
x
y
y
M
y
T
x
τ
T
τ
T
σ
M
σ
M
Figure 3.7: Schematic of a segment of a shaft under combined stress.
Solution:
In this example, there is no dimension-dependent parameter.
(1) e Von-Mises stress.
As shown in Figure 3.7, the critical points on this critical section will be the points A
and B because there are the maximum values of bending stress. Stress elements at points A and
B are shown in Figures 3.7b,c, where
M
and
T
are the bending stress due to the bending
moment M
y
and the shear stress due to the torque T
x
, respectively. e Von Mises stress [1, 2]
at points A and B for the loading case in this example are the same. e point A is used to run
the calculation. At point A, we have:
x
D
M
D
32M
y
d
3
;
y
D
z
D 0;
xy
D
T
D
16T
x
d
3
;
yz
D
zx
D 0: (a)
e Von Mises in this case is:
von
D
s
32M
y
d
3
2
C 3
16T
x
d
3
2
D
16
d
3
q
4M
2
y
C 3T
2
x
: (b)
166 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
(2) e limit state function.
e limit state function of this example by using the distortion energy theory [1] is:
g
S
y
; T
x
; M
y
; d;
D S
y
16
d
3
q
4M
2
y
C 3T
2
x
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(c)
ere are four normally distributed random variables in this limit state function. e mean and
standard deviations of the bending moment and the torque can be calculated per Equation (1.2).
e standard deviation of the diameter d can be calculated per Equation (1.1). eir distribution
parameters for this example are listed in Table 3.35.
Table 3.35: e distribution parameters for the limit state function (c)
S
y
(psi) T
x
(klb.in) M
y
(klb.in) d (in)
μ
S
y
σ
S
y
μ
T
x
σ
T
x
μ
M
y
σ
M
y
μ
d
σ
d
34.5 3.12
2.5 0.045 4.6 0.085
μ
d
0.00125
(3) Use the modified H-L method to determine the dimension.
Following the modified H-L method procedure discussed in Section 3.2.3 and the
flowchart shown in Figure 3.2, we can make a MATLAB program for the limit state func-
tion (c).
e iterative results for this problem are listed in Table 3.36.
Table 3.36: e iterative results for the strength issue in this problem
Iterative #
S
y
*
T
x
*
M
y
*
d
*
|∆d
*
|
1 34.5 2.5 4.6 1.144977
2 27.35034 2.503722 4.632582 1.239629 0.094652
3 27.30947 2.502936 4.625846 1.23973 0.000102
4 27.30941 2.502938 4.625835 1.239731 1.2E-07
According to the result obtained from the program, the mean of the diameter with the
reliability 0.99 is
d
D 1:240
00
: (d)
3.3. DIMENSION OF A COMPONENT UNDER STATIC LOADING 167
erefore, the diameter of the shaft with the required reliability 0.99 under the specified loading
is
d D 1:240 ˙ 0:005
00
:
Example 3.16
Schematic of a thin-cylindrical vessel is depicted in Figure 3.8. e vessel has an inner diameter
d D 50
00
˙ 0:125
00
. e internal pressure is p D 350 ˙30 (psi). e vessel material is ductile. e
yield strength S
y
of this material follow a normal distribution with a mean
S
y
D 34;500 (psi)
and a standard deviation
S
y
D 3120 (psi). Use the maximum shear stress (MSS) theory with
the modified H-L method to design the thickness t of the vessel with a reliability 0.999 when
the thickness d has a tolerance ˙0:030
00
.
(a) Schematic of a in-Wall Vessel
(b) Stress Element of a Critical Point
σ
h
σ
l
p
Figure 3.8: Schematic of a thin-wall cylindrical vessel.
Solution:
In this example, there is no dimension-dependent parameter.
(1) e maximum shear stress.
Stress element of this vessel at the critical point is shown in Figure 3.8b, where p is the
internal pressure,
l
is the longitudinal normal stress and
h
is the normal stress in the hoop
direction.
l
and
h
can be calculated by using the following equation:
h
D
pD
2d
;
l
D
pD
4d
; (a)
where D is the inner diameter of the vessel, and d is the wall thickness.
Since
h
,
l
, and p are three principal stresses and are arranged as
h
>
l
> p in this
case, the maximum shear stress [1, 2] in this case will be:
max
D
h
C p
2
D p
D
4d
C
1
2
: (b)
168 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
(2) e limit state function of the vessel.
e limit state function of this vessel by using the maximum shear stress theory [1] is
g
S
y
; p; D; d
D
S
y
2
p
D
4d
C
1
2
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(c)
ere are four normally distributed random variables in this limit state function. e mean
and standard deviation of the internal pressure p can be determined per Equation (1.2). e
mean and standard deviation of the inner diameter D can be determined per Equation (1.1).
e standard deviation of the wall thickness d can be determined per Equation (1.1). eir
distribution parameters are listed in Table 3.37.
Table 3.37: e distribution parameters of random variables in Equation (c)
S
y
(psi) p (psi) D (in) d (in)
μ
S
y
σ
S
y
μ
p
σ
p
μ
D
σ
D
μ
d
σ
d
34,500 3,120 350 7.5 50 0.03125 50
0.0075
(3) Use the modified H-L method to determine the dimension.
Following the modified H-L method procedure discussed in Section 3.2.3 and the
flowchart shown in Figure 3.2, we can compile a MATLAB program for the limit state func-
tion (c). e iterative results for this problem are listed in Table 3.38.
Table 3.38: e iterative results for Example 3.16
Iterative #
S
y
*
p
*
D
*
d
*
|∆d
*
|
1 34,500 350 50 0.256223
2 25,543.6 355.1015 50.00061 0.352448 0.096226
3 25,132.85 353.8936 50.00047 0.357054 0.004606
4 25,122.04 353.8485 50.00047 0.357164 0.00011
5 25,121.78 353.8474 50.00047 0.357166 2.68E-06
According to the result obtained from the program, the mean of the thickness with a
reliability 0.999 is
d
D 0:358
00
: (d)
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