probabilistic r5v2/3 assessments rick bradford peter holt 17 th december 2012
TRANSCRIPT
Probabilistic R5V2/3 Assessments
Rick Bradford
Peter Holt
17th 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 10-7pry 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• Introduction 9:30 – 10:00• Computing platform 10:00 – 10:15• Methodology
– Deterministic 10:15 – 10:45– Probabilistic 10:45 – 12:30
• Lunch 12:30 – 13:00• Input Distributions
– Materials Data (316) 13:00 – 14:30– Loading / Stress 14:30 – 14:45– Plant History 14:45 – 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
,,,,,, TT
Z
E
dt
dc
Rrefccc
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 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
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
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
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
ity
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
on
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!
• 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 VariablesHere 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 VariablesDeadweight 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 2541 = 2 x 1057 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, 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.
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
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
• “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 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
on
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
iati
on
Ra
te p
er
Tri
al
Case 3
Case 3b
Initiation Predictions for Repeat Cases50,000 trials (restricted)
250,000 trials (unrestricted)
No. of Initiations by 2024
Case Restricted Tubes
Unrestricted Tubes
Total
12 0.2 0.6 0.8
12b 0.2 0.6 0.8
12c 0.2 0.46 0.7
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
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
Pro
ba
bili
ty p
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., 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 550oC
• This latter test consistent with continuous hardening at 550oC
• A further 9 tests at 550oC reviewed• All these much closer to primary reset than
continuous hardening• The O’Donnell 550oC 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 550oCTest 62401 at 550 deg.C: Relaxations v Time
Continuous Hardening, RCC-MR Cx7, Z=1 cf Test Data
110
115
120
125
130
135
0 500 1000 1500 2000 2500 3000 3500 4000
Total Dwell Time, Hours
Str
es
s, M
Pa
RCC-MR Cx7
test data (end of dwell)
start of dwell
Example: Test 62401 at 550oCTest 62401 at 550 deg.C: Relaxations v TimePrimary Reset, RCC-MR Cx7, Z=1 cf Test Data
110
115
120
125
130
135
0 500 1000 1500 2000 2500 3000 3500 4000
Total Dwell Time, Hours
Str
es
s, M
Pa
RCC-MR Cx7
test data (end of dwell)
start of dwell
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
• 550oC 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 Creep Test 62431 - 550C
0
50
100
150
200
250
0 1000 2000 3000 4000 5000 6000
Time (hrs)
Str
es
s (
MP
a)
Test (End of Dwell)
Test (Start of Dwell)
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 62401: Development of Cyclic Hardening - First 69 Cycles
-300
-200
-100
0
100
200
300
-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4
Strain (%)
Str
ess
(MP
a)
Cycles 1 to 10Cycles 11 to 20Cycles 22-30Cycles 32 to 40Cycles 41 to 50Cycles 51 to 69
Test 12161 Stress-Strain Hysteresis
Loops up to Cycle 100 Test 12161 (Cast 69431) Evolution of Cyclic Hardening: First 100 Cycles
-400
-300
-200
-100
0
100
200
300
400
-8.00E-01 -6.00E-01 -4.00E-01 -2.00E-01 0.00E+00 2.00E-01 4.00E-01 6.00E-01 8.00E-01
Strain (%)
Str
es
s (
MP
a)
Cycles 1 to 10
Cycles 11 to 20
Cycles 21 to 30
Cycles 31 to 40
Cycles 41 to 100
Comparison of Saturated Cycle Peak Stress from Cast 69431 Creep-Fatigue Tests with R66 Expectations
Comparison of Saturated Cycle Peak Stress from Cast 69431 Tests with R66 Expectations
200
250
300
350
400
200 250 300 350 400 450 500
R66 Peak (Half Stress Range)
Te
st
Pe
ak
Str
es
s
R66 Lower Bound
R66 Best Estimate
R66 Upper Bound
Line of Equality
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
Creep-Fatigue Test 613Z , 550 deg.C: Start-of-Dwell Stress v Cycle
0
50
100
150
200
250
300
0 500 1000 1500 2000 2500 3000
Cycles #
Ma
x s
tre
ss
MP
a
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.
12 nC1
c t100
C
nc C
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: log10(7) = 0.85
Log10(Ratio of NIMS Test Strain Rates to RCC-MR)
-1.20
-0.80
-0.40
0.00
0.40
0.80
1.20
480 500 520 540 560 580 600 620
Temperature, deg.C
Lo
g10
of
Rat
e R
atio
(N
IMS
/RC
C-M
R)
Rate Ratio at 100 hours
Rate Ratio at 1000 hours
Uniaxial Creep Ductility• R66 implies a log-normal distribution• At 550oC, median 10.7%, 98%CL 2.6%• Mean of Log10(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!
Log-Normal Distribution for R66 Uniaxial Creep Ductility of 316H
at 500-550oC: Median 10.7%, 98%CL 2.6%
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 5 10 15 20 25 30 35 40
creep ductility (%)
Cu
mu
lati
ve
Pro
ba
bili
ty
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
Pro
ba
bili
ty D
en
sit
y (
pe
r 1
% s
tra
in)
Cumulative Log-Normal Probability
Log-Normal PDF (per 1% strain range)
Comparison of RCC-MR with NIMSNIMS Monotonic Creep Test Ductility Data
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
480 500 520 540 560 580 600 620
Temperature (C)
Lo
g1
0(C
ree
p D
uc
tilit
y (
%))
NIMS DataNIMS Data Linear FitNIMS Data Fit 98% CL'sR66 MeanR66 98% CL's
Comparison of RCC-MR with GLIMGLIM Monotonic Creep Test Ductility Data
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
480 500 520 540 560 580 600 620
Temperature (C)
Lo
g10
(Cre
ep
Du
cti
lity
(%
))
GLIM DataGLIM Data Linear FitGLIM Data Fit 98% CL'sR66 MeanR66 98% CL'sNIMS Data Linear FitNIMS Data Fit 98% CLs
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 600oC 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
Test Matrix Temperatures v Stresses
0
50
100
150
200
250
300
350
400
450
500
480 500 520 540 560 580 600
Temperature, deg.C
Str
ess,
MP
a
GLIM Data
NIMS Data
Likely Bounding Plant Regime
Is Ductility Correlated with Stress – or with Temperature – or Not?
NIMS Creep Ductility cf Creep-Fatigueand Spindler, Ref.[13]
0
10
20
30
40
50
60
0 50 100 150 200 250 300 350 400 450 500
Stress, MPa
Cre
ep S
trai
n a
t F
ailu
re o
r In
itia
tio
n (
%)
500 deg.C
550 deg.C
575 deg.C
600 deg.C
creep-fatigue tests (550 degC)
Spindler, Ref.[13]
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 550oC 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
Loading Distributions
• Does the load correlate with some known parameter, e.g.,– Thermal load with temperature– Deadweight load with some dimension– System load with range of assumptions in a
pipework model, etc
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