# Probabilistic R5V2/3 Assessments Rick Bradford Peter Holt 17 th December 2012.

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Probabilistic R5V2/3 Assessments Rick Bradford Peter Holt 17 th December 2012

Introduction & Purpose Why probabilistic R5V2/3? If deterministic R5V2/3 gives a lemon 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? (All 316H) 2009/10: HPB/HNB Bifs – R5V4/5 (Peter Holt) 2011: HYA/HAR Bifs – Creep Rupture (BIFLIFE) 2012: HYA/HAR Bifs – R5V2/3 (BIFINIT) Today only R5V2/3

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 -7 pry to 10 -5 pry (maybe) But the would be assessment the same!

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

Aim of Today Probabilistic “How to do it” guide For people intending to apply – I hope Knowledge of R5V2/3 assumed We’ve only done it once So everything based on HAR bifurcations experience (316H)

Limitation 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

Agenda Introduction9:30 – 10:00 Computing platform10:00 – 10:15 Methodology –Deterministic 10:15 – 10:45 –Probabilistic10:45 – 12:30 Lunch12:30 – 13:00 Input Distributions –Materials Data (316)13:00 – 14:30 –Loading / Stress14:30 – 14:45 –Plant History14:45 – 15:00

Computing Platform So far we’ve used Excel Latin Hypercube add-ons available RiskAmp / “@Risk” ?? 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?

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

Deterministic Methodology R5V2/3 Issue 3 (Revision 1 2013 ?) Will include new weldments Appendix A4 BUT when used for 316H probabilistics, we advise revised rules for primary reset NB: Deterministic assessments should continue to use the ‘old’ Manus O’Donnell rules E/REP/ATEC/0027/AGR/01

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 – HANDOUT (from BIFINIT-RB) Remember that the dwell stress cannot be less than the rupture reference stress

Talking to the HANDOUT A brief run-through the elements of R5V2/3 Appendix A7 hysteresis cycle construction methodology using the hysteresis cycle on the next slide as the basis of the illustration

Illustrative Hysteresis Cycle A B C D E F G H J

Weldments R5V2/3 Appendix A4 Initially use parent stress-strain data WSEF used in hysteresis cycle construction – not FSRF WER – leave nucleation cycles out of parent fatigue endurance Factor dwell stress by ratio of weld:parent cyclic strength (unless replaced by rupture reference stress) For 2mm thick we assumed weld = parent

Creep Dwell Creep relaxation, and hence damage, by integration of forward creep law Prohibits relaxation below rupture reference stress Strain hardening – both terms evaluated at the same strain Both evaluated at same point in scatter, h

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 550 o C Do not reset primary creep at or below 550 o C… …use continuous hardening instead (creep strain accumulates over cycles)

Primary Reset Issue: 316H For probabilistic assessments we advise the use of primary reset at all temperatures 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

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

A Brief Reminder of the Basics: PDFs versus cumulative distributions, definitions of mean, median, variance, standard deviation, CoV, correlation coefficient. Illustrative graphs. DO VIA HANDOUT Illustrate number of standard deviations against probability for normal distribution, e.g., 1.65 sigma = 95%, etc.

Non-Standard Distribution: Elastic Follow-Up

Non-Standard Distribution: Overhang

Non-Standard Distribution Thermal Transient Factor wrt Reference Trip

How Many Distributed Variables Generally – lots! 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

How Many Distributed Variables Here are those used for the HYA/HAR bifurcations (PJH distribution types given)… Bifurcation inlet sol inner radius (PERT) Bifurcation inlet sol outer radius (PERT) Boiler tube sol inner radius (PERT) Boiler tube sol outer radius (PERT) Inner radius oxidation metal loss k parameter (Log- Normal) Off-Load IGA at bore (Log-Normal) Chemical clean IGA at bore (Log-Normal or Uniform) Outer radius metal loss (Log-Normal)

How Many Distributed Variables Deadweight moment (Normal) Bifurcation thermal moment (Normal) Unrestricted MECT (not used by PJH) Gas temperature (Normal) Steam temperature (Normal) Metal temperature interpolation parameter (Normal) Follow-up factor, Z (PERT) Carburisation allowance (Log-Normal of Sub-Set) Number of restricted tubes post-clean (not used by PJH)

How Many Distributed Variables Restriction after 3rd clean (not used by PJH) Overhang distribution (actual overhangs used) Tube thermal moment (Normal) Bifurcation 0.2% proof stress (Log-Normal) Tube 0.2% proof stress (Log-Normal) Zeta (  ) (Log-Normal) Ramberg-Osgood A parameter (Log-Normal) Young's modulus (Log-Normal)

How Many Distributed Variables The weld fatigue endurance (Log-Normal) The bifurcation parent fatigue endurance (Log- Normal) The boiler tube fatigue endurance (Log-Normal) The minimum differential pressure in hot standby (Uniform distribution of a set of cases) The start-up peak thermal stress (Ditto) The trip peak thermal stress (Ditto) The minimum temperature during hot standby (Ditto)

How Many Distributed Variables Creep strain rate (weld) (Log-Normal) Creep strain rate (bifurcation) (Log-Normal) Creep strain rate (boiler tube) (Log-Normal) Creep ductility (weld) (Log-Normal) Creep ductility (bifurcation) (Log-Normal) Creep ductility (boiler tube) (Log-Normal)

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 41 distributed variables That’s 25 41 = 2 x 10 57 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

Options for Sampling: (2)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, P N, N = number of variables

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

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 B N 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 B N 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 1234 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 4123 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.

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, This leads to “symmetrisation on average” a = 0 symmetrises every cycle a = 0.93 is believed reasonable

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 “Failure” (crack initiation) conceded when any one location “fails” 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… HANDOUT

Terminology: “Trial” A trial is an assessment of one component for one particular possible plant history It covers all assessment locations for one component It covers the whole of life, hence all load cycles Have achieved 0.15 seconds per trial for 3 assessment locations, ~260 cycles over 260,000 hours life, on core i5 PC (41 distributed variables)

How Many Trials Do You Need? It will depend on the application The smaller the “failure” probability, the larger the number of trials needed to calculate it by Monte Carlo If there is a large number of components per reactor, the number of trials needed to get a good reactor-average “failure” probability will be much smaller than required to resolve “failure” probabilities for individual components across the whole reactor

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 Predictions for Repeat Cases 50,000 trials (restricted) 250,000 trials (unrestricted)

Outputs Cumulative number of crack initiations by now and by each future year May be fractional –Reactor average –Average per component –Optional: Prediction for individual components Annual probability of crack initiation – plot as graph against time

Annual Probability Showing Upturn

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., restricted tubes – temperatures not as important as stress

Learning Point? Probabilistics implies stress is the dominant issue (for this application), and yet… –It was approx operating year 26 before we commissioned FE models of the tailpipes! –Contrast with the huge sums spent on chemical cleaning over last 12 years –(Although this is also a boiler stability issue) Deterministic assessment did not reveal the relative importance of the stress analysis (and R5 methodology)

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

Materials Data: 316H Parent Bradford & Holt E/REP/BBAB/0022/AGR/12 Reviews 316H material data specifically for the purposes of BIFINIT There’s a lot of unpublished data around for 316H – hence the need for a review BIFINIT task has taken all year More than half time on materials review

Review for Your Application? You may not have the time, or the data, to make a similar review possible If yours is 316H parent then E/REP/BBAB/0022/AGR/12 should be a good guide For our thin (<2mm) sections we assumed weld = HAZ = parent You cannot do this for thick sections Bottom line: such a comprehensive review is not essential to benefit from probabilistics

Cyclic Creep Relaxation Under constant load 316H material shows a primary behaviour in which the strain rate drops with accumulated time/strain. Under a relaxing load, this behaviour is usually modelled as dependent on accumulated strain. Under a cyclic load application the relaxation would then be expected to show the following pattern Stress Time

Cyclic Stress Relaxation On the other hand, if the cycle unloading phase induces reverse plasticity, the creep deformation behaviour may revert to that at zero creep strain (so called ‘Primary reset’). Then the relaxation would then be expected to show the following pattern. Stress Time

Data Analysis Behind Primary Reset and Zeta Factor O’Donnell E/REP/ATEC/0027/AGR/01 used only three cyclic tests, only one of which was at 550 o C This latter test consistent with continuous hardening at 550 o C A further 9 tests at 550 o C reviewed All these much closer to primary reset than continuous hardening The O’Donnell 550 o C test appears exceptional rather than typical

Data Analysis Behind Primary Reset and Zeta Factor Analysis consists of calculating relaxation using (a) continuous hardening, and, (b) primary reset Comparing with experimental relaxation data Calculations adjusted the forward creep deformation behaviour according to where the cast in question lay in the scatter band Example for one test…

Example: Test 62401 at 550 o C

Data Analysis Behind Primary Reset and Zeta Factor Can’t defend continuous hardening based on available data Primary reset much closer but retains some conservatism Hence factor creep strain increase over the dwell by zeta 12 data points give mean ln(zeta) = -0.26 (median zeta = 0.77) standard deviation of ln(zeta) = 0.32 Cap zeta at 1

Is Primary Reset x Zeta Pessimistic? Creep-fatigue tests generally have dwells of 24 hours or shorter Plant has dwells of typically ~1000 hours The short dwells in lab tests will greatly accentuate the primary creep part So the recommendation may be very conservative But there’s no other evidence at present (Need long dwell creep-fatigue tests)

Cyclic Stress-Strain? Same cyclic tests used to compare with R66 cyclic Ramberg-Osgood fits 550 o C only Cyclic hardening – and softening – also seen in the data. Saturated cyclic data compare reasonably with R66 Lie between lower bound and mean (NB: Upper bound is probably most onerous)

Cyclic Stress-Strain? Example of Hardening/Softening Behaviour

Cyclic Stress-Strain? Hence we just used R66 Ramberg- Osgood Together with log-normal distribution of A parameter With standard deviation equivalent to quoted +/-25% error i.e., CoV = 0.176

Test 62401 Stress-Strain Hysteresis Loops up to Cycle 69

Test 12161 Stress-Strain Hysteresis Loops up to Cycle 100

Comparison of Saturated Cycle Peak Stress from Cast 69431 Creep-Fatigue Tests with R66 Expectations

Transition to Saturated Cycle? R66 §8.7 advice (based on 5% of the mean fatigue cycle endurance) looks wrong – don’t use it. (CR for ATG to review) The above tests generally achieved 95% of the final cyclic hardening by about cycle 40 (one at cycle 70 and one at just over 100 cycles) and symmetrisation much sooner. cf. ~260 plant cycles over life So we ignored the transition period and used fully hardened from the start of life

Start-of-Dwell Stresses versus Cycle Number

Long Dwell Softening Advice in R66 §8.6 – but this also seems anomalous (because it is based purely on the dwell in the previous cycle, whereas softening appears accumulative). (CR for ATG to review) Nevertheless we applied it as a sensitivity study It made little difference in our assessments

R66 Forward Creep RCC-MR Primary Secondary Scatter defined by factoring C1 parameter by x7.0754 and C by x6.583 for upper 95% CL Or reciprocals of these for lower 95% CL Hence a Log-Normal Distribution is appropriate. Standard deviation on ln(C) is 1.146 (CoV of about 1.77) NB: These factors apply to C and C1, not to the strain rate. This makes a big difference in strain hardening.

How Good is RCC-MR? Comparisons with test data have been done before J.Taylor et al E/REP/BBGB/0066/AGR/10 Wang SERCO/E004792/001 Conclusion: RCC-MR with R66 95%CL scatter bands is representative Our review against large database bought from NIMS came to the same conclusion

NIMS Data cf RCC-MR, NB: log 10 (7) = 0.85

Uniaxial Creep Ductility R66 implies a log-normal distribution At  550 o C, median 10.7%, 98%CL 2.6% Mean of Log 10 (ductility,%) 1.029 Standard deviation 0.299 Comparison made with NIMS dataset And also with a large dataset from various sources (referred to as GLIM) R66 advice looks representative So we just used R66 (log-normal) But remember the multiaxial factor!

Comparison of RCC-MR with NIMS

Comparison of RCC-MR with GLIM

What’s Tricky about Ductility? The test do not cover the plant conditions of temperature and stress Low ductilities occur only for stresses above about 260 MPa Lower stresses are used in monotonic creep tests only at higher temperatures (generally 600 o C plus) So, is ductility really inversely correlated with stress… …or is this just an artefact of the bias in the test matrices?

Creep Tests Do Not Address Plant Conditions

Is Ductility Correlated with Stress – or with Temperature – or Not?

Ductility in Creep-Fatigue Is the effective ductility in creep-fatigue better than implied by monotonic creep test data? Lowest creep strain at initiation from 7 creep-fatigue tests at 550 o C was 9% (and this test was fatigue dominated) But creep-fatigue data looks compatible with NIMS Too few creep-fatigue data to deduce lower bound And would expect longer dwells to produce poorer ductility

Loading Distributions Loadings can be distributed because, –They vary in time –They vary from one component to another –They are of uncertain magnitude Can use one distribution to represent all these Or separate distributions

Converting Load to Stress Given the load and dimensions, the stress may be regarded as determinate Or there may be some intrinsic uncertainty in the stress analysis But the stress is most likely to be distributed only as a consequence of the underlying distributions of load and dimensions

Dimension Distributions Start of life dimensions – drawing tolerances Thickness may vary over life due to corrosion – hence bringing temperature and chemistry uncertainties into play Dimensional distributions may be used to represent deterministic differences across the different components Or this can be addressed by running the code for individual components separately treating these quantities as deterministic

Load Cycles First need to identify cycle types, e.g., –Reactor cycles to cold shutdown, –Reactor cycles to hot standby –Boiler cycles Decide whether each cycle to be modelled will be, –Chosen randomly, or, –Taken from a pre-determined sequence Numbers of cycles obviously needed Dwell times – deterministic or sampled Need to predict future numbers of cycles of each type, and dwells

Transient Loads The peak transient loads usually define the peaks of the hysteresis cycles, and so are particularly important Try to get plant data Transients will generally be particularly variable For example, distributions of peak start-up and peak trip thermal loads can be important

Transient Loads Golden Rule: Everything best estimate Including the uncertainties Don’t make the mistake of assuming that over-estimating a transient load is conservative It may be optimistic if it leads to a smaller dwell stress Remember ALL this is for normal / typical / representative conditions – not faults

Initial Residual Stress Of course the secondary part of the dwell stress is a residual stress! R5 does require that damage due to any initial welding residual stress be included But notice that, as soon as there is an elastic-plastic hysteresis cycle due to service loading, the initial residual stresses will be modified – and probably replaced by the shakedown residual stress So don’t over-cook the damage due to residual stresses One crude method: see HANDOUT

Elastic Follow-Up Ideal to have non-linear FEA to derive Z for a range of assumptions Possibly a range of different plant geometries Hence distribution of Z But this may be a luxury you cannot afford Good news is that results are often insensitive to Z

THE END