why need probabilistic approach? rain probability how does that affect our behaviour? ?
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Why need probabilistic Why need probabilistic approach?approach?
Rain probabilityRain probability
How does that affect our behaviour?How does that affect our behaviour?
? ?
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Uncertainties in Uncertainties in EngineeringEngineering
Natural HazardsNatural Hazards
Material PropertiesMaterial Properties
Design ModelsDesign Models
Construction ErrorsConstruction Errors
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Absolute Safety Not Absolute Safety Not GuaranteedGuaranteed
Engineers need to:Engineers need to:
model, analyze, update model, analyze, update
uncertaintiesuncertainties
evaluate probability of failureevaluate probability of failure
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QuestionsQuestions
What is acceptable failure What is acceptable failure probability?probability?
-- stadium vs shedstadium vs shed
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QuestionsQuestions
Should one want to be Should one want to be conservative if a perfectly safe conservative if a perfectly safe system is possible?system is possible?
-- overbooking in airlinesoverbooking in airlines
-- parking permitsparking permits
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QuestionsQuestions
Should one minimize risk if Should one minimize risk if money is not a problem?money is not a problem?
-- system consideration – system consideration –
e.g.dame.g.dam
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Trade-off Decision AnalysisTrade-off Decision Analysis
Risk vs. consequenceRisk vs. consequence
System riskSystem risk
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Formal analysis of Formal analysis of uncertainties and probabilityuncertainties and probability
• Not all problems can be solved by analysis Not all problems can be solved by analysis of dataof data
• Set TheorySet Theory
• Sample spaceSample space: collection of all possibilities: collection of all possibilities
• Sample pointSample point: each possibility: each possibility
• EventEvent: subset of sample space: subset of sample space
• Probability TheoryProbability Theory
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Union: Union: eithereither E E1 1 or Eor E2 2 occuroccur
EE11∪∪EE22
Intersection:Intersection:bothboth E E1 1 and Eand E2 2 occuroccur
EE11∩∩ E E22 or E or E11 E E22
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ExamplesExamples
A B
No communication between A and B = E1E2
CB
A
No communication between A and B = E3∪∪E1E2
1
2
EE11 = road 1 closed = road 1 closedEE22 = road 2 closed = road 2 closed
EE33 = road 3 closed = road 3 closed1
23
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Example - pair of footingsExample - pair of footings
1 2
EE11 = 1 settles = 1 settles
ĒĒ11 = 1 does not = 1 does not
settlesettle
EE22 = 2 settles = 2 settles
ĒĒ22 = 2 does not = 2 does not
settlesettle
Settlement occurs = E1∪∪E2
Tilting occurs = E1ĒĒ2∪∪ĒĒ1E2
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de Morgan’s rulede Morgan’s rule
EE1 1 = pipe 1 breaks= pipe 1 breaks
EE22 = pipe 2 breaks = pipe 2 breaks
1 2
Water Supply
E = failure in water supply = EE = failure in water supply = E11∪∪EE22
no failure in water supply = no failure in water supply = ĒĒ = =
EE11∪E∪E22
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2121EEEE
nEEEEEE n ....... 2121
Event of “no failure”
Extension to n events
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de Morgan #2de Morgan #2
2121 EEEE
nn EEEEEE ........ 2121
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Basis of Probability Basis of Probability EstimationEstimation
a)a) Subjective assumption e.g. P(Q) = 1/2Subjective assumption e.g. P(Q) = 1/2
b)b) Relative frequency e.g. Relative frequency e.g.
P(Q)=502/1000P(Q)=502/1000
c)c) Bayesian (a)+(b) Bayesian (a)+(b) judgment + limited observation judgment + limited observation
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Probability of UnionProbability of Union
in generalin general
E1 E2
1 2( )P E E
1 2 1 2( ) ( ) ( )P E P E P E E
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Using de Morgan’s ruleUsing de Morgan’s rule
)(1)( 321321 EEEPEEEP
)(1 321 EEEP
P (intersection) conditional
probability
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( ) ( | ) ( )P AB P A B P B ( ) ( | ) ( )P AB P B A P A
( ) ( | ) ( )P ABC P A BC P BC
( | ) ( )P B C P C
or or
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Statistical independence
if E1 and E2 are s.i.
2 1 2( ) ( )P E E P E
1 2 1( ) ( )P E E P Eor
s.i.
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Example:
E2 = flood in 廣西 on June E3 = flood in 哈爾濱 on June
E1 = flood in 廣東 on June
P(E1) = 0.1; P(E2)=0.1; P(E3) = 0.1
121 3.0| EPEEP E1 and E2 are not s.i.
1.0| 31 EEPE1 and E3 are s.i.
1EP
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1 2 1 2 2( ) ( ) ( )P E E P E E P E
1( )P Eif E1 and E2 are s.i.
1 2 3 1 2 3( ) ( ) ( ) ( )P E E E P E P E P Eif all are s.i.
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s.i. and m.e. (mutually exclusive)
if E1 and E2 are m.e.
1 2( ) 0P E E
1 2 1( ) ( )P E E P E
if E1 and E2 are s.i.
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B
C
A
2
1
3
P(E1)=2/5
P(E2)=3/4
P(E3)=2/3
P(E3|E2)=4/5
P(E1|E2E3)=1/2
a) P(go from A to B through C)
32EEP 2 3( ) ( )P E P E
5
3
4
3
5
4
3 2 2( ) ( )P E E P E
E1 : ① is openP 2.15
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b)
P(go from A to B)
132 EEEP
132132 EEEPEPEEP
1 2 3 2 3
3 2( | ) ( )
5 5P E E E P E E
7.0 1/2 3/5
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T.O.T (Theorem of Total Probabilities)
Bayes theorem
AP
EPEAPAEP jj
j
||
P(A) = P(A|E1)P(E1)+P(A|E2)P(E2)+…+P(A|En)P(En)
Ei’s are m.e. and c.e.
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7.0GP 3.0GP
good enough for construction
2.0|;8.0| GTPGTP
9.0|;1.0| GTPGTPpositive
E 2.30 aggregate for construction
engineer's judgment based on geology and experience
crude test
reliability (or quality) is as follows:
not a perfect test
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After 1 successful test, what is P(G)?
TP
GPGTPTGP
||
3.01.07.08.0
7.08.0
95.0
( | ) ( )
( | ) ( ) ( | ) ( )
P T G P G
P T G P G P T G P G
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After another successful independent test, P(G)?
GPGTPGPGTP
GPGTPTGP
||
||
22
22
05.01.095.08.0
95.08.0
993.0
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What if the two tests were performed at the same time?
)()|()()|(
)()|()|(
2121
2121
GPGTTPGPGTTP
GPGTTPTTGP
1 2
1 2 1 2
( | ) ( | ) ( )
( | ) ( | ) ( ) ( | ) ( | ) ( )
P T G P T G P G
P T G P T G P G P T G P T G P G
0.7
0.30.7
3.1.1.7.8.8.
7.8.8.
993.0
P(G)
UST
HKU
0.7
0.3
After 1 test
0.95
0.77
After 2 tests
0.993
0.965
… 5
1.0000
0.9999
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Random variables
A device to:
a) formalize description of event
b) facilitate computation of
probability
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PMF PX(x) FX(x)
PDF fX(x) FX(x)
CDF
( )( ) X
X
dF xf x
dx
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Main descriptors of R.V.
The PMF or PDF completely define the r.v.
Descriptors give partial information about the r.v.
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Mean value
Define = E(X)
= expected value of X or mean value of X
( )i X ii
x P xa measure of central tendency a measure of central tendency
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Measure of spread
Standard deviation X
X dimensionless %
range
( )Var X
X
X
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Expected value of function
( ) ( )XE X xf x dx
recall
( ) ( ) ( )XE g X g X f x dx
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Recall2( ) [ ( )] ( )XVar X x E X f x dx
2 2( ) ( ) ( )Var X E X E X
After some algebra,
2 ( )Xx f x dx
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0.00
0.05
0.10
0.15
0.20
0.25
-5 0 5 10 15 20
fX(x)
x
Normal distribution
= 5
X : N (, )
N (5, 2)
x
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Effect of varying parameters ( & )
fX(x)
x
A
B
C
for C
for B
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S: N (0,1)
Standard normal distribution
0.00
0.10
0.20
0.30
0.40
0.50
-6 -4 -2 0 2 4 6
fX(x)
x
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(0) 0.5 (1) 0.8413
(2) 0.97725 ( 1) 1 (1) 0.1587
(4) 1 0.00003167 0.9999683
-5 -4 -3 -2 -1 0 1 2 3 4 5
21
21( )
2
a sa e ds
a
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Page 380 Table of Standard Normal Probability
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Example: retaining wall
x
F
Suppose X = N(200,30)
(230 260)
260 200 230 200
30 30
(2) (1)
0.977 0.841 0.136
P x
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If the retaining wall is designed such that the reliability against sliding is 99%,
How much friction should be provided?
1
( ) 99%
200 2000.99
30 30
200(0.99)
30
P x F
F
F
1200 30 (0.99) 270F
2.33
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Lognormal distribution
Parameter
0 2 4 6 8
fX(x)
x
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21ln
2
2
2 22
ln 1 ln 1
Parameters
for 0.3,
ln mx
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Probability for Log-normal distribution
ln( )
aP X a
If a is xm, then is not needed.
ln ln( )
b aP a X b
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Other distributions
Exponential distribution Triangular distribution Uniform distribution Rayleigh distribution
p.224-225: table of common distribution
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Exponential distribution
x
fX(x)
( ) xXf x e x 0
1( )E X
21
( )Var X
100%X
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Beta distribution1 1
1
( ) ( ) ( )( )
( ) ( ) ( )
q r
X q r
q r x a b xf x
q r b a
a x b
0.0
0.1
0.2
0.3
0 2 4 6 8 10 12x
fX(x)q = 2.0 ; r = 6.0
a = 2.0 b = 12
probability
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0
1
2
3
4
0 0.2 0.4 0.6 0.8 1
Standard beta PDF
q = 1.0 ; r = 4.0
q = r = 3.0 q = 4.0 ; r = 2.0
q = r = 1.0
x
fX(x)
(a = 0, b = 1)
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Bernoulli sequence
Discrete repeated trials 2 outcomes for each trial s.i. between trials Probability of occurrence same for all
trials
SF
p = probability of a success
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SF
x = number of successp = probability of a success
P ( x success in n trials)
= P ( X = x | n, p) (1 )
( 0,1,2,..., )
x n xnp p
x
x n
Binomial distribution
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Examples
Number of flooded years
Number of failed specimens
Number of polluted days
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Example:
Given: probability of flood each year = 0.1
Over a 5 year period
1 45( 1) 0.1 (0.9) 0.328
1P X
P ( at most 1 flood year) = P (X =0) + P(X=1)
= 0.95 + 0.328
= 0.919
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P (flooding during 5 years)
= P (X 1)
= 1 – P( X = 0)
= 1- 0.95
= 0.41
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For Bernoulli sequence Model
No. of success binomial distribution
Time to first success geometric
distribution
E(T) =1/p = return period
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Significance of return period in design
Suppose a bldg is expected to last 100 years and if it is designed against 100 year-wind of 68.6 m/s
P (exceedence of 68.6 m/s each year) = 1/100 = 0.01
P (exceedence of 68.6 m/s in 100th year) = 0.01
Service life
design return period
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P (1st exceedence of 68.6 m/s in 100th year)
= 0.99990.01 = 0.0037
P (no exceedence of 68.6 m/s within a service life of 100 years)
= 0.99100 = 0.366
P (no exceedence of 68.6 m/s within the return period of design) = 0.366
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If it is designed against a 200 year-wind of 70.6 m/s
P (exceedence of 70.6 m/s each year) = 1/200 = 0.005
P (1st exceedence of 70.6 m/s in 100th year)
= 0.995990.005 = 0.003
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P (no exceedence of 70.6 m/s within return period of design)
= 0.995200 = 0.367
P (no exceedence of 70.6 m/s within a service life of 100 years)
= 0.995100 = 0.606 > 0.366
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How to determine the design wind speed for a given return period?
Get histogram of annual max. wind velocity
Fit probability model Calculate wind speed for a design
return period
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N (72,8)Example
V100
0.01
Design for return period of 100 years:
p = 1/100 = 0.01
100( ) 0.99P V V
100 720.99
8
V
V100 = 90.6 mph
Annual max wind velocity
Frequency
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Suppose we design it for 100 mph, what is the corresponding return period?
( 100)
100 721
8
1 (3.5)
0.000233
P V
T 4300 years
Alternative design criteria 1
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Pf = P (exceedence within 100 years)
= 1- P (no exceedence within 100 years)
=1- (1-0.000233)100 = 0.023
Probability of failure
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1x n xn t t
x n n
P ( x occurrences in n trials)
= limn
( )
!
xtt
ex
x = 0, 1, 2, …
Poisson distributionPoisson distribution
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P 3.42
Service stations along highway are located according to a Poisson process
Average of 1 station in 10 miles = 0.1 /mile
P(no gasoline available in a service station)
( ) 0.2P G
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(a) P( X 1 in 15 miles ) = ?
0 1
0 1.5 1 1.5
( 0) ( 1)
( ) ( )
0! 1!
(1.5) (1.5)
0! 1!0.223 0.335
0.558
t t
P X P X
t e t e
e e
No. of service stations
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(b) P( none of the next 3 stations have gasoline)
0 3
( 0 | 3, )
( 0 | 3,0.8)
3(0.8) (0.2)
0
0.008
P Y p
P Y
binomial
No. of stations with gasoline
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(c) A driver noticed the fuel gauge reads empty; he can go another 15 miles from experience.
P (stranded on highway without gasoline) = ?
P (S)( | 0) ( 0) ( | 1) ( 1)
( | 2) ( 2) ......
P S X P X P S X P X
P S X P X
No. of station in 15 miles
binomial Poisson
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x P( S| X = x ) P( X = x ) P( S| X = x ) P( X = x )
0 1 e-1.5 = 0.223 0.223
1 0.2 1.5 e-1.5 = 0.335 0.067
2 0.22 1.52/2! e-1.5 = 0.251 0.010
3 0.23 1.53/3! e-1.5 = 0.126 0.001
4 0.24 1.54/4! e-1.5 = 0.047 0.00007
Total = 0.301
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Alternative approach
Mean rate of service station = 0.1 per mile
Probability of gas at a station = 0.8
Mean rate of “wet” station = 0.10.8 = 0.08 per mile
Occurrence of “wet” station is also Poisson
P (S) = P ( no wet station in 15 mile)0
0.08 15 1.2(0.08 15)0.301
0!e e
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Time to next occurrence in Poisson process
Time to next occurrence = T is a continuous r.v.
( ) ( ) ( ) 1 ( )T T
d d df t F t P T t P T t
dt dt dt
( )P T t = P (X = 0 in time t)te
( ) 1 t tT
df t e e
dt
Recall for an exponential distribution
( ) xXf x e
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T follows an exponential distribution with parameter =
E(T) =1/
If = 0.1 per year, E(T) = 10 years
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Bernoulli Sequence
Poisson Process
Interval Discrete Continuous
No. of occurrence Binomial Poisson
Time to next occurrence Geometric Exponential
Time to kth occurrence Negative binomial Gamma
Comparison of two families of occurrence models
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Significance of correlation coefficient
= +1.0 = -1.0
y: strength
x: Length
GlassGlass
y: elongation
x: Length
SteelSteel
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= 0 0< <1.0
y: ID No
x: height
y: weight
x: height
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Functions of Random Variable Functions of Random Variable (R.V.)(R.V.)
In general, Y = g(X)
Y = g(X1, X2,…, Xn)
If we know distribution of X distribution of Y?
M = 4X + 10
X5
22
M1)
X
b
b = Xtan 2)
cost of delay = aX23) where X – length of delay
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Consider M = 4X + 10Consider M = 4X + 10
Observation: the distribution of y depends on
(1) Distribution of X
(2) g(X)
X654
0.6
0.2 0.2
M343026
0.6
0.2 0.2
X654 M343026
wider distribution
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2
11
2
1
2
2 3 2
1 1 2 3 3exp 2
2 22 2
1e 0,1
2
, 0,1
y x x
y
dgf y f g f y
dy
y
N
xif x N y N
Y X X
X YX
Eq. 4.6Eq. 4.6
E4.1
See E4.1 in text for details
3
2
xy
,
3, 2x N
1 2 3 2dx
g x ydy
X
21
21
2
x
xf x e
X
where
1
1y x
dgf y f g
dy
Y X
For monotonic For monotonic function g(X)function g(X)
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if X1 and X2 are Poisson with mean rates 1 and 2 respectively Z is Poisson with z = 1 + 2
(see E4.5 on p. 175)
1 2Z X X
if X is N(,) Y is N(a + b, a)Y aX b ( 0)a
if X is N(,) Y is N(a, a)
if X is LN(,) Y is LN(ln a + , )Y aX
( 0)a
Summary of Common ResultsSummary of Common Results
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Summary of Common Results Summary of Common Results (Cont’d)(Cont’d)
if X1 and X2 are N(,) and N(,) respectively Z is N(z,z)where
if X1 and X2 are s.i. 12 = 0
1 1 2 2Z a X a X
2 2 2 21 1 2 2 1 2 12 1 22Z a a a a
if Xi = N(i,i) ; i = 1 to n Z is N(z,z)where1 1 2 2
... n n
Z a X a X
a X
1 1 2 2 ...Z n na a a
2 2Z i i
i
a correlation terms
1 1 2 2Z a a
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E4.8E4.8
S = D + L + W
Column with capacity, R
D = 4.2 , D = 0.3 , D = 7%L = 6.5 , L = 0.8 , L = 12%W = 3.4 , W = 0.7 , W = 21%
S = total load = D + L + W
S = D + L + W = 14.1S = 0.32 + 0.82 + 0.72 =1.1
a) P(S > 18) = 1 –
= 1 – (3.55) = 0.000193
18 14.1
1.1
Assume D, L, W Assume D, L, W are s.i.are s.i.
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E4.8 (Cont’d)E4.8 (Cont’d)P(failure) = P(R < S)
= P(R – S < 0)= P(X < 0)
R = N(R, RR)where R = 1.5S
= 1.5 x 14.1 = 21.15R = 0.15 R = N(21.15, 3.17)
X = -14.1 + 21.15 = 7.05X = 1.12 + 3.172 = 3.36
P(F) = = (-2.1) = 0.0180 7.05
3.36
X = R – SX = R – S
RR – design capacity – design capacity
1.5 – design safety factor, SF1.5 – design safety factor, SF
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If the target is P(F) = 0.001 R = ? and assume R = 0.15
Recall:X = R - 14.1
X = (0.15R)2 + 1.12
= -1(0.001) = – 3.09
0.0225R2+1.21 = 20.8 – 2.95R + 0.105R
2
0.0825R2 – 2.95R +19.59 = 0
R = 8.812 or 26.9
2 2
14.1
0.15 1.1R
R
2
22 2 14.1(0.15 ) 1.1 4.56 0.324
3.09R
R R
Set: P(F) = = 0.001 2 2
14.1
0.15 1.1R
R
Since R should be larger than 21.5R = 26.9
Check R = 8.812
P(F) = = (3.09) 1.0
8.812 14.1
1.7197
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W = weight of a truck = N(100, 20)We are interested in the total weight of 2
trucks
or
1.
Total weight = T = 2W
Normal with
T = 2W = 200
T = 2W = 40
Total weight = T = W1 + W2
Normal with T = 200
T =
=
=
if s.i. and 1 = 2 = 20
2 21 220 20 2
2 220 2020 2
2.
?
T
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E4.10E4.10
Sand Footing
P
S
Sand Property – M Footing Property – B and I
PBIS
M
Assume P, B, I, and M are s.i. and log-normal with parameters P, B, I, M and P, B, I, M, respectively
c.o.v.
P B I
M
1.0 6.0 0.6
32.0
0.10 0
0.10 0.15
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Central Limit TheoremCentral Limit Theorem
S will approach a normal distribution regardless the individual probability distribution of Xi if N is large enough
0.5
0.7
0.4
0.2
0.6
0.3
0.1
-1 1
P N = 1 S = X1
0.5
0.7
0.4
0.2
0.6
0.3
0.1
-2 0 2
P N = 2 S = X1 + X2
0.5
0.7
0.4
0.2
0.6
0.3
0.1
2 -4 -2 0 4 8 6 10
P N = 10 S = X1 + X2 +…+X10
P
0.5
0.7
0.4
0.2
0.6
0.3
0.1
18 16 14 12 10 8 6 4 2 0 -2 -4
N = 20 S = X1 + X2 +…+ X20
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First order approximation:
2 2
X X X
X
X
X X
dgE g X g E X g
dX
dg dgVar g X Var X Var X
dX dX
g(X)
g’(X) (X – X)
X
g(X)
X X
Taylor Series ApproximationTaylor Series Approximation
X X X
2X
g(X) g( ) g ( ) X-
"...
2 X
gX
x - x
X X
dgg(X) g( ) X-
dXX
Var(X)
…
is known and are known valuesX
X
dgg X g
dX
2 2
X X X
X
X
X X
dgE g X g E X g
dX
dg dgVar g X Var X Var X
dX dX
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Observe validity of linear approx Observe validity of linear approx depends on:depends on:
1) Function g is almost linear, i.e. small curvature
2) x is small, i.e. distribution of X is narrow
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UsesUses
1. Easy calculations
2. Compare relative contributions of uncertainties – allocation of resource
3. Combine individual contributions of uncertainties
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Reliability ComputationReliability ComputationSuppose R denotes resistance or capacity
S denotes load or demandSatisfactory Performance = {S < R}
PS = P(S < R) and Pf = 1 - PS
Case 1: If R, S are normal
where Z = S – R and Z = S2 + R
2
0 0
0 Z
Z
P S R P S R P Z
Case 2: If R, S are lognormal
where Z = S – R and Z = S2 + R
2
ln11 1 Z
Z
SP S R P P Z
R
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Case 3: If R is discrete, S is continuous
i ii
S i R ii
P S R P S R R r P R r
F r P r
Example on Case 3
S = N(5, 1)
5 5 6 6 7 7P S R P S P R P S P R P S P R
5 5 6 5 7 50.1 0.3 0.6
1 1 1
0 0.1 1 0.3 2 0.6
0.5 0.1 0.84 0.3 0.98 0.6
0.889
r
P(R = r)
5 6 7
0.1
0.3
0.6
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Reliability – Based DesignReliability – Based Design
Observe for Case 1 in which R and S are both Normal
2 2
R SS
S R
P
If PS Reliability
= Reliability index = -1(PS)
Design Design RR = = SS + + SS22 + + RR
22
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Example:S = N(5, 2)R = N(R, 1)
R = ?
Require Pf = 0.001 or PS = 0.999 = -1(0.999) = 3.1
Design R = 5 + 3.122 + 12 = 11.93