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Lecture 3: Probabilistic Design

Uncertainty in Engineering Systems and Risk Managements

Professor CHOI Hae-Jin

Contents

• Error Propagation

• Decision-making under Uncertainty

• Probabilistic Design

Conventional Approach

• Conventional engineering design uses a deterministic approach. It disregards the fact that material properties, the dimensions of the parts, and the externally applied loads vary statistically.

• In conventional design, theses uncertainties are handled by applying a factor of safety

Probabilistic Approach

• In critical design situation, such as aircraft, space, and nuclear applications, it is often necessary to use a probabilistic approach for quantifying uncertainty and increasing reliability of a system.

Error Propagation

• When working with random variables, it is necessary to propagate error (variability) through systemic equations (or models).

• For example, we need to know the variability of the deflection of the cantilever beam with a given variability of the load

3

3

L

EI

P

???δP

Error Propagation

• For normal distributions, following equations are a method for estimating propagated errors.

1 2

1/ 22

2

1

when an output is ( ),

the mean of is ( , ,..., ), and

the standard deviation of is

ny x x x

n

y xi

i ix

1 2 ny x , x , ..., x

y

y (5.6)

Error Propagation

• The estimate of the mean of a function relationship comes from substituting the mean values of the random variables.

• The estimate of the variance of a function relationship is simply the weighted variances of the constitutive variances, the weighting factors being the squares of the partial derivatives evaluated at the means

Mean and standard deviations for simple algebraic operations (x, y : independent

random variables)

Function Mean Standard Deviation

a a 0

x x x

ax x a x

ax xa xa

x y x y

1/ 22 2

x y

x y x y

1/ 22 2

x y

xy x y 2 2 2 2 1/ 2( )x y x y x yC C C C

/x y /x y 1/ 2

2 2 2

/ / 1x y x y yC C C

1/ x 211 x

x

C

21xx

x

CC

x 211

8x xC

21

12 16

x

x xC C

nx 2( 1)

12

n

x x

n nC

22( 1)

14

n

x x x

nn C C

Where, Cx =σx

μx Coefficient of variation of

the random variable. x

Table 5.3

Error Propagation by Simulation

• In following situation, it is very difficult to estimate error propagation by the method.

– Distribution of random input is not a normal distribution (such as, lognormal and Weibull)

– Function is not a form of mathematics, but computer simulation or experiments.

• We can employ Monte Carlo simulation for propagating the error (variability).

Error Propagation by Simulation

• Procedure of the Monte Carlo simulation1. Define a domain of possible inputs.

2. Generate an instance of inputs randomly from the domain using random number generator.

3. Perform a deterministic computation using the instance.

4. Repeat step 2 and 3 to collect enough amount of data

5. Aggregate the results of the individual computations into the final result to estimate a statistical distribution

Example of Error Propagation

• Example 6.1: the load of the cantilever beam varies as P ~ N(100, 10) N, what are the mean and standard deviation of deflection?

Deterministic parametersE = 200 GPaI = 1000 mm4

L = 500 mm

P

L

Example of Error Propagation

• δ= (L3/3EI)P = AP where

• Deterministic parameter

• A=L3/3EI =

• µP = 100 and σP2 = 10

• From the Table 5.3 or Eq. 5.6, µδ = AµP and σδ=AσP

• µδ = AµP = 0.20833 x 100 = 20.833 (mm)

• σδ= AσP = 0.20833 x 3.16 = 0.6587 (mm)

-3 3 3

9 12

(500 10 ) 5 10(m/N)

3 200 10 1000 10 24

0.20833 (mm/N)

Probabilistic Decision-making

• Example 6.2: The requirement of the cantilever beam deflection is less than 22 (mm). Does this beam design satisfy the requirement with 99% chance?

Find the deflection limit (critical point) of 99% percentile.

0.99

δ=??µδ = 20.833 (mm)

Probabilistic Decision-making

lim,0.99 lim,0.99

lim,0.99

x=2.326 at =0.01 (i.e.,1- 0.99)

20.833x= =2.326

0.6587

20.833 2.326 0.6587 22.37 22

From Table 5.2, the critical point is

or

Therefore,

This beam design is NOT acceptable. 0.99

δlim,0.99µδ = 20.833 (mm)

x

1-Ф(x)=α

=2.326

=0.010.99

Probabilistic Design Approach

• Procedure of probabilistic design approach

1. Identify sources of uncertainty and system constraints (such as yield strength, deflections, etc.),

2. Establish system function (model),

3. Categorize system parameters: random variables, deterministic parameters, and design variables,

4. Find the distribution of system output by error propagation, and

5. Determine the values of design variables

Probabilistic Design Approach

• Example 6.3: the load of the cantilever beam varies as P ~ N(100, 10) N, determine the maximum length of beam so that the deflection of the beam is less than 22(mm) with 99% chance.

P

L

Deterministic parametersE = 200 GPaI = 1000 mm4

Probabilistic Design Approach

• STEP 1: Identify sources of uncertainty and system constraints

• Source of uncertainty is in loading, P ~ N(100,10) N

• System constraints: Deflection < 22 (mm)

• STEP 2: Establish system function (model)

• Deflection δ = PL3/(3EI)

• STEP 3: Categorize system parameters

• Random variables: P

• Deterministic parameters: E, I

• Design variable: L

Probabilistic Design Approach

• STEP 4: Find the distribution of output by error propagation

δ= (L3/3EI)P = (1/3EI) PL3 = A PL3

Where the deterministic parameters, Α =1/3EI =

1/(3*200*10e9*1000*10e-12) = 1/600

Mean of propagated error, µδ = A µP L3

Standard deviation of propagated error, σδ = A σP L3

Therefore, the estimated distribution of deflection

δ ~ N(µδ, σδ2)

Probabilistic Design Approach

• STEP 5:Determine the values of design variables

3

lim,0.99lim,0.99

3

3 3 -3

lim,0.99

-3 -3

3

x=2.326 at =0.01 (i.e.,1- 0.99)

x= =2.326

2.326 22 10

22 10 22 10L

A( 2.326 ) (1/

p

p

p p

p p

A L

A L

or A L A L

From Table 5.2 the critical point is

Therefore,

3 0.4973600)(100 2.326 3.162)

497.3( )mm