T73S04 (R5V2/3) Session 38BProbabilistic Assessments

Rick Bradford

Full procedure here http://rickbradford.co.uk/ProbabilisticProcedureForR5V2_3.pdf

Exclusion Clause

• I am quite ignorant about statistics

• Probabilistic assessment is not statistical analysis

• My spin: probabilistic assessment is deterministic assessment done many times with suitable probability weighting of the inputs – “Monte Carlo”

Introduction & Purpose

• Why probabilistic R5V2/3?– If deterministic R5V2/3 gives a lemon– If components have already failed– If you want to know about lifetime / reality

• Trouble with ‘bounding’ data is,– It’s not bounding– It’s arbitrary– Most of the information is not used

What have we done? (mostly 316H)

• 2008-2010: HPB/HNB Bifs – R5V4/5 E/REP/BBAB/0009/AGR/08, Rev.001

• 2011: HYA/HAR Bifs – Creep Rupture (BIFLIFE) E/REP/BBAB/0019/HAR/11 and E/REP/BBAB/0021/HAR/11

• 2012: HYA/HAR Bifs – R5V2/3 (BIFINIT) E/REP/BBAB/0023/AGR/12 and E/REP/BBAB/0025/HAR/12

• 2013: HPB/HNB Bifs – R5V2/3 with carburisation E/REP/BBAB/0027/AGR/13

• 2013-14 Various DNB FPU probabilistics (R6)• 2015 in progress Spine Welds 12.3 R5V4/5

Psychology Change

• Best estimate rather than conservative

• Including best estimate of error / scatter

• The conservatism comes at the end…– ..in what “failure” (initiation) probability is

regarded as acceptable…– …and this may depend upon the application

(safety case v lifetime)

So what is acceptable?

• Will vary – consult Customer

• For “frequent” plant might be ~0.2 failures per reactor year (e.g., boiler tubes)

• For HI/IOGF plant might be 10-7pry to 10-5 pry (maybe)

• The assessment would be the same, but the latter is more dependent on the uncertain tails of the distributions

What’s the Downside?• MINOR: Probabilistic assessment is more work

than deterministic

• MAJOR: Verification– The only way of doing a meaningful verification of a

Monte Carlo assessment is to do an independent Monte Carlo assessment!

– Or is it…..?

• Learning Points: Can be counterintuitive– Acceptance by others– Brainteaser

Limitation re R5V2/3 Applications

• Only the crack initiation part of R5V2/3 addressed

• Not the “precursor” assessments– Primary stress limits– Stress range limit– Shakedown– Cyclically enhanced creep

• Complete job will need to address these separately

Computing Platform

• So far we’ve used Excel• Latin Hypercube add-ons available• RiskAmp / “@Risk” Being developed• Most coding in VBA essential• Minimise output to spreadsheet during

execution• Matlab might be a natural platform• I expect Latin Hypercube add-ons would

also be available – but not checked• Develop facility within R-CODE/DFA - No

Run Times

• Efficient coding crucial • Typically 50,000 – 750,000 trials• (Trial = assessment of whole life of one

component with just one set of randomly sampled variables)

• Have achieved run times of 0.15 to 0.33 seconds per trial on standard PCs (~260 load cycles)

• Hence 2 hours to 3 days per run

Methodology

• We shall assume Monte Carlo

• Monte Carlo is just deterministic assessment done many times

• So the core of the probabilistic code is the deterministic assessment

Hysteresis Cycle Construction• R5V2/3 Appendix A7• Always sketch what the generic cycle will

look like for your application• Helpful to write down the intended

algorithm in full as algebra• Recall that the R5 hysteresis cycle

construction is all driven by the elastically calculated stresses

• Example – see http://rickbradford.co.uk/HysteresisCycleConstructionMethodologyForInteractingCycles.pdf

• Remember that the dwell stress cannot be less than the rupture reference stress

• R5V2/3 Appendix A7• Always sketch what the generic cycle will

look like for your application• Helpful to write down the intended

algorithm in full as algebra• Recall that the R5 hysteresis cycle

construction is all driven by the elastically calculated stresses

• Example – see http://rickbradford.co.uk/HysteresisCycleConstructionMethodologyForInteractingCycles.pdf

• Remember that the dwell stress cannot be less than the rupture reference stress

Non-Closed Cycles

• Actual plant cycle sequences may not produce cycles whch are closed even in principle

• e.g., cold – operating – hot standby

• My approach described in detail here, http://rickbradford.co.uk/HysteresisCycleConstructionMethodologyForInteractingCycles.pdf

• e.g., symmetrisation: find reverse stress datum for left and right hand half-cycles separately and use the average.

Primary Reset Issue: 316H

• Is creep strain reset to zero at the start of each dwell – so as to regenerate the initial fast primary creep rate?

• Existing advice is unchanged for deterministic assessments…

• Reset primary creep above 550oC

• Do not reset primary creep at or below 550oC…

• …use continuous hardening instead (creep strain accumulates over cycles)

Primary Reset Issue: 316H

• For probabilistic assessments I advise the use of primary reset at all temperatures (for justification see E/REP/BBAB/0022/AGR/12)

• But with two alleviations,– Application of the zeta factor, z– Only reset primary creep if the previous

unload caused significant reverse plasticity

• “significant” plasticity in this context has been taken as >0.01% plastic strain, though 0.05% may be OK

The zeta factor

tt

t

cc tdTtt ,,,~c

tt

t

cRref

Pc tdTt ,,,~

c

cc ~ Pc

Pc ~

fccD /

Pccsodeod Z

E

Probabilistics

• Is it all just normal distributions?

• No

• Also Log-normal, also…

• All sorts of weird & wonderful pdfs

• Or just use random sampling of a histogram…

Normal and Log-Normal PDFs

• Normal pdf

• Log-normal is the same with z replaced by ln(z)

• Integration measure is then d(ln(z))=dz/z

2

2

2exp

2

1

zz

zP

Go tohttp://

rickbradford.co.uk/PeterHoltNotesOnPDFs.pdf

Non-Standard Distribution:Elastic Follow-Up

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 2 4 6 8 10

Z

Pro

ba

bil

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pe

r u

nit

Z

Non-Standard Distribution: Overhang

actual plant overhang distribution

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.15 0.25 0.35 0.45 0.55 0.65 0.75 0.85

overhang (m)

fra

cti

on

of

bif

urc

ati

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s

Non-Standard Distribution Thermal Transient Factor wrt Reference Trip

0

50

100

150

200

250

300

0.50

0.53

0.56

0.59

0.62

0.65

0.68

0.71

0.74

0.77

0.80

0.83

0.86

0.89

0.92

0.95

0.98

1.01

1.04

1.07

1.10

System Load Factor

Fre

qu

en

cy

How Many Distributed Variables

• Generally – lots! (20 to 40)

• If a quantity is significantly uncertain…

• …and you have even a very rough estimate of its uncertainty…

• …then include it as a distributed variable.

• The Latin Hypercube can handle it

Where are the pdfs?

• “But what if no one has given me a pdf for this variable”, I hear you cry.

• Ask yourself, “Is it better to use an arbitrary single figure – or is it better to guestimate a mean and an error?”

• If you have a mean and an error then any vaguely reasonable pdf is better than assuming a single deterministic value

How is Probabilistics Done?

• (Monte Carlo) probabilistics is just deterministic assessment done many times

• This means random sampling (i.e. each distributed variable is randomly sampled and these values used in a trial calculation)

• But how are the many results weighted?

Options for Sampling: (1)Exhaustive

(Numerical Integration)

• Suppose we want +/-3 standard deviations sampled at 0.25 sd intervals

• That’s 25 values, each of different probability.

• Say of 20 distributed variables• That’s 2520 ~ 1028 combinations• Not feasible – by a massive factor

Options for Sampling: (2)Unstructured Combination

• Each trial has a different probability

• Range of probabilities is enormous

• Out of 50,000 trials you will find that one or two have far greater probability than all the others

• So most trials are irrelevant

• Hence grossly inefficient / results vary wildly from run to run (no convergence)

Options for Sampling: (3)Random but Equal Probability

• Arrange for all trials to have the same probability

• Split all the pdfs into “bins” of equal area (= equal probability) – say P

• Then every random sample has the same probability, PN, N = number of variables

Equal Area “Bins” Illustrated for 10 Bins (More Likely to Use 10,000 Bins)

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-3.000 -2.000 -1.000 0.000 1.000 2.000 3.000

Bins

mean of last bin (1.755)

Bins v Sampling Range

• 10 bins = +/- 1.75 standard deviations (not adequate)

• 300 bins = +/- 3 standard deviations (may be adequate)

• 10,000 bins = +/- 4 standard deviations (easily adequate for “frequent”; not sure for “HI/IOGF/IOF”)

Optimum Trial Sampling Strategy

• Have now chosen the bins for each variable

• Bins are of equal probability

• So we want to sample all bins for all variables with equal likelihood

• How can we ensure that all bins of all variables are sampled in the smallest number of trials?

• (Albeit not in all combinations)

Answer: Latin Hypercube

• N-dimensional cube

• N = number of distributed variables

• Each side divided into B bins

• Hence BN cells

• Each cell defines a particular randomly sampled value for every variable

• i.e., each cell defines a trial

• All trials are equally probable

Latin Hypercube

• A Latin Hypercube consists of B cells chosen from the possible BN cells such that no cell shares a row, column, rank,… with any other cell.

• For N = 2 and B = 8 an example of a Latin Hypercube is a chess board containing 8 rooks none of which are en prise.

• Any Latin Hypercube defines B trials which sample all B bins of every one of the N variables.

Example – The ‘Latin Square’

• N=2 Variables and B=4 Samples per Variable

1 2 3 4

1

2

3

4

•B cells are randomly occupied such that each row and column contains only one occupied cell.

•The occupied cells then define the B trial combinations.

Generation of the Latin Square

• A simple way to generate the square/hypercube

4 123

3

1

4

2

•Assign the variable samples in random order to each row and column.

•Occupy the diagonal to specify the trial combinations.

•These combinations are identical to the ones on the previous slide.

Homework

• Given N and B, how many different Latin hypercubes are there?

Range of Components

• Modelling just one item – or a family of items?

• Note that distributed variables do not just cover uncertainties but can also cover item to item differences, – Temperature– Load– Geometry– Metal losses

Plant History• A decision is required early on…

• Model on the basis of just a few idealised load cycles…

• …or use the plant history to model the actual load cycles that have occurred

• Can either random sample to achieve this

• Or can simply model every major cycle in sequence if you have the history (reactor and boiler cycles)

• Reality is that all cycles are different

Cycle Interaction• Even if load cycles are idealised, if one or

more parameters are randomly sampled every cycle will be different

• Hence a cycle interaction algorithm is obligatory

• And since all load cycles differ, the hysteresis cycles will not be closed, even in principle

• This takes us beyond what R5 caters for

• Hence need to make up a procedure

Unapproved Cycle Interaction• “Symmetrisation” of the hysteresis cycles

has no basis when they are not repeated• Suggested methodology is,

• http://rickbradford.co.uk/HysteresisCycleConstructionMethodologyForInteractingCycles.pdf

• This leads to “symmetrisation on average”

= 0 symmetrises every cycle = 0.93 is believed reasonable

symrevi

revi

revi

,1 1

i

symrevi

i

revi

,

Multiple Assessment Locations

• In general you will need to assess several locations to cover just one component

• E.g., a weld location, a stress-raiser location, and perhaps a second parent location

• Crack initiation conceded when any one location cracks

• So need to assess all locations in parallel at the same time

Correlations Between Locations

• Are the material property distributions the same for all locations?

• Even if they are the same distributions, is sampling to be done just once to cover all locations? (Perfect correlation)

• Or are the properties obtained by sampling separately for each location (uncorrelated)

• Ditto for the load distributions

Time Dependent Distributions

• Most distributed variables will be time independent

• Hence sampled once at start of life, then constant through life

• But some may involve sampling repeatedly during service life

• E.g., transient loads are generally different cycle by cycle

Time Dependent Distributions• Cycle-to-cycle variations in cyclic loading

may be addressed…– Deterministically, from plant data– Probabilistically but as time independent

(sampled just once) – not really right– Probabilistically sampled independently on

every cycle

• Latter case can be handled outwith the Latin Hypercube but must be on the basis of equal probabilities

• Combination of the above for different aspects of the cyclic loading

Imposing Correlations

• Correlations can be extremely important to the result

• Proprietary software will include facilities for correlating variables

• Input the correlation coefficient

• If writing your own code, here’s how correlation may be imposed…

•

Imposing Correlations

• For correlation between variables x and y, invent a new variable v, sampled independently of x, and then set y to,

• This will generate the desired correlation Cxy between x and y.

• For mutual correlations between three or more variables, see Appendix A of http://rickbradford.co.uk/ProbabilisticProcedureForR5V2_3.pdf

vCxCy xyxy 21

Results• If n cases of crack initiation occur in a run of N

trials, then your estimate of the crack initiation probability is P = n/N

• If n relates to the period from now to the end of life, then so does P

• If n relates to a specified year, then so does P• If n relates to a specified individual component,

then so does P• If the N trials randomly sample all the

components in the reactor, then P is the reactor-average cracking probability per component

Results

• If there are C components per reactor, the probability of cracking per reactor is then CP where P is the reactor-average cracking probability per component provided that CP << 1

• More precisely, for any CP, the probability of one or more cracks per reactor is,

1 – (1 – P)C

How Many Trials Do You Need?

• It will depend on the application

• The smaller the probability of cracking, the larger the number of trials needed to calculate it by Monte Carlo

• You cannot justify a cracking probability of 10-4 per component in 1000 trials of single components. You would need ~100,000 single-component trials

How Many Trials Do You Need?

• Suppose you want to justify a cracking probability of 0.1 per reactor year for a reactor consisting of 1000 components.

• Suppose you run for 10 future years, then you need to justify 1 crack per 1000 components, i.e., 10-3

• So 10,000 single-component trials would be appropriate

How Many Trials Do You Need?

• If there is a large number of components per reactor, the number of trials needed to get a good reactor-average probability of cracking will be much smaller than required to resolve probabilities of cracking for individual components across the whole reactor if these individual components differ (e.g., different temperatures, stresses, geometry, etc).

How Many Trials Do You Need?

• Convergence should always be checked in two ways…

• Converging to stable probability in real time as the run proceeds

• Repeat runs with identical input to confirm reproducibility of result

Convergence of Initiation Rate

Initiation Results - Restricted Tubes

0.0000

0.0005

0.0010

0.0015

0.0020

0.0025

0.0030

0.0035

0.0040

0.0045

0 10000 20000 30000 40000 50000

Trial No.

Init

iati

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Rat

e p

er

Tri

al

Case 1

Case 1b

Convergence of Initiation Rate

Initiation Results - Unrestricted Tubes

0.0000

0.0005

0.0010

0.0015

0.0020

0.0025

0 50000 100000 150000 200000 250000

Trial No.

Init

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Case 3

Case 3b

Annual Probability Showing Upturn

Annual Crack Initiation Probabilities versus Year (All Tubes)

0.0000

0.0200

0.0400

0.0600

0.0800

0.1000

0.1200

1985 1990 1995 2000 2005 2010 2015 2020 2025

Year

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er

Ye

ar

Output Correlations

• Look at correlation between cracking probability and certain distributed variables – to identify the significant variables

• Can be salutary

• Factors which seem important in deterministic assessments may not be so important in the probabilistics

• E.g., HYA/HAR restricted tubes – temperatures not as important as stress

Specific Interesting Outputs

• Track the proportion of cycles which deploy primary reset – does this correlate with cracking? (Likely)

• Track the proportion of cycles with dwell stresses above the rupture reference stress – does this correlate with cracking? (Likely)

INPUT DISTRIBUTIONS

Beyond the scope of this session, but for 316H see the full presentation from December 2012 here, http://

rickbradford.co.uk/StructuralIntegrity.html

and also the report,

E/REP/BBAB/0022/AGR/12

THE

END