222 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
e limit state function (c) is a nonlinear equation with four normal distributed ran-
dom variables. 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
Table 4.46. e reliability index ˇ and corresponding reliability R of this component are:
ˇ D 2:65702 R D ˆ
.
2:65702
/
D 0:99606:
Table 4.46: e iterative results of Example 4.22 by the H-L method
Iterative #
S
y
*
σ
x
*
σ
y
*
τ
x
*
y
β
*
|∆β
*
|
1 50.5 15.2 17.5 19.24162 2.885932
2 42.52771 16.21553 21.07022 10.22716 2.642539 0.243393
3 43.32856 16.22218 21.09359 11.0102 2.656992 0.014453
4 43.28253 16.22313 21.09694 10.96119 2.656799 0.000193
5 43.28351 16.22336 21.09776 10.96125 2.656979 0.00018
6 43.28302 16.22343 21.09801 10.96049 2.65702 4.04E-05
4.10.2 RELIABILITY OF A COMPONENT OF BRITTLE MATERIAL
UNDER COMBINED STRESSES
e maximum normal stress (MNS) theory and the Brittle Coulomb–Mohr (BCM) theory are
two failure theory for a component of brittle material when it is subjected to combined stresses
due to static loadings [5].
e MNS theory states that failure occurs whenever one of the three principal stress equals
to or exceeds the strength.
1
,
2
, and
3
are three principal stresses at the critical point which
are arranged in such a way
1
2
3
. According to the MNS theory for brittle materials,
the limit state functions of a component under combined stresses due to static loadings are as
follows.
When
1
2
3
0,
g
.
S
ut
;
1
/
D S
ut
1
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.48)
where S
ut
is the ultimate tensile strength of brittle material.
1
will be determined per case and
is the function of loading and component geometric dimensions.
4.10. RELIABILITY OF A COMPONENT UNDER COMBINED STRESSES 223
When 0
1
2
3
,
g
.
S
uc
;
3
/
D S
ut
j
3
j
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.49)
where S
uc
is the ultimate compression strength of brittle material.
3
will be determined per case
and is the function of loading and component geometric dimensions.
e MNS theory is not recommended to be used when the principal stresses at the critical
point have both tensile stress and compressive stress.
BCM theory is a failure theory for a component of brittle material under plane stress due
to static loading. Let
A
and
B
represent two principal stresses of a component of brittle mate-
rial at the critical point under a plane stress condition. Let also assume
A
B
. According to
the BCM theory and the different conditions of
A
and
B
, which can be calculated per Equa-
tion (4.43), three different versions of limit state functions can be established for a component
of brittle material in a plane stress condition.
When
A
B
0,
g
.
S
ut
;
A
/
D S
ut
A
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(4.50)
When
A
0
B
,
g
.
S
ut
; S
uc
;
A
;
b
/
D 1
A
S
ut
C
j
B
j
S
uc
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(4.51)
When 0
A
B
,
g
.
S
uc
;
B
/
D S
uc
j
B
j
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(4.52)
e limit state functions (4.48)–(4.52) can be used to calculate the reliability of a compo-
nent of brittle material under combined stresses due to static loadings.
We will use two examples to demonstrate how to calculate the reliability of a component
of brittle material under combined stresses.
224 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
Example 4.23
A schematic of the cross-section of a column is subjected to a compressive force and bending
moment as shown in Figure 4.21a, where the tensile force F
x
is along the x-axis and through the
centroid of the cross-section, and the bending moment M
z
is about the neutral z-axis. e tensile
force is F
x
D 800 ˙ 60 (lb). e bending moment is M
z
D 4000 ˙ 320 (lb/in). e cross-section
of the column is shown in Figure 4.21b, where L D 1:25 ˙ 0:010
00
and t D 0:25 ˙ 0:010
00
. e
column is made of brittle material. Its ultimate tensile strength S
ut
follows a normal distribu-
tion with a mean
S
ut
D 22,000 (psi) and standard deviation
S
ut
D 1800 (psi). e ultimate
compression strength S
uc
follows a normal distribution with a mean
S
uc
D 82,000 (psi) and a
standard deviation
S
uc
D 10,500 (psi). Calculate the reliability of the component by using the
MNS theory.
x
y
y
t
ba
L
L
A o
B
z
zt
M
z
F
x
(a) Schematic of Axial Loading and
Bending Moment
(b) Schematic of the Cross-section
and Neutral Axis Z
Figure 4.21: Schematic of a segment of a column under compressions and bending.
Solution:
1. e maximum tensile normal stress and normal compressive stress.
e centroid of the cross-section, as shown in Figure 4.21b, will be:
a D
3L C t
4
; b D
3t C L
4
: (a)
e area A and the moment of inertia I of the cross-section are:
A D 2Lt (b)
I D Lt
b
t
2
2
C
Lt
3
12
C Lt
a
L
2
2
C
tL
3
12
D
5Lt
3
C 5tL
3
C 6t
2
L
2
24
: (c)
e critical points on the cross-section in this example will be the points on the outer
layers, that is, point B and point A.
4.10. RELIABILITY OF A COMPONENT UNDER COMBINED STRESSES 225
On point B, the normal stress will be the combined stress of the tensile stress due to the
bending moment M
z
and the tensile stress due to the tensile force F
x
.
B
D
M
z
b
I
C
F
x
A
D
6M
z
.3t C L/
5Lt
3
C 5tL
3
C 6t
2
L
2
C
F
x
2Lt
: (d)
On point A, the normal stress will be the combined stress of the compression stress due
to the bending moment M
z
and the tensile stress due to the tensile force F
x
A
D
M
z
a
I
C
F
x
A
D
6M
z
.3L Ct/
5Lt
3
C 5tL
3
C 6t
2
L
2
C
F
x
2Lt
: (e)
If we use the mean values to calculate normal stresses at point A and B, we will have
A
D 23296 (pis) and
B
D 13568 (psi). erefore, the normal stress at point A will be
compression stress, and the normal stress at point B will be tensile.
2. e limit state functions.
Per Equation (4.48), the limit state function of the column at the critical point B by using
the maximum normal stress theory will be
g
.
S
ut
; t; L; F
x
; M
z
/
D S
ut
6M
z
.
3t C L
/
7Lt
3
C 7tL
3
C 6t
2
L
2
F
x
2Lt
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(f)
Per Equation (4.49), the limit state function of the column at the critical point A by using
the maximum normal stress theory will be:
g
.
S
uc
; t; L; F
x
; M
z
/
D S
uc
6M
z
.3L C t/
7Lt
3
C 7tL
3
C 6t
2
L
2
C
F
x
2Lt
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(g)
Geometric dimensions can be treated as normal distributions. eir distribution param-
eters can be determined per Equation (4.1). e loading F
x
and M
z
can be treated as
normal distributions. eir distribution parameters can be determined by Equation (4.2).
eir distribution parameters are listed in Table 4.47.
Table 4.47: e distribution parameters of random variables in Equations (f) and (g)
S
ut
(psi) S
uc
(psi) t
(in)
L
(in)
F
x
(lb)
M
z
(lb/in)
μ
S
ut
σ
S
ut
μ
S
uc
σ
S
uc
μ
t
σ
t
μ
L
σ
L
μ
F
x
σ
F
x
μ
M
z
σ
M
z
22,000 1,800 82,000 10,500 0.25 0.0025 1.25 0.0025 800 15 3200 60
226 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
3. Reliability of the column under combined stresses.
We can use the Monte Carlo method to calculate the reliability of this example based
on two limit state functions (f) and (g). e Monte Carlo method has been discussed
in Section 3.8. We can follow the procedure of the Monte Carlo method and the pro-
gram flowchart in Figure 3.8 to create a MATLAB program. Since this problem is no
very big and complicated, we will use the trial number N D 1,598,400 from Table 3.2 in
Section 3.8. e estimated reliability R of this column at the critical point B is:
R D 0:9983:
e estimated reliability R of this column at the critical point A is:
R D 1:0000:
erefore, the reliability of this complement in this example will be
R D 0:9983:
Example 4.24
A plane stress element of a component of a brittle material at the critical point is shown in
Figure 4.22. e normal compressive stress
x
(ksi), the normal tensile stress
y
, and the shear
stress
xy
follows normal distributions. eir distribution parameters are listed in Table 4.48.
Table 4.48: e stresses in a plane stress element
σ
x
(ksi) σ
y
(ksi) τ
xy
(ksi)
μ
σ
x
σ
σ
x
μ
σ
y
σ
σ
y
μ
τ
xy
σ
τ
xy
31.2 2.51 1.80 0.085 15.0 2.31
e component’s ultimate tensile strength S
ut
follows a normal distribution with a mean
S
ut
D 22:00 (ksi) and standard deviation
S
ut
D 1:80 (ksi). Its ultimate compression strength
S
uc
follows a normal distribution with a mean
S
uc
D 82:00 (ksi) and standard deviation
S
uc
D
10:50 (ksi). Calculate the reliability of the component by using the BCM theory.
Solution:
1. e two principal stresses in a plane stress.
As shown in Figure 4.22,
x
is normal compressive stress,
y
is normal tensile stress, and
xy
is negative shear stress. In the following calculations,
y
,
x
, and
xy
will be the values
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