Download Probabilistic R5V2/3 Assessments

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
• Computing platform
• Methodology
– Deterministic
– Probabilistic
• Lunch
• Input Distributions
– Materials Data (316)
– Loading / Stress
– Plant History
9:30 – 10:00
10:00 – 10:15
10:15 – 10:45
10:45 – 12:30
12:30 – 13:00
13:00 – 14:30
14:30 – 14:45
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
C
G
F
D
E
B
H
A
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


d
E
R


   c  ,  c , T ,    c  ref ,  c , T ,
dt
Z

• 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 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
~c 
t  t
 c  t ,  c t , T , dt 
~cP 
t
t  t

t
 c  ~c
 cP  ~cP
Dc   c /  f
 eod   sod

E

 c   cP
Z


R



 c ref ,  c t , T , dt 
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
Pz  
2

1
 z  z  

exp
2 


2
2  


• 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
0.35
Probability per unit Z
0.3
0.25
0.2
0.15
0.1
0.05
0
0
2
4
6
Z
8
10
Non-Standard Distribution:
Overhang
actual plant overhang distribution
fraction of bifurcations
0.600
0.500
0.400
0.300
0.200
0.100
0.000
0.15
0.25
0.35
0.45
0.55
overhang (m)
0.65
0.75
0.85
Non-Standard Distribution
Thermal Transient Factor wrt Reference Trip
300
250
150
100
50
System Load Factor
10
1.
07
1.
04
1.
01
1.
98
0.
95
0.
92
0.
89
0.
86
0.
83
0.
80
0.
77
0.
74
0.
71
0.
68
0.
65
0.
62
0.
59
0.
56
0.
53
0.
50
0
0.
Frequency
200
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 (LogNormal)
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 (LogNormal)
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)
0.45
Bins
0.4
0.35
0.3
0.25
0.2
0.15
mean of last bin
(1.755)
0.1
0.05
-3.000
-2.000
-1.000
0
0.000
1.000
2.000
3.000
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
1
2
3
4
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
3
3
1
4
2
4
2
1
•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,

rev
i
 
rev
i 1
 1   
rev, sym
i
• This leads to “symmetrisation on average”

i
rev
i


rev, sym
i
i
• 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 Results - Restricted Tubes
0.0045
Case 1
Initiation Rate per Trial
0.0040
Case 1b
0.0035
0.0030
0.0025
0.0020
0.0015
0.0010
0.0005
0.0000
0
10000
20000
30000
Trial No.
40000
50000
Convergence of Initiation Rate
Initiation Results - Unrestricted Tubes
0.0025
Case 3
Initiation Rate per Trial
Case 3b
0.0020
0.0015
0.0010
0.0005
0.0000
0
50000
100000
150000
Trial No.
200000
250000
Initiation Predictions for Repeat Cases
50,000 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
0.1200
Annual Crack Initiation Probabilities versus Year (All Tubes)
0.1000
Probability per Year
0.0800
0.0600
0.0400
0.0200
Year
0.0000
1985
1990
1995
2000
2005
2010
2015
2020
2025
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
135
Test 62401 at 550 deg.C: Relaxations v Time
Continuous Hardening, RCC-MR Cx7, Z=1 cf Test Data
RCC-MR Cx7
test data (end of dwell)
start of dwell
130
Stress, MPa
o
550 C
125
120
115
Total Dwell Time, Hours
110
0
500
1000
1500
2000
2500
3000
3500
4000
Example: Test 62401 at 550oC
135
Test 62401 at 550 deg.C: Relaxations v Time
Primary Reset, RCC-MR Cx7, Z=1 cf Test Data
RCC-MR Cx7
test data (end of dwell)
start of dwell
Stress, MPa
130
125
120
115
Total Dwell Time, Hours
110
0
500
1000
1500
2000
2500
3000
3500
4000
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
250
Stress (MPa)
200
150
100
50
Test (End of Dwell)
Test (Start of Dwell)
0
0
1000
2000
3000
Time (hrs)
4000
5000
6000
Cyclic Stress-Strain?
• Hence we just used R66 RambergOsgood
• 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
300
Test 62401: Development of Cyclic Hardening - First 69 Cycles
200
Stress (MPa)
100
0
Cycles 1 to 10
Cycles 11 to 20
Cycles 22-30
Cycles 32 to 40
Cycles 41 to 50
Cycles 51 to 69
-100
-200
-300
-0.4
-0.3
-0.2
-0.1
0
Strain (%)
0.1
0.2
0.3
0.4
Test 12161 Stress-Strain Hysteresis
Loops up to Cycle 100
400
Test 12161 (Cast 69431) Evolution of Cyclic Hardening: First 100 Cycles
300
100
-8.00E-01
-6.00E-01
-4.00E-01
-2.00E-01
Stress (MPa)
200
0
0.00E+00
Strain (%)
2.00E-01
4.00E-01
6.00E-01
-100
-200
-300
-400
Cycles 1 to 10
Cycles 11 to 20
Cycles 21 to 30
Cycles 31 to 40
Cycles 41 to 100
8.00E-01
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
400
R66 Lower Bound
R66 Best Estimate
R66 Upper Bound
Line of Equality
Test Peak Stress
350
300
250
200
200
R66 Peak (Half Stress Range)
250
300
350
400
450
500
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
300
250
Max stress MPa
200
150
Creep-Fatigue Test 613Z , 550 deg.C: Start-of-Dwell Stress v Cycle
100
50
0
0
500
1000
1500
Cycles #
2000
2500
3000
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
c 
C1 C 2 n 1
t 
100
 c  C n
• 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: log10(7) = 0.85
Log10(Ratio of NIMS Test Strain Rates to RCC-MR)
Rate Ratio at 100 hours
Rate Ratio at 1000 hours
Log10 of Rate Ratio (NIMS/RCC-MR)
1.20
0.80
0.40
0.00
480
-0.40
-0.80
-1.20
Temperature, deg.C
500
520
540
560
580
600
620
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%
1
0.08
0.9
0.07
Cumulative Probability
0.06
0.7
0.6
0.05
Cumulative Log-Normal Probability
Log-Normal PDF (per 1% strain range)
0.5
0.04
0.4
0.03
0.3
0.02
0.2
0.01
0.1
creep ductility (%)
0
0
5
10
15
20
0
25
30
35
40
Probability Density (per 1% strain)
0.8
Comparison of RCC-MR with NIMS
NIMS Monotonic Creep Test Ductility Data
2.5
NIMS Data
NIMS Data Linear Fit
NIMS Data Fit 98% CL's
R66 Mean
R66 98% CL's
Log10(Creep Ductility (%))
2.0
1.5
1.0
0.5
0.0
480
500
520
540
560
-0.5
Temperature (C)
580
600
620
Comparison of RCC-MR with GLIM
GLIM Monotonic Creep Test Ductility Data
2.5
GLIM Data
GLIM Data Linear Fit
GLIM Data Fit 98% CL's
R66 Mean
R66 98% CL's
NIMS Data Linear Fit
NIMS Data Fit 98% CLs
Log10(Creep Ductility (%))
2.0
1.5
1.0
0.5
0.0
480
500
520
540
560
-0.5
Temperature (C)
580
600
620
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
500
Test Matrix Temperatures v Stresses
450
GLIM Data
NIMS Data
Likely Bounding Plant Regime
400
Stress, MPa
350
300
250
200
150
100
50
0
480
500
520
540
Temperature, deg.C
560
580
600
Is Ductility Correlated with Stress – or with
Temperature – or Not?
Creep Strain at Failure or Initiation (%)
60
500 deg.C
550 deg.C
575 deg.C
600 deg.C
creep-fatigue tests (550 degC)
Spindler, Ref.[13]
NIMS Creep Ductility cf Creep-Fatigue
and Spindler, Ref.[13]
50
40
30
20
10
Stress, MPa
0
0
50
100
150
200
250
300
350
400
450
500
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