Download The use of probabilistics with R5V2/3

T73S04 (R5V2/3) Session 38B
Probabilistic 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
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
~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
Go to
http://rickbradford.co.uk/Peter
HoltNotesOnPDFs.pdf
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! (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)
0.45
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.
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,

•
rev
i
 a
rev
i 1
 1  a 
rev, sym
i
http://rickbradford.co.uk/HysteresisCycleConstructionMethodologyForInteractingCycles.pdf
• This leads to “symmetrisation on average”
 irev   irev, 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
• 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,
y  C xy x  1  C  v
2
xy
• 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
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 reactoraverage 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
0.0045
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
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., 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