structural reliability

2013 Spring Structural Reliability Analysis 1 2013 Spring 2013 Spring

Structural Reliability Chapter I. Introduction

Jung J. Kim, Ph.D. Adjunct Professor Department of Civil & Environmental Engineering Sejong University

2013 Spring Structural Reliability Analysis 2 2013 Spring

Office: Chungmu-gwan 706

e-mail: [email protected]

Office hour: Mon/Tue 10:00~11:00 am

Mid-term (30%), Final (30%), Homework (40%)

Exams: Open book

Course information


Schedule

week Date Contents Etc.

1 Mar 06 Introduction/ Probability applications to Engineering

2 Mar 13 Probability density functions, histogram

3 Mar 20 Cumulative probability functions, fitting Probability curve

4 Mar 27 Limit state function, Probability of failure and Reliability index

5 Apr 03 Integration at (G < 0), direct integration

6 Apr 10 Random Sampling

7 Apr 17 Monte Carlo method

8 Apr 24 Mid-term

9 May 01 FOSM

10 May 08 Calibration (1)

11 May 15 Calibration (2)

12 May 22 Models of Load and resistance

13 May 29 System reliability

14 June 05 Examples of application

15 June 12 Presentation

16 June 19 Final


1. COURSE LAYOUT

• Design Philosophy

• Deterministic vs. Stochastic

• Reliability Analysis

• Partial Safety Factors


Design Philosophy

R, S

R (resisting strength) > S (acting force)

R S

R, S f R g S

Engineer’s confidence to uncertainties,

reliability

Uncertainties in acting force and resisting strength

are considered in design by partial safety factors

(load factors and resisting factors)

f R > g S Wider gap, more confidence to designers

Narrower gap, less cost for contractors


Deterministic vs. Stochastic

R, S R S

probability

Make acting force and resisting strength stochastic variables,

then calculate the probability that acting force is bigger than

resisting strength.

)( SRPp f


Reliability analysis

A tool that can give theoretical background to consider

uncertainties of design variables in engineering problems.

)( SRPp f

General procedure

1. Determine the distributions of variables

2. Define Limit State Function

PDF, CDF

3. Calculate the probability of ‘violation of

the Limit State Function’

0 SRG

4. Check the probability with the target

probability of failure

Generally,

0.003%, (=4)

0.135%, (=3)

: Reliability index


Partial Safety Factors

Although it is desirable to do reliability analysis for each design

case, it will be more practical to use partial safety factors for

most designers.

)( SRPp f gf

General procedure

1. Determine the distributions of variables

2. Define Limit State Function

PDF, CDF

3. Calculate the probability of ‘violation of

the Limit State Function’

0 SRG gf

4. Check the probability with the target

probability of failure (reliability index)

Generally,

0.003%, (=4)

0.135%, (=3)

5. Iteration to find f and g combination, which always give

over the target reliability index.


2. PRERIMINARIES

• Events and Sets (Domain)

• De Morgan’s Rule

• Probability in Engineering

• Dependent Events and Sets

• Bayes’ Theorem


Events and Set (Domain)

The formulation of a probabilistic problem starts with the

identification of the event of interests for a set.

Problems of “happen or not”

E11: Failure of link 1, E12: Failure of link 2

Parallel system Serial system

Event: failure.

Set: two set of link1 and link 2

Eij i: Event number

j: Set number


De Morgan’s Rule

When does the following system fail?

E11: Failure of link 1, E12: Failure of link 2, E13: Failure of link 3, E14: Failure of link 4

Find when the system operates (gets connected).

Then, apply De Morgan’s Rule.

(Prob. 1-1) For the event “connected”, find the probability that system fails.


Probability of events

P(E): probability that the event “E” occurs.

When two events E11 and E21 are mutually exclusive event in a set S,

the probability of “E11 or E21”, P(E11 U E21) is P(E11)+P(E21) and

the probability of “E11 and E21”, P(E11 ∩ E21) is 0 (impossible).

e.g) En1: occurrence of number n for a dice (n=1,2,…,6)

P(E11 U E21) = 1/6+1/6 =1/3

P(E11 ∩ E21) =0, mutually exclusive event.

However, for different two sets S1, S2,

P(E11 U E22) = 1/6+1/6 =1/3

P(E11 ∩ E22) =1/6(1/6)=1/36, mutually exclusive set.


Probability of events in Engineering

The probabilities of events in engineering problem are not such

random as previous examples, those are already engineered.

Sometimes, the events and sets are not mutually exclusive.

e.g.) Event 1: fracture, Set: link 1 and link 2 P(E11) = 0.02, P(E12) = 0.03

1. Engineered: the event is no more about “happen or not, 1/possible

cases problem”. The probability of the event was already engineered

or designed for preventing or holding the event.

2. The selection of events in a set is not necessary to be mutually exclusive.

e.g.) Event 1: fracture, Set: link 1 and link 2 from same manufacturer.

2111 EE

2212 EE

e.g.) Event 1: fracture, Event 2: Damage, Set: link 1

3. The selection of sets is not necessary to be mutually exclusive.

dependency


Conditional Probability

Conditional probability: For some occasions when the probability of an

event depends on the occurrence of another event.

A For the case that event A and B

are in one set,

)(

)()|(

B

BABA

EP

EEPEEP

B

For the case that event A and B

are in different set, A B dependency

pEEP BA )|(

For this case, will be given such as “When rain falls (B), the probability to sell umbrella (A) is 70%.”

)()|( ABA EPEEP

If the probability that rain falls, what is the probability to sell umbrella? 56.0)8.0(7.0)()|( BBA EPEEP

If there is no dependency between A an B in different sets,

)()()( BABA EPEPEEP

)|( BA EEP


Dependent events in a set

Intersect part needs to be omitted.

When the events in a set are not mutually exclusive,

2111 EE


)()()( 21112111 EPEPEEP

)()()()( 211121112111 EEPEPEPEEP

)()( 112111 EPEEP )()( 212111 EPEEP

Fracture

Damage

No damage

Link 1


Dependent event in different sets

Intersect part needs to be omitted.

When the selected sets are not mutually exclusive,


)()()( 12111211 EPEPEEP

)|()()()(

)()()()(

1211121211

121112111211

EEPEPEPEP

EEPEPEPEEP

Fracture

Link 1

Link 2

Fracture

If there is no dependency,

)()()()()( 111212111211 EPEPEPEPEEP

If there is total dependency (so, if link 1 fractures then link 2

also fractures.), 1)|( 1211 EEP

)()()()( 1212111211 EPEPEPEEP

(Prob. 1-2) Which case will be better for engineers? Why? (Put numbers and

check.)


Total Probability

For this case, will be given such as “When rain falls (B), the probability to sell umbrella (A) is 70%.”

If the probability that rain falls, what is the probability to sell umbrella? 56.0)8.0(7.0)()|( BBA EPEEP

)|( BA EEP

Remind

Is 56% the probability to sell umbrella?

The total probability to sell umbrella should include when rain does not fall.

Therefore, if 3.0)|( BA EEP

62.0)2.0(3.0)8.0(7.0)()|()()|( BBABBA EPEEPEPEEP

Generally, )()|()(

1

i

n

i

i EPEAPAP


Bayes’ Theorem

The Bayes’ theorem (Bayesian update) provides a valuable and useful tool for revising

or updating a calculated probability as additional data or information becomes

available.

)()( ii EAPAEP )()|()()|( APAEPEPEAP iii

)()|()(1

i

n

i

i EPEAPAP

)(

)()|()|(

AP

EPEAPAEP ii

i

)()|(

)()|()|(

1

j

n

j

j

iii

EPEAP

EPEAPAEP

Previous slide

Updating

)|()|(

)|(

)|()|(

)(

)|(

1

p

j

n

jp

j

j

u

p

ip

i

i

u

u

i

AEPAEP

EAP

AEPAEP

EAP

AEP

For the previous probability of event Ei for sample size Ap,

the updated probability of event Ei for the new sample size

Au can be calculated as

Previously find )()|( i

p

i EPAEP ?)|( u

i AEP


Bayes’ Theorem Example 1

The fraction defective of a product has been 5% after observing 100 samples.

Additional observation by an excellent expert (assume he/she would not miss detecting

any significant flaw) detected a defective product at the 3rd product. Show the

sequence of the fraction defective using Bayesian update.

1st observation: No defective

Put Event 1: Defective, Event 2: No defective

0495.0

95.095.0

101/9605.0

05.0

101/5

05.005.0

101/5

)|()|(

)|(

)|()|(

)(

)|(

1

p

j

n

jp

j

j

u

p

ip

i

i

u

u

i

AEPAEP

EAP

AEPAEP

EAP

AEP

2nd observation: No defective

049.0

9505.09505.0

102/970495.0

0495.0

102/5

0495.00495.0

102/5

)|(

u

i AEP

3rd observation: defective

0583.0

951.0951.0

103/970495.0

0495.0

103/6

049.0049.0

103/6

)|(

u

i AEP


Bayes’ Theorem Example 2

For the previous example, if the reliability of expert is only 90%, what is the updated

fraction defective when the first observation detect a defective product.

321.0)95.0)(1.0()05.0)(9.0(

)05.0)(9.0(

)()|(

)()|()|(

2

1

111

j

j

j EPEDP

EPEDPDEP

(Prob. 1-3) What is the updated fraction defective when the first observation is not a defective product.

006.0)95.0)(9.0()05.0)(1.0(

)05.0)(1.0(

)()|(

)()|()|(

2

1

111

j

j

j EPENP

EPENPNEP

(Prob. 1-4) What happens when the reliability of expert is only 50%. Discuss the results.


Structural Reliability Chapter II. Stochastic Variables

Jung J. Kim, Ph.D. Adjunct Professor Department of Civil & Environmental Engineering Sejong University


1. UNCERTAIN VARIABLES

• Stochastic variables and others

• Probability Density Function (PDF)

• Generation of PDFs


Evidence theory

belief, plausibility

Probability theory

probability

Possibility theory

necessity, possibility

Modeling of Uncertain Variables

There exist many methodology to model uncertain variables according to the types of

uncertainties such as likelihood, non-specificity, ambiguity and so on.

Fuzzy logic

membership

Fuzzy set Crisp set


Distributions of variables

What is the probability that the event x1 occurs?

For },...,,,{ 10321 xxxxX

How about that of x3?

10321 ... xxxx

Uniform distribution,

if there are no intentions, no

constraints and so on.

p

Engineering properties are

designed to have an

intended value, such as

fy = 400 MPa, fck = 25 MPa.

0

2

4

6

8

15 20 25 30 35 Over

Fre

qu

ency

Strength (MPa)


Discrete vs. Continuous

Discrete set Continuous function },...,,,{ 321 nxxxx

},...,,,{ 321 npppp

2

2

1

2

1)(

cx

d expe.g.)

Easy to understand the concepts of probability.

For practical use, it is necessary for the probability concepts to be extended to continuous domain

0

0.02

0.04

0.06

0.08

0.1

15 20 25 30 35 Over

Fre

qu

ency

den

sity

Strength (MPa)

Histogram

0

0.02

0.04

0.06

0.08

0.1

0 10 20 30 40 50

Pro

ba

bil

ity

den

sity

Strength (MPa)

Normal distribution function


• Probability Density Function (PDF)


Probability Density Function (PDF)

Normalized distribution of a continuous variable x.

The types of PDFs Uniform, Normal, Lognormal, Gamma, Gumbel (Extreme type I),

Frechet (Extreme type II), Weibull(Extreme type III), Poison distributions

… discussed later!

Central Limit Theorem

If the number of samples for a PDF are infinity, the PDF approaches

a normal distribution.

Normalized means that the area under the distribution is 1.0.

1)pdf(

dxx

This can be the basis of normal approximation of

distributions.

11

N

i

ipDiscrete set


Cumulative Density Function (CDF)

Definition of CDF of a continuous variable x.

dxx

)pdf()cdf(

Is the inverse of CDF available? How about PDF?

i

k

ki pcp1

Discrete set

)(X)cdf( xprobx where X is a continuous set.


Moments of PDF

Moments

The first moment

The r-th moment r is defined as dxxpxr

r

)(

Expected value

The r-th central moment r is defined as dxxpx r

r

)()( 1

dxxxp

)(1

The second central moment

dxxpx

)()( 2

12 Variance

e.g.)

With the uniformly distributed assumption

},...,,{ 21 NxxxX For the discrete set of

Nppp N

1...21

cxN

xpxN

i

i

N

i

ii 11

1

1)(

2

1

2

2 )(1

N

i

cxN

average square of the standard deviation

Those are the first two moments, which are most useful.


Characteristics of PDF

Coefficient of Variation (COV)

c

1

2/1

2COV

Coefficient of Skewness (COS)

2/3

2

3COS

Coefficient of Kurtosis (COK)

2

2

4COK

Figure

(+) COS

Right

(-) COS

Left

COK <3.0

flatter COK >3.0

narrower


Types of PDF (uniform)

Uniform PDF

CDF

The uniform distribution is the most uncertain distribution.

(Mathematically, the minimum information and the maximum

entropy principle)

otherwise0

1

)(bxa

abxpU

2

ba

12

)( 22 ab

(Prob. 2-1) Derive the equation for 2 (The second central moment)

Figure


Normal PDF

Types of PDF (Normal or Gauss)

CDF

2

2

1

2

1)(

x

N exp

The most important distribution in structural reliability theory.

(Central limit theorem)

dexPx

N

2

2

1

2

1)(

Figure

No closed-form solution

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

10 20 30 40 50

Pro

ba

bil

ity d

en

sity

Strength (MPa)

PDF (Normal distribution)


Standard Normal PDF Normal PDF for = 1, = 0

Types of PDF (Standard Normal)

CDF

)(2

1)(

2

2

1

zezpz

SN f

)()( zzPSN

)(1)( zz


PDFs are standardized to the Standard Normal PDF, not only for the normal PDFs, but also for the other type of PDFs based on Second Moment (SM) approximation, which is that most distributions can be described by the first two moments, and .

Standard form

XZ ZX

xxZprobxZprobxP )()(

then Put

X

2

2

Z

-2 -1 0 1 2

Moving and scaling

f

xx

dx

dxP

dx

dxp

1)()(

X

2

2

Z

-2 -1 0 1 2


Only defined in the positive region, which gives reasons to be used

for most physical quantities.

Types of PDF (Lognormal)

2ln

2

1exp

2

1)(

xxpLN

Lognormal PDF

Normal PDFs for ln(X)

2)ln(

2

X

2

XX

2 /1ln

2exp

2X CDF 1)exp( 22 XX

For COVX <20%

Xln

XCOV


Normal vs. Lognormal

Normal and lognormal Distributions for (, ) =(30, 3), (30,9) MPa.

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

10 20 30 40 50

Pro

ba

bil

ity

den

sity

Strength (MPa)

PDF s(Normal and Lognormal distributions)

(Prob. 2-2) Draw figures


Types of PDF (Binomial)

The probability that no bar fails under 210 MPa

The probability that a steel bar fails under 210 MPa by tension is

0.05. If we do tensile strength tests of 5 steel bars, what is the

probability that at least one bar fails under 210 MPa.

Describe the probability of occurrence of x among n trials for the

occurrence probability of p.

nxppx

nxp xnx

BN ,...,2,1,0for)1()(

x

k

knk

EX ppk

nxP

0

)1()(

PDF

CDF

774.0)05.01()05.0)(1()1(0

5)0( 50050

pppBN

Therefore, the probability that at least one bar fails under 210 Mpa is then

226.0774.01)0(1 BNp

(Prob. 2-3) The probability that no more than two bars fails under 210 Mpa (CDF)

Bernoulli Sequence


Types of PDF (Poisson)

(Prob. 2-4) The probability that such large earthquakes occurs at least two times in the next 5 years.

Describe the probability of random occurrence at any given time or

space interval. e.g.) earthquake, tsunami…

,...2,1,0forexp

!)( xvt

x

vtxXp

x

tPSPDF

x is the average number of occurrences in a time [0, t].

v is the mean occurrence rate (time/number of event).

In the last 50 years, there were three large earthquakes (with M > 6)

in Southern California. So, the mean occurrence rate is 3/50=0.06

per year. The probability that such large earthquakes in the next 10

years is then

Poisson Sequence

329.0)10(06.0exp

!0

)10(06.0)0(

0

10 XpPS

671.0329.01)0(1)1( 1010 XpXp PSPS


Types of PDF (Exponential)

No occurrence of event during time t

vtvt

vtXp tPS expexp

!0)0(

0

The probability that an event occurs in the next t years (CDF).

Poisson Sequence

)exp(1)0(1)1( vtXpXp tPStPS

The probability that an event occurs at t year is then (PDF).

)exp()0(1)1( vtvXpXp tPStPS

For the previous example, the probability that such large

earthquakes in the next 20 years is then

699.0)]20(06.0exp[1)20( EXp

)exp(1)( vttpEX

)exp()( vtvtpEX

(Prob. 2-5) What are the differences between Poisson and Exponential distributions?


Types of PDF (Geometric)

Describe the probability of occurrence at n-th trial for the occurrence

probability of p.

integerfor)1()( 1 nppnp n

GMPDF

(Prob. 2-6) The probability that the first occurrence of 50-yr wind velocity within 5 years (sum).

Mean Recurrence Time (Return period)

In design code, design wind is defined as such a way that wind

velocity with return period 50yr (return period).c

yearper02.050

11

Tp

For a reference time (completion of building), the probability that the

first occurrence of 50-yr wind velocity on the second year is

196.0)98.0(02.0)1()2( 12 pppGM


Gamma PDF

X

2

X 2

k

k

Types of PDF (Gamma)

Usually used for modeling of sustained live load.

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 1 2 3 4 5 6 7 8 9 10

Pro

ba

bil

ity

den

sity

Sustained live load (kN/m2)

Gamma PDFs

1,1 k

1,2 k

1,3 k1,1 k 0forexp)( xxxpG

1,2 k 0forexp)( xxxxpG

1,3 k 0for

2

exp)(

2

xxx

xpG

0for

)(

exp)()(

1

xk

xxxp

k

G

0integerfor)!1()(0

1

kkduuek ku

(Prob. 2-7) Draw figures for various parameters (2)


(x )

X (x) exp (x ) for xuf u e

Extreme Type I PDF

X

22

X 2

0.577

6

u

Types of PDF (Gumbel)

Extreme values, wind loads



1

X (x) exp for 0 xx x

k kk u u

fu

Extreme Type II PDF

X

2 2 2

X

11 for 1

2 11 1 for 2

where ( ) ( 1)!

u kk

u kk k

y y

Types of PDF (Frechet)

Extreme values, the maximum seismic load

Prob. 12: Draw figures for

various parameters (2)



Extreme Type III PDF

Types of PDF (Weibull)

Important distribution for most research works.

Used to simulate a lifetime event of a structure

Bathtub curve: Summation of three phase curves


structural reliability

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