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An Introductory Talk on Reliability
Analysis
Jeen-Shang Lin
University of Pittsburgh
With contribution from Yung Chia HSU
May 2007
Hua Fan University, Taipei
Supply vs. Demand
 Failure takes place when
demand exceeds supply.
 For an engineering system:
– Available resistance is the
supply, R
– Load is the demand, Q
– Margin of safety, M=R-Q
 The reliability of a system
can be defined as the
probability that R>Q
represented as:
  PM  0  PR  Q
Risk
 The probability of failure, or risk
pF  P( M  0)  1  P( M  0)  1  
 How to find the risk?
– If we known the distribution of M;
– or, the mean and variance of M;
– then we can compute P(M<0) easily.
Normal distribution: the bell curve
For a wide variety of conditions, the
distribution of the sum of a large number of
random variables converge to Normal
distribution. (Central Limit Theorem)
  0  1
1
( x   )2
f ( x) 
e( 
)
2

2 
x
F ( x) 


1
( x   )2
e( 
)
2

2 
  0  1
When
x
F ( x) 


1
( x   )2
1
2
e( 
)

e
(

x
)  ( x )
2


2 
 2
x
IF M=Q-R is normal
pF  P( M  0)  (
 M
M
)
Because of symmetry
(  x )  1  ( x )
Define reliability index
M
pF  1  (
)
M
  M /  M
Example: vertical cut in clay
M  c  H / 4
c  100  c  30 kPa
  20   2 kN / m3
c  0.5
H  10
If all variables are normal,
M  50  M  27.83   1.796 pF  3.62 102
Some basics
 y  a1x  a2 x
y  a1x1  (a2 ) x2
y  a1x1  a2 x2
y  a1x  (a2 )x
1
2
1
 y 2  E[( y   y ) 2 ]
 E[ y 2 ]   y
2
2
y  ai xi
2
2
 y 2  E[( y   y ) 2 ]
 E[ y 2 ]   y
 a1  x1  a2  x2  2a1a2 12 x1 x2
2
Negative
coefficient
2
2
 a1  x1  a2  x2  2a1 (  a2 ) 12 x1 x2
2
2
2
2
y  ai x
i
 y 2  ai 2 x 2  ai a j ij x  x
i
i
j i
i
j
Engineers like Factor of safety
 F=R/Q, if F is normal
pF  P( F  1)  (
1  F
F
)  1  (  )
F  1
reliability index  
F
Lognormal distribution
 The uncertain variable can increase without
limits but cannot fall below zero.
 The uncertain variable is positively skewed, with
most of the values near the lower limit.
 The natural logarithm of the uncertain variable
follows a normal distribution.
F is also often treated as lognormal
In case of lognormal
Ln(R) and ln(Q) each is normal
1
ln[x]   )2
f ( x) 
e(
)
2

2 
First order second moment method
 The MFOSM method assumes that the
uncertainty features of a random variable
can be represented by its first two moments:
mean and variance.
 This method is based on the Taylor series
expansion of the performance function
linearized at the mean values of the random
variables.
First order second moment method
 Taylor series
expansion
g  g ( x1, x2 ,.....xn )
g  g( X )  ( x  X )
g
g ( x1 , x2 ,...,xn )  g ( x1 , x 2 ,...,xn )   ( xi  xi )
xi
g  g( x1, x 2 ,...,xn )
2

2
g
 g 
g g
    xixj xi xj
   xi 
xi x j
i i j
 xi 
2
dg
dx
Example: vertical cut in clay
F
4c
H
c  100  c  30 kPa
  20   2 kN / m3
c  0.5
H  10
If all variables are normal,
F
4

c H
F  4c
 2
  H
F  1

F
F  2  F  0.5292   1.8896 pF  2.94 102
1-normcdf(1.8896,0,1) MATLAB
Slope stability
 cx
n
FS 
i 1
i
n
f R
FS 
 Q
-55


 Wi  u i xi  tani  1

M
(

)
i


-35
W sin 
i 1
i
25
i
2 (H): 1(V) slope
with a height of 5m
5
-15
5
-15
-35
25
45
Reliability Analysis
 The reliability of a system can be defined as the
probability that R>Q represented as:
  PQ  R
 cx  W
n
FS 
i 1
i
i


 u i xi  tani  1

 M i ( ) 
n
W sin 
i 1
i
i
FS FS   FS 

X i
2m X i
FS contour
  10   2
c  10  c 2 kPa
4
2.17
2
2.24
2
05 8
2.1
7
3
37
2.0
1.6
94
9
9
94
1.6
8
68
1.9
19
83
1.
20
FS  1.436
1.626 4
2.2
42 7
1.558
1
2
FS
 0.056

3
4
5
6
7
8
2.
1
2. 05 8
17
42
1
2. 1 .900
03 .9 3
7 3 68 8
5
89
1.4
1
1. .763 4
83
19
1.
1.6 626
94 4
9
5
0
2.1742
1.626 4
2.1058
2.0373
1.968 8
3
9
58
1.5
58
1.5
34
76
1.
1.8
1.9 31 9
1.9 00 3
68
8
FS
 0.089
c
5
1.489 5
10
0
1.90
1. 48
9
1
1.83
1.6949
0.21.
9
94
1.6
4
26
1.6
15
4
63
1.7
1.558
,
1.6
2
1.6 6 4
94
9
,
1.7
63
4
03
90
1.
1.
76
34
25
  2.079
2
11 79 748 1 6 6
2.32.32.42.51
9
 FS  0.21
10
pF  0.0188
First Order Reliability Method
Hasofer-Lind (FORM)


Probability of failure can be found
obtained in material space
Approximate as distance to Limit
state
Distance to failure criterion
 If F=1 or M=0 is a straight line
 Reliability becomes the shortest
distance
Constraint Optimization:Excel
May get similar results with FOSM
  1.796
FOSM
1-normcdf(1. 796,0,1)=0.0362 MATLAB
Monte Carlo Simulation
correlation=0
Monte Carlo=0.0495
Monte Carlo Simulation
correlation=0.5
{Y * }  [CY ]1/ 2 [T ]T {X *}
 1/ 2 1/ 2 
[T ]  

 1 / 2 1 / 2 
FORM=0.0362
0 
1  
[CY ]  
1   
 0

1
( X  X )T C X ( X  X )
min
XLimit state
3
2
FS=1.0
* 
 

1
Mean
FS=1.549
Mean + - S.D.
FS=1.259
FS=1.613
0
FS=1.436

-1
FS=1.324
-2
1
-3
y
2
x
-4
-5
-5
-4
-3
-2
-1
0
c* 
1
cc
c
2
3
4
5

FS=1.0 (M=0)
The matrix form of the
Hasofer-Lind (1974)
  c  c
UNSAFE
Region

FS<1 or
M<0
 
c
The matrix form of the
Hasofer-Lind (1974)
Soil properties
FOS=1
Soil properties>0
25
25
2.2
2.1
5. 25
1.8
5
4.8
2
1.9
3.
75
1.7
1.8
20
3.3
5
3.15
3.3
5
1.6
1. 7
1.6
5
3.1
1.48
5
3.3
3. 15
2.2
2.75
5
2.9
1.7
1.8
1.6
1. 65
5
3.7
2.1
15
4. 85
5
2.9
2
1.9
4.55
4.15
5
1.6
1.55
5.25
1.
7
1.65
20
15
5
4.5
5
4.1
3.75
1. 55
1
2
3
4
4.15
5
2.9 15
3.
35
3.
5
5
0
2.7
5
5.
25
2.1
2.
2
8
1.
3.
15
3.7
5
1.9
2
1.6
4.
1
4.84.55 5
5
4
4.8 .55
5
1.65
1.48
7
1.
1.55
2. 75
1.6
2.1 2
2.55
5
2.5
95
2.
1.
65
5
3.3
5
3.7
7
1.
8
1.
1.
9
10
1.48
1.55
10
5
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10

FS=1.0
c
c
Correlation=.99


UNSAFE
Region
FS<1 or
Correlation=-.99
Correlation=0
M<0
c
 The distance
  (c* )  ( * ) 
2
2
FS  1
2
2
(FS / c)2  c  (FS /  )2  
FOSM maybe wrong
1
2
Y
Soil 1
c1
X
Soil 2
c2
2
9.10 m
6.14 m
 FOSM
A projection Method
( xi ) p  X i 
( FS  1)(FS / X i ) X
n
2
(

FS
/

X
)

i X
i 1
2
I
2
i
 Check the FOSM
 Use the slope, projected to where the
failure material is
 Use the material to find FS
 If FS=1, ok
May 2007
Hua Fan University, Taipei