4.10. RELIABILITY OF A COMPONENT UNDER COMBINED STRESSES 215
4.10 RELIABILITY OF A COMPONENT UNDER
COMBINED STRESSES
4.10.1 RELIABILITY OF A COMPONENT OF DUCTILE MATERIAL
UNDER COMBINED STRESSES
e maximum-shear-stress (MSS) theory and the distortion-energy (DE) theory are two widely
accepted failure theory for a component of a ductile material when it is subjected to combined
stresses due to static loadings [5]. ese failure theories can be used to establish the limit state
function of a component under combined stresses due to static loading.
e MSS theory predicts that yielding begins whenever the maximum shear stress in
any element equals or exceeds the maximum shear stress in a tensile-test specimen of the same
material when that specimen begins to yield. According to the MSS theory for a ductile material,
the limit state function of a component under combined stress due to the static loadings is:
g
S
y
;
max
D
S
y
2
max
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.41)
where S
y
is the yield strength and
max
is the maximum shear stress at the critical point.
max
will be determined per case and will be the functions of static loading and component geometric
dimensions.
If three principal stress
1
,
2
, and
3
at the critical point are known and are arranged in
such a way
1
2
3
; the maximum shear stress in such case will be:
max
D
.
1
3
/
2
: (4.42)
If the stress status of a component at the critical point is plane stress and when two normal stress
x
,
y
and shear stress
xy
are known, the two principal stress
A
and
B
can be calculated and
are arranged in such a way
A
B
:
A
D
x
C
y
2
C
r
x
y
2
2
C
2
xy
B
D
x
C
y
2
r
x
y
2
2
C
2
xy
:
(4.43)
For a plane stress,
A
and
B
are two principal stresses. Another principal stress is 0. e maxi-
mum shear stress for such case will be:
max
D
8
ˆ
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
ˆ
:
A
2
when
A
B
0
A
B
2
when
A
0
B
B
2
when 0
A
B
:
(4.44)
216 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
e DE theory predicts that yielding occurs when the distortion strain energy per unit
volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension
or compression of the same material. According to the DE theory for a ductile material, the limit
state function of a component under combined stresses due to static loadings is:
g
S
y
;
von
D S
y
von
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.45)
where S
y
is the yield strength and
von
is the Von-Mises stress of the component at the critical
point.
von
will be determined per case and will be the function of static loadings and component
geometric dimensions
If three principal stresses
1
,
2
, and
3
at the critical point are known, the Von-Mises
stress
von
is:
von
D
s
.
1
2
/
2
C
.
2
3
/
2
C
.
3
1
/
2
2
: (4.46)
If the stress status of the critical point is known, the Von-Mises stress
von
is:
von
D
s
x
y
2
C
y
z
2
C
.
z
x
/
2
C 6.
2
xy
C
2
yz
C
2
zx
/
2
: (4.47)
Limit state functions (4.41) and (4.45) can be used to calculate the reliability of a component of
a ductile material under combined stresses due to static loading.
We will use three examples to demonstrate how to calculate the reliability of components
of a ductile material at the critical point under combined stress due to static loadings.
Example 4.20
A segment of a solid shaft with a diameter d D 1:125 ˙ 0:005
00
is shown in Figure 4.18. e
resultant internal torsion and the resultant internal bending moment of a shaft at the critical
section are T D 1400 ˙ 120 (lb/in), and M D 3500 ˙ 450 (lb/in). e yield strength S
y
of the
shaft’s material follows a normal distribution with the mean
S
y
D34,500 (ksi) and the standard
deviation
S
y
D 3120 (ksi). Calculate the reliability of the shaft by using the DE theory.
Solution:
1. e Von-Mises stress.
As shown in Figure 4.18, the critical points on this critical section are point A and point B
because there are the maximum values of bending stress. Stress elements at points A and B
are shown in Figure 4.18b and c, where
M
and
T
are the bending stress due to the bending
moment M and the shear stress due to the torque T , respectively. Per Equation (4.46), the
4.10. RELIABILITY OF A COMPONENT UNDER COMBINED STRESSES 217
(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 4.18: Schematic of a segment of a shaft under combined stress.
Von-Mises stress at points A and B in this example are the same. e point A is used as
an example 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)
Per Equation (4.46), we have:
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)
2. e limit state function.
By using the DE theory, the limit state function of this example per Equation (4.44) is:
g
S
y
; d; T
x
; M
y
D S
y
16
d
3
q
4M
2
y
C 3T
2
x
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(c)
ere are four random variables in this limit state function. e diameter d can be treated
as a normal distribution. Its distribution parameters can be determined per Equation (4.1).
e bending moment and the torque will be treated as normal distributions. eir distri-
bution parameters can be determined per Equation (4.2). eir distribution parameters
are listed in Table 4.40.
3. e reliability of the shaft under combined stress.
e limit state function (c) is a nonlinear equation with four normal distributed ran-
dom variable. 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 are listed in
218 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
Table 4.40: e distribution parameters of random variables in Equation (c)
S
y
(psi)
d (in)
T
x
(lb/in)
M
y
(lb/in)
μ
S
y
σ
S
y
μ
d
σ
d
μ
T
x
σ
T
x
μ
M
y
σ
M
y
34500 3120 1.125 0.00125 1400 30 3500 90
Table 4.41. From the iterative results, the reliability index ˇ and corresponding reliability
R of this shaft are:
ˇ D 2:51570 R D ˆ
.
2:51570
/
D 0:99406:
Table 4.41: e iterative results of Example 4.20 by the H-L method
Iterative #
S
y
*
d
*
T
x
*
M
y
*
β
*
|∆β
*
|
1 34500 1.125 1400 4667.651 2.539208
2 26737.03 1.124885 1401.118 3533.743 2.515704 0.023504
3 26800.86 1.124912 1401.433 3543.36 2.515703 1.04E-06
Example 4.21
Schematic of a thin-cylindrical vessel is depicted in Figure 4.19. e vessel has an inner diame-
ter d D 48
00
˙ 0:125
00
and a thickness t D 3=8
00
˙ 0:060
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 distri-
bution with the mean
S
y
D 34500 (psi) and the standard deviation
S
y
D 3120 (psi). Calculate
the reliability of this vessel by using the MSS theory.
Solution:
1. e maximum shear stress.
Stress element of this vessel at the critical point is shown in Figure 4.19b, 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
2t
;
l
D
pd
4t
; (a)
4.10. RELIABILITY OF A COMPONENT UNDER COMBINED STRESSES 219
(a) Schematic of a in-Wall Vessel
(b) Stress Element of a Critical Point
σ
h
σ
l
p
Figure 4.19: Schematic of a thin-wall cylindrical vessel.
where p is the internal pressure. d is the inner diameter of the vessel, and t 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 per Equation (4.41) will be:
max
D
h
C p
2
D p
d
4t
C
1
2
: (b)
2. e limit state function of the thin-cylindrical vessel.
Per Equation (4.41), the limit state function of this vessel by using the maximum shear
stress theory is
g
S
y
; t; p; d
D
S
y
2
p
d
4t
C
1
2
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(c)
ere are four random variables in this limit state function. e geometric dimensions d
and t can be treated as a normal distribution. eir distribution parameters can be deter-
mined per Equation (4.1). e internal pressure p can be treated as a normal distribution
too. Its distribution parameter can be determined per Equation (4.2). eir distribution
parameters are listed in Table 4.42.
Table 4.42: e distribution parameters of random variables in Equation (c)
S
y
(psi) t (in) p (psi) d (in)
μ
S
y
σ
S
y
μ
t
σ
t
μ
p
σ
p
μ
d
σ
d
34500 3120 0.375 0.015 350 7.5 48 0.03125
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