Lecture 9. Failure probabilities and safety indexes
Jesper Rydén
Department of Mathematics, Uppsala University
[email protected]
Statistical Risk Analysis • Spring 2011
Failure probability
Often the failure probability Pf = P(B) is formulated in terms of
a function of uncertain values (random variables) exceeding some
critical level u crt
Pf = P(B) = P(h(X1 , X2 , . . . , Xn ) > u crt )
Introducing Z = h(X1 , X2 , . . . , Xn ), we may write
Pf = P(Z > u crt ) = 1 − FZ (u crt )
Finding the distribution of Z is in general difficult.
Failure probability
In some situations, the distribution of Z can be found.
If knowledge of the distributions of Xi are too uncertain, safety
indexes are used. These can be related to costs and consequences.
Approximations of safety indexes can be found by employing
Gauss’ formulae. These can also be used to approximately
compute confidence intervals (the so-called delta method).
Linear functions: Load and strength
Often a function can be formulated on the form supply versus
demand, i.e. the (supply) capacity of a system must meet certain
(demand) requirements.
Simple example: One single random strength R and one load S.
System fails when
Z =R −S <0
i.e. when R < S. This probability can be computed by means of
numerical integration if distributions of R and S are completely
known:
Z
Z
P(R < S) = P(R < s)fS (s) ds = FR (s)fS (s) ds
Alternatively: simulate random numbers ri , si and estimate the
frequency of cases ri < si .
Common situation, reliability of structures: R is Weibull
distributed, S is Gumbel distributed.
Example: Crack propagation
Consider crack growth in some specimen. The time to failure, T ,
due to cracking is the sum of two times, T = T1 + T2 . An integral
expression can be formulated for this sum.
T1 =
“Time to initiation of a microscopic crack”
T2 =
“Time for the crack to grow a fixed distance”
If the component is supposed to be used for a period of time t0 ,
failure occurs if T < t0 .
Example: Hooke’s law
By Hooke’s law, the elongation of a fibre is is proportional to the
force F , that is, = K −1 F or F = K . Here K , called Young’s
modulus, is uncertain and modelled as a random variable with
mean m and variance σ 2 .
Consider a wire containing 1000 fibres with individual independent
values of Young’s modulus Ki . A safety criterion is given by ≤ 0 .
Linear function
In the case where
Z = X1 + · · · + Xn ,
the exact distribution can be found in some cases. Assume Xi to
be independent and from the same distribution F (x) with
E[X ] = µ, V[X ] = σ 2 .
I
Sum of normal variables: Normal.
I
Sum of gamma variables: Gamma.
I
Sum of Poisson variables: Poisson.
Lognormal distribution
A variable Z such that
ln Z ∈ N(m, σ 2 )
is called a lognormal variable.
ln z − m .
σ
One can prove that for a lognormally distributed variable
ln Z ∈ N(m, σ 2 ),
FZ (z) = P(Z ≤ z) = P(ln Z ≤ ln z) = Φ
E(Z ) = em+σ
,
2σ 2
2
· (e − eσ ),
p
p
2
D(Z ) = em e2σ2 − eσ2 = em+σ /2 · eσ2 − 1.
V(Z ) = e
2m
2 /2
Lognormal distribution
Note that the coefficient of variation R(Z ) =
a function of σ 2 .
p
exp(σ 2 ) − 1 is only
Solving for σ 2 , we obtain σ 2 = ln(1 + R(Z )2 ). With σ 2 known, m
can be computed if E(Z ) is given. However, m is much easier to
find in the case when the median of Z is specified.
For a normal variable Z ∈ N(m, σ 2 ) the parameter m is both mean
and median, thus
0.5 = P(ln Z ≤ m) = P(Z ≤ em ),
and hence the median of Z is exp(m).
Example: Waste-water treatment
Suppose that spill water in a chemical factory is treated before it is
dumped into a nearby lake.
Let X denote the concentration of a pollutant feeding into the
treatment system, and Y denote the concentration of the same
pollutant leaving the system. Suppose that for a normal day, X
has a lognormal distribution with median 4 mg/l and coefficient of
variation R[X ] = 0.2.
Because of the erratic nature of biological and chemical reactions,
the efficiency of the treatment system is unpredictable. Hence the
fraction of pollutant remaining untreated, denoted by K , is also a
random variable. Assume K is lognormal with median of 0.15 and
coefficient of variation R[K ] = 0.1. We also assume that X and K
are independent.
Example: Waste-water treatment
(a) What is the distribution of Y = K · X ?
(b) Suppose that maximal concentration of the pollutant
permitted to be dumped into the lake is specified to be 1
mg/l. What is the probability (failure probability) that on a
normal day this specified standard will be exceeded?
(c) On some working days, owing to heavy production in the
factory, the influent X will have higher median of 5 mg/l. All
other parameters remain unchanged. Suppose that such a
heavy-work day happens only in average 1 per 10 days of
production. Then, on a given day selected at random, what is
the probability that the standard 1 mg/l will be exceeded?
(What is the probability of failure of the treatment system?)
Minimum – Weibull variables
Weakest-link principle in mechanics: the strength of a component
is equal to the strength of its weakest part.
In other words we may say that “failure” occurs if the minimum
strength of some components is below a critical level ucrt :
min(X1 , . . . , Xn ) ≤ ucrt .
If Xi are independent with distributions Fi , then
P(min(X1 , . . . , Xn ) ≤ ucrt ) = 1 − P(min(X1 , . . . , Xn ) > ucrt )
= 1 − P(X1 > ucrt , . . . , Xn > ucrt )
= 1 − (1 − F1 (ucrt )) · . . . · (1 − Fn (ucrt )).
The computations are particularly simple if Xi are Weibull
distributed.
Safety index: Cornell’s index
Consider again Z = R − S. Special case: R and S are independent
and R ∈ N(mR , σR2 ) and S ∈ N(mS , σS2 ).
2
Then Z
q∈ N(mZ , σZ ), where mZ = mR − mS and
σZ = σR2 + σS2 , and thus
Pf = P(Z < 0) = Φ
0 − mZ = Φ(−βC ) = 1 − Φ(βC ),
σZ
where
βC =
mZ
σZ
is the so-called Cornell’s safety index.
Safety index: Cornell’s index
The index measures the distance from the mean mZ = E(Z ) > 0
to the unsafe region (that is zero) in the number of standard
deviations.
Example: βC = 4 means that the unsafe region is four standard
deviations away from the mean, which is often chosen as a “typical
value”.
If Z is normally distributed then we can find from tables of the
normal distribution that for βC = 4 the failure probability
Pf = 1 − Φ(βC ) ≈ 3/10 000.
Safety index: Cornell’s index
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
−2
−1
0
1
2
3
4
5
6
Illustration of safety index. Here: βC = 2. Failure probability
Pf = 1 − Φ(2) = 0.023 (area of shaded region).
Safety index: Hasofer–Lind index
In practice, βC is seldom used; for example, it depends on the
choice of function h.
The Hasofer–Lind index βHL measures distance from expectations
of the strength and load variables to the unsafe region in a way
that is independent of a particular choice of the h function.
The Hasofer–Lind safety index is commonly used in reliability
analysis, although one needs quite advanced computer software to
compute it.
In the special case when h is a linear function, the Hasofer-Lind
index βHL is equal to Cornell’s index βC .
Safety index: Hasofer–Lind index
For βHL , approximately Pf ≈ Φ(−βHL ). Higher value of the safety
index implies lower risk for failure, but also a more expensive
structure.
In order to propose the so-called target safety index, both costs
and consequences must be considered. Possible classes of
consequences are:
Minor Consequences This means that risk to life, given a failure, is
small to negligible and economic consequences are small
or negligible (e.g. agricultural structures, silos, masts).
Moderate Consequences This means that risk to life, given a failure, is
medium or economic consequences are considerable (e.g.
office buildings, industrial buildings, apartment buildings).
Large Consequences This means that risk to life, given a failure, is
high or that economic consequences are significant (e.g.
main bridges, theatres, hospitals, high-rise buildings).
Safety index: Hasofer–Lind index
Target reliability indexes (safety index for a particular failure mode
will have the target value).
One considers here the so-called “ultimate limit states”, which
means failure modes of the structure. This kind of failure concerns
mainly the maximum load-carrying capacity as well as the
maximum deformability.
Relative cost of
safety measure
Minor conseq.
of failure
Moderate conseq.
of failure
Large conseq.
of failure
Large
Normal
Footnotesize
βHL = 3.1
βHL = 3.7
βHL = 4.2
βHL = 3.3
βHL = 4.2
βHL = 4.4
βHL = 3.7
βHL = 4.4
βHL = 4.7
Recommendations proposed by Joint Committee on Structural
Safety.
Return periods and safety index
The safety index βHL = 3.1 implies that the intensity of accidents
is 1/1000 [year−1 ], or equivalently, the return period is 1000 years.
Similarly, the corresponding return periods corresponding to the
other values of βHL in the previous table can be found:
Safety index βHL
Return period (years)
3.1
103
3.3
2 · 103
3.7
104
4.2
105
4.7
106
Gauss’ Approximations
Gauss’ approximation, one variable. Let X be a random
variable with E(X ) = m and V(X ) = σ 2 . Further, let h be a
function with continuous derivative. Then
E(h(X )) ≈ h(m)
and
V(h(X )) ≈ (h0 (m))2 σ 2 .
Gauss’ approximation, two variables
Gauss’ approximation, two variables. Let X and Y be
independent random variables with expectations mX , mY ,
respectively. For a smooth function h, the following
approximations are called Gauss’ formulae:
E(h(X , Y )) ≈ h(mX , mY ),
2
2
V(h(X , Y )) ≈ h1 (mX , mY ) V(X ) + h2 (mX , mY ) V(Y ),
where
h1 (x, y ) =
∂
h(x, y ),
∂x
h2 (x, y ) =
∂
h(x, y ),
∂y
Gauss’ approximation, two variables
For correlated variables X and Y , the formulae are as follow:
E(h(X , Y ))
≈
V(h(X , Y ))
≈
h(mX , mY ),
2
2
h1 (mX , mY ) V(X ) + h2 (mX , mY ) V(Y )
+2h1 (mX , mY )h2 (mX , mY ) C (X , Y ).
Example: Gauss’ approximation
Consider a beam of length L = 3 m. A random force P with
expectation 25 000 N and standard deviation 5 000 N is applied at
the midpoint of such a beam. The modulus of elasticity E of a
randomly chosen beam has the expectation 2 · 1011 Pa and the
standard deviation 3 · 1010 Pa. All beams share the same second
moment of (cross-section) area I = 1 · 10−4 m4 . Then the vertical
displacement of the midpoints is
U=
PL3
.
48EI
Give approximately E(U) and V(U).
Example: Gauss’ approximation
Rating life of ball bearings. In an earlier example, 22 lifetimes
of ball bearings were presented. We will assume a parametric
model for the distribution and study the uncertainty of the
parameter estimates. In particular, we will study the so-called
rating life, L10 , a statistical measure of the life which 90% of a
large group of apparently identical ball bearings will achieve or
exceed. In other words, L10 satisfies
P(X ≤ L10 ) = 1/10.
We will assume a Weibull distribution for the distribution of the
lifetime:
c
FX (x) = 1 − e−(x/a) , x ≥ 0.