000000000001000311-Guidelines for Performing Probabilistic Analyses of Boiler Pressure Parts

March 27, 2018 | Author: danish873 | Category: Fracture Mechanics, Creep (Deformation), Fatigue (Material), Fracture, Elasticity (Physics)
Share
Report this link


Comments



Description

Guidelines for Performing Probabilistic Analyses of Boiler Pressure PartsEffective December 6, 2006, this report has been made publicly available in accordance with Section 734.3(b)(3) and published in accordance with Section 734.7 of the U.S. Export Administration Regulations. As a result of this publication, this report is subject to only copyright protection and does not require any license agreement from EPRI. This notice supersedes the export control restrictions and any proprietary licensed material notices embedded in the document prior to publication. M AT E R I A L N LICE SED WARNING: Please read the License Agreement on the back cover before removing the Wrapping Material. Technical Report Guidelines for Performing Probabilistic Analyses of Boiler Pressure Parts 1000311 Final Report, December 2000 EPRI Project Manager R. Tilley EPRI • 3412 Hillview Avenue, Palo Alto, California 94304 • PO Box 10412, Palo Alto, California 94303 • USA 800.313.3774 • 650.855.2121 • [email protected] • www.epri.com INCLUDING ANY PARTY'S INTELLECTUAL PROPERTY.O. OR (B) ASSUMES RESPONSIBILITY FOR ANY DAMAGES OR OTHER LIABILITY WHATSOEVER (INCLUDING ANY CONSEQUENTIAL DAMAGES. OR (II) THAT SUCH USE DOES NOT INFRINGE ON OR INTERFERE WITH PRIVATELY OWNED RIGHTS. ANY MEMBER OF EPRI. All rights reserved. INCLUDING MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE. ELECTRIFY THE WORLD is a service mark of the Electric Power Research Institute. APPARATUS.DISCLAIMER OF WARRANTIES AND LIMITATION OF LIABILITIES THIS DOCUMENT WAS PREPARED BY THE ORGANIZATION(S) NAMED BELOW AS AN ACCOUNT OF WORK SPONSORED OR COSPONSORED BY THE ELECTRIC POWER RESEARCH INSTITUTE. THE ORGANIZATION(S) BELOW. Inc. INC. METHOD. Inc. (I) WITH RESPECT TO THE USE OF ANY INFORMATION. EXPRESS OR IMPLIED. Box 23205. ANY COSPONSOR. Inc. Electric Power Research Institute and EPRI are registered service marks of the Electric Power Research Institute. CA 94523. P. EPRI. NEITHER EPRI. Pleasant Hill. ORGANIZATION(S) THAT PREPARED THIS DOCUMENT Engineering Mechanics Technology ORDERING INFORMATION Requests for copies of this report should be directed to the EPRI Distribution Center. . EVEN IF EPRI OR ANY EPRI REPRESENTATIVE HAS BEEN ADVISED OF THE POSSIBILITY OF SUCH DAMAGES) RESULTING FROM YOUR SELECTION OR USE OF THIS DOCUMENT OR ANY INFORMATION. 207 Coggins Drive. (EPRI). NOR ANY PERSON ACTING ON BEHALF OF ANY OF THEM: (A) MAKES ANY WARRANTY OR REPRESENTATION WHATSOEVER. OR (III) THAT THIS DOCUMENT IS SUITABLE TO ANY PARTICULAR USER'S CIRCUMSTANCE. PROCESS. Copyright © 2000 Electric Power Research Institute. OR SIMILAR ITEM DISCLOSED IN THIS DOCUMENT. APPARATUS. METHOD. PROCESS. (800) 313-3774. OR SIMILAR ITEM DISCLOSED IN THIS DOCUMENT. CA: 2000. Palo Alto.CITATIONS This report was prepared by Engineering Mechanics Technology 4340 Stevens Creek Blvd. 1000311. iii . Suite 166 San Jose. The report is a corporate document that should be cited in the literature in the following manner: Guidelines for Performing Probabilistic Analyses of Boiler Pressure Parts. CA 95129 Principal Investigator D.. Harris This report describes research sponsored by EPRI. EPRI. . These results are generally expressed as failure probabilities (or failure rates) versus time. maintenance. Utilities can now use these options to improve decisions on boiler component inspections. The time for remedial action can be keyed to the time at which failure rate v . However. It provides guidelines on the generation and use of such results in maintenance and inspection planning. Such a realistic basis also couples with economic parameters to allow utilities to make better overall decisions in their efforts to reduce operating and maintenance costs. Objectives N To review probabilistic methodologies for use in boiler component life management N To provide guidelines on the generation and use of such results in maintenance and inspection planning Approach Using a probabilistic approach. because of the emerging competitive environment and more financially oriented management.REPORT SUMMARY Using probabilistic methodologies for life assessment of boiler components provides a more realistic basis for managing the inspection. and replacement actions for those components. maintenance. repair. Recent EPRI analytical models such as the Boiler Life Evaluation and Simulation System (BLESS) incorporate options to allow probabilistic analyses. Probabilistic results are generated by running a series of analyses in which key input parameters are varied to reflect the actual variation occurring in the population of similar components. more accurate assessments of risks and benefits must be incorporated into utility decision making. and replacement. maintenance decisions and corresponding expenditures could be based on engineering analyses using approximate models for component damage mechanisms and adding conservative safety factors to account for both model and data inaccuracies. utilities are finding that such conservative approaches are non-optimum in balancing costs and benefits. In the competitive environment for power generation. rather than providing a deterministic failure time. the probability of failure within a certain time range can be estimated. Decisions need to be justified from an economic point of view that better incorporates risks of equipment failure. Probabilistic techniques have been used in other industries to provide such a risk-based bridge to economic decision making. repair. Probabilistic techniques are a proven way to refine the basis for such decisions. This document reviews some life prediction methodologies and discusses relevant statistical principles. Background In the past. The deterministic result uses a single set of input variables and is used with a safety factor to account for inaccuracies in the model and its input. Keywords Probabilistic analysis Boiler components Life assessment vi . but much of the discussion is readily applicable to other components that degrade due to material aging. Special attention is paid to probabilities associated with crack initiation and growth. this report provides guidance for the use of probabilistic approaches in managing boiler component life. The probabilistic approach reviewed in this document is based on an underlying mechanistic model of lifetime. which are of great use in component life management EPRI Perspective As utilities come under increased pressure to reduce costs and extend the lifetime of plant components. which are leading causes of material degradation and component failure. The development of probabilistic models of component lifetimes is also discussed. This document concentrates on boiler pressure parts. the following points need to be kept in mind: N N N N N Lifetime models are available Scatter and uncertainties in inputs to the models usually preclude accurate deterministic results Probabilistic lifetime models can be obtained by quantifying the scatter and uncertainty and incorporating them into the underlying deterministic lifetime model Numerical procedures for generation of failure probabilities are available.becomes excessive. and numerical results can usually be obtained using a personal computer Probabilistic results can be used in analyses of expected future operating costs. EPRI and other organizations have facilitated the use of risk-principles to prioritize maintenance actions. Results When using probabilistic methodologies for component life management. Statistical background information is provided in the area of probabilistic structural analysis. Representative lifetime models are reviewed with a focus on boiler pressure parts. Building on software tools such as the BLESS code and processes such as those developed for extending intervals between turbine maintenance outages (TURBO-X). This information can then be incorporated into the component cost-benefit models to optimize overall costs. interest has increased in procedures for rationally planning inspection and maintenance. ..................................................................................1............................. 2-6 2..........................EPRI Licensed Material CONTENTS 1 INTRODUCTION...............2.......................................2............................................ 2-8 2...........................................................1 3......4 Oxide Notching .....3................................................................. 3-1 Fitting Distributions .....................................................3 Simple Example Problems........................................................................... 2-1 2....................................3 Creep/Fatigue Crack Initiation.........................1 Analytical Methods............................................................................2 Creep Damage in a Thinning Tube ................................................... 2-13 2........................................................................3...................................................................... 2-18 2...............2 Creep Crack Initiation .....................................................................5 Creep Crack Growth ...... 3-1 3................................ 3-4 Combinations of Random Variables ..........3 Monte Carlo – Confidence Intervals ........................................ 3-16 3.............................................1........2 Monte Carlo Simulation – Principles .2................................................................. 2-15 2........................... 2-3 2...................... 2-8 2....6 Creep/Fatigue Crack Growth ......................................................................................1 Crack Tip Stress Fields...............2..... 2-10 2............................2 3................1 Crack Initiation...............2......... 2-1 2.................................................................. 3-12 3.3................................................................................................................................................................................................. 1-1 2 REVIEW OF DETERMINISTIC LIFE PREDICTION PROCEDURES FOR BOILER PRESSURE PARTS ................................2...................................................1 Fatigue Crack Initiation ................................. 2-18 2.................... 2-1 2..........................1......... 2-14 2...................3 Calculation of Critical Crack Sizes ............ 3-11 3....... 2-7 2.................1............4 Monte Carlo Simulation – Importance Sampling ........................3..2 Crack Growth ..................... 2-13 2................3 Probability Density Functions ... 2-19 3 SOME STATISTICAL BACKGROUND INFORMATION......................................................... 3-19 vii ................................................................... 3-11 3........4 Fatigue Crack Growth ...........................................................3..1 Fatigue of a Crack in a Large Plate.........3....................2 Crack Driving Force Solutions.............. ............................................................................................................................................................... 3-21 3............1... 5-12 6 USE OF PROBABILISTIC RESULTS........ 4-4 4.............. 3-27 4 DEVELOPMENT OF PROBABILISTIC MODELS FROM DETERMINISTIC BASICS.2 Target Hazard Rates ................2..........2 5......... 7-1 A DETAILS OF BLESS EXAMPLE ..................................5 First Order Reliability Methods – Basics................... 5-3 5..................................................................................................................................... 5-1 5...............1 Gathering the Necessary Information ........................... 5-6 Combining Data.................. 5-1 5................................................3.........................1 4.................................... 4-4 4...............EPRI Licensed Material 3.......................................................................1........................................................................................2 Discussion................................................. 6-5 7 CONCLUDING REMARKS .....................B-1 viii ................................................2......................................................................................................A-1 B REFERENCES ... 5-1 5..................................................................1 6.......................3.......................2 Operating Conditions ..........................1 Fatigue Crack Growth in a Large Plate ....................3 Inspection Detection Probabilities................................... 4-1 Simple Example Problems..... 6-1 Economic Models ..................... 4-15 4............... 4-1 4................................................................................................................. 4-11 4...2 Creep Damage in a Thinning Tube ...........................................................3 Hazard Rates............................3 Performing the Analysis................................. 4-18 5 EXAMPLE OF A PROBABILISTIC ANALYSIS ...............................................................................................................2................................6 First Order Reliability Methods – General .................................................................................................................................... 6-1 6.............1 Component Geometry and Material ...... ........................................................................ 2-6 Figure 2-4 Crack-Like Defect Initiated by Oxide Notching........................................................................ 2-19 Figure 2-12 Time to Failure as a Function of the Wall-Thinning Rate for the Creep Rupture Example Problem (1 Mil/Year=25...................................................... 2-2 Figure 2-2 Creep Rupture Data for 1 ¼ Cr ½ Mo With Curve Fit [From Grunloh 92] (1 ksi=6............................................... 2-12 Figure 2-9 Fatigue Crack Growth as a Function of the Cyclic Stress Intensity Factor for 2 ¼ Cr 1 Mo at Various Temperatures [Drawn From Viswanathan 89]........................................................................ 2-11 Figure 2-8 Single Edge-Cracked Strip in Tension With J-Solution ................. 3-15 Figure 3-5 Values of Factors in Table 3-5 Divided by the Number of Failures .............................................................................................................. and the Most Probable Failure Point Is Indicated..................................................................... ..... 3-9 Figure 3-4 Cumulative Probability of the Sum of Two Lognormals as Computed by Numerical Integration and Monte Carlo With 20 and 500 Trials ...4 µm/yr)................................................ 2-17 Figure 2-11 Half-Crack Length as a Function of the Number of Cycles to 20 ksi (137.............................................................. 2-14 Figure 2-10 Creep and Creep/Fatigue Crack Growth Data and Fits.................................. 2-7 Figure 2-5 Depiction of Procedure for Determination of Oxide Thickness for a Time at Tlo Followed by a Time at Thi ...............EPRI Licensed Material LIST OF FIGURES Figure 2-1 Strain Life Data for A106B Carbon Steel in Air at 550°F (290°C) With Median Curve Fit [From Keisler 95] .............................. The Origin Is at the Upper Right Corner.......................... 2-4 Figure 2-3 Creep/Fatigue Damage Plane Showing Combinations Corresponding to Crack Initiation............................ 2-21 Figure 3-1 Plot of Data of Table 3-2 on Log-Linear Scales ............................... 2-9 Figure 2-7 Through-Crack of Length 2a in a Large Plate Subject to Stress σ.................... 3-19 Figure 3-6 Cumulative Probability of the Sum of Two Lognormals as Computed by Numerical Integration and Monte Carlo Simulation With 20 Trials.........................................................895MPa) .................... 3-22 Figure 3-8 Plot of the Performance Function in Reduced Variate Space for the Example Problem of Two Lognormals for z=2.................................................... 3-24 ix .............................................................................................. 3-8 Figure 3-3 Normal Probability Plot of Data of Table 3-2........................................................................... 2-8 Figure 2-6 Coordinate System Near a Crack Tip .................................................................. With and Without Importance Sampling...... Left Figure Is for Constant Load and Right Figure Is for Cyclic Load With Various Hold Times [From Grunloh 92]..................................... 3-21 Figure 3-7 Pictorial Representation of Joint Density Function in Unit Variate Space Showing Failure Curve and Most Probable Failure Point (MPFP) ............... 3-8 Figure 3-2 Lognormal Probability Plot of Data of Table 3-2 .......9 MPa) for Example Fatigue Problem ................................................. ........................................................ Points are Results From Rackwitz-Fiessler.................................................... 3-25 Figure 3-10 Example of a Performance Function With the Vector Normal to the Axis of One of the Variables Showing the Insensitivity of to That Variable............................. 3-26 Figure 3-11 The Direction Cosines of x and y for the Example Problem of the Sum of Two Lognormals ............................................................ 3-31 Figure 4-1 Probabilistic Treatment of Strain-Life Data for A106B Carbon Steel in Air at 550°F (288°C) Showing Various Quantiles of the Keisler Curve Fit [From Keisler 95].................... 4-18 Figure 4-14 Nondetection Probability as a Function of Crack Depth Divided by Plate Thickness for Fatigue Cracks in Ferritic Steel for Three Qualities of Ultrasonic Inspection .............. 4-15 Figure 4-12 Hazard Function vs.......................... tR in Hours) ............ 4-11 Figure 4-9 Creep Rupture Data for 1 ¼ Cr 1/2 Mo as Obtained From Van Echo 66 With Least Squares Linear Fit (Stress in ksi...... 4-8 Figure 4-6 Cumulative Failure Probabilities in the Lower Probability Region for the Fatigue Crack Growth Example Problem.............................................................. 3-27 Figure 3-12 Diagrammatic Representation of a Procedure for Finding Most Probable Failure Point .................. 4-17 Figure 4-13 Failure Rate as a Function of Time for the Thinning Problem Only......................................................................................... Solid Line is Monte Carlo With 106 Trials.................................................................................... .......................................................................... 4-13 Figure 4-11 Results of Example Problem of Creep Damage in a Thinning Tube as Obtained by Monte Carlo Simulation With 104 Trials ................ 4-7 Figure 4-5 Cumulative Failure Probabilities in the Lower Probability Region of the Fatigue Crack Growth Example Problem With Two Distributions of the Fracture Toughness (106 Trials)....................................................................................... 4-6 Figure 4-3 Lognormal Probability Plot of the Failure Probability as a Function of the Number of Cycles for the Fatigue Example Problem (104 Trials)......... 4-3 Figure 4-2 Cumulative Distribution of Critical Crack Size for the Fatigue Crack Growth Example Problem (104 Trials) .................................................. 4-13 Figure 4-10 Cumulative Distribution of the Random Variable A Used to Describe the Scatter in the Larson-Miller Data for 1 ¼ Cr 1/2 Mo Steel .................................................................................................................................................... Cycles for the Fatigue Crack Growth Example Problem........................................................EPRI Licensed Material Figure 3-9 Cumulative Probability of the Sum of Two Lognormals as Computed by the First Order Reliability Method and Numerical Integration ............................ 4-7 Figure 4-4 Plot on Lognormal Scales of the Distribution of Cycles to Failure for the Example Fatigue Crack Growth Problem and the Same Problem With the Mean and Standard Deviation Divided by Two (106 Trials) ........ Points are From Rackwitz-Fiessler.............................000 Failure Times in 106 Trials........................................ 4-10 Figure 4-8 Direction Cosines for the Fatigue Crack Growth Example Problem as a Function of the Number of Cycles to Failure .................... ............ Histogram Results Are for the Smallest 2............. 4-9 Figure 4-7 Cumulative Failure Probabilities in the Lower Probability Region for the Fatigue Crack Growth Problem as Obtained Using Importance Sampling With 1000 Trials With Different Shifts in Parameters of Input Random Variables...................................................................................... 4-21 Figure 4-15 Failure Probability as a Function of Time for the Inspection Example Problem ..... the Line Is for the Closed Form Result............................ 4-22 x ....................... Ta in Degrees Rankine.................................. ...................................................................... 5-9 Figure 5-6 Comparison of Hazard as a Function of Time as Obtained From the BLESS Output Data and the Fitted Lognormal Relation ................................................................................. ...................................... 5-7 Figure 5-4 Lognormal Probability Plot of BLESS Results for the Header Example Problem (Probability in Percent) ........................... 5-2 Figure 5-2 Summary of Header Geometry Analyzed (Length Dimensions in Inches)....... 5-10 Figure 5-7 Results of Header Example Problem With Varying Shifts and Number of Trials............000 Trials) ............................................................................................ The Solid Line is the Curve Fit Based on the Estimated Lognormal Distribution With No Shifts (the Line of Figure 5-6)........................... 5-12 Figure 6-1 Log-Log Plot of Frequency Severity Data for Boiler Components From Table 6-2 ................ 5-7 Figure 5-5 Lognormal Hazard Function for the Header Example Problem ........................................... 4-23 Figure 5-1 Schematic Representation of Typical Header With Illustration of Segment Considered in Analysis ..... 6-4 xi ................................... 5-3 Figure 5-3 Log-Linear Plot of Probability of Leak as a Function of Time for Oxide Notching Initiation and Creep Fatigue Crack Growth in Header Example Problem (2..........................................................................................................................................................EPRI Licensed Material Figure 4-16 Comparison of Fatigue Example Problem of Infinite Plate With Finite Thickness Results With No Inspection Using PRAISE Code...................... . ....... 3-10 Table 3-4 Results of Monte Carlo Example With 20 Trials................................ 3-32 Table 3-9 Intermediate Steps in Iterative Procedure for Finding the MPFP for the Example Problem With z=2 ................ 3-18 Table 3-7 Coordinates and Direction Cosines of the MPFP for Example Problem With z = 2 ........................................................................................... 4-5 Table 4-2 Steps in Estimating the Hazard Function for the Fatigue Crack Growth Example Problem ............................................ 6-3 Table 6-3 Calculation of Expected Cost of Continuing Operation of Example Header for Another 20 Years.................... 3-7 Table 3-3 Information on Parameters of Distribution of Cf Data of Table 3-2 ... 2-12 Table 3-1 Characteristics of Some Commonly Encountered Probability Functions Used to Describe Scatter and Uncertainty ............................................................................................................. 5-8 Table 6-1 Examples of Hazards of Common Activities as Measured by Fatality Rate ...................... 3-33 Table 4-1 Random Variables for the Fatigue Crack Growth Example Problem................ 4-16 Table 4-3 Parameters of the Equation Describing the Non-Detection Probability ......................................................................................................................................... 3-14 Table 3-5 Summary of Some of the Statistics from the Monte Carlo Trials ............. 6-2 Table 6-2 Partial List of Boiler Component Failure Rate and Consequences........................................................EPRI Licensed Material LIST OF TABLES Table 2-1 Table of the Function h1(α................................................................................................................................................................................................................................................................. 4-20 Table 5-1 Summary of Time-Variation of Operating Conditions .......................... 6-8 xiii ............................. 3-3 Table 3-2 Values of Cf for Fatigue Crack Growth Data .....n) in the J-Solution for an Edge Cracked Plate in Tension for Plane Strain [From Shih 84] ................................................................................................................................. 3-15 Table 3-6 Confidence Upper Bounds on Ntrpf for Various Numbers of Failures ......................................................... 6-7 Table 6-4 Calculation of Expected Cost of Replacing Header Now and Then Operating for Another 20 Years..................................................... 5-5 Table 5-2 Summary of Initiation Time for Various Operating Scenarios.. 3-25 Table 3-8 Intermediate Steps in Iterative Procedure for Finding the MPFP for the Example Problem with z=3 .......... . Guidance on exercising the resulting probabilistic models is given and interpretation of the results is discussed.EPRI Licensed Material 1 INTRODUCTION Deregulation of the electric utility industry has lead to increased competition in the generation of electrical power. Due to uncertainties and inherent randomness in parameters that determine the component life. The points to be made are that: • • • • Lifetime models are available There are scatter and uncertainties in inputs to the models that usually preclude accurate deterministic results Probabilistic lifetime models can be obtained by quantifying the scatter and uncertainty and incorporating them into the underlying deterministic lifetime model Numerical procedures for generation of failure probabilities are available. Use of probabilistic approaches provides a more useful way of accounting for analysis uncertainties. It is not economical to replace them earlier than necessary. Using a probabilistic approach. nor is it economical to run them until they cause an unscheduled outage or safety problem. which are of great use in component life management • This document reviews some life prediction methodologies. To account for inherent inaccuracies. rather than providing a deterministic failure time. The availability of probabilistic information can be viewed as a plus. the probability of failure within a certain time range can be estimated. All of this information is available elsewhere. including the analysis of a header using the EPRI BLESS software. Hence. and personal computers are becoming so fast and cheap that generation of numerical results is usually not a problem The probabilistic results generated (failure probability as a function of time) are of direct use in analyses of expected future operating costs. The purpose of this document is to review probabilistic methodologies for use in component life management and to provide guidelines on the generation and use of such results in maintenance and inspection planning. it is desired to define an optimum time range for remedial action. the precise time of failure can rarely be accurately defined. Examples of probabilistic analyses are provided. conservative safety factors are often applied to deterministic results. but not in a single convenient document. The use of the failure probability results in a run/retirement decision is demonstrated. which has led to increasing need for systematic means of inspection and maintenance planning. followed by a discussion of relevant statistical principles. because this can lead directly to application of risk-based concepts in 1-1 . Power plant components are subject to aging due to a variety of mechanisms and must occasionally be replaced or repaired. Another advantage of using failure probability is that it can be used to estimate the future expected cost of failure. which can be an important component of expected future operating cost. one component of risk (the failure probability) is a direct outcome of probabilistic analyses. This document concentrates on boiler pressure parts. Risk is conventionally defined as the product of the probability of failure and the consequences of failure. These expected costs can be incorporated into financial models to optimize equipment life management over an entire group of plants. but much of the discussion is readily applicable to other components that degrade due to material aging. The time for remedial action can be keyed to the time at which failure rate becomes excessive – excessive being based on level of risk. Items with a high failure rate but low consequences do not require the level of attention that would occur if only failure rate was used in the decision process. Including consequences in the process allows attention to be focused on the items of importance.EPRI Licensed Material Introduction run/replace/retire decisions. These costs are expressed in terms that are readily communicated to utility management. 1-2 . Hence. such models are fundamental to the approach. and economizer). this is about 800°F (427°C). Crack initiation in headers by oxide notching is an important degradation mechanism. Viswanathan 89 and 92 provide information on these topics. creep. creep crack initiation occurs due to time spent in the stress and temperature range where time-dependent material response (creep) is important. oxide notching or a combination of these.1.1 Crack Initiation Crack initiation can occur due to fatigue. the procedures to be employed for crack initiation are quite different than for crack growth. stress corrosion cracking. Also. 2. oxidation and erosion can also be problems. In the steels used in electrical power generating plants. Fatigue crack initiation occurs due to cyclic loading and may occur in the absence of time-dependent material response. In contrast to this. The crack growth methodologies discussed here are based on fracture mechanics and are applicable when the lifetime is controlled by the behavior of a single (or a few) dominant cracks. and pipes. reheater. but will not be covered here. The life prediction methodologies for creep and fatigue are quite different. This is the degradation mechanism most familiar to 2-1 . Material degradation in boiler pressure parts is mostly due to creep and/or fatigue. Stress corrosion will not be discussed here. and this section will review some representative lifetime models. Boiler pressure parts are considered to be headers (superheater. tubing (superheater and reheater). Boiler pressure parts are subject to material degradation by a wide variety of mechanisms.1 Fatigue Crack Initiation The initiation of fatigue cracks is due to cyclic loading and can occur at temperatures well below the range where creep is important. 2. Oxidation and fire-side corrosion can be troublesome in tubing. pitting. This can involve crack initiation and/or crack growth. Oxide notching is a problem that is aggravated by load cycling. Corrosion. Viswanathan 89 provides a comprehensive summary of degradation mechanisms and life prediction approaches. Creep generally is not a problem in metals at temperature less than about 1/3 to 1/2 of the absolute melting temperature. Hence.EPRI Licensed Material 2 REVIEW OF DETERMINISTIC LIFE PREDICTION PROCEDURES FOR BOILER PRESSURE PARTS The probabilistic approach reviewed in this document is based on an underlying mechanistic model of lifetime. and the model used in BLESS [Grunloh 92] is discussed and considered in an example problem. with his Chapter 5 being devoted to boiler components. which corresponds to A=0. The data is in terms of the strain amplitude (one-half the peak-to-peak value). 2-1 The line in Figure 2-1 is a plot of the best fit obtained for this data by Keisler and Chopra [Keisler 95]. which is from Keisler 95 and is for A106 carbon steel in air at 550°F (290°C). and the desirability of a probabilistic approach is immediately apparent. and expressed as an “S-N” curve. The following functional form is often used to represent fatigue data: ε a = BN − b + A Eq. B=27.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts engineers. The amount of scatter in the data is usually large.11. which corresponds approximately to a 3 mm deep crack. The cyclic lifetime is measured in the laboratory as a function of the cyclic stress (or strain) level. and b=0. Figure 2-1 provides an example. rather than cycles to initiation.47. The data in Figure 2-1 are actually the cycles for a 25% load drop in the test.534. Figure 2-1 Strain Life Data for A106B Carbon Steel in Air at 550°F (290°C) With Median Curve Fit [From Keisler 95] 2-2 . The data is most often in terms of the cycles to failure. so the damage for n cycles of amplitude ε is n/N(ε). which is defined as LMP = TA [C + log(t R )] Eq. except for the presence of the crack. and there are L different amplitudes of cycling. C is evaluated as part of the curve fitting procedure. The number of time periods to failure is the number of time periods to reach D=1.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts In cases where the cyclic stress level varies during the lifetime. the presence of a crack. A plot of the best fit is also shown. the cycles to failure is generally computed by use of Miner’s rule. The damage per cycle of strain of amplitude ε is taken to be equal to 1/N(ε). but a considerable amount of scatter is again observed. The accumulated fatigue damage does not change the stresses in a complex body. Fracture mechanics principles can be used to then grow the crack once it initiates. Figure 2-2 is a typical plot of creep rupture data for 1 ¼ Cr ½ Mo steel. if a given time period consists of ∆ni cycles of strain amplitude εi. then the fatigue damage for this time period is Df = ε ∑ N∆ ( ∆ε ) L i i =1 i Eq. 2-3 . The total damage is then the sum for each of the contributors. such as is usually the case due to different loads. and failure is taken to occur when the damage totals to unity. For instance. 2-3 TA is the absolute temperature. and the usefulness of a probabilistic approach is apparent.2 Creep Crack Initiation Creep cracks can initiate after a period of time under steady loading. The rupture lifetimes are often then plotted as function of a so-called Larson-Miller parameter. 2-2 The value of N(∆εi) is obtained from Equation 2-1. or. 2. a crack of about 3 mm in size. Failure can be final failure of a “specimen”. in the particular case of the Keisler data. The lifetimes of uniaxial tensile specimens are typically measured for a range of temperatures as a function of the applied stress.1. with C in Equation 2-3 equal to 20: LMP = 42869 − 5146[log σ ] − 956[log σ ]2 Eq. tR(σ. The time to rupture for varying stress conditions is often evaluated by the creep counterpart of Miner’s rule. which is called Robinson’s rule.T). 2-4 The log is to the base 10. T ) L ti Eq. the temperature is in degrees Rankine (1°R=5/9°K). the creep damage is evaluated by use of the expression Dc = ∑ t (σ . Equations 2-3 and 2-4 can be considered to provide a function that gives the rupture time as a function of stress and temperature.895MPa) The following expression is the curve fit shown in Figure 2-2. the stress is in thousand pounds per square inch (ksi). For a set of times ti spent at a stress σi and temperature Ti. and the rupture time is in hours. 2-5 i =1 R i i 2-4 .EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Figure 2-2 Creep Rupture Data for 1 ¼ Cr ½ Mo With Curve Fit [From Grunloh 92] (1 ksi=6. Creep damage can be included in the stress-strain rate relation in the following way = Ae − Q / T ε LM σ OP N1 − ω Q n Eq. Failure is the rupture of a laboratory specimen. 2-5 . Failure occurs when =1. can be considered to be the initiation of a creep crack. Such factors are readily treated by continuum creep damage mechanics. In the simplest case of uniaxial tension.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Failure is considered to occur when the creep damage reaches unity. which has the advantage of being easily expanded to complex geometries and stress/temperature histories. The creep damage enters into the relation between the creep strain rate and the stress and temperature. 2-8 The only additional material constant is . or. Unlike fatigue damage. which can be evaluated from data on the tertiary creep characteristics of the material (the increase in strain rate that occurs as failure is approached). Finite element procedures can easily be implemented.3. An alternative procedure for considering creep damage has been suggested that is usually associated with the names of Kachanov and Rabotnov. which accumulates according to the relation dω 1 = dt (1 + γ )t R (σ . T ] Eq. Equation 2-8 can be used to evaluate the time to failure by separating variables and integrating. 2-9 This is the counterpart of Robinson’s rule (Equation 2-5) expressed as an integral rather than a sum. This is the continuum creep damage mechanics methodology. An example of the use of continuum creep damage mechanics to a simple problem is provided in Section 2. The following form contains a term to account for temperature variations and describes the minimum strain rate (which is also known as the steady state or secondary creep rate). This leads directly to the following relation z tR 0 dt =1 t R [σ (t ). Skrzypek and Hetnarski [Skryzpek 93] provide a recent summary. T is the absolute temperature. For constant stress and temperature conditions this corresponds to failure at tR for that stress and temperature. The discussion here is limited to simple stress conditions. = Ae − Q / T σ n ε Eq. The initial damage is 0 and failure occurs when =1. T )(1 − ω ) γ Eq. the creep rate is often expressed by a so-called Norton relation. 2-7 The term is the creep damage. in a larger component. For stress and temperature that vary in a known fashion. creep straining and associated damage can result in appreciable redistribution of stresses relative to initial elastic conditions.2. 2-6 The parameter Q is the activation energy for creep divided by the gas constant. 3 Creep/Fatigue Crack Initiation Creep/fatigue crack initiation is based on the fatigue and creep damage expressed by Equations 2-2 and 2-5. A. Such representations are beyond the scope of these guidelines.1. at least over a limited range of temperature. Q. It is tempting to assume that failure (crack initiation) will occur when the sum of the creep and fatigue damage is equal to unity. Figure 2-3 Creep/Fatigue Damage Plane Showing Combinations Corresponding to Crack Initiation 2-6 . respectively. They are considered to be independent of temperature. Figure 2-3 schematically shows the usual treatment. The first term is the primary creep strain rate. and n are material constants that are obtained from curve fits to creep strain—time test data. and is often included as another term in the creep rate—stress relation. p. it has been experimentally observed that there is some interaction between the damage mechanisms. 2. and failure is considered when the creep and fatigue damage conditions fall outside a line on a creepfatigue damage plot. 2-10 No creep damage is included in this expression. B. Such representations are much more convenient to include in finite element computations for life prediction of complex geometries. the following relation is employed: = Be − Q / T (1 + p) ε 1/(1+ p ) σm + Ae − Q / T σ n p /(1+ p ) p t (1 + ) Eq. primary creep can be important. However. There are ways to include both of these terms in one expression [Stouffer 96] and include creep damage at the same time. Typically. m.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts In some instances. and the second term is the secondary creep strain rate. If the metal temperature decreases below Tlo-ox. and the crack depth is incremented by an amount equal to the increment in the oxide thickness since the last time it cracked. 2-7 . The value of Tlo-ox is 700ºF (371ºC) in BLESS and the depth of notching at which fracture mechanics principles takes over is 0. Section 4. Simple procedures for evaluation of the oxide thickness when temperature varies are given by Grunloh 92 and are depicted in Figure 2-5 for a time ∆t1 at T1 and ∆t2 at T2. The oxide thickness is taken to continuously increase as long as the adjacent metal temperature is greater than Tlo-ox. the growth of the steam-side oxide under constant temperature conditions is expressed as hox = C1e C2 / T t C3 Eq.4 Oxide Notching Crack-like defects can be initiated by the growth and cracking of oxide layers.76 mm). the oxide is assumed to crack. Harris 93]. Figure 2-4 Crack-Like Defect Initiated by Oxide Notching This initiation mechanism is considered in a deterministic fashion in the BLESS software [Grunloh 92. The oxide thickness is then rezeroed and grown during subsequent times above Tlo-ox.1 of Grunloh 92 provides the details of the oxide notching model in BLESS.2.030 inches (0.2. 2-11 The values of C1 – C3 are taken to be deterministically defined. Once the oxide notch crack depth reaches a specified depth. it is considered to be an initiated crack that then grows by fracture mechanics principles. A corresponding probabilistic treatment is not available.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts 2. Figure 2-4 is a photomicrograph of a crack-like defect that has initiated due to repeated cracking of the oxide scale during start-stop cycles. as discussed in Section 2. In this case.1. The temperature is in absolute degrees. Similarly.1 Crack Tip Stress Fields Fracture mechanics principles are most often based on the analysis of the stresses near a crack tip. which.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Figure 2-5 Depiction of Procedure for Determination of Oxide Thickness for a Time at Tlo Followed by a Time at Thi 2. This is readily applied to primary creep also. Hence. Fracture mechanics procedures can be used to estimate the remaining time to grow the crack to the point where it will pose a significant risk to continued operation. 2-8 . 2-12 When n=1 and D=E. cracks may initially be present in a component. If the strain is a rate. rather than a strain directly. then this can represent nonlinear elastic behavior which is the same as plasticity as long as no unloading occurs. Equation 2-12 can be used to describe a variety of material responses. 2. When n ≠ 1. The stresses depend on the stress strain relation of the material. for uniaxial tension. can often be expressed as Fσ I ε=G J H DK n Eq.2. and fracture mechanics is again called for. The strain is then the plastic strain. This is the Ramberg-Osgood representation of plasticity. then this can represent the secondary creep relation of Equation 2-6. this is the familiar Hooke’s law of linear elasticity. and ε is the elastic strain.2 Crack Growth The initiation of a crack most often does not mean that the component has reached the end of its useful life. and iii) the magnitude of the field (for a given D and n) is controlled by the single parameter J. Dimensional considerations require that J has the units of Dr. The parameter J is Rice’s J-integral. 2-13 ~ . which is the value of the strain energy release rate with respect to crack area (joules/m2. Anderson 95]  JD n σ ij =   I r  n  n +1 ~   σ ij (θ . ii) the deformation field (for a given n) always has the same spatial variation. 2-14 2-9 .EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Figure 2-6 shows the coordinate system near a crack tip. Figure 2-6 Coordinate System Near a Crack Tip The deformation field near a crack tip in a homogeneous isotropic body whose stress-strain relation is given by Equation 2-12 is characterized by the so-called Hutchinson-Rice-Rosengren singularity and is given as [Kanninen 85.2.).2. Kumar 81. etc. which is (stress)x(length) or (F/L). Specific examples of J solutions are given in Section 2. and Equations 2-13 can be written explicitly as follows: σx = θ θ 3θ cos (1 − sin sin ) 2 2 2 2πr K θ θ 3θ σy = cos (1 + sin sin ) 2 2 2 2πr K θ θ 3θ τ xy = sin cos cos 2 2 2 2πr K Eq. n)  n 1  J  n +1 ~ ε ij =   DI r   ε ij (θ . in-lb/in2. n)  n Eq. n)  n   J ui =   DI  n  n +1 n1 +1 ~   r u i (θ . This case is of particular interest. The case of linear elasticity is when n in Equation 2-12 is equal to 1.ε ~ and u ~ are dimensionless tabulated functions [Shih 83] and I is a dimensionless where σ ij ij i n constant [Anderson 95. These equations show that i) the stresses and strains are large as r approaches zero. Kanninen 87] that depends on n and whether the conditions are plane stress or plane strain. J is a measure of the crack driving force. If the strain in Equation 2-12 is replaced by a strain rate. such as is shown in Figure 2-7. if n=1. which is denoted as C*. K must vary linearly with the applied loads. For the linear problems to which K is applicable. and the field is controlled by a single parameter which is the rate analog of the J-integral. as in Equation 2-10. 2-10 . K must be proportional to σ a . which are referred to here as the crack driving forces. As Equation 2-14 shows. Equations 2-13 still describe the stress and strain rate field near the crack tip. have an influence on the crack tip fields. If primary creep is also included. The magnitude and type of loading. (E is Young’s modulus and ν is Poisson’s ratio). the stresses are controlled by a single parameter.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts As expected. 2-15 plane strain Equation 2-14 shows that K has the units of (stress)x(length)1/2. but K is related to J by Equation 2-15.2. and this influence enters into the expression for J or K. which is denoted as K and is called the stress intensity factor. Specific examples of K solutions are discussed in Section 2. From dimensional considerations. the stress-strain rate relation is as given in Equation 2-6. the stress intensity factor K has dimensions of (stress x square root of length). then C* is applicable to the secondary creep and there is another parameter to account for the primary creep.2 Crack Driving Force Solutions Equation 2-13 shows that the stress-strain-displacement field near a crack tip is controlled by a single parameter J. This * parameter is referred to as Ch . and is the parameter controlling the crack tip stress field when primary creep is dominant.2. then the parameter K controls the field. because a is the only length available. For a through-crack in a large plate.2. as well as the geometry of the cracked body. K and J are related to one another by the expression R K | |E J=S | K (1 − ν ) | T E 2 2 2 plane stress Eq. 2. The proportionality constant turns out to be π1/2. n) Eq. If the crack driving force is expressed in terms of J. Tada. 2-16 There are numerous such K-solutions available for a wide range of crack configurations and loadings.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Figure 2-7 Through-Crack of Length 2a in a Large Plate Subject to Stress σ. If the material is creeping. The expression for J is given in the figure. stress intensity factor solutions can be written as K = σ πa F (geometry) Eq. consider the edge-cracked strip in tension shown in Figure 2-8.n) has been determined by finite element computations and tabulated [Shih 84]. As an example of a J-solution. the crack driving force can be written as J = σεa G ( geometry . Paris and Irwin [Tada 00] is an example of such a compendium. J-solutions are tabulated in Kanninen 87. then J is . and n=1. 2-11 . G=π. Table 2-1 is the table for plane strain. The function h1(α. 2-17 For the simple problem of Figure 2-7. Anderson 95 replaced by C* and ε is replaced by ε and Kumar 81. as obtained from the limiting case of the elasticity solution for stresses near the tip of an elliptical hole in a plate. which has units of in-lb/in2 or Joules/m2. In general. 62 2.452 2-12 .01 4.43 5 12.13 1.243 0.45 2.09 5.70 1.38 0.62 3 9.290 0.375 0.373 0.585 0.188 0.89 2.93 2 7.420 0.2 1.142 0.1 1.125 0.812 13 27.0952 0.20 3.111 0.86 1.00 10 21.744 0.0710 0.88 2.7 4.409 0.754 0.0391 0.500 0.875 n=1 5. R | κ =S | T1.225 0.18 7 16.750 0.3 3.563 20 45.87 1.186 0. n) σ n +1 a n D (1 − α ) n (κη) n +1 α =a/h η= 1 − 2α + 2α 2 − α 1− α plane strain plane stress 1455 .48 2.54 1.55 1.250 0.90 0.16 1.858 1.02 1.7 3.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts J= h1 (α .28 0.11 1.22 1.672 16 34.92 1.17 5.396 0.16 2.48 0.42 3.02 0.3 2.38 1.704 0.97 3.625 0.072 Figure 2-8 Single Edge-Cracked Strip in Tension With J-Solution Table 2-1 Table of the Function h1(α.n) in the J-Solution for an Edge Cracked Plate in Tension for Plane Strain [From Shih 84] a/h⇓ 0. such as very slow (<~10-7 in/cycle) or quite rapid (>~10-3 in/cycle). because the procedures involved (usually finite elements) are more straightforward and linear superposition is applicable. The material is 2 ¼ Cr 1 Mo steel at various temperatures. This has been observed to be the case for a wide variety of metals and conditions. and the following relation is often found to provide a good fit to data da = C f ∆K m dN Eq. that is. which is equal to Kmax – Kmin. however. The critical value of J or K is usually called the fracture toughness. See for instance. There are many complications. the failure criterion can be expressed in terms of the stress intensity factor. JIc. and more complex representations are appropriate. These complications will not be considered here. The room temperature fit is also shown. This suggests the use of a probabilistic treatment. Tada 00 is an example of a compendium of stress intensity solutions. the cyclic parameter is ∆K. however. Kanninen 87.2. including the influence of non-singular terms on the stresses and strains. Many more crack cases have been analyzed for the linear problem. and Anderson 95. Kumar 81. At extremes of crack growth rates.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts A variety of other geometries have been analyzed with solutions analogous to the one shown in Figure 2-8. For linearly elastic bodies. Both of the lines in Figure 2-9 are of the form of Equation 2-18.3 Calculation of Critical Crack Sizes Since the stresses and strains surrounding a crack tip are controlled by the value of the J-integral. and the dashed line is a least squares curve fit to the 1100°F (593°C) data. The 1100°F (593°C) data is considerably above the data for the lower temperatures. The value of JIc is obtained in a test. The figure is drawn from Viswanathan 89. Figure 2-9 is an example of fatigue crack growth data. The Forman relation is widely used in such instances. as well as well as increasing resistance of the material to crack growth as the crack extends. In the case of conditions where the body remains substantially elastic. the crack growth behavior can deviate from this relation. This figure shows that the fatigue crack growth rate is not a strong function of temperature.2. especially for the 1100°F (593°C) data. The values of Cf and m for the lines in Figure 2-9 are as follows: 2-13 . The amount of scatter in the data is seen to be quite large. a reasonable failure criterion is that failure occurs when the applied value of J equals some critical value. with data for 700°F (371°C) being comparable to the room temperature line. Anderson 95 provides details. 2-18 This is the so-called Paris relation and is a suitable representation under a wide variety of conditions. see for instance Henkener 93. 2. with a critical value being denoted as KIc. 2. This criterion is often valid and has been widely used. it is reasonable to presume that the rate of growth of a crack in a body subject to cyclic loading (da/dN) is controlled by the cyclic value of the crack tip stress parameter. the Paris relation.4 Fatigue Crack Growth Since the stress-strain field near the tip of a crack is controlled by a single parameter. elastic and primary creep strain rates must be negligible. Under the case of secondary creep dominating.41x10 8.2. Figure 2-9 Fatigue Crack Growth as a Function of the Cyclic Stress Intensity Factor for 2 ¼ Cr 1 Mo at Various Temperatures [Drawn From Viswanathan 89] 2. 2-19 2-14 . there is still an elastic response that must be considered. but many complicating factors arise. Equation 2-19 has been found to provide a good representation of data da = Cc C * q dt Eq. it would be reasonable for the creep crack growth rate (da/dt) to be controlled by the value of C*.099 MPa-m1/2) and da/dN is in inches per cycle.95 The values of Cf are applicable when K is in ksi-in1/2 (1 ksi-in1/2=1.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Cf Room temperature 1100°F (593°C) 1.85 2.07x10 -11 -10 m 3. which is the time analog of the J-integral. The primary restriction is that the body must be fully in the steady-state creep range. This has been observed to be the case. Even if the material exhibits no primary creep.5 Creep Crack Growth The stresses near a crack tip in a body that is undergoing secondary creep according to Equation 2-6 are controlled by the parameter C*. Hence. and the third line is the steady-state creep. and secondary creep. 2-20 + C* The first line is the elastic transient that occurs on initial loading. and Bloom and Malito [Bloom 92]. As time becomes very large. The parameter C*h is the primary creep analog of the steady state parameter C*. The approach of Bloom is to consider a timedependent crack driving force that has terms corresponding to elastic.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts When the strain rates consist of elastic. The crack driving force is referred to as Ct(t) and is given as Ct (t ) = C *2 /[(1+ p )( m−1)] LM (1 − υ ) K OP N E (n + 1)t Q 2 2 1− 2 (1+ p )(1− m ) + n + p +1 * 1 Ch t (n + 1)( p + 1) FG IJ HK p /(1+ p ) Eq. the second line is the secondary creep. It is obtainable n from the J-solution by replacing (1/D ) with B(1 + p)e − Q / T 1/(1+ p ) The creep crack growth rate is then considered to be related to Ct(t) by use of Equation 2-19 with C* replaced by Ct(t). the situation becomes more complex.ave t h q Eq. such as described in Riedel 87 . the increment of crack growth per cycle has been found to be related to the average value of Ct(t) during the cycle plus a fatigue contribution.6 Creep/Fatigue Crack Growth When cyclic loading occurs at temperatures and cycling rates where creep is important. The growth per cycle is given by the expression ∆a|cycle = C f ∆K mf Eq. primary and secondary creep rates.2. the third line dominates. Saxena 98. primary creep. A variety of procedures have been suggested. 2-21 + Cc Ct . This provides the relation da = Cc [Ct (t )]q dt 2. 2-22 The average value of Ct is given by 2-15 . 1 kip/inch-hour = 1.75x10 J/m -hr). 2-16 . The line in the figure is best fit to the data and corresponds to Cc = 0. and the right-hand part of the figure is for cyclic loading with various hold times. no load cycling). The data is for 1 ¼ Cr ½ Mo steel 1000°F (538°C). such as the rise time of the loading.825 in Equation 2-21. th is the duration of the time at load and to is a small time. 2-23 to In this expression. The left-hand part of the figure is for creep crack growth (i. (The value of Cc is for Ct(ave) in kips/inch-hour and 5 2 crack growth rates in inches per hour.0246 and q = 0.e. which suggests the usefulness of a probabilistic approach. There is a considerable amount of scatter observed in Figure 2-10.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Ct . Figure 2-10 provides an example of creep crack growth and creep/fatigue crack growth data.ave = z th Ct (t )dt Eq. EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Figure 2-10 Creep and Creep/Fatigue Crack Growth Data and Fits. 2-17 . Left Figure Is for Constant Load and Right Figure Is for Cyclic Load With Various Hold Times [From Grunloh 92]. The initial half-crack length is ao.3 Simple Example Problems Two simple problems are presented to serve as demonstration of the procedures involved in life prediction. ac. da = C f ∆K m = C f ∆σ πa dN m = C f ∆σ m π m / 2 a m / 2 This equation can be solved by separating variables and integrating. is 1 K Ic ac = π ∆σ LM OP N Q 2 Eq. KIc. the cycles to failure are not strongly influenced by the critical crack size if the initial size is small. Nf. independent of critical crack size. Nf = 1 m− 2 Cf 2 ∆σ π m m/ 2 LM Na 1 ( m − 2 )/ 2 o − 1 ac ( m − 2 )/ 2 OP Q Eq.3.1 Fatigue of a Crack in a Large Plate Consider a through crack in a large plate. 2-24 The fatigue crack growth relation is the Paris relation of Equation 2-18. Their probabilistic counterparts are included in Section 4. Figure 2-11 is a plot of a vs N for a ∆σ of 20 ksi (137.050 inches (1. and Cf and m for the room temperature line in Figure 2-9. This is a consequence of the nearly vertical slope of the line in Figure 2-11 as a exceeds about 2 inches (50 mm).000. 2. If the critical crack size is larger than about 4 inches (100 mm). Failure occurs when the maximum applied stress intensity factor is equal to the critical value. but once the crack starts to grow appreciably. 2-18 . Hence the cyclic stress intensity factor is given by the expression ∆K = ∆σ πa Eq.300. then the cycles to failure is nearly 1. The problems in this section are deterministic.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts 2. 2-26 As an example of the above relations. it quickly becomes long. for a given initial crack size ao. such as shown in Figure 2-7.9 MPa).2. and the plate is subject to a stress that cycles between 0 and ∆σ. not much happens for a long period. 2-25] A differential equation for the crack length as a function of the number of fatigue cycles is obtained by inserting the relation for ∆K into the Paris relation for crack growth rate. thereby providing the following end result for the cycles to failure.27 mm). Also. The results of Figure 2-11 are fairly typical of fatigue crack growth problems with initial cracks that are quite small. The critical crack size. an initial crack half-length of 0. This example problem could be representative of a superheater/reheater tube. for this example consider the maximum principal stress to be governing. so the integration can not be performed. and a thickness that decreases with time according to h(t ) = ho − t Eq. in which case the stresses are dominated by pressure and easily estimated.9 MPa) for Example Fatigue Problem In most practical situations. Another complicating factor is that the fatigue crack growth relation is usually more complex than the Paris relation. and the hydrostatic stress. In general. Although the stress that controls creep rupture can be a combination of the principal stress. but realistic problems usually require numerical procedures for computation of crack sizes and lifetimes.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Figure 2-11 Half-Crack Length as a Function of the Number of Cycles to 20 ksi (137. As a simple example. Consider a tube with constant internal pressure p. Stress intensity factor solutions for realistic problems are more complex than in this example. if the curve fit in Figure 2-2 is taken to be a straight line. The above example demonstrates the principles involved. rather than the second order relation of 2-19 . a closed form expression can not be obtained for the crack size as a function of the number of cycles. This is the hoop stress due to the pressure. The above equation can be used along with Equation 2-9 to obtain the time to failure. equivalent stress. numerical integration is necessary. mean radius R.2 Creep Damage in a Thinning Tube An example of creep rupture with wall thinning is presented as an example of creep life prediction. which is given by σ (t ) = pR pR = h(t ) ho − t Eq. 2-27 The stress strain rate relation is given by Equation 2-7 and the damage kinetics by Equation 2-8.3. 2-28 Consider temperature to be fixed. 2. the following values are obtained -4 β = 1/BT = 1/(1460x1.453 CR = 10-C+A/(BT) = 10-20+6. T ) = 10 − C + A /( BT ) σ 1/ BT Eq. consider a 2. this can be rearranged to give the following expression for the rupture time t R (σ .0001538) = 2. 2-31 As an example. Using the above definitions.55 MPa) pressure. 2-30 Use the following definitions: β = 1 / BT C R = 10 − C + A / BT t T = time for wall to thin to zero = h / pR σ o = hoop stress at initial thickness = ho t C = time for creep rupture at intial stress = 10 − C + A / BT C R = β / BT σ1 σo o The time to rupture with wall thinning and creep damage is then obtained by using Equation 2-30 for the rupture time and Equation 2-28 for the stress in conjunction with Equation 2-9.615 − 1538 .0625/0. 2-29 Using the definition of LMP from Equation 2-3. x10 −4 LMP Eq.125 inch (54 mm) diameter 1 ¼ Cr 1/2 Mo tube with a wall thickness of 0.4 inches (10. then a closed form relation for the rupture time can be obtained.460°R = 811.000°F (1.1°K) and 2400 psi (16.4x1.88x109 σo = 2.38 ksi (44.7 years 2-20 .0 MPa) tC = 7. Using the above definitions and grinding through the algebra leads to the following relation for the rupture time tR: tR = 1− tT LM1 + (β − 1) t OP t Q N C T 1 1/( β −1) Eq.0 MPa) log σ = A − B( LMP) = 6.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Equation 2-4.2 mm) operating at 1.615/(1460x0.4 = 6.51x105 hours = 85. The following is a good linear representation of the data of Figure 2-2 for stresses less than 20 ksi (137.538x10 ) = 4. Although such models are available. and detailed lifetime evaluations often require finite element computations. etc. but first some statistical background information is provided in Section 3. so the stress analysis is simple. It is seen that there is a strong interaction between the creep and thinning degradation mechanisms.4 µm/yr) In most practical situations. temperature. because of more complex stress and temperature histories and more complex geometries. and derivation of model constants from test data. Probabilistic models account for these uncertainties and scatter.EPRI Licensed Material Review of Deterministic Life Prediction Procedures for Boiler Pressure Parts Figure 2-12 provides a plot of the failure time for various wall thinning rates . because the failure time is much smaller than if only one mechanism is acting. including scatter in material properties. 2-21 . uncertainty in service conditions (pressure. a closed form expression for the creep lifetime can not be obtained. This simple example problem serves to show the principles involved. Figure 2-12 Time to Failure as a Function of the Wall-Thinning Rate for the Creep Rupture Example Problem (1 Mil/Year=25.). and are discussed in Section 4. The above discussion provides examples of deterministic lifetime models. In fact. The stresses in the above example are statically determined. results obtained in their application to real plant components are subject to many sources of uncertainty. creep strain and damage can result in large changes in stress in complex bodies. . If a set of x data is available. p(x). x. Hahn 67. Mathematically. If N is large. commonly denoted as σ. The following basic references are suggested for additional information [Ang 75. 3-1 Various characteristics of random variables are of use. it doesn’t make much difference. this is expressed as probability that x falls between x and x + dx = p( x )dx Eq. if N=1 then σ is said to be biased.1 Probability Density Functions For a continuous random variable.EPRI Licensed Material 3 SOME STATISTICAL BACKGROUND INFORMATION Some background information on statistics is provided in areas of particular interest in probabilistic structural analysis. Ang 84. 3-4 ( x − x ) 2 p( x )dx −∞ 3-1 . If N-1 is used. the probability of x falling within a range of values is described by the probability density function. Wolstenholme 99] 3. and those familiar with statistics can proceed directly to Section 4. The values of the mean and variance can be obtained from the probability density function (pdf) by use of the following x= σ2 = z z ∞ −∞ xp( x )dx ∞ Eq. 3-3 The square of the standard deviation is called the variance. No attempt is made to be comprehensive. then the value of σ is called unbiased. is 1 σ = N −1 2 ∑ N i =1 1 ( xi − x ) = N −1 2 LM x MN∑ N i =1 2 i −Nx 2 OP PQ 1/ 2 Eq. the mean is given by the expression 1 x= N ∑x N i =1 i Eq. the most common ones being the mean (or average) and the standard deviation. Ayyub 97. Sometime the N-1 in the denominator in Equation 3-3 is replaced by N. 3-2 and the standard deviation. and is often referred to. Mathematically. This is known by various names. and sometimes the reliability or survivor function. Their usefulness here is in describing the scatter or uncertainty in the variables entering into the lifetime model. is equal to the standard deviation divided by the mean. if sufficient data is available. The complementary cumulative distribution is equal to one minus the cumulative distribution. The most commonly used distribution is the normal. distribution. this is expressed as 1 = 2 z x50 p( x )dx or P( x50 ) = −∞ 1 2 Eq. or personal taste. 3-7 There are many different probability density functions. This is the value that is exceeded with a 50-50 chance. The cumulative distribution function is obtained from the probability density function by the equation P( x) = z x p( y )dy Eq. h(x). The median of a random variable is also often of interest. This is referred to as the cumulative distribution function and is denoted as P(x). 3-2 . Any random variable that is the sum of many other random variables is normally distributed. Just about any function that integrates to unity can be used as a pdf. This is a consequence of the “central limit theorem. The most compelling reason is fits to data. theoretical considerations. This is also called the hazard function.EPRI Licensed Material Some Statistical Background Information The coefficient of variation. and is related to the pdf and cumulative by the expression h( x ) = p( x ) 1 − P( x ) Eq. The probability that x is less than some value is often also of interest. or Gaussian. The type of distribution to be used in a given situation can be selected based on fits to data. 3-6 Another item of interest is the failure rate. the maximum value of P(x) is 1 and Equation 3-5 implies that z ∞ p( x )dx = 1 −∞ The probability that x is greater than some value is also of interest. convenience. This is the probability of failure between x and x+dx given that failure has not already occurred. regardless of the distributions of the individual variables.” Table 3-1 summarizes several of the distributions most often encountered in probabilistic lifetime analysis. cov. 3-5 −∞ Since probabilities are always between 0 and 1. usually the complementary cumulative distribution. R | | F 1I L F 1 I O U Γ G 2 + J − MΓ G 1 + J P V . c=2 is Rayleigh 1. x50 is equal to exp[mean of ln(x)]. ∞ x50 e µ 2 /2 x50 e 2 µ − e µ 2 2 log(x) is normally distributed. erf(x) commonly tabulated normal or Gaussian x−x 1 1 + erf 2 σ 2 2 LM N FG H IJ OP KQ I OP JK PQ −∞.EPRI Licensed Material Some Statistical Background Information Table 3-1 Characteristics of Some Commonly Encountered Probability Functions Used to Describe Scatter and Uncertainty Probability Density Function Cumulative Distribution Function Range of Random Variable Name uniform Mean Standard Deviation Comments simplest 1 b 1 − x/λ e λ − 1 e σ 2π ( x− x )2 2σ 2 x−a b 1− e − x/λ a. which is widely tabulated (see. has constant hazard function. simple cumulative distribution. µ is the standard deviation of ln(x) 2. x50 is 1 the median c x b b FG IJ HK c −1 e − ( x /b )c 0. a + b 0. for S H K H K | T c N cQ| W 2 3-3 . Weibull with c=1 most commonly encountered. ∞ x σ lognormal 1 µx 2π Weibull LM ln( x / x ) OP e N µ 2 Q − 50 ln( x / x50 ) 1 1 + erf 2 µ 2 1 − e − ( x /b ) c LM MN F GH 0. Γ is the gamma function. Γ function tabulated. ∞ a+ b 2 b 12 exponential simple. ∞ bΓ 1 + FG H 1 c IJ K see note 2 a favorite of many. standard deviation of a Weibull variate is b instance. Abramowitz 64). c=1 is the exponential distribution. The parameters are evaluated by finding their values that will maximize the likelihood.2 Fitting Distributions A key part of developing probabilistic lifetime models is the definition of the distributions of the random input variables. They also provide curve fits for inverse error functions. This method has some theoretical advantages but is often more difficult to implement. 3. as shown in Table 3-1. The standard normal variate is often tabulated in statistics texts. It is one minus the error function.EPRI Licensed Material Some Statistical Background Information A function that is often encountered is the so-called error function. 3-12 3-4 . 3-11 For instance. That is erfc( x ) = 1 − erf ( x ) Eq. and is denoted as erfc(x). c) = ∏ FGH IJK N i =1 c xi b b e − ( xi /b ) c Eq. parametersg N i i =1 c −1 Eq. 3-8 0 The complementary error function is often also of use. 3-10 The error function is widely tabulated. which is the cumulative distribution for a normally distributed variate with zero mean and a standard deviation of unity. are often referred to as the parameters of the distribution. Using the Φ( x ) notation. Another way to evaluate the parameters of the distribution that entails the distribution type is the method of maximum likelihood. either as erf(x) or Φ( x ) . the expression for the likelihood would be L(b. Abramowitz and Stegun [Abramowitz 64] provide convenient tables and curve fits. if a Weibull distribution is assumed. which is expressed as L( parameters) ≡ ∏ pb x . This is most often done by use of data. the relationship to the error function is Φ( x ) = x 1 1 + erf 2 2 LM N OP Q Eq. and is denoted as P(x) or Φ( x ) . such as b and c for the Weibull distribution. It is useful because it conveniently expresses the cumulative normal distribution. The constants. and Table 3-1 then used to get the constants in the distribution functions. Equations 3-2 and 3-3 can be used to compute the mean and standard deviation from the data. 3-9 The error function is related to the cumulative unit normal variate. erf(x). The error function is defined as erf ( x ) ≡ 2 π z x e − y dy 2 Eq. If maximum likelihood is used. 3-14 and this will be used herein. Alternatively. 3-15 3-5 . The data. 3-13 Many relations similar to Equation 3-13 have been suggested as better approximations for P. this can be written as x = λ ln FG 1 IJ H1− PK so that if ln[1/(1-P)] is plotted versus x. P(x). The -x/ simplest example is for an exponential distribution. Once the values of si (sorted xi) are known. Rearranging and taking logarithms. a Pi – si plot is made on paper whose scales are transformed in such a manner that the cumulative distribution will plot as a straight line if the data is of the distribution type that the paper was constructed for. In this case. Just knowing the parameters is not sufficient to define the distribution—the distribution type must also be known. It is most often not easy to solve Equation 3-12. The values of P are then estimated by a relation such as Pi = i N Eq.EPRI Licensed Material Some Statistical Background Information The values of b and c that maximize L would be selected. lognormal. Similar transformations can be made for other distribution types. Such paper is available for selected distribution types. then the distribution type has already been assumed. including normal. such as lognormal or Weibull. but it is still not known how well the data has been represented. then the data will plot as a straight line with slope related to the parameter . and x is exponentially distributed. a histogram of the data can be constructed. by first sorting the data in ascending order. Denote this sorted list as N values of s. can be used to obtain the cumulative distribution. but this paper can be constructed by appropriate transformations. such as normal probability paper. and Weibull. the cumulative distribution can be written as ln( x ) = ln( x50 ) − µ 2erfc -1 (2 P) Eq. For the lognormal distribution. and numerical methods must be employed. P(x)=1-e . A good way to assess the fit to the distribution type is to plot the distribution on a standard paper. One widely used relation is Pi ≈ i− N 1 2 Eq. say N values of x. 3-16 As an example. fix m at 2. that is if erfc()= then -1 erfc ()=. erfc-1(x) is the inverse complementary error function. The counterpart of Equation 3-15 for a normal variate is x = x − σ 2erfc -1 (2 P) Eq. P is evaluated by use of Equation 3-14 and Y=erfc-1(2P).95. 3-6 . which can be used to define a curve fit to the inverse complementary error function. Abramowitz and Stegun contain convenient curve fits to the inverse complementary unit normal variate (see page 933 of Abramowitz 64). if erfc-1(2P) is plotted versus x on a log scale.95 and evaluate Cf for each of the 25 data points in Figure 2-9. respectively. In this table. There are 25 data points. The linear least squares curve fit to the data gave a Cf and m (in Equation 2-18) of 8. the data will plot as a straight line if it is lognormally distributed. This provides the data in Table 3-2.07x10-10 and 2.EPRI Licensed Material Some Statistical Background Information In this equation. consider the 1100ºF (593ºC) fatigue crack growth data in Figure 2-9. Hence. In order to characterize the scatter in the 1100ºF (593ºC) data. 7840 7.3200 10.5690 8.7640 .4547 .0350 7.5458 -.6880 7.5870 8.1435 .1000 -1.5000 .9300 -20.5458 .9800 X=ln(Cf) Sorted -21.2920 6.3000 .4600 .3170 9.7000 P .0210 7.7640 -.1900 -21.3400 .8600 .0100 -21.6472 .1970 8.4200 .9700 -20.5250 8.7300 -202.0500 -21.4540 1.2163 .9200 -20.0022 -.6600 -20.6880 7.5300 7.1900 -21.2580 6.0700 -21.9900 -20.5250 7.9400 -20.3707 -.9450 10.7870 8.2163 -.9450 10.0724 -.5400 8.7140 9.0100 -21.5400 .7000 7.5300 7.1000 .EPRI Licensed Material Some Statistical Background Information Table 3-2 Values of Cf for Fatigue Crack Growth Data (values of Cf are for crack growth in inches per cycle and ∆K in ksi-in ) 1/2 i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 10 Cf 7.3170 9.3730 10.5400 9.7140 7. Both Figures 3-2 and 3-3 provide a good fit to the data. The 3-7 .0200 .1800 .5690 7.2100 7.8200 .2100 7.3800 .2200 . and Figure 3-3 is the corresponding normal probability scale plot.9066 .6000 9.2870 7.5870 9.9400 .4540 The data in Table 3-2 is plotted on log-linear scales in Figure 3-1.6472 -.1490 8.1660 7.2870 7.7870 6.1435 -.2580 6.9066 -1.4740 7.1660 7.4547 -.6900 -20.7900 -20.4740 9.1500 -21.9000 .0600 -21.7840 7.3200 10. The straight lines shown in these figures are the result of linear least squares curve fits on the transformed scales of the figures.6200 .8800 -20.3707 .6000 10.7800 .0350 7.3730 9.9800 -20.1400 .0400 -21.1000 .7400 .2920 6.0000 -20. Figure 3-2 is the same plot on lognormal probability scale.7900 -20. The solid line in these figures is the linear least squares fit to the corresponding X-Y data.7700 -20.6600 .5800 .7000 .2917 .9700 -20.0600 .2600 .2917 -.6600 Y 1.1970 10 Sorted 10 10 Cf 6.0210 8.1490 7.0724 -. and the interest in probabilistic structural models is often in extrapolations beyond the data. D’Agostino 86. Such goodness-of-fit tests are restricted to the range of data. for instance. Figure 3-1 Plot of Data of Table 3-2 on Log-Linear Scales Figure 3-2 Lognormal Probability Plot of Data of Table 3-2 3-8 .EPRI Licensed Material Some Statistical Background Information lognormal fit (Figure 3-2) appears to be somewhat better than the normal fit (Figure 3-3). Which of these is better can be judged by the “goodness of fit”. see. which is an entire topic in itself. The values from the data are for lognormal fit-0. and the lognormal distribution would be selected for further use. 3-17 The value of quantifies how well the x-y data is represented by a linear relation.977 for normal fit -0. unless the normal fit was substantially better than the lognormal. In fact. In selecting between the normal and lognormal fits of Figures 3-2 and 3-3. this would be the basis for selection of the lognormal. This is preferred. A value of +1 or –1 means that the data is fitted perfectly. as observed in Figure 2-9. because if the data on a log da/dN-log∆K plot are evenly distributed above and below the best fit line. the fit to the lognormal is somewhat better. then Cf would be expected to be lognormally distributed rather than normal. 3-9 . the corresponding coefficient of linear correlation is of interest.EPRI Licensed Material Some Statistical Background Information Figure 3-3 Normal Probability Plot of Data of Table 3-2 Another way to see how well the data is being fitted within the range is the coefficient of linear correlation. and a value of 0 means that there is no linear relation between them (that is not to say that there may be some nonlinear relation between them). which can be evaluated from the data by the expression ρ= R U | | x − x) y − y) V ( ( S ∑ ∑ | | T W N N 2 2 i i i =1 i =1 ∑ (x − x) ( y − y) N 2 i i i =1 2 1/ 2 Eq.966 Hence. We would much rather have a good representation in the tails with a poor agreement with the mean of the data than the other way around. the least squares line on probability paper is highly influenced by a few data points out in the tail. The fit on probability paper generally provides a better representation in the tails.160 -10 The parameters for this particular set of data.78x10 8. and parameters based on such a fit are preferred to a good fit in the central region of the distribution. However. the mean and standard deviation evaluated from the data by Equations 3-2 and 3-3 do not agree well with a fit on probability paper. there is a finite probability that Cf will be negative.153 0. agree well with one another.07x10 -10 1. Quite often. and the related curve fits.160 and a median of 8. which is physically unrealistic (fatigue cracks do not usually get shorter). so this is not really a serious objection. the probability of having a negative Cf is about 10-10. The parameters as evaluated by various procedures do not always agree as well as in the above example. 3-10 . For the mean and standard deviation based on the data and included in Table 3-3. A µ of 0. as estimated by various means summarized in Table 3-3.07x10 -10 -10 -10 8.28x10 0. Since the goal of probabilistic analyses is usually to define the reliability out where the failure probability is small.07x10 -10 7.EPRI Licensed Material Some Statistical Background Information Another reason for selecting the lognormal is that if Cf is normally distributed. the information in Table 3-3 on the parameters of the distribution of Cf is obtained. it is important to have a good representation of the random variables out in the appropriate tails. Based on the data.16x10 8. Table 3-3 Information on Parameters of Distribution of Cf Data of Table 3-2 (values of Cf are for crack growth in inches per cycle and ∆K in ksi-in1/2) least squares fit on log-log da/dN-∆K median of data mean of Cf exp[mean(ln Cf)] median from Fig 3-2 fit (Y=0) standard deviation of Cf (from data) stdev[ln(Cf)] µ from slope of Fig 3-2 fit 8. This is because the mean and standard deviation evaluated by Equations 3-2 and 3-3 provide good representations near the mean and are not affected by a few data points well away from the mean (out in the tails of the distribution).07x10-10 would be selected. and cyclic stress as inputs. p( z ) = 1 e −z/λ − e − z/µ µ−λ c h (Great care must be used in the limits of integration). If some of the input variables are random. then it turns out that z is also normally distributed. and are denoted as F(x) and G(y). respectively. there may be an equation for the lifetime that contains several inputs. but numerical procedures may be required.EPRI Licensed Material Some Statistical Background Information 3. Equation 2-26 is an example. and the parameters describing the distribution of z are readily obtained from the parameters of x and y. Then find the distribution of z = x + y. a closed form expression for the distribution of Nf is not obtainable (as far as known to the author) regardless of the distribution types of the random variables—unless most of them are deterministic. In more realistic problems. The cumulative distribution of z is then obtained from the equation P( z < u) = probability that z is less than u = = ∑ (probability that x is between x and x + dx)bprobability that y < u − xg z x x f ( x )G (u − x )dx If x and y are normally distributed. but z is no longer exponentially distributed. the probability of failure prior to a given number of cycles is just the cumulative distribution of Nf. This is a simple problem. For instance. with the initial crack size. Using Equation 2-26 as an example. If x and y are lognormally distributed with parameters x50. Closed form expressions can be obtained only in exceptional circumstances. the reliability as a function of the number of load cycles is known if the distribution of Nf is known. let x and y be random variables with known probability density functions f(x) and g(y). (respectively). then Nf is a random variable. About the simplest case would be for x and y to be exponentially distributed with parameters and µ. fatigue crack growth coefficient. Hence. The probability density function of z is given by the following expression. there is still a way to compute the lifetime for a set of inputs. The corresponding cumulative distributions of x and y are also known. µ and y50. it is desired to obtain the distribution of Nf for a given set of random input variables. then the cumulative distribution of z is given by the expression 3-11 .3. there is often an underlying deterministic model with some of the inputs being random variables.3 Combinations of Random Variables In probabilistic lifetime models. critical crack size. 3. for the simple example of Equation 2-26.1 Analytical Methods As an example of analytical methods. In fact. and numerical procedures are most often required. A Weibull to some power is a Weibull. 3-18 The integral in this equation can not be evaluated in closed form. then there are other more straightforward procedures. They can provide closed form representations that are of use in some instances that provide good checks on numerical results. then z is a Weibull with parameters b 1/ p and c / p . such as Monte Carlo simulation. However. the distribution of the sum of two lognormals can not be expressed in closed form.i and second parameters µi. Hence. if z = K ∑ x and the x have means x and variances σ then K 2 i i i i i =1 2 the mean of z is ∑ x and variance σ = ∑ σ K i i =1 i =1 2 i • The product of lognormals is lognormal. to the author’s knowledge. In these days of fast. inexpensive PCs. This is the counterpart of the above relation for normals. means that the product of lognormals to powers is itself lognormally distributed with easily evaluated parameters. This list is probably not complete. • The sum of normals is normal. then z is lognormally distributed with median x50 and second parameter pµ. • A lognormal to some power is lognormal. 3. The following provides a brief summary of such cases. The point is that analytical methods are of limited use. if z = ∏ x and the x have medians x K i i 50. It is simple to understand and implement and can be made as accurate as desired by taking sufficient samples. but if numerical techniques are called for. along with the immediately above case. which is not near the problem that it was formerly when computer time was much more expensive. then z is lognormally distributed with median i =1 ∏x K i =1 50 .3.i and second parameter µ = 2 ∑µ K i =1 2 i . There are a few cases where analytical methods are useful.2 Monte Carlo Simulation – Principles This is a commonly used method of obtaining results from a probabilistic model. if z = x p and x has median and second parameter p x50 and µ. 3-12 . if z = x p and x is a Weibull with parameters b and c.EPRI Licensed Material Some Statistical Background Information P( z) = 1 2 µ 2π z z 0 1 − 2 µ2 G H ln x50 JK erfc 1 ln y50 e x z−x λ 2 1 x F I 2 LM N FG H IJ OPdx KQ Eq. it can be burdensome in computer time. • Analytical methods are of limited use in all but the simplest of component reliability models. This. The integral can be easily evaluated by numerical integration. so you do not know in advance. This is like a cumulative probability and you want to get the value of the random variable that corresponds to this cumulative. a value of the lifetime is calculated using the deterministic lifetime model. The probabilistic model then involves characterizing the distributions of the random input variables and evaluating the failure probability. Nearly all computers have random number generators built into them (of varying quality).EPRI Licensed Material Some Statistical Background Information Monte Carlo simulation is a useful tool. Monte Carlo is one alternative. 3-19 Eq. Once a sample of each of the random input variables is drawn. Typically the random number is for a uniform distribution of a random variable that is between 0 and 1. analytical tools are seldom suitable. Another way to look at it is that the lifetime “data” from the trials can be sorted into a histogram and the resulting histogram normalized to define the probability density function of the lifetime. Each of these lifetime calculations is called a trial. it provides the relation between the input variables and the failure time. much like finite elements. For a Weibull variate. However. Monte Carlo simulation involves drawing a sample from the distribution of each of the random input variables. x = x + σ cos(2πu1 ) −2 ln(u2 ) x = x50 exp µ cos(2πu1 ) −2 ln(u2 ) (normal) (lognormal) Eq.” This depends on the answer. The cumulative distribution of the lifetime is the failure probability as a function of time. Before further discussions on the number of trials to take and the resulting confidence in the results. A frequently asked question is “how do I know how many trials to take. an example of Monte Carlo simulation will be provided. the more trials you need. Once a deterministic model of lifetime is available. Equation 2-26 is a simple example for fatigue crack growth. The sampling of an input variable is analogous to rolling a dice that is shaped like the distribution of the random variable. Call the sampled random number u. there are faster ways than using inverse error functions. It is immediately obvious that the better you want to define the failure probability and the smaller the failure probability that you want to consider. Each trial provides a value of the lifetime. 3-20 3-13 . A series of trials is then sorted (like the Cf data of Table 3-2) and the distribution of the lifetime is constructed. you can always make a problem bigger and computer time can become of concern. As discussed immediately above. The following then appear to be normal and lognormal variates. the sampled value is L F 1 IJ OP x = b MlnG N H 1 − uK Q 1/ c (Weibull) For normal and lognormal variates. Abramowitz 64 provides the trick that if u1 and u2 are two samples drawn from unit uniform random numbers then cos(2πu1 ) −2 ln(u2 ) appears to be a unit normal. First you need to shape the die that you are rolling. 042 242.660 27.333 1.675 .161 . Since the results in Figure 3-4 do not fall on a straight line.600 P .580 14.282 5.097 .685 2.380 19.509 4.355 .806 240.140 7.660 52. This is easy to program in FORTRAN.350 34. As an example.212 6. Figure 3-4 also shows the result of numerical integration of Equation 3-18 obtained by use of the numerical integrator in MATHCAD [MATHCAD 91].876 26.180 93.275 .059 4.900 19.075 .701 48.375 .390 14.225 .140 12.010 6.840 97.270 z 11. MATHCAD 91.267 9. Table 3-5 summarizes some of the statistics from the Monte Carlo trials. and y have a median of 4 and a µ of 3.982 8. the die are shaped.900 32.649 24.267 6.475 .525 .580 6.212 52.150 1.473 3. apply Monte Carlo simulation to compute the distribution of the sum of two lognormals (which is the problem that provided the complicated integral in Equation 3-18). Let z = x + y.326 6.440 34.325 .135 7.982 3.EPRI Licensed Material Some Statistical Background Information Thus.708 3.042 11.925 .400 13. or on a spreadsheet—once the underlying deterministic model is available.649 8.575 .000 6.235 1.579 17.900 .625 .175 . Let x have a median of 3 and a µ of 0.131 15.100 9.440 97.025 .666 4.517 9.975 3-14 .840 24.708 1.510 12.775 .125 .320 2.725 .390 34.699 1.875 . they are not lognormally distributed.526 2.825 .150 19.057 1.010 9.437 .350 Sort z 1.513 6.978 7.511 4.071 5. Table 3-4 Results of Monte Carlo Example With 20 Trials i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x 1.8.090 y 9.780 32.600 1.064 29.780 1.510 27. Table 3-4 provides some intermediate results for the case of 20 trials.780 242.110 32.425 . Figure 3-4 provides a plot on lognormal scales of the cumulative distribution of z as obtained by 20 and 500 trials.353 2.135 34.830 5. 357 7.131 360. To the right of the vertical line in Table 3-4 are the sorted values of z and the value of the probability (obtained by use of Equation 3-14).254 9. which are smoother than the results for 20 trials.898 26. These left two columns are plotted as crosses in Figure 3-4.914 3.1 5.129 331.8 3 0.782 1.951 13.2 Median 3 4 4.215 0.06 335. Table 3-5 provides some of the statistics from the trials. the 3-15 .42 4.929 Figure 3-4 Cumulative Probability of the Sum of Two Lognormals as Computed by Numerical Integration and Monte Carlo With 20 and 500 Trials Table 3-4 provides each of the 20 randomly sampled values of x and y along with the resulting value of z.844 3. Note that with 20 trials. Figure 3-4 also shows the results for 500 trials.52 32.177 µ 0.36 2.EPRI Licensed Material Some Statistical Background Information Table 3-5 Summary of Some of the Statistics from the Monte Carlo Trials Mean Inputs x y 20 Trials x y z 500 Trials x y z 4.989 1.178 1. In addressing the number of trials to be performed and what is adequate. We will concentrate on the case where the failure probability (proportion of failures in the trials) is small. and her treatment will be followed here. if there is one failure in 10. The probability of seeing f or fewer failures is the cumulative binomial distribution and is equal to 3-16 . Wolstenholme 99 provides a good discussion on confidence intervals on the results of Monte Carlo simulations. The agreement between the 500 Monte Carlo trials and the numerical integration is good. there are ways to estimate the confidence intervals of the results. For instance.3. This is because the integration got progressively less accurate as the probability increased beyond this point. it would be burdensome to improve the numerical integration but it would be easy for Monte Carlo simulation—just perform more trials. Some random number generators have a seed that allows you to control where they start. The situation improves with 500 trials. For example. (Large means say 100). because it is discreet (that is. which is discussed in Section 3. then the probability of observing f failures in n trials is binomially distributed with a probability “density” given by the expression pf ( f ) = n! p f (1 − p) n − f (n − f )! f ! (pf is not really a probability density function. and Ayyub 95. (or ( f − 1 / 2) / n ) where f is the number of failures and n is the number of trials. and the results are not plotted. If p is the probability of a failure in one trial. Let p be the probability of failure (which is what we want). just make more trials.4.3 Monte Carlo – Confidence Intervals The above example shows how the results change as more trials are taken. (The outcome of the Monte Carlo simulation is itself a random number. Making more trials can lead to excessive computer time. such as importance sampling. and the number of trials is large. and they will then give you the same sequence of random numbers. The estimate of p will be f/n. and will not always give the same answer each time you do the problem. To define the lower tail better. If greater accuracy was desired. the estimated failure probability will be below this number 90% of the time. which is relevant to the question of how many samples to take. This gives the probability of seeing exactly f failures in n trials if p is the failure probability in one trial. which is a big number for many applications. Probabilistic analyses are usually concerned with low failure probabilities. so that you will always get the same answer for the same problem. The scale in Figure 3-4 only goes down to 10-4. but is an artifact of the use of a pseudo-random number generator). it would be useful to say that I am 90% confident that the true value lies below some number that I can evaluate. There are tricks that can be employed. Figure 3-4 also shows the results of numerical integration.95. f is an integer)). This is equivalent to saying that if I perform the Monte Carlo simulation very many times. so we need to define the lower tail of the distribution.000 trials.EPRI Licensed Material Some Statistical Background Information median and µ of x and y do not agree well with the inputs.3. This is helpful in checking. It is then of interest to address the confidence interval on this estimate. a point estimate of the failure probability would be 10-4. 3. Latin hypercube and stratified sampling are also often useful and are discussed in Mahadevan 97a. but the numerical integration is shown only up to a probability of about 0. because we are interested in values of f that are not too big. The root of the equation can be found by MATHCAD [MATHCAD 91] with Table 3-6 summarizing the results for 90 and 95% confidence levels.1 for a 90% confidence). If p is the true value. because of numerical overflow problems with n!. 3-21 where 100(1-α)% is the desired confidence level. Equation 3-21 is difficult to employ. 3-17 . When n is large. α=0. is provided by the Poisson distribution p f ( f ) = e − pn ( pn) f f! The corresponding cumulative distribution is Pf ( f ) = e − pn ∑ f k =0 ( pn) f f! Then the 100(1-α)% upper confidence upper limit on p is the value of p that will give f or fewer failures in n trials. (For instance. A convenient approximation to the binomial density function when p is small and n is large. This is given by the root of the expression e − pn ∑ f k =0 ( pn) k =α k! Eq.EPRI Licensed Material Some Statistical Background Information Pf ( f ) = ∑ (n − nf!)! f ! p (1 − p) f k k =0 n−k n−k An upper limit on p (with a given confidence) can be obtained by finding the value of p that is the solution to the equation ∑ (n −nk!)!k ! p (1 − p) f k k =0 =α Eq. 3-22 Results obtained by use of this equation can be evaluated. there would be f failures in n trials fewer that 100(1-α)% of the time. this value is 1.962 23.5 19. which is well below the 95% confidence limit.339 114. The confidence level (90 vs 95%) does not have a large influence.5 14. The 95% upper confidence limit on the probability of seeing the value 1. Table 3-5 shows an order of magnitude difference (0.303 3.075 95% 2.994 10.322 6.5 1.5 99.708 and the point estimate of the probability is 0. Figure 3-4 shows that the more accurate results for 500 trials and the numerical integration both gave a probability of about 0.97 29.435 16. the upper confidence limit is about 15% above the point estimate.05 (5%) for z = 1. 3-18 . normalized with respect to the number of failures. This also shows the big difference between the point estimate and the 95% confidence limit when there is only one failure.287 118.045 60.996 4.292 27.079 Figure 3-5 provides a plot of the results in Table 3-6.5 7.774/20 = 0.842 14.403 21.5 5.754 9.296 7.5 3.5 49. Even with 100 failures.995 15.708 is 4. (This is not a large number of trials. As an example of the application of these results.744).532 12.708.025. There must be about six failures before the 95% confidence limit is within a factor of two of the point estimate.EPRI Licensed Material Some Statistical Background Information Table 3-6 Confidence Upper Bounds on Ntrpf for Various Numbers of Failures Number of Failures 0 1 2 3 4 6 8 10 15 20 50 100 0.890 5. which shows how the upper limit confidence limits approach the point estimate as the number of failures increases.5 9.154 11.5 vs 4. but the above tabulation should be adequate for one failure.239.5 2.681 7.5 Point Estimate Confidence Value 90% 2.744 6.) From Table 3-4.062 63. consider the first value of z in the Monte Carlo example with 20 trials in Table 3-4. Importance sampling involves sampling from the portions of the distributions of the input random variables that are more likely to lead to failure. The failure probability can be written as 3-19 . and more rapid procedures are desirable.2 discusses Monte Carlo procedures that involve direct sampling from the distributions of the random variables and calculation of the value of the outcome. Let g(xi) be the function that describes failure.5 discusses an alternative procedure that can lead to substantial economies.4 Monte Carlo Simulation – Importance Sampling Section 3. and Latin hypercube sampling. the large number of trials required can lead to excessive computer time. Such techniques include stratified sampling. but are not efficient for evaluation of low probabilities. Section 3.3. with g(xi)<0 meaning failure occurs. In some instances. Such procedures as particularly straightforward. Mahadevan 97a and Ayyub 95 provide discussions on procedures that can be used to speed up Monte Carlo simulation. g(xi) is called the performance function and is a function of the n random variables xi.EPRI Licensed Material Some Statistical Background Information Figure 3-5 Values of Factors in Table 3-5 Divided by the Number of Failures 3. but there are ways to employ Monte Carlo techniques to provide much quicker results than those obtained from standard Monte Carlo simulation.3 discusses the number of trials that must be performed in order to confidently evaluate low probabilities. importance sampling. and the need for large numbers of trials. this is best described by first looking at an alternate means of combining random variables. Importance sampling is discussed in this section.3. and then compensating the results to account for not having sampled from the appropriate original distributions. Section 3. Following Mahadevan 97a. such as the lifetime.3. The failure probability (at a given time) can be viewed as the integral of the joint probability density function of the random variables over the volume in which the combination of random variables could lead to failure (within that time).3. . 3-27 Consider h(xi) as a new set of density functions.. The altered probability density functions. Ig(xi) allows Equation 3-23 to be rewritten as in Equation 3-25 0 R | I (x ) = S | T1 g i if g ( xi ) > 0 Eq... etc. must be selected to provide more failures.. For the purposes of importance sampling.. which can be written as 1 Pf ≈ N ∑ I (x ) i Eq. One way to do this is to use the same type of density function but increase (or decrease) the mean. h(xi). 3-25 −∞ In the case of simple Monte Carlo.. ∞ f ( xi )dxi Eq. reducing the mean fracture toughness. Importance sampling can lead to significant reduction in computer time when computing small probabilities.... A suggested procedure is to 3-20 . but each failure counts as f(xi)/h(xi) rather than one. Equation 3-26 also shows that the failure probability is the mean value of the indicator function (see Equation 3-24). This is simply the mean value of Ig(xi). 3-26 where N is the number of trials... 3-23 g ( xi ) < 0 Defining the following indicator function. but too large a shift can introduce inaccuracies in the results. I g ( xi ) −∞ −∞ f ( xi ) h( xi )dxi h( xi ) Eq... which can be estimated by Monte Carlo simulation as 1 Pf ≈ N ∑ N i =1 I g ( xi ) f ( xi ) h( x i ) Eq. rewrite Equation 3-25 as Pf = z z ∞ ∞ . such as increasing the mean initial crack size.. 3-28 The density function h(xi) is chosen so as to lead to more failures during the N Monte Carlo trials. 3-24 Pf = z z . −∞ N g i =1 if g ( xi ) < 0 ∞ I g ( xi ) f ( xi )dxi Eq. The failure probability is then equal to the mean value of I g ( xi ) f ( xi ) / h( xi ) .. the failure probability is estimated as the number of failures divided by the number of trials.EPRI Licensed Material Some Statistical Background Information Pf = z z .. The amount of the increase or decrease influences the number of failures that are obtained for a given shift.. EPRI Licensed Material Some Statistical Background Information increase (or decrease) the mean by an amount scaled by the standard deviation. Such shifts can be made to all of the random variables or only to the dominant ones. As an example, consider the problem of the sum of two lognormal variates, as discussed in Section 3.3.2 with the previous results summarized in Figure 3-4. Figure 3-6 shows corresponding results as obtained with 20 trials and importance sampling. The medians of x and y were both shifted downward by a factor of 2 in order to obtain more small values of z in the 20 trials. The previous results with 20 trials and no importance sampling are also shown along with the values obtained by use of numerical integration. The agreement between the three sets of results in Figure 3-6 would be improved by the use of more trials, the 20 used in this example is a small number of trials. Figure 3-6 shows results down to probabilities of 0.005 and z of 0.401. This is in contrast to a probability of 0.025 and a z of 1.708 (see Table 3-4) obtained by conventional Monte Carlo simulation with the same number of trials. This demonstrates the lower probabilities that are obtainable with importance sampling. In order to get to the same probabilities using conventional Monte Carlo, some 200 trials would be required, which is an order of magnitude more than used here. Even greater benefits can be obtained in many situations. Figure 3-6 Cumulative Probability of the Sum of Two Lognormals as Computed by Numerical Integration and Monte Carlo Simulation With 20 Trials, With and Without Importance Sampling 3.3.5 First Order Reliability Methods – Basics Monte Carlo simulation techniques can sometimes involve excessive computer time, and faster methods can become of interest. Specialized sampling techniques can be helpful [Mahadevan 97a, Ayyub 95], as mentioned above. There is an alternative procedure that has seen widespread use. [Ang 84, Haldar 95, Mahadevan 97b] They are known by a variety of names and involve various “wrinkles”, but the method discussed here is one of the simplest and is often referred to 3-21 EPRI Licensed Material Some Statistical Background Information as the first order reliability method (FORM). This method relies on a couple of observations, which are first most easily visualized for the case of two random variables. Let x1 and x2 be two normally distributed random variables. Then define the transformed random variables, u1 and u2 ui = xi − xi σi Eq. 3-29 These are referred to as unit normal variates; they have a zero mean and a standard deviation of 1. The joint probability density function of u1 and u2 is an axisymmetric unit normal variate. Denote the condition that is to be evaluated as g(ui)=0. For instance, in the example problem of the sum of two variables that is discussed in Section 3.3.1, if we are interested in the probability that z<4, then the function g is written as g ( x i ) ≡ x1 + x 2 − 4 = 0 The function g(xi) is called the performance function. If the xi are normally distributed, then the probability that z<4 is the volume under the joint probability density function of the unit normals outside the line in u1-u2 space that is a plot of the performance function. This is shown pictorially in Figure 3-7. Figure 3-7 Pictorial Representation of Joint Density Function in Unit Variate Space Showing Failure Curve and Most Probable Failure Point (MPFP) The above performance function is a straight line in xi space. If the xi are normally distributed, then the performance function is a straight line in the reduced variate space, ui. The volume under the joint density function outside a line in reduced variate space turns out to be related to 3-22 EPRI Licensed Material Some Statistical Background Information the closest distance (in reduced variate space) from the origin to the line. This distance is called β. The volume is equal to the probability of interest. The relation is P = volume = Φ( − β ) = β 1 erfc 2 2 FG IJ H K Eq. 3-30 Hence, in order to obtain the probability, all that is required is the distance from the origin in reduced variate space to the line corresponding to the performance function. As shown in Figure 3-7, if the performance function is not a straight line in the reduced variate space, then procedure involves locating the point of closest approach of the line to the origin, and the performance function is then approximated as a straight line passing through this point. The above relation still then provides an approximation to the failure probability. This is also shown in Figure 3-7. The point of closest approach, which is a distance β away from the origin, is where the joint density function above the performance function is a maximum and is called the most probable failure point (MPFP). It is also sometimes called the design point. If the random variables are not normal, then they can be converted to an equivalent normal. This conversion depends on the location of the point where the conversion is being made and uses an equivalent normal, which is a normal with the mean and standard deviation set to give the same value of the probability density function and cumulative at that point as the function being converted. For lognormal variates, the log of the variate is normally distributed, so the performance function is written in terms of the logarithms of the variables. Using Equation 3-29, the reduced normal variates are ui = x ln( xi ) − ln( xi ) mean ln( xi ) − ln( x50 ) 1 = = ln i x50 µ µi ln( xi ) stddev FG IJ H K Eq. 3-31 As an example, consider the sum of two lognormals discussed above. The performance function in the xi-space is (take x1 as x and x2 as y) x1 + x 2 − z = 0 The reduced variates are given in Equation 3-31, and the equation for the performance function in xi-space leads to the following equation in ui-space. x50 e µ x u1 + y50 e µ y u2 −z=0 The example problem has a linear performance function in xi-space, but a nonlinear one in uispace. Consider the case of z=2, Figure 3-8 is a plot of the performance function in ui-space. The location of the MPFP is shown. The distance from a point on the line to the origin in reduced variate space is denoted as S and is given by the expression 3-23 and finding the smallest value of S. Using Equation 3-30. This is easy to do for two dimensions (two random variables). Figure 3-9 shows the results on lognormal probability scales along with the results of the numerical integration included in Figure 3-4.EPRI Licensed Material Some Statistical Background Information 2 S 2 = u12 + u2 Eq. calculating the value of S for each set of xi and ui.098.368. the location of the MPFP was found by making up a table of xi and ui. It is relatively straightforward to find the MPFP when there are only two random variables. 3-32 The MPFP is the point of closest approach of the performance function to the origin. This defines the location of the MPFP. and the smallest value of S is β. The Origin Is at the Upper Right Corner. 3-24 . this corresponds to a probability of 0. Good agreement is observed. Figure 3-8 Plot of the Performance Function in Reduced Variate Space for the Example Problem of Two Lognormals for z=2. but not a good way to do it for many random variables. In this case. and the Most Probable Failure Point Is Indicated. The value of β in Figure 3-8 is 1. Similar calculations were carried out for a range of z. 0 3.EPRI Licensed Material Some Statistical Background Information Figure 3-9 Cumulative Probability of the Sum of Two Lognormals as Computed by the First Order Reliability Method and Numerical Integration The direction cosines of the angles between the vector from the origin and the MPFP are of interest in studying the influence of the random variables on the probabilities.8 1.7593 -0.7225 2 y 4.41 -0.0 0. They are given by the expression αi = ui β Eq.0 0.7936 -0.6913 3-25 . 3-33 Table 3-7 summarizes the information on the MPFP for the example problem with z=2.59 -0. Table 3-7 Coordinates and Direction Cosines of the MPFP for Example Problem With z = 2 Identification Name median µ coordinates in x-y space coordinates in u1-u2 space direction cosines 1 x 3. Figure 3-11 presents the direction cosines as functions of z for the example problem of the sum of two lognormals. y has the larger values of both these parameters. which has a small median and µ. and then x2 (y) becomes more influential as z becomes large. This switches at about z=2. then the probability is relatively sensitive to the value of that random variable.EPRI Licensed Material Some Statistical Background Information The negative values of ui in Table 3-7 means that the values of the random variables at the MPFP are less than the median. then the corresponding direction cosine is also negative. Figure 3-10 Example of a Performance Function With the Vector Normal to the Axis of One of the Variables Showing the Insensitivity of to That Variable 3-26 . If ui is negative. its value will be only slightly influenced by adding on the value of x. it will do so on the basis of the random variable that has a large median and standard deviation (µ). This means that the value of the random variable xk has little or no influence on the probability. then the vector is normal to the uk coordinate. and the insensitivity of the location of the MPFP to the value of u2 is apparent. they are at the low end of their distributions. If the direction cosine αk is zero. In Figure 3-8 (and Table 3-7) the direction cosines are nearly the same. If the absolute value of the direction cosine is relatively large. so each of the random variables has about the same influence. Hence. This is because as z becomes large. Figure 3-10 shows a case with a zero α2. The direction cosines serve as measures of the sensitivity of the probability to the random variables. Figure 3-11 shows that the value of x (x1) is more influential on the probability of z when z is small. When z is large. This is an example of the use of the direction cosines to assess the relative influence of the random variables. g(ui)=0. It is inefficient to just make up a table of the distance to points on the performance function for all combinations of the random variables and then look for the minimum. The performance function becomes a hypersurface in hyperspace. It considered two random variables that were independent and of a normal type. Numerical procedures are available for accomplishing this. the dimension of the hyperspace being the number of random variables. First. is used to define the distance to the origin β = 2 ∑u N i =1 2 i Eq. there can be many random variables. more efficient methods have been devised.6 First Order Reliability Methods – General The previous section gave a basic discussion of the first order reliability methods. 3-27 . The following equation. 3-34 where N is the number of random variables. things are not this simple. Instead. if they are normal. and restricting the discussion to two dimensions allowed a plot to be drawn of the performance function and the location of the MPFP. It served to show the concepts involved.EPRI Licensed Material Some Statistical Background Information Figure 3-11 The Direction Cosines of x and y for the Example Problem of the Sum of Two Lognormals 3. which is the multidimensional analog of Equation 3-32. as was done in the above example problem. The desired end result is to minimize the distance to the origin (in reduced variate space) subject to the condition of being on the performance function. i.e. The performance function can not then be drawn as a line on a sheet of paper. Equation 3-23 is still used to define the reduced variates. Mathematically.3. the problem is to minimize subject to the constraint that the point is on the performance function. This is the point of closest approach of the performance function to the origin of the reduced variate space. In real problems. The process still involves finding the MPFP. The procedure is often called the Rackwitz-Fiessler algorithm. This results in the set of N+1 equations for the N+1 unknown and ui. these are the values of ui on the plane defined by Equation 3-35 at which g=0 and δ = ∑ cu − u h N i i =1 N 0 i i i =1 * 2 i is minimized. Mathematically. Denote the values at the guessed location of the MPFP with a * superscript. find the point (values of ui) on the hyperplane that is closest to the guessed point and at which g=0. The procedure shown in this figure is based on linearizing the performance function in the reduced variate space (xi) at the guessed MPFP (this is why the derivatives are needed). This involves guessing the MPFP. Mahadevan 97b. This defines a hyperplane in hyperspace. 3-37 The above set of equations can be solved by taking the derivative of L with respect to and ui and setting each of them to zero. The linearized performance function is obtained by expanding the performance function in a Taylor series about the guessed point and retaining only the linear terms. Once the guess no longer changes. after Rackwitz 78. to update our guess. 3-36 Now. an approximation to the location of the MPFP has been found. and defining L = δ + λg = F I u − u h + λG A + Au J c ∑ H ∑ K N i * 2 i i =1 Eq. This can be accomplished by the use of a Lagrange multiplier. and then updating the guess. and uses a Newton-type recursive formula to search for the MPFP.EPRI Licensed Material Some Statistical Background Information see. Figure 3-12 depicts one way of searching for the MPFP. 3-35 where the coefficients A are given by F ∂g IJ A =G H ∂u K i i * A0 = g (u ) − * i ∑Au N i i =1 * i Eq. ui ∑ (u − u ) N i i =1 N + λAi = ui + λAi = 0 δ * 2 i A0 + ∑Au = 0 i i i =1 3-28 . for instance Ang 84. This results in the following expression FG ∂g IJ cu − u h = A + A u g (u ) = g (u ) + ∑ H ∂u K ∑ N * N i * i i * i 0 i i =1 i i =1 i Eq. . EPRI Licensed Material Some Statistical Background Information From the first of these equations. Inserting this into the second equation A0 + ∑ A u = 0 = A + ∑ A (−λδA ) = A − λδ ∑ A N N N i i 0 i i 0 i =1 i =1 i =1 N 2 i i =1 2 i Hence. This is alluded to in Figure 3-11. Mahadevan 97b]. The equation P=(-) is for independent normal variates. λδ = A0 / ∑ A . The process in Figure 3-12 is continued until the value of changes by only a small amount (ε) between steps. and sometimes it diverges away from the solution. Equation 3-33 still provides the direction cosines. For a particular guess of the location of the MPFP. The equation P=(-) is limited to normal distributions. Once the location of the MPFP is available. It must be kept in mind that the equation P=(-) is exact only for independent normal variates with a linear performance function in reduced variate space. and the performance function had better be smooth. Ang 84. The use of this relation for nonlinear performance functions involves replacing the performance function with a hyperplane that is 3-29 . Then P=(-). so that the derivatives can be accurately and economically evaluated. It is said that the procedure usually converges rapidly. and Mahadevan 97b describe the process and give several examples. Ayyub 97. the values of ui are known. and Equation 3-30 still gives the probability. equivalent normals are used. The value of is then the value in Equation 3-34 obtained by using the MPFP ui. 3-38 2 i Equation 3-38 provides the updated values of the coordinates of the guessed MPFP. the actual non-normal variate is replaced by a normal with a mean and standard deviation that are adjusted to have the same value of the probability density function and cumulative distribution function as the non-normal variate at that same point. A second complication involves random variables that are not normal (or lognormal). if it converges at all. which leads directly to the following expression for the updated u Ai i ui = − λδAi = − A0 ∑A N i =1 Eq. and the value of g at the candidate MPFP is close to zero. Problems with lack of independence can be treated by a rotation of coordinates using a correlation matrix or by use of a so-called Rosenblatt transformation [Ang 84. Sometimes it bounces between two points. ui = − λδAi . This process can be tricky. Harris 95a and 97a provide examples of the use of these procedures with fracture mechanics problems. To circumvent this problem. A third complication occurs if the random variables are not independent. EPRI Licensed Material Some Statistical Background Information tangent to the performance function at the MPFP. but this involved finding the coordinates of the MPFP by making up a big table of for points on g=0. The change in and the value of the performance function at the guessed MPFP (g0) both got progressively smaller. which summarizes the results of the intermediate steps for z=3 with an initial guess of x and y at their median values (ui=0). additional approximations are involved in the use of equivalent normal distributions. apply the above procedure to the example problem of the sum of two lognormal variates. The table shows how the procedure converged in seven steps. For non-normal variates. 3-30 . A variation of FORM was used in the above discussion. such as in obtaining Figure 3-8. This is not efficient for problems with more random variables. Applying the procedure of Figure 3-12 provides the results shown in Table 3-8. As an example of the use of FORM and the Rackwitz-Fiessler algorithm. EPRI Licensed Material Some Statistical Background Information Figure 3-12 Diagrammatic Representation of a Procedure for Finding Most Probable Failure Point 3-31 . As mentioned above. After about 13 steps. Table 3-9 provides intermediate steps for the case of z=2.5251 -. This would probably work for the example problem with z=2. which is where convergence problems would be more likely. neither of which is correct.0000E+00 1.3235E-03 1.6615 .6617 new .5.3417E-03 2.3670E+00 2.3292 . the result is bouncing between two points. As an example of bouncing between two points. or can diverge away from the solution.7995 g0 4.3810 -.3737 -. which largely defeats the purpose of using the Rackwitz-Fiessler approach.5410 -.6182 -.1972 -. 3-32 .6082 -.4492 -.6612 .5876 .0649 -.6578 .8250 -.4066 -.3228 -.4023 -.5876 .098.5250 -.5652 -. so the conditions are not at low probabilities and out in the tails of the distributions.3975 u2 -.6615 . Line 5 is fairly close to the answer.6612 .3292 . The process has not converged after 20 steps and the value of g0 has not approached zero. The known answer is =1. The agreement of the direction cosines in Table 3-8 with the results in Figure 3-11 for z=3 is good. but would involve numerous calls to the performance function.5455 -. Ang 84 suggests a procedure that involves solving a nonlinear equation for that involves roots of the nonlinear performance function. but further steps wander away from the known answer from Table 3-7.368.0234E-03 The validity of the final results in Table 3-8 can be verified by comparing them with the results obtained by the look-up table approach discussed in Section 3.5701E-01 1. For instance. which corresponds to a probability of 0.6617 .5758 -.5170 -.9804 -.0000 .6578 . the iterative procedure often converges quickly. but can sometimes jump between two points.7860 -.2640 -.8934 -.7938 -.EPRI Licensed Material Some Statistical Background Information Table 3-8 Intermediate Steps in Iterative Procedure for Finding the MPFP for the Example Problem with z=3 i 1 2 3 4 5 6 7 old .5290 α1 -.5062E-02 3. There are other variations of the theme.6007 α2 -. neither of which is correct.3.6617 u1 -.8176 -. 4639 -.7919 -.9833 .46.1205E-01 6.4656 -.3236 -.0138 .3296 -.8883 -.8861 -.8732 -.2847E-01 5.9833 .8927 -. can take an appreciable amount of computer time.2743E-01 5.6289 -.9285 -.4562 -. One problem with FORM is that you can not account for actions taken during the lifetime.0771 1.4575 -.4515 -.3680 -.8383 -.3714 -.4560 -.8446 -.8697 -.9009 u1 -.0000 .9581E-01 4.9005 .EPRI Licensed Material Some Statistical Background Information Table 3-9 Intermediate Steps in Iterative Procedure for Finding the MPFP for the Example Problem With z=2 i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 old .3300 u2 -.1972 -. As computer time has become cheaper and computers faster.8858 -.8850 -. such as second order reliability methods and using Monte Carlo simulation to sample about the MPFP in order to improve the accuracy.9009 .3613 -.9834 .8261 -.9834 .9306 -.6813E-01 2.7899E-02 8.8713 -.4116 .8392 -.8711 -.8716 -.4593 -.8530 -.9305 -.0000E+00 2.4116 .2719E-01 5.2436E-01 6.9830 .7775 -.0626 1. such as inspections and repairs.9305 g0 5.1232E-01 6.0682E-01 5.9097 .9837 .9827 .4562 -.0149 1.8859 -.0138 .9005 .1232E-01 6.3379 -.9025 . FORM (and its related procedures) is ideally suited to such problems.5344 -.6576 -.3321 -.4639 -. 3-33 .5457 -.8380 -.9834 .8712 -.9305 -.9837 .7281 -.4563 -.8843 -.4035 -.4640 -.8383 α1 -.2594E-01 8.6683 -.9010 .3664 -.8350 -.9025 .9834 new .3663 -.9306 -. Sometimes a single call to the performance function.9804 -.8866 -.1203E-01 6.4128E-01 6.8614 1.0811 -.8955 . such as in problems involving nonlinear finite element analysis.8383 -.8614 1. Monte Carlo simulation is returning from disfavor.4981 -.0653 1.3661 -.8381 -.9324 -.1205E-01 The first order reliability method is finding wide use and provides great economies in computer time.2270E-01 5.3301 -.0626 1.8955 .4689 -.4626 -.0771 1.8452 -.3300 -. There are variations on the theme.4908 -.3663 α2 -. This can be treated as a natural part of Monte Carlo simulation.04 -.9009 .8988 -. which is analogous to one Monte Carlo trial.9097 .8859 -.9827 .0037E+00 5.9010 .5813 -.4253E-01 5.9830 .0149 1.4635 -.2260E-02 1.8712 -.0653 1. and Monte Carlo simulation can be impractical in such situations.6106 -. . using the lifetime model as the performance function in FORM to obtain the failure probability as a function of time). Defining the values of the deterministic inputs. in order to provide simple results. The simple deterministic example problems of Section 2. 4-1 Equation 2-26 is an example for fatigue crack growth lifetime.3 are analyzed probabilistically. we are now ready to discuss the development of probabilistic models of component lifetime. Nf = 1 m− 2 Cf 2 ∆σ π m m/ 2 LM Na 1 ( m − 2 )/ 2 o − 1 ac ( m − 2 )/ 2 OP Q Eq. This equation is repeated here for convenience.EPRI Licensed Material 4 DEVELOPMENT OF PROBABILISTIC MODELS FROM DETERMINISTIC BASICS Having discussed underlying deterministic lifetime models and statistical background.1 Discussion Once a deterministic lifetime model is available. 2. although there are advantages to this approach. It is important to remember that. This can often be done based on data from the literature. it can be cast into a probabilistic model by simply: 1. Characterizing the statistical distributions of the input random variables (defining the distribution type and the parameters of the distributions). Analyzing the set of failure times to obtain the failure probability as a function of time. or a testing program can be performed. The deterministic model provides a relation such as lifetime = function (input variables) Eq. 2-26 4-1 . Using the deterministic lifetime model in conjunction with Monte Carlo simulation to generate a set of failure times (Alternatively. Defining which of the inputs are deterministic and which are probabilistic. The underlying deterministic bases were discussed in Section 2 and statistical background was provided in Section 3. 5. and then a more realistic example problem is provided in Section 5. 4. 3. 4. there are also pitfalls—as discussed in the Introduction and the Summary sections. then this equation can be used to compute a precise number of cycles to failure. then take it to be deterministic. once a distribution type was assumed. This characterizes the scatter in the fatigue data and facilitates a probabilistic analysis of fatigue crack initiation. the inputs are seldom known precisely.. There are various types and sources of uncertainty. In this discussion. Usually.. The lack of confidence can be expressed in terms like “I am 90% certain that the cyclic stress is less than 30 ksi. obtained from the literature.16. The following quote is from Ang 84 (page 384): “Uncertainties in engineering may be associated with physical phenomena that are inherently random or with predictions . there may be uncertainty in the value of an input due to lack of data.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics This equation is particularly simple.. Figure 2-1 provided the data and the best fit to it. To transform the deterministic model to a probabilistic one. ao. 4-2 .” This could be used to define the parameters of the distribution of the cyclic stress. and some of them may be subject to inherent randomness—such as scatter in fatigue crack growth data. If an input is well known or if its value does not have a strong influence on the lifetime. ∆σ and m) are precisely known (and the underlying model is correct). (Let’s not get into the discussion of what is sufficient data. and an average (best guess) would be about 23 ksi (158. This figure is based on fatigue data that was characterized by Keisler and Chopra [Keisler 95]. going out and getting more data will not alter the distribution of the crack growth rate.. the lifetime model can not be written explicitly and will involve numerical calculations for its evaluation. it must be decided which of the inputs are to be considered as random. and then to characterize the statistical distributions of these random inputs. ac. hopefully enough data to also provide guidance on the distribution type. If all of the inputs (Cf. The decision on what to treat as random depends on the inputs that are subject to the greatest scatter and/or uncertainty.07x10-10 (crack growth rate in inches per cycle and cyclic stress intensity factor in ksi-in1/2) and a µ of 0. performed under conditions of incomplete or inadequate information. Now the curve fits to the various quantiles fitted by Keisler and Chopra [Keisler 96] are plotted. An example was provided in Section 3. However. Once sufficient data has been obtained..) Alternatively. Fatigue data is an example. The statistical distributions of the input random variables are preferably determined by the use of data.” (italics added) Some inputs are inherently random. perhaps. these sources of scatter/uncertainty will be treated the same. In that case Cf was found to be well represented by a lognormal distribution with a median of 8.6 MPa). Figure 4-1 provides another example of the characterization of the randomness of a material input variable.2 of fitting the distribution of the fatigue crack growth for a given ∆K. Such data can be generated as a test program or. which is an important input.3. can be based on estimates on what could escape detection.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Figure 4-1 Probabilistic Treatment of Strain-Life Data for A106B Carbon Steel in Air at 550°F (288°C) Showing Various Quantiles of the Keisler Curve Fit [From Keisler 95] Returning to the example of Equation 2-26. or on simulation of the fabrication process. Monte Carlo simulation is the most straightforward and suitable. The distribution of the fracture toughness could be characterized by a set of replicate tests to provide a set of values that could be analyzed to define the distribution type and the corresponding parameters. or on a set of observations. and could be of a random amplitude. in addition to the fatigue crack growth coefficient. with the deterministic lifetime model serving as the performance function. This can be accomplished by a variety of means. Most often. The distribution of ∆σ could be obtained from strain gage data. Examples are provided in the following. The initial crack size. with Khaleel 99 providing curve fits of parameters of crack depth distributions obtained from simulations of the welding process in carbon and austenitic steels.2. The work of Chapman [Chapman 93] is an example of this latter process. The remaining step is to get results (failure probability as a function of time).2. critical crack size.2. As discussed in Section 3. and cyclic stress could also be taken as random. The critical crack size is a function of the fracture toughness and the peak stresses (see Equation 2-25). The fatigue crack growth example is discussed further in Section 4. Alternatively. 4-3 . FORM methods can be used. the initial crack size. Then it is only necessary to program a Monte Carlo simulator. this involves sampling each of the random variables on each trial and calculating a value of the failure time. Khaleel. It is lognormal with a median of 8.. This is a direct.2.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics 4. or base metal. is taken to be lognormal with median 0. The stress is considered to cycle between zero and a maximum value equal to the cyclic stress. In a realistic situation. et al. take 1/2 1/2 the toughness to be Weibull distributed with a mean of 120 ksi-in (131. way to estimate the distribution. The importance of the fracture toughness on the failure probability can be ascertained by sensitivity studies or by direction cosines if FORM is used. cov) is given by the expression 4-4 . The value of the fracture toughness will not have a large influence on the results. Table 3-1 shows that for a Weibull variate. ao.) The distribution of initial crack length is an important input to the problem. Equation 2-26 (which was repeated above) provides the deterministic lifetime model.160. [Khaleel 99] provide results that can be used for estimation of the crack size distribution.100°F (593°C) which is cycled at a high enough rate that creep is not a contributor to crack growth. this ratio (which is called the coefficient of variation. 4. Consider the material to be 2 ¼ Cr 1 Mo steel at 1. The distribution of the fatigue crack growth coefficient.0519 inches (1. Using their results for a 1 inch (25. heat affected zone.9 MPa-m ) and a 1/2 1/2 standard deviation of 20 ksi-in (22. so assume these parameters to be applicable. The values of the parameters of the Weibull distribution that correspond to the assumed mean and standard deviation of the fracture toughness can be found by trial and error.5665.4 mm) thick weld in a ferritic steel. More complex problems only involve more complex lifetime models. The cyclic stress is assumed to be normally distributed with a mean of 20 ksi (137. MATHCAD is useful in this regard. This distribution could be estimated by sectioning of welds and gathering statistics on the observed crack sizes.100°F (593°C) from the literature. Cf. then more resources could be spent to better define the distribution. the toughness distribution could be assumed (as here).07x10-10 and a µ of 0. and a series of replicate tests to generate data would be expensive – and of dubious value. As an example.6 MPa). the initial crack size. The above assumption gives a ratio of the standard deviation to the mean of 1/6.1 Fatigue Crack Growth in a Large Plate Once again.2 Simple Example Problems Two simple example probabilistic problems will be presented to demonstrate the procedures involved. is plotted in Figure 3-1. and at 1100°F (593°C) the material will be very tough (high fracture toughness). the example of fatigue of a large plate will be considered. µ.32 mm) and second parameter. (Fatigue crack growth properties are not strongly influenced by whether the crack is in a weld.0 MPa-m ). and if this turned out to have an important influence.9 MPa) and a standard deviation of 4 ksi (27. It would be difficult to obtain data on the toughness of 2 ¼ Cr 1 Mo at 1. In this case consider a weld in the plate. but tedious and expensive. of 0. 036. b is easily found to be 128. The distribution of the critical crack size. the value of corresponding value of the critical crack size is 1 120 π 20 LM OP N Q 2 = 1146 . it is not expected to be an important factor. Using the mean values of the fracture toughness and cyclic stress. Once c is known. µ=0. std dev =4 ksi b=128.07x10-10. as obtained by Monte Carlo simulation with 104 trials using MATHCAD is plotted in Figure 4-2.5665 median =8.25 ksi-in1/2.29 m) Since this is so much larger than the median crack depth (0. c=7. It can be obtained by use of Equation 2-24. Table 4-1 summarizes the input random variables for the example problem. µ=0.32 mm).0519 inches.25 ksi-in1/2 from the expression mean = b(1+1/c).EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics cov = R | | F 1I L F 1I O U Γ G 2 + J − MΓ G 1 + J P V S H K H K | T c N cQ| W F 1I Γ G1 + J H cK 2 MATHCAD has a built in gamma function and a root finding capability that allows the value of c to be obtained. Table 4-1 Random Variables for the Fatigue Crack Growth Example Problem Distribution Type lognormal lognormal normal Weibull Name initial crack size fatigue crack coefficient cyclic stress fracture toughness Symbol ao Cf ∆σ KIc Parameters median=0. 4-5 . inches (0.036 The distribution of the critical crack size is of interest in its own right.0519 inches = 1.160 mean=20 ksi. The resulting value is c=7. However. This figure shows that the failure probability is not strongly influenced by the value of b once the number of cycles exceeds about 5x104. 4-6 . The results in this figure were obtained with 106 trials.050 inches (1. which corresponds to a failure probability of about 10-3. which is a consequence of the central limit theorem [Ang 75.25/2 ksi-in (140.25 ksi-in (140.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Figure 4-2 Cumulative Distribution of Critical Crack Size for the Fatigue Crack Growth Example Problem (104 Trials) This figure shows that the critical crack size is nearly lognormally distributed. Hahn 67. the interest is usually in lower failure probabilities. regardless of the distribution-types of the random variables. Figure 4-3 is a plot of the cumulative failure probability as a function of the number of fatigue 4 cycles as obtained with Monte Carlo simulation with 10 trials. Ayyub 97] which states that the sum of random variables will tend to be normal. for component reliabilities.4 mm) is 10-4. and the median initial crack size is about 0.27 mm). A corollary to this is that the product of random variables to powers will tend to be lognormal.95/2 MPa-m1/2). Figure 4-4 is a plot on lognormal scales of the cumulative failure probability for the base case of the Weibull parameter 1/2 1/2 1/2 b=128.95 MPa-m ) along with the case of b=128. regardless of the type of distribution of the random variables. The probability that the critical crack size is less than 1 inch (25. in which case the distribution of the fracture toughness can be important. So at cumulative -3 failure probabilities exceeding 10 the toughness has a negligible influence. EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Figure 4-3 Lognormal Probability Plot of the Failure Probability as a Function of the Number of Cycles for the Fatigue Example Problem (104 Trials) Figure 4-4 Plot on Lognormal Scales of the Distribution of Cycles to Failure for the Example Fatigue Crack Growth Problem and the Same Problem With the Mean and Standard Deviation Divided by Two (106 Trials) 4-7 . EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Figure 4-5 presents the results with 106 trials at lower failure probabilities. and greater attention should be paid to the appropriate failure criterion. Monte Carlo simulation with 10 trials took about 100 minutes on a 400MHz Pentium II pc and gave 137 failures. The problem ran quickly and gave results to lower failure probabilities than shown in Figure 4-5. 4-8 . 107 trials would have been necessary to obtain similar results with Monte Carlo. the flattening out of the results at low cycles in Figure 4-6 is due to the influence of the finite critical crack size. then additional information on the statistics of the fracture toughness would be worthwhile to gather. This gives a point estimate of the failure probability of 1. The implications of this on the hazard function are discussed in Section 4. Figure 4-6 provides results to probabilities that are an order of magnitude lower than the Monte Carlo. but Monte Carlo is easy to perform. If the interest is in lifetimes of tens of thousands of cycles.000 cycles there is over two orders of magnitude difference in the failure probabilities. which involves only calculating σ πa and comparing it to Kc. The data points are seen to 4 flatten out below 10 cycles. Hence.25 ksi-in1/2 (140. whereas the higher toughness result appears to be far below 10-7. -6 the lower toughness result is a failure probability of about 8x10 . Rackwitz-Fiessler and numerical integration were attempted with no success. The failure probability at N=0 can be evaluated by Monte Carlo simulation. Hence. They could probably be made to apply to 9 this problem. which is probably similar to the flattening seen in Figure 4-5 for the results with b=128.25/2.2.37x10-7. The distribution of the fracture toughness is seen to have a large effect at small numbers of cycles.95 MPa-m1/2) as obtained by the Rackwitz-Fiessler procedure depicted in Figure 3-12.3. Figure 4-5 Cumulative Failure Probabilities in the Lower Probability Region of the Fatigue Crack Growth Example Problem With Two Distributions of the Fracture Toughness (106 Trials) Figure 4-6 provides results for the problem with b=128. At 10. At very few cycles. Figure 4-7 summarizes the results obtained with 1000 trials while employing the following shifts initial crack size and fatigue crack growth coefficient cyclic stress mean shifted up 1 or 2 standard deviations µ held constant mean shifted up 1 or 2 standard deviations standard deviation held constant b shifted down 1 or 2 standard deviations c held constant fracture toughness Figure 4-7 also presents the Rackwitz-Fiessler results from Figure 4-6. Thus. the fatigue example problem was run with 1. except at small numbers of cycles. It is known that this flattening is real. The Rackwitz-Fiessler results are seen in Figure 4-6 to agree well with 6 the results Monte Carlo simulation with 10 trials and no importance sampling. Points are Results From Rackwitz-Fiessler As an example of the use of importance sampling. because of the Monte Carlo simulation without importance sampling to obtain 137 failures in 109 trials. Hence. the Monte Carlo simulation with and without importance sampling agree well with one another. but the flattening of the curve at small numbers of cycles is not captured.000 trials and various shifts in the distributions of the random input variables. The results with a shift of two standard deviations provides results to very low failure probabilities. Hence. comparable results were obtained with importance sampling with only one-thousandth as many trials. which demonstrates the substantial savings that can be made with importance sampling.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Figure 4-6 Cumulative Failure Probabilities in the Lower Probability Region for the Fatigue Crack Growth Example Problem. it is apparent that importance sampling can 4-9 . It was hoped that this would serve to better identify the failure probability for small numbers of cycles. Solid Line is Monte Carlo With 106 Trials. Figure 4-7 shows good agreement between the importance sampling results and the Rackwitz-Fiessler result. EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics provide substantial economies, but may fail to capture some important features. In this particular instance, this shortcoming is probably due to not really having many failures in the simulation that were influenced by being initially close to failure, a feature that was not captured by the shifts in the importance sampling. Figure 4-7 Cumulative Failure Probabilities in the Lower Probability Region for the Fatigue Crack Growth Problem as Obtained Using Importance Sampling With 1000 Trials With Different Shifts in Parameters of Input Random Variables. Points are From Rackwitz-Fiessler. Returning to the FORM results, this method also gives the direction cosines, which serve as measures of the sensitivity of the failure probability to the values of the random variables. Figure 4-8 provides a plot of the direction cosines for the fatigue crack growth example problem. This figure shows many interesting features. For small number of cycles to failure, the problem is dominated by the fracture toughness, and the fatigue crack growth coefficient has no influence (its direction cosine is close to zero). The only negative direction cosine is for the fracture toughness. This is because in order to get failure in a relatively few cycles, the toughness must be low (below the mean, so the direction cosine is negative), and the cyclic stress, crack size, and fatigue crack growth coefficient must be large (above the mean, so the direction cosine is positive). At larger numbers of cycles, the direction cosine for the toughness becomes close to zero, which is a consequence of the insensitivity of the lifetime to the critical crack size, as discussed in Section 2.3.1 in conjunction with Figure 2-11. The initial crack size and cyclic stress are always important. 4-10 EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics The example of fatigue of a crack in a very large plate demonstrates the principles involved in a probabilistic analysis of lifetime, even though it is a particularly simple fracture mechanics problem. In more complex problems, the calculation of the critical crack size is not so easy, and the lifetime relation (the counterpart of Equation 2-26) is not as simple. Most often, the lifetime evaluation itself involves numerical calculations, but the principles are no more complicated. It’s all in the performance function. Figure 4-8 Direction Cosines for the Fatigue Crack Growth Example Problem as a Function of the Number of Cycles to Failure 4.2.2 Creep Damage in a Thinning Tube The deterministic basis of this superheater/reheater tube problem was covered in Section 2.3.2, where a closed form expression for the lifetime, tR, was derived. This relation was tR = 1− tT LM1 + (β − 1) t OP t Q N C T 1 1/( β −1) Eq. 2-31 In this relation, tC is the time for creep rupture if no thinning occurred, and tT is the time for the wall thickness to go to zero if thinning is the only degradation mechanism. For a wall thinning rate of and initial thickness ho, tT is ho/ . For a given stress and temperature, and for the linear LMP relation of Equation 2-29, the expression for tC is t C (σ , T ) = 10 − C + A /( BT ) σ 1/ BT Eq. 2-30 4-11 EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics In this expression, C comes from the definition of the Larsen-Miller parameter in Equation 2-3, and is equal to 20. A and B come from the linear fit to creep rupture data, Equation 2-29. The value of in Equation 2-31 is 1/(BT). Since there is considerable scatter in creep rupture data, as shown in Figure 2-2, and there would be expected to be great uncertainty and scatter in wall thinning rates, a probabilistic analysis is appropriate. The random variables to include in the analysis are the thinning rate, , and some “constant” associated with the creep rupture. Take C=20, because this is the value used in the original analysis employing the Larsen-Miller parameter, then take the slope of the fit to the data to be constant. This fixes B. Then consider A to be a random variable with distribution type and parameters obtained from data. It would be possible to use the characterization of the scatter employed in Grunloh, et al., [Grunloh 92], but this example will proceed independently of this and use data available from the literature. This serves as an example of developing statistical distributions from information available in the literature, and will lead to a result somewhat different than obtained by Grunloh, et al. Van Echo 66 contains a tabulation of creep rupture data for chromium-molybdenum steels, and data for wrought 1 ¼ Cr 1/2 Mo steel with stresses less than or equal to 30 ksi was obtained from this reference. This provides 207 data points, which are temperature, stress and time to rupture. This data is plotted in Figure 4-9 in terms of log10(σ) (σ in ksi) and LMP=Ta[20+log10(tR)]/1000 (Ta is degrees Rankine, stress in ksi, and tR is the rupture time in hours.) Also shown in Figure 4-9 is the least squares linear fit to the data. This fit is log(σ ) = A − B( LMP) with A=5.645 and B=0.1264. The scatter in Figure 4-9 can be characterized by fixing B and calculating a value of A for each data point. This will provide 207 data points for evaluation of the distribution of A. Figure 4-10 presents a normal probability plot of the cumulative distribution of A. The straight line corresponds to the normal distribution with the mean and standard deviation as obtained from the data. A good fit to the data is observed. Figure 4-10 shows that A is well represented by a normal distribution with mean and standard deviation obtained directly from the data. These values are 5.645 and 0.0579, respectively. Note that this is a “tight” distribution; the cov is 0.01. Since the random variable A appears as an exponent in the expression for tC, as seen in Equation 2-30, and A is normally distributed, then tC is lognormally distributed. 4-12 tR in Hours) Figure 4-10 Cumulative Distribution of the Random Variable A Used to Describe the Scatter in the Larson-Miller Data for 1 ¼ Cr 1/2 Mo Steel 4-13 . Ta in Degrees Rankine.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Figure 4-9 Creep Rupture Data for 1 ¼ Cr 1/2 Mo as Obtained From Van Echo 66 With Least Squares Linear Fit (Stress in ksi. 88x1010/6. so assume that the coefficient of variation (cov) is large. Since the thinning failure time (tt) is so much smaller than the creep rupture time (tR) the problem will be dominated by the distribution of the thinning rate—especially at small times.3.13 mm) for 3x104 hours at 1. but not on the median. From Table 3-1. This variable will be dependent on the problem at hand. From Figure 13 of Viswanathan 92 a mean wall loss of 5 times the steamside oxide thickness appears reasonable. This means that the standard deviation of the thinning rate is 2 times the mean.2 mm) operating at 1. although the data information is for 2 ¼ Cr 1 Mo.42 median CR = 10-C+A50/(BT) = 3. with this as a median value. The following provides some intermediate results σo=6. Alternatively. OD oxidation and OD erosion. [Viswanathan 92] shows the wall loss to be a multiple of the steamside oxide thickness.40 inches (10. All of the inputs are now available to address the problem analyzed deterministically in Section 2. Figure 4 of Viswanathan 92 provides a mean line that gives a thickness of about 5 mils (0.2.400 psi 16. such procedures are especially easy to apply to this problem because analytical derivatives of the performance function 4-14 . The scatter in the thinning rate is large. Hence.385.38 ksi (44 MPa) (same as in Section 2.005 inches = 0.88x1010 tC (median properties) = 3. Wall thinning can occur due to ID oxidation. a very large variance could be expected in wall thinning rates. and can be subject to wide variation depending on.2.7x10-7 inches per hour or 1.125 inch (54 mm) diameter 1 ¼ Cr 1/2 Mo tube with a wall thickness of 0.19 mm/yr).42 = 1. Figure 4 of this same reference shows considerable scatter in the steamside oxidation rate.8°C) and 2.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics The other random variable is the wall thinning rate .68x106 hours = 192 years tT (median properties) = ho/ 50 = 0. the thinning mechanism.0075 = 53. To include fireside effects. Some estimate of a mean rate can be found from Viswanathan 92. This corresponds to a thinning rate of 1. it is seen that the cov for a lognormal variate depends on µ. which gives 7. FORM procedures could be used.3. Assume the thinning rate to be lognormally distributed. The results obtained here will differ from those in Section 2. because different constants in the Larson-Miller curve fit are used. multiply this by 5.5 mils/year (38µm/yr). Figure 13 of Viswanathan. The cov in terms of µ for a lognormal distribution is given by the expression µ = ln(1 + cov 2 ) This results in a value of µ of ln 5 which equals 1.5 mils per year (0.3.2) -3 =1/BT=1/(1460x0.3 years. et al. with large amounts of scatter.27.1264x10 )=5. Consider again a 2. and a thinning rate that is not constant in time.000°F (537°C).000°F (537. Monte Carlo simulation can be easily used to evaluate the failure probability of the tube as a function of time. among other things.5 MPa).40/0. say 2. 1.1 that a lognormal variate to some power is also lognormally distributed). the failure probability is constant at about the value obtained from the Monte Carlo simulation.2. For fewer cycles. the hazard function is 10-7 per cycle on the first cycle. This suggests that in this example problem.3 Hazard Rates As discussed in Section 3. Starting with the fatigue crack growth problem. given that failure has not already occurred.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics (Equation 2-31) are available and the random variables are of the normal type (lognormal). and is then essentially zero until about 104 cycles. which is the probability of failure within the next time increment. the probabilistic information often of most use is the hazard or failure rate. Figure 4-6 provides the cumulative failure probability as a function of the number of fatigue cycles and shows that the Rackwitz-Fiessler data points provide a good representation for cycles greater that about 10. The hazard functions for the example problems are developed here. because the two lines in the figure are close to one another. The deviation from linearity for the thinning-only result is due to the finite number of samples taken. Figure 4-11 provides the results of a Monte Carlo simulation with 104 trials plotted on lognormal probability scales. This is because the thinning rate ( ) is lognormally distributed and tT=h/ . 10-7. more care should be taken in the definition of the distribution of the thinning rate than in the distribution of the creep rupture properties.000. This is especially true at the low failure rates generally of interest. Hence. The hazard rate as obtained by Monte Carlo simulation is usually much less continuous than the cumulative probability results that have been presented in the above examples. 4-15 .3. Figure 4-11 Results of Example Problem of Creep Damage in a Thinning Tube as Obtained by Monte Carlo Simulation With 104 Trials 4. because numerical differentiation or binning into histograms is involved. so tT is lognormally distributed (recall from Section 3. because tT is lognormally distributed. This figure shows the expected behavior of the failure probability at short times (and low probabilities) in that it is dominated by the thinning rate. 5 25 4.249 2.485 12.42 1.04x10-7 1.5 45 1. This slope is most suitable at the mid-point between the points.30x10-8 0.5 20 8.57x10-7 4.356 6. kc 10 Pf 3.14x10-8 0.55x10-6 5.905 17.94x10-4 5.907 22.11x10-10 1.87x10-5 2.637 37.57x10-4 4. and a good estimate of the slope of this curve is just a straight line between the data points.35x10-5 3.559 8.25x10-9 0.5 40 9.06x10-8 L(1) 0.83x10-4 4.cyc-1 4-16 .638 27.00 2.197 32.537 0.993 42.38x10-3 8.53x10-4 4.29x10-7 0. This procedure appears suitable to this problem.5 15 8.08x10-6 2.66x10-8 0.440 2.44x10-4 5.295 1.288 47. The hazard function can then be estimated by the approach outlined in Table 4-2.46x10-11 N(2) ∆L P(3) h(4).5 35 4.5 50 NOTES (1) L=log10(108Pf) (2) Midway point between line above and line below (3) Linear interpolation on a log scale [10-8 x 10(Labove + Lbelow)/2] (4) See text 3.5 30 1.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Figure 4-6 shows that the cumulative failure probability is a smooth curve on the log scale employed.59x10-3 1.33x10-4 4.61x10-4 2. and its applicability to other problems should be assessed before it is used elsewhere.731 1.73x10-9 1.27x10-5 9. Table 4-2 Steps in Estimating the Hazard Function for the Fatigue Crack Growth Example Problem N. 4-2 The slope of the cumulative probability on a log scale is approximately ∆L/∆N. 4-3 Equation 4-2 in conjunction with Equation 4-3 and the values in Table 4-2 provide the values of the hazard function in the right-hand column of the table.000 cycles. Cycles for the Fatigue Crack Growth Example Problem This figure shows the interesting feature of a high hazard on the first cycle. d(logP)/dN~∆L/∆N and dP/dN can be estimated from the following expression: d (log P ) dP dP d (log P) d (log P) ∆L dN = = =P ≈P d (log P ) dN d (log P) dN dN ∆N dP Eq. Figure 4-12 Hazard Function vs. with a much lower hazard in succeeding cycles.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics The hazard function. which includes the value of the hazard function on the first cycle.) The above wall thinning example is also useful in this discussion of hazard rates. where ∆L is given in the above table. (This is why we perform leak and proof tests on new equipment under well-controlled and careful conditions. the problem in dominated by the thinning and is therefore lognormally distributed 4-17 . and ∆N is 5. It takes 50. h. Hence. is given in Equation 3-7 as h( N ) = p( N ) dP / dN = 1 − P( N ) 1 − P( N ) Eq.000 cycles before the hazard again reaches the value it had on the first cycle. These results are plotted in Figure 4-12. At low failure probabilities. To represent their results. ultrasonic. this is the crack depth (not the surface length). the nondetection probability to be used in a given application must be carefully selected. 95. median of 53. Very good agreement is observed. eddy current.27).3 years and a µ of 1. Such databases are not numerous. Khaleel 94.000 Failure Times in 106 Trials. etc). All of these things remaining constant. Hence. See for instance. Histogram Results Are for the Smallest 2.000 failure times from 106 trials.] The nondetection probability is a very strong function of the inspection technology employed. Hence. and the probability of not-detecting a crack as a function of its size is considered. and radiography are presented by Rummel. but the general applicability of their results must be kept in mind. [Rummel 89]. the material and the crack morphology (tightness. et al. This discussion concentrates on inspections for cracklike defects. This is denoted as PND(a) [a is the crack dimension that most strongly influences the value of the stress intensity factor.3 Inspection Detection Probabilities Once a deterministic model has been cast into a probabilistic form. For surface cracks. The Pacific Northwest National Laboratory (PNNL) has investigated nondetection probabilities for ultrasonic inspections of austenitic and ferritic steels commonly used in power plant piping. Figure 4-13 presents the closed form failure rate along with the results the smallest 2. Example nondetection probabilities for dye penetrant. the hazard function is known in closed form at low failure probabilities. Figure 4-13 Failure Rate as a Function of Time for the Thinning Problem Only. roughness. The influence of inspection enters through the probability of detecting a defect as a function of its size.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics with known parameters (i. they use the following functional form 4-18 . the Line Is for the Closed Form Result. there will still be wide variations between individual inspectors. it is possible to address issues that can not be handled in a deterministic context—such as the influence of inspection and proof testing on component reliability. The best way is to have a database for your inspection procedure and material. 4.e. outstanding. (This performance level implies advanced technologies and/or improved procedures that could be developed in the future. and marginal. The PNNL model allows for performance levels possible with future technology and procedures. 4-19 . good. for various qualities for inspection. 95] in Equation 4-4 for the nondetection probabilities for the various qualities. According to Khaleel 95.) • • • Using the three-tier model of Khaleel et al. The four tiers are as follows: • Marginal: a detection performance described by this curve would represent a team using given equipment and procedures that would have only a small chance of passing an Appendix VIII (of the ASME Boiler and Pressure Vessel Code) performance demonstration. and ν controls the slope of the PND – a curve. Good: a detection performance described by this curve corresponds to the better performance levels in the PNNL round robin. equipment. Results are suggested by Khaleel et al.. and procedures that have passed an Appendix VIII-type of performance demonstration.ε )erfc ν ln a* 2 LM N OP Q Eq. Advanced: this curve describes a level of performance significantly better than expected from present-day teams. materials and types of cracks are summarized in Table 4-3. 4-4 In this relation. ε is the probability of missing a very large crack (which will not be zero). PNNL experts use a four-tier model to judge inspection qualities. The value of a* scales with the thickness. the constants [Khaleel 94. h.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics PND (a ) = ε + a 1 (1. Very good: this curve corresponds to a team with given equipment and procedures that significantly exceed the minimum level of performance needed to pass an Appendix VIII performance test. a* is the crack depth that has about a 50% chance of being found. Since is the same for all qualities.6 0.6 0.10 0.10 0. which makes the results a straight line when PND>>ε.6 0.4h 1.15h 1.05h 1.15h 1.6 0.25 Figure 4-14 provides a plot of the of the nondetection probability for fatigue cracks in a ferritic plate as defined by the constants in Table 4-3.65h 1.6 0.02 0.05h 1.05h 1.6 0. the lines are parallel except when the value of ε starts to have an influence.005 0. The parameter controls the slope of the nondetection curve when PND>>ε.4h 1. 4-20 .6 0.6 0. The plot is on lognormal probability scales.6 0.4h 1.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Table 4-3 Parameters of the Equation Describing the Non-Detection Probability Ferritic Fatigue Cracks Austenitic Fatigue Cracks SCC Cracks Inspection Qualities/Parameters Outstanding a* ν ε Good a* ν ε Marginal a* ν ε 0.005 0.10 0.02 0.005 0. This problem is simple enough that the results presented there were generated by Monte Carlo simulation using a program written in MATHCAD.2. In order to demonstrate the type of results obtainable when inspections are employed. When inspections are performed. because it will tend to remove the larger cracks.1. As an example of the influence of inspection on calculated failure probability. and it is assumed that all detected cracks are removed. For instance. The only difference is that now a very long surface crack in a plate of 4-21 . then the probability density function of cracks after the inspection is given by the expression ppost (a ) = z p pre (a ) PND (a ) ∞ Eq.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Figure 4-14 Nondetection Probability as a Function of Crack Depth Divided by Plate Thickness for Fatigue Cracks in Ferritic Steel for Three Qualities of Ultrasonic Inspection The nondetection probabilities are used in a probabilistic fracture mechanics analysis to “remove” cracks that are detected. consider the fatigue crack growth problem of Section 4.1. The benefit of the inspection depends on the nondetection probability and when the inspections are performed. It is therefore suitable for an example that is very similar to the fatigue example in Section 4.) The inspection has a beneficial effect. PRAISE was originally developed for nuclear reactor piping [Harris 81.92a].2. the PRAISE code. but it is applicable to fatigue crack growth of semi-elliptical surface cracks in materials that follow the Paris-type relation. if ppre(a) is the probability density function of cracks before an inspection with nondetection probability PND(a). the logic becomes more complex. 4-5 p pre (a ) PND (a )da 0 (The term in the denominator is merely a normalizing constant so that the probability density function for post-inspection crack depths integrates to unity. use will be made of existing software. so the use of MATHCAD and starting from scratch would not be effective. This is because the finite thickness is large relative to the small size of cracks that require so many cycles to grow to failure. 4-22 .EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics finite thickness is considered (PRAISE does not consider the simple problem of a through crack in a very large plate. The nondetection probability parameters for a “good” inspection were employed. Failure is the growth of a crack to become through-thickness or exceedance of the fracture toughness. the fracture toughness will be taken to be deterministic with a value of 120 ksi-in1/2 (131. that K-solution is not built into PRAISE). This figure shows remarkably similar results when the number of cycles exceeds about 50. For small numbers or cycles. the finite thickness results of Figure 4-15 are higher than the infinite plate results. Figure 4-15 presents the results. Also. pre-service inspection and in-service inspections at 25 and 50 thousand cycles were obtained. fatigue crack coefficient and cyclic stress as in the fatigue example problem. which is expected.9 MPa-m1/2). Calculations were performed for a one-inch-thick plate with the same distributions of initial crack size (a).000. Results with no inspection. Figure 4-15 Failure Probability as a Function of Time for the Inspection Example Problem A comparison of the results with no inspection in Figure 4-15 with the results for an infinite plate in Figure 4-3 is included in Figure 4-16. This means that the hazard rate is small following the inspections. where a decision tree is applied to estimate the cost of candidate inspection schedules. 4-23 . Such results can not be obtained by the use of FORM. Results such as included in Figure 4-15 are useful in inspection planning.000 cycles. The economic models discussed in Section 6. Studies on the influence of the quality and inspection frequency on the failure probability can be performed and inspection strategies optimized. The effects of the in-service inspections at 25 and 50 thousand cycles are noticeable in that the cumulative failure probability remains flat for a period following the inspection. Results such as in Figure 4-15 account for actions taken as a result of “observations” made during a Monte Carlo trial (that is. The effects of an inspection are most pronounced immediately following the inspection. One advantage of Monte Carlo is that it can consider the effects of remedial actions taken during the lifetime. The pre-service inspection has a pronounced effect below about 50. because the performance function can not be written to include inservice inspection. the inspection has only a small influence. For large numbers of cycles (time) beyond the inspection. a crack is removed if it is detected). and diminish with increasing cycles (time). This is discussed in Harris 92b.2 are also useful in this regard. but the pre-service and no inspection results are very similar beyond that point.EPRI Licensed Material Development of Probabilistic Models From Deterministic Basics Figure 4-16 Comparison of Fatigue Example Problem of Infinite Plate With Finite Thickness Results With No Inspection Using PRAISE Code The results of Figure 4-15 are typical of those obtained for various inspection schedules. . Harris 93]. Although specialized software is used in this example. it is important to remember that the technical basis is really no more complex than the simple example problems already discussed. the models for prediction of reliability are already available in the BLESS software [Grunloh 92. and the stress intensity and C* solutions for cracks in headers are not so simple. A header is selected for the example. but basically the same principles are involved. [Harris 98] will be used for this example. the principles are the same simple ones discussed in the previous examples. The superheater outlet header considered by Harris..1 Component Geometry and Material A header analysis by use of the BLESS code will be used as an example of a probabilistic analysis of a boiler component. but. Statistical background was provided. The lifetime analysis itself now requires numerical calculations. 5. which shows that the header is idealized as a repeating pattern of tube penetrations. and then cast this model into a probabilistic form by the five steps outlined at the beginning of Section 4. Harris 93] is used for the evaluation of the reliability. once again. et al.EPRI Licensed Material 5 EXAMPLE OF A PROBABILISTIC ANALYSIS We are now ready to provide a realistic example of a probabilistic analysis of a boiler pressure component. Of course. A typical geometry of a header is depicted in Figure 5-1. 5. because the stresses are more complex. the volume of the Monte Carlo simulations requires a computer. and the BLESS software [Grunloh 92. 5-1 . Previous sections discussed background on how a deterministic lifetime model can be constructed and then placed into a probabilistic format for evaluation of component reliability. and it is only necessary to identify the material and geometry along with the operating history.1.1 Gathering the Necessary Information It is necessary to perform the following steps when performing a probabilistic analysis of a given component: • • • • Identify the material Identify the degradation mechanism Define the component geometry Define the operating history Once the above steps have been completed. In the case of a high-temperature header. it is necessary to construct a deterministic lifetime model applicable to the degradation mechanism. spacing. This picture is generated as a standard feature in BLESS. The appendix provides portions of the drawing of the example header. which shows the tube size. Figure 5-1 Schematic Representation of Typical Header With Illustration of Segment Considered in Analysis 5-2 .EPRI Licensed Material Example of a Probabilistic Analysis The component material and the required geometric information can be obtained from a drawing of the header. and geometry of connections to the header. the header is idealized as 47 slices each having the configuration shown in Figure 5-2. The available information often appears to be scanty. Based on the information in the drawing in the Appendix. The material is identified in the drawing as SA336 (Cr 1 ¼) which corresponds to 1 ¼ Cr 1/2 Mo steel. but the necessary information can usually be obtained from an old drawing. Also required is the number of other operating excursions (such as load swings) and the corresponding conditions. information on the operating transients that occur is required. The case of headers is different. because the temperature gradients within the thick header wall can lead to appreciable thermal stresses.) In addition to the nominal operating conditions.2 Operating Conditions The history information that is required is the number of cold starts in the plant history and/or projected into the future. pressure and flow rate conditions during typical transients. because pipes are thin enough that radial temperature gradients do not lead to significant thermal stresses.EPRI Licensed Material Example of a Probabilistic Analysis Figure 5-2 Summary of Header Geometry Analyzed (Length Dimensions in Inches) 5.000 lbs/hour of steam translates to 600.000 kg/hr) of steam.1. along with the number of occurrences and sequencing of each of the transient types. In turn.122 lbs/hour (2783 kg/hr) for each tube.500 psi (10.000°F (538°C) operating temperature. (The flow rate in the tube is an input to BLESS. For the example.000/98 = 6.000 lbs/hour (272.3 MPa) operating pressure. In the case of pipes. this consists of typical temperature. The nominal operating conditions for the header are usually readily available. the details of the temperature history are not important. and 600. The 600. because it influences the temperature in the header when steam temperature transients occur. Ideally. 1. this influences the stress and creep rate. these are 1. 5-3 . but no leaks were reported. These charts show a typical day of main steam pressure and temperature history. Based on the data in the charts and similar data for a heat-up and cool-down. All that can be hoped for is to construct some representative repeating pattern. The information in that table leads to an average of 43 cold starts per year with an average of 6. Data on the typical conditions during a heat-up and cool-down should be gathered. The cracks were observed to grow during this period. and will be used to construct the operating history in the BLESS example. but was continued in operation for another 4 years. Such information is often in the form of charts. as well as information on the conditions during a typical day of operation.EPRI Licensed Material Example of a Probabilistic Analysis It is impossible to reconstruct the operating history in great detail. Table A-1 in the Appendix summarizes information for the example plant. Studies have shown that fairly large variations in operating conditions do not necessarily have an appreciable influence on the predicted reliability [Harris 97b]. 5-4 . The example header was observed to be extensively cracked during an inspection at year 34. the data of Table 5-1 is compiled.618 yearly hours of operation. examples of which are shown in Figure A-3 in the Appendix. The number of plant cold starts is usually readily available. and variations in operation undoubtedly occur from year to year and even from day to day. 0 3.5 24.5 1. the user should perform some sensitivity studies to see if these observations are applicable to his case.5 9. For a particular application.5 2. psi Steam Temp.5 3.0 Steady Operation Pressure.5 Cool-Down 0.. klb/hr Time. do not have a noticeable influence on the predicted lifetime or reliability. °F 100 300 1000 1200 1400 1500 1500 1500 96 196 294 390 588 600 600 600 200 350 500 550 850 920 970 1000 1500 1500 1300 1000 100 600 600 400 200 96 1000 960 680 520 200 1500 1500 950 300 300 900 1500 600 600 500 400 400 600 600 1000 1000 850 670 670 820 1000 1500 600 1000 It has generally been found that details of the operating history.0 4.0 8. Similarly.0 21. Hours Heat-up 0.0 0.0 Daily Cycle 0. variations in the flow rate have been found to not be influential.0 2.0 1.EPRI Licensed Material Example of a Probabilistic Analysis Table 5-1 Summary of Time-Variation of Operating Conditions Header Flow Rate. Hence.0 11.0 1. such as minor short duration temperature excursions. 5-5 .0 13.5 11. extreme accuracy in these inputs is not necessary.7 22. There is no need to precisely define operating histories if wide variations do not influence the results. and a sensitivity study performed to see if this has an appreciable effect. (The corresponding time for bore-hole to bore-hole growth was 500.031°F (528 to 555°C).EPRI Licensed Material Example of a Probabilistic Analysis The temperature of the steam in the tubes is often different than the temperature of the steam in the header. along with the results included in Figure 5-3.42 days. so the remaining discussion will consider ID to OD growth.000 hours of service. and an analysis for this mechanism was also performed. (As mentioned above. Figure 5-4 presents the results for this run on lognormal probability scales.290 hours. which is nearly 20 times the length of time that the header has been in service. the BLESS oxide notching model is deterministic. A BLESS analysis of lifetime with oxide notching for initiation with creep/fatigue crack growth from ID to OD predicted a median time to leak of 499. 5. Therefore.2 Performing the Analysis The information from Section 5. This model predicted an initiation time of 5. A BLESS run was also made for 20. The BLESS software was used to predict the header failure times.020°F (549°C). and the temperature often varies from tube to tube. Hence oxide notching is predicted to result in crack initiation in a very short time.1. This plot is as directly generated by BLESS. measurements showed tube temperatures during steady operation that varied from 982 to 1. and these procedures are combined to define operating histories.433 hours. The predicted median time for 6 initiation by creep/fatigue was 4.) BLESS considers oxide notching as an alternative crack initiation mechanism. Based on the averages. In the example problem. Temperature is a very important input. extensive cracking was observed in the header at the 34th year. 6 days of daily cycle. say 1.1 and the Appendix is input to the BLESS software through the Windows interface. with the current time being 244. consider the history to be a heat up. Therefore a probabilistic lifetime analysis for oxide notching initiation and creep/fatigue crack growth is considered. which is just twice the reported hours of service to date. an average year consists of 6. which corresponds to 236.000 trials and a maximum evaluation time of 100. such as with thermocouples. The data of Table 5-1 is input as operating procedures.4. This is one cold start every 276/43 = 6. A typical tube temperature can be selected.800 hours.618 hours (276 days) of operation with 43 cold starts.000 hours. and a cool down. As discussed in Section 2. 5-6 .855 hours (Table A-1).25x10 hours. Figure 5-3 provides a log plot of the cumulative leak probability as a function of time for the example problem. This will be repeated over the time span of interest.) The oxide notching initiation and creep/fatigue crack growth analysis provided results more in line with observations of header performance than corresponding evaluations considering initiation by creep/fatigue. so it is therefore advisable to make some tube temperature measurements. EPRI Licensed Material Example of a Probabilistic Analysis Figure 5-3 Log-Linear Plot of Probability of Leak as a Function of Time for Oxide Notching Initiation and Creep Fatigue Crack Growth in Header Example Problem (2.000 Trials) Figure 5-4 Lognormal Probability Plot of BLESS Results for the Header Example Problem (Probability in Percent) 5-7 . The predicted time for oxide notching should increase if the minimum temperature during the daily cycle was increased above 700°F (371°C).000 hours. This combining of the data uses each set for the times at which it is most accurate 5 (most failures).000 trials for times beyond 40. the oxide layer is considered to crack whenever the temperature falls below Tlo-ox of 700°F (371°C). Table 5-2 summarizes the oxide initiation time under various operating scenarios and shows that increasing the minimum temperature during the daily cycle has a large influence.EPRI Licensed Material Example of a Probabilistic Analysis The dominance of oxide notching and the short time for oxide notching initiation suggest that little can be done to increase the reliability of the header except for aspects that reduce the oxide initiation time or reduce the creep/fatigue crack growth.810. About the only thing that could be done to increase the oxide notching initiation time is to decrease the number of times the temperature falls below this value. it would be necessary to extend the growth portion of the lifetime. and the lower part shows the short time results out to 50.000 hours was performed. Table 5-2 Summary of Initiation Time for Various Operating Scenarios Cold Starts per Year 41 41 41 Notching Initiation Time. The upper part of the figure shows results out to 106 hours.4. This could be accomplished by decreasing the operating pressure and/or temperature.000 trials for times beyond 100. Hours 5290 13575 14240 Cycles Between Cold Starts daily. in order to increase the predicted component reliability. The linearity of the results of Figure 5-4 suggests that the lifetime is lognormally distributed. Figure 5-5 provides plots of the corresponding hazard function.17x10 hours and a µ of 0. As discussed in Section 2. This resulted in a fit with a median failure time of 6. this results in reduced plant output with a corresponding economic penalty.1. Tmin=720°F (382°C) daily. 5-8 . of course. Tmin=720°F (382°C). A linear least-squares fit that included the BLESS results generated with 2. tube temp=1010°F (543°C) none none 41 20 8631 11609 Hence. but that the initiation time is short for all cases considered. Tmin=670°F (354°C) daily. but. Table 5-1 shows that this occurs during the daily cycle.000 hours and the results with 20. 000 hours.000 hours along with a histogram corresponding to the BLESS results of Figure 5-4 with 20.000 trials. which is 2. 5-1 dt is the time between the tabulated data points in the BLESS output. The histogram values are calculated from the relation h( t ) ~ P (t + dt ) − P(t ) dt[1 − P (t )] Eq. 5-9 . Good agreement between the BLESS histogram and the fitted lognormal is seen in Figure 5-6.EPRI Licensed Material Example of a Probabilistic Analysis Figure 5-5 Lognormal Hazard Function for the Header Example Problem Figure 5-6 presents a plot of the fitted hazard function out to 100. EPRI Licensed Material Example of a Probabilistic Analysis Figure 5-6 Comparison of Hazard as a Function of Time as Obtained From the BLESS Output Data and the Fitted Lognormal Relation Section 6 discusses the application of these results. The times considered in the example are from the current time (244.9 thousand hours) to 20 years into the future (end time of 377.2 thousand hours). Figures 5-4 and 5-6 show that the BLESS run provides accurate results within this time frame. The failure rate within the time frame of interest was generated by BLESS using standard Monte Carlo without excessive computer run times. The smallest cumulative failure probability in Figure 5-4 is 10-4. In many instances, lower failure probabilities than this are of concern, and it is of interest to generate results to lower probabilities. As discussed in Section 2, tools are available to conveniently accomplish this, such as first order reliability methods and importance sampling. Importance sampling has recently been incorporated into BLESS (Version 4.2), and an example of its use in the sample header problem is provided. The importance sampling is defined by altering the distributions of selected random variables. The dominant random variables in headers have been found to be the creep rate coefficient (A1000, which is Ae-Q/T in Equation 2-6, with T=1,000°F=538°C) and the coefficient in the creep crack growth relation (Cc in Equation 2-21). Both of these are lognormal. The altered distribution is defined by increasing the median of the distribution by adding on a multiple of the standard deviation, but leaving the second parameter alone (µ of the lognormal density function, see Table 3-1). A shift of one increases the median used in the Monte Carlo sampling to one standard deviation above its nominal value. A shift of zero corresponds to no shift, and a positive shift increases the median. The shift selected should be in the direction to increase the number of failures. (The shift should be towards the most probable failure point). In the current case, the shifts should be greater than zero (increases in A1000 and Cc result in more trials that produce failures). 5-10 EPRI Licensed Material Example of a Probabilistic Analysis The selected shifts should be held to a minimum and the number of trials maximized, while still maintaining the computer time within acceptable limits. As always, trade offs are involved. The shifts in each of the variables do not have to be the same, and should be larger for the more influential variables. The shifts should be gradually increased in successive runs to see the trends. Large shifts are to be avoided, unless they are gradually approached. For example, Figure 5-7 shows the results of eight BLESS runs with varying numbers of trials and multipliers, including no shifts. The solid line is the line defined by the median and µ evaluated for the unshifted results in Figure 5-4 (t50=6.17x105, µ=0.810), and shows that this line agrees quite well with most of the results generated by the importance sampling. The table with Figure 5-7 identifies the shifts employed, the number of trials taken, the computer execution time in hours and the minimum failure probability obtained in the run. If no importance sampling was employed, it would take some 1400 years of computer time to obtain results down to 10-14, whereas employing importance sampling with shifts of 1 and 3 with 20,000 trials provided results down to nominally this value with only 27 hours of computer time. However, care must be taken in defining the shifts; the triangles and diamonds data do not agree well, and differ by about two orders of magnitude at the lower failure probabilities. The diamonds are believed to be more accurate, because a smaller shift and larger number of trials were used. Also, the diamonds agree better with the solid line. The diamond data took nearly five times as long to generate, however. 5-11 EPRI Licensed Material Example of a Probabilistic Analysis Figure 5-7 Results of Header Example Problem With Varying Shifts and Number of Trials. The Solid Line is the Curve Fit Based on the Estimated Lognormal Distribution With No Shifts (the Line of Figure 5-6). These results demonstrate the usefulness of importance sampling in generating results at lower failure probabilities, but that some care must be exercised in using this feature. In practical problems that require estimates involving very low failure probabilities, importance sampling could be used to good advantage while maintaining reasonable computer execution times. 5.3 Combining Data The above discussion covers the estimation of component reliability by use of deterministically based models. This can provide the failure probability as a function of time. Sometimes there is data or reliability estimates from other sources, and it is desired to somehow factor this other information into the estimates from the component reliability model. Potential other sources of 5-12 such as the use of so-called conjugate functions that simplify the integrations involved. The selected standard deviation is usually subjective.EPRI Licensed Material Example of a Probabilistic Analysis information are generic failure rate data (such as the NERC-GADS data based discussed in Section 6. the distribution of the parameter can be considered in the analysis (for 5-13 . that is. denote the distribution as pA50(A50). The likelihood of observing the set of failure times is L( A50 ) = ∏ p (t n f i =1 f .i. Ayyub 97. 5-2 prior p A50 ( A50 ) L( A50 )dA50 0 The term in the denominator is merely a normalizing constant that assures that the posterior distribution integrates to unity. expert opinion and the results of interviews with plant personnel. Equation 5-2 can then be simply stated as posterior distribution = (prior distribution)(likelihood)(normalizing constant) An updated prediction of the reliability can then be made by using the median from the posterior distribution. The values of the parameters are then “updated” using other information. i=1. rather than any set of data. Procedures for combining these types of information are discussed in ASME 00. such as a set of observations.n.2). as a probability density function must. This is denoted as pf(tf|A50). This is the prior distribution. Using the header of Section 5. the mean value of the parameter describing the secondary creep rate (A of Equation 2-7 or 2-10) could itself be considered to be described by a statistical distribution with a mean equal to the mean currently in use and a standard deviation selected to describe the uncertainty in this value. Alternatively. Considering the median value of A as being the parameter that is to be updated. The use of conjugate functions simplifies the evaluation of the integral involved in the normalization.i | A50 ) The posterior distribution of A50 is then given by the following expression. The distribution type is usually selected for mathematical convenience. specific failures or inspection results. Wostenholm 99]. with the spread in the distribution representing uncertainty in the value of the parameters. The engineering model of component reliability is then used to evaluate the probability density function of failure times given a value of A50. This will presumably lead to a better estimate than the use of one source by itself. and EPRI 98 If more than one source of failure probability information is available. The distribution of A50 is then updated (to give the posterior) based on some observations. One means of combining information is to construct a model similar to those discussed above. Bayesian procedures conventionally involve considering some of the parameters (mean and standard deviation) of the random input variables to themselves be statistically distributed.2 as an example. it is important to combine them in some fashion. it is set by the user based on judgment. such as a set of n observed failure times. which is an expression of Bayes theorem posterior p A50 ( A50 ) = z prior p A50 ( A50 ) L( A50 ) ∞ Eq. tf. and then to use Bayesian procedures for updating some inputs to the model based on some additional information [Ang 75. Harris 97a provides such an example.EPRI Licensed Material Example of a Probabilistic Analysis instance. The use of Bayesian updating requires a set of observations. one from a reliability model (such as concentrated upon in this document) and a set of estimates based on interviews. 5-14 . ASME 00 and EPRI 98 suggest a procedure that they refer to as “Bayesian-like” updating. A set of observations for use in the updating procedure is usually difficult to obtain in practical cases of interest here. Another example would be a set of crack sizes found in an inspection and using this to update the parameters of the initial crack size distribution in a probabilistic fracture mechanics model. To overcome this difficulty and to still be able to combine sets of data. The procedure is appealingly simple. but is not based on any known (to the author) mathematical principle. the value of the parameter can be sampled from its posterior distribution for use in each trial). the two sets are combined by multiplying their probability density functions together and normalizing so that the product integrates to unity. Using the example of two sets of failure probability predictions. such as a set of failure times in the above example. This provides the combined probability density function of lifetime that is presumably a better estimate than either set of information by itself. if Monte Carlo simulation is employed. which combines probabilistic engineering and economic risk assessment for optimal outage planning.2. An estimate of an acceptable failure rate (or hazard rate) can be obtained by looking at past failures of similar components or by following the discussion in Harris 95b. can be set relative to the frequency of events that occur 6-1 . 94. which can use riskprinciples to prioritize inspection and maintenance. and state-of-the-art computer model predictions of remaining component life as inputs to a decision analysis process. Since one measure of risk is the product of the failure rate times the consequences. and are in References ASME 91. “The Turbo-X program utilizes component history data from NERC-GADS (North American Electric Reliability Council). but is a secondary goal of this document. an inspection could be performed and the initial conditions for the probabilistic evaluation reset.” The emphasis herein is on the estimation of component failure probability by state-of-the-art computer models. Alternatively. The application of these results to decisions of run/replace/retire is certainly of interest. As utilities come under increasing pressure to reduce costs and extend the lifetime of plant components. maintenance personnel interviews. The question of acceptable risk is a complex one. Some discussion and examples in these areas are included in this section. such as (if no cracks are found during the inspection) resetting the crack size to that size which would be readily detected by the inspection. a target level of the hazard rate. EPRI’s Turbo-X process [EPRI 98] is another example of an economic approach that uses probabilistic information as an input.EPRI Licensed Material 6 USE OF PROBABILISTIC RESULTS The emphasis in this guideline is on performing probabilistic analyses based on underlying deterministic lifetime models to obtain estimates of component reliability as a function of time. interest has increased in procedures for rationally planning inspection and maintenance. this can result in a somewhat constant risk of operation of components.1 Target Hazard Rates One way of using the results of probabilistic lifetime analyses is simply to replace components that are operating with a hazard rate that is above a selected target value. 98. ASME 94 is specific to fossil-fired power plants and contains an example of a multi-component economic optimization. 92. and will not be dwelled upon here. Following first the approach in Harris 95b. ASME 00 is another volume of the series that is nearing completion and provides a handbook approach to risk-based methods for equipment life management. which is a subset of the efforts in Turbo-X. The use of failure probability in economic models is demonstrated in the simple examples of Section 6. the lower the target hazard rate is set. h(t). The higher the consequences of a failure. 6. As stated in EPRI 98. One such procedure is risk-based (or risk-informed) inspection. Several volumes on the topic have been published by the ASME. The numbers in Table 6-1 are largely based values in Accident Facts [National Safety Council 92] for the year 1990.700 46. per Hour 4.000 hours of exposure per year.400.43x10-6 per hour. The other approach to setting a target hazard rate is to look at past experience with related components. The severity is in lost megawatt hours.000 Fatality Rate.EPRI Licensed Material Use of Probabilistic Results and are evidently acceptable to society. Considering a life span of 80 years. This could be used as the target level for purposes of discussion for events that are likely to lead to a fatality. or fatalities in the work place. As an example of a high fatality rate.2x10 per year. the average frequency can be thought of in simple terms as 1/80 years.000 — ~400 Category work (all industries) work (trans & util) motor vehicles Annual Fatalities 10. which translates to 1.500 1.0x10 * ~5x10 -7 -7 *Accident Facts fatality rate of 2. consider life in general.000. which suggests that a hazard rate of 2x10-8 per hour (~2x10-4 per year) is accepted by society. The North American Electric Reliability Council (NERC) of Princeton.300 -4 Persons Exposed 117. with 2. Other examples are summarized in Table 6-1. The fatality rate in Table 6-1 is simply the fatalities per year divided by the product of the number of persons exposed timed the annual exposure per person.5x10 -8 1. For instance. maintains a database on failures in components in electric power stations. including the frequency and severity of failures. Table 6-1 Examples of Hazards of Common Activities as Measured by Fatality Rate Annual Hours of Exposure per Person 2. Information is provided on a wide variety of components. Table 6-2 (which is NERC data reported in ASME 94) summarizes the failure rate and consequences from generic NERC data for boiler components. The frequency is in terms of fatalities per hour of exposure. New Jersey. Such events could include fatalities in commercial air travel or automobile travel.000 — 250. 6-2 . 66 x10 1. based on generic data.EPRI Licensed Material Use of Probabilistic Results Table 6-2 Partial List of Boiler Component Failure Rate and Consequences Failure Rate.90 x10 -6 8.42 x10 -7 -6 Figure 6-1 provides a log-log plot of the frequency-severity data in Table 6-2. followed closely by 103 (superheater) and 104 (reheater-first).07 x10 5.00 x10 -4 -5 5.60 x10 5. these would be the boiler components to concentrate on.07 x10 2. 6-3 .65 x10 5.37x10 -5 1.00 x10 -6 -5 1. If risk is defined as the product of the frequency times the severity. and such lines are shown. The figure shows that. then lines of constant risk are straight lines on theses scales. per Hour 1.59x10 -4 Component Code and Name 101 waterwalls 102 generating tubes 103 superheater 104 reheater-first 105 reheater-second 106 economizer 107 air preheater-tubular 108 air reheater-regenerative 109 induced draft fans 110 forced draft fans 111 recirculating fans 112 desuperheaters and attemperators Average MWh Lost per Forced Outage 21951 21991 15765 21866 30351 16137 35107 24722 11331 8865 26412 11064 1. Hence. Component 101 (waterwalls) has the highest risk.38 x10 -6 -6 -5 8. and the use of 10-4 per hour would be more representative of past experience with headers. This is not a long time. which is much higher than the fatality rate of 2x10-8 per hour in Table 6-1.EPRI Licensed Material Use of Probabilistic Results Figure 6-1 Log-Log Plot of Frequency Severity Data for Boiler Components From Table 6-2 Table 6-2 suggests typical failure rates that are much higher than the 2x10-8 per hour that is based on the rates from Table 6-1. This is well above the fatality rates in Table 6-1. but still below the failure rates based on component failures given in Table 6-2. the use of 2x10-8 per hour as a target hazard is conservative. and Figure 5-7 shows that the cumulative leak probability at this time is about 0. the risk of future operation of the header is not excessive based on past experience. The header has been operated for nearly 245. Hence. -8 Using the target hazard rate of 2x10 per hour based on Table 6-1. Hence. the predicted hazard rate increases rapidly. but the data of Table 6-2 do suggest that typical component failure rates are well above the rates of Table 6-1. Figure 5-8 shows that the predicted hazard rate for this header is below 1. and such failures are evidently most often leaks that are not life threatening. Evidently.10.5x10-6 per hour for times below 106 hours (114 years). 6-4 . This is consistent with the observations of extended cracking but no leak at this time. This seems high. the risk of future operation of the example header may not be excessive. Since the leaking of the header will most likely not result in a fatality.000 hours (3.000 hours (Table 5-1). which suggests a superheater failure about once a year. and the predicted failure rate is within the range of experience for boiler parts.4 years) of operation. At times beyond this. Economic considerations of the cost of header failure may be the driving factor. The failure rate for superheaters given in Table 6-2 is about 10-4 per hour. superheater failure seldom cause fatalities. from Figure 5-8 the example header can be operated at a hazard below this for the first 30. The failure rate for superheaters in Table 6-2 is about 10-4 per hour. 6-1 Thus. which provides a good representation of the BLESS results in the time frame of interest (the next 20 years). and can constrain the analysis so that failure rates will not exceed a defined value—regardless of the cost.2 Economic Models Another way to address the header example problem is to consider the expected costs of various scenarios. 6-5 . This is related to the hazard function but is most conveniently expressed in terms of the cumulative failure probability by use of the following relation Pf ( year n) = 1 − Pf ( year n) z year ( n +1) f year n p (t )dt = Pf [ year (n + 1)] − Pf [ year n] 1 − Pf [ year n] Eq. the cost of failures in the future can be cast in terms of current dollars. The header failure lifetime will be taken to be lognormally distributed with a median of 6. Such models can involve many components distributed over several plants. This is because $1 now is worth $1. The expected cost of operating a component for a year that has a failure probability Pf in that year (given that it did not fail before that year) and a cost of a failure of Cf is PfCf. These values of t50 and µ correspond to the solid line in Figure 5-7. the effects of inflation. So a failure that occurs one year from now that will cost $1 at that time will cost only $0. taxes.943 in current dollars. This allows safety-related constraints to be imposed that will over-ride economic considerations.2. and the probability of failure in the future can be obtained from the reliability model. This involves the use of economic models that incorporate the failure probability as an input to the cost estimates.EPRI Licensed Material Use of Probabilistic Results 6. If r is the interest rate and n is the number of years into the future. the net present cost of a failure that costs Cfo at that time is given in Equation 6-1 Cf = C fo (1 + r ) n Eq. and the net present value of money must be accounted for. what would be the expected costs over the next 20 years if the header was continued in operation. As projections are made over time.06 one year from now if the interest rate is 6%.810. 6-2 Returning to the superheater header example. In this simple discussion. This allows the failure probability within each year to easily be computed. and EPRI 98. This discussion concentrates on the simpler problem of estimating the expected cost of candidate scenarios for a single component without constraints on failure rate. versus replacing it now and maybe even replacing it again 10 years from now? This can be addressed using the above relations and the failure probability of the header as evaluated in Section 5. inflation will be assumed to be zero and the effects of taxes will be ignored. The failure probability within one year. Such models have been discussed in the current context in Mauney 93. If the cost of a failure at the present time is Cf. ASME 00. then a failure in the future is lower in terms of net present value.17x105 hours and a µ of 0. given that it has not failed prior to that year. is the appropriate probability to use. ASME 94. From Table 5-1. to be $1 million. the current time is 244. The actual values are not important to the discussion. The cost of lost capacity varies greatly. but should be within an order of magnitude of 1¢ per kilowatt hour. 6-6 . and the cost of a catastrophic header failure to be $10 million. This seems low and is probably applicable to failures that are readily fixed rather than requiring replacement of the header. The failure probability in each of the future 20 years and the net present cost of failure are obtainable from results discussed immediately above.000 per superheater failure. so the results shown in Table 6-3 can be obtained. consider the cost of a new header.EPRI Licensed Material Use of Probabilistic Results The cost of a header failure is an input to the analysis. This results in an estimated lost capacity of about $150. First consider just continuing operation for the next 20 calendar years in the same mode as in the past. This is consistent with the high failure frequency for superheaters in Table 6-2. including installation. For discussion purposes. Table 6-2 suggests that this would be the cost of the header (and installation) plus the cost of the lost generating capacity of 15. and one calendar year of operation typically consists of 6618 hours.765 MWh.885 hours. since the desire is to demonstrate the principles involved. 63 51.78 73.2 Incremental Hazard. because it has not been 6-7 .1 264. Years 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Cost of Failure. but its failure probability will be different.60 61. Eq.36 — Cumulative Time.8 304.EPRI Licensed Material Use of Probabilistic Results Table 6-3 Calculation of Expected Cost of Continuing Operation of Example Header for Another 20 Years Time Into Future.33 44.73 58.83 35.93 34. which is $1.009474 0.4 278.5 344.64 70.95 76.009611 0.007995 0.10 32.4 364. The results in Table 6-3 were obtained by use of MATHCAD.7 271. 6-1.009844 The expected cost of operating the header for the next 20 years is the sum of the right-hand column in Table 6-3.11 48. 6-2 0. The handbook approach of ASME 00 utilizes spreadsheets for the type of calculation included in Table 6-3.9 337.009734 0.9 251.0 370.7 324.2 297.009322 0.1 317.25 53.009545 0.008757 0.009400 0. but now the failure rate for a new header is used.4 311.55 64.008864 0.3 330.6 377. The new header is assumed to be the same as the old header.8 357. k$ 10000 9434 8900 8396 7921 7473 7050 6651 6274 5919 5584 5268 4970 4688 4423 4173 3936 3714 3503 3305 3118 Incremental Cost of Failure.2 350.07 41.0 284. k$ 79.008645 0.67 46.5 258.008138 0.83 37. but could easily be obtained by the use of a spreadsheet.008404 0.000.90 39.009061 0. khrs 244.009674 0.008274 0.008528 0.6 291.008965 0.56 67.009791 0. Similar results for the cost of replacing the header and operating for the next 20 years are obtainable.075. Eq.009152 0.009239 0.95 56. 081 10.3327 0.9 112.942 8.465 4. Eq.08903 0.96x10 -8 -6 Incremental Cost of Failure.94 59.80 79. Years 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Cost of Failure.0007342 0. k$ 0.001827 0.56 66. Table 6-4 Calculation of Expected Cost of Replacing Header Now and Then Operating for Another 20 Years Time Into Future. 6-1.k$ 10000 9434 8900 8396 7921 7473 7050 6651 6274 5919 5584 5268 4970 4688 4423 4173 3936 3714 3503 3305 3118 Cumulative Time.507 2.001243 0.09 39.85 26.002136 0.00x10 3.051 9.61 12. This provides the results included in Table 6-4.004305 6-8 . The use of the diamond results in Figure 5-7 would reduce the incremental hazard below the values used here.004008 0.003705 0. 6-2 1.5 119.04x10 1.606 5.0009767 0.18 72.002450 0.71 46.65 99.13 12.01 10.0001082 0.54 12.8027 1.0005210 0. which may somewhat overestimate the probabilities at the shorter times now of interest (see Figure 5-7).0001013 0.83 11.0002016 0.009837 0.EPRI Licensed Material Use of Probabilistic Results operated in the past.27 105.618 13.25 — -5 -5 0.002766 0.08x10 1.42 86.0003421 0.1 125.781 6.33 52. Eq.412 3. khrs 0. but only for about the 10.47 33.003082 0.000 6.03 92.003396 0.000 hours. The failure probability is evaluated using the same lognormal parameters as used above.24 19.001528 0.98 13.4 Incremental Hazard.7 132. 000 + $126.000 = $2.001 Thus.EPRI Licensed Material Use of Probabilistic Results Reducing the incremental hazard during the first 10.004 0.075. Inspection schedules and type of inspection could be optimized by the use of results such as included in Figure 4-15 when the failure probabilities are combined with the costs of inspections and failure.019. which is $1. it is seen that the least expensive scenario is to continue operating the existing header. This results in an expected cost of 2($1.000.400 $2.126.400 = $1. The failure probability results play the same role as they did in the above header example for run/retire decisions.000 hours would have a minimal influence on these results.019.000 Max Annual Pf 0. Expected Cost $1.000. 6-9 . The relative costs of the scenarios would change with the relative costs of header failure versus header replacement. the “replace header now” would be selected.000) + $19.126. If a constraint of the annual failure probability not exceeding 0.010 0. because the incremental costs associated with these short times are low. and the cost benefit of inspection would result from the reduced hazard rate following an inspection.000.400 If the header is replaced again in 10 years. the cost for the next 20 years is the cost of two headers plus the sum of the right-hand column of Table 6-4 for the first 10 years.000 $1. and the cost would not increase much.005. but this also has the highest maximum annual failure probability. The cost of replacing the header and operating for the next 20 years is the sum of the right-hand column plus the cost of the new header. The estimated expected costs for the three scenarios considered are summarized below: Scenario Continue operation Replace header now Replace header now and at 10 years. . Sections 5 and 6 of this document illustrate the application and potential use of these procedures. A good model can be misused.EPRI Licensed Material 7 CONCLUDING REMARKS Damage models are available for many degradation mechanisms operating in boiler pressure parts. Probabilistic results generated by such models can be used as inputs to economic models to optimize the operation. replacement. Some cautions are required in model selection and execution. The probabilistic approach described here should play an ever-increasing role in component life management. These optimizations can be done on a plant-wide basis or even on a utility-wide basis. and the best that you can do with the methodologies concentrated upon here is obtain the failure probability given the dominance of the damage mechanism modeled. and inspection of components. a good lifetime model can still give inappropriate results if the distributions of the inputs are not correct. 7-1 . such as using too few Monte Carlo trials or excessive shifts in importance sampling. There are other areas of caution. Perhaps of even more concern is the use of an inappropriate lifetime model. Cautions in this regard are provided. and these can be cast into a probabilistic format by considering some of the inputs to the model to be random variables to describe the scatter and/or uncertainty in these inputs. You must be sure that you have modeled the damage mechanism of importance. . A DETAILS OF BLESS EXAMPLE Figures A-1 and A-2 are examples of the type of information required for definition of the header geometry. These are provided as examples of the type often available. Figure A-1 Portion of Header Drawing Showing Configuration of Tube Penetration . and Figure A-2 provides details of the tube penetrations. Figure A-1 provides the overall definition of the tubes connected to the header. there is no offset. it can be verified that the tubes connected to the header are 2inch OD. and that 94 of the 98 tubes of this size are in rows A and F.e.EPRI Licensed Material Details of BLESS Example Figure A-2 Portion of Header Drawing Showing Details of Tube Penetration From the drawing in Figure A-1. The tubes penetrate into the header along a radial line (i. The axial distance between tubes (axial pitch) is 6 inches and there are 24° between these tube rows. A-2 .) The analysis will concentrate on the tubes in Rows A and F. Table A-1 Summary of Cold Starts and Operating Hours for Example Header Year 1-13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 Starts 60 4 0 12 5 2 5 7 8 18 31 27 34 78 138 149 46 187 180 123 109 72 118 113 78 Cumulative Starts 60 64 64 76 81 83 88 95 103 121 152 179 213 291 429 578 624 811 991 1114 1223 1295 1413 1526 1604 Operating Hours 8485 8739 8639 7936 8439 8735 7739 8100 8228 7418 7395 6345 6794 5638 4178 3251 2451 4977 2211 5898 2487 4091 2638 ~2500 ~3600 Cumulative Hours 106728 115467 124106 132069 140508 149244 156983 165083 173311 180729 188125 194470 200964 206602 210780 214031 216482 221460 223671 229569 232056 236147 238785 ~241285 ~244885 A-3 .EPRI Licensed Material Details of BLESS Example Table A-1 summarizes the information on the number of past cold-starts of the unit. 12000 600. Table A-2 BLESS Output File for Example Header Problem—2.000 500.000 6.000 970.00 1000.00 1020.000 350.00000 588.000000 100.000 6.000000 24.00 600.000 6.000 350.00 1500.000 Trials With Maximum Time of 200.344 1.096 2 Ref Angle .000000 .000 Steady PI 1500.500 6.000 1000.00 Flow Rate Header Tube 96.000 hours.000 1000.00 1000.00 Number of Tubes = Tube Row a b Steam Pressure Steam Temperature Tube Temperature Header Flow Rate Tube Flow Rate Material Type Code Offset .50000 2.00000 196.EPRI Licensed Material Details of BLESS Example Table A-2 provides the BLESS output for 2.000 300.00 1500.50000 3.000000 .000 1.000 6.00 1020.00 1000.000 3.12000 600.000 970.000 2.800000 Number of Operating Procedures Number 1 Time .12000 # of Times 2 Steam Temperature Header Tube 1000.000 500.50000 Lig Heat PI 100.00000 3.000 200.00 Flow Rate Header Tube 600.1 Analysis Title: guidelines example Units Selected: US Customary Crack and Header Dimensions in inches Pressure in psi Stress in ksi Temperature in degrees F Crack growth rates in in/hour.000 Hours ** Inputs were read from file: C:\BLESS41\BMARK.000 920.INB BLESS-Headers 4.000 550.000 550.000000 Penetration .00 1500.00 1500.00000 600.00 A-4 .12000 Number 2 Time .12 2 4 # of Times 8 Steam Temperature Header Tube 200.12000 600.00000 294.250 10.500000 1.00 1400. in/cycle Flow rate in klbs/hr Time in hours Ligaments/Pipe = L Outlet Header Header OD Header ID Axial Pitch Tube OD Tube ID Bore Hole 15.0000 1500.000 6.00000 2.00 6.0000 1.000 trials and a maximum evaluation time of 200.800000 .000 920.00 1020.000 2.000 6. 0000 Lig Cool PI 1500.66000 .000 1000.000 5.820000 82432.00000 4.000000 8.7000 22.120000E-13 1. coeff.000 870.12000 600.00000 # of Times 5 Steam Temperature Header Tube 1000.00 1000.000 4.50000 9.12000 400.0000 5.00 1020.000 200.594000 .00 1300.946000 80.00000 26.00 1020.50000 11. UTS-virgin n (stress-strain curve) D Jic Flow stress Tearing Modulus (Tmat) = = = = = = = = = = = = = = = 1.00 1000.25Cr-0.000 520.000 6.000 Number 4 Time .0 .000000E-17 8.000 960.00000 11.923000 -.000 600.000 6.000 670.5Mo Base & Weld ) A1000 (median) B1000 (median) n (creep) m p Q-prime (Secondary Creep) A1000 (second parameter) B1000 (second parameter) A1000-B1000 corr.00000 96.0000 Daily Cycle (8-24-89) Flow Rate PI Header Tube 1500.00 960.0000 1000.00000 300.00 600.00000 5.00000 69.000 6.000 840.000 6.00 600.0000 5.000 500.10000 200.00 Number of Operating Histories 100 per year Op# 1 2 3 daily cycle Op# 4 steady Op# 2 Op Description steady Op Description daily cycle (8-24-89) Op Description lig heat steady lig cool down 4 # of Operating Procedures Repeats 1 1 1 # of Operating Procedures Repeats 1 # of Operating Procedures Repeats 1 # of Operating Procedures Repeats 1 6 1 3 1 1 3 daily cycle with cold start every 6 days Op# 1 4 3 Op Description lig heat daily cycle (8-24-89) lig cool down Material ID = 2 (1.000 4.000 820.000 680.000 400.00000 300.00000 1500.00 100.000 400.0000 13.11000 1500.000 4.5000 24.00 600.0000 21.00 1500.000 2.000 520.000 650.000 670.00 A-5 .000 6.000 870.00000 900.000000 1.00 1020.11000 # of Times 7 Steam Temperature Header Tube 1000.EPRI Licensed Material Details of BLESS Example Number 3 Time .000 690.11000 950.0000 1.00 1020.000 200.000 6.00 850.000 Flow Rate Header Tube 600. ) fatigue exp. C (median) creep coeff.00000(Mean) OPERATING HISTORY: daily cycle with cold start every 6 days 3 Std.030000 .5Mo Base & Weld) SOLUTION DESCRIPTION: Creep-fatigue crack initiation & growth CRACK GROWTH DIRECTION: ID to OD Crack Depth = .000000(Mean) a/w for terminating calculations = .EPRI Licensed Material Details of BLESS Example fatigue coeff.000000 Oxide cracking Std.300000E-01 Time (hrs) 5289.4 Crack Depth .4 37536.460000E-02 1.00000 4 **** Deterministic results using median properties **** Estimated time to crack initiation (hours) Crack initiation controlled by oxide cracking At initiation: oxide crack depth (inches) .00000 Random Number Seed = 9275 Max.038288 .500000 .0 Number of Simulations = 2000 Temp.) = .Damage diagram mid-point TplQ-Prime (Primary Creep) = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 5.14200 -1.80000 Initiation Model: Fast Initiation calculation mode selected LMP-age = . Dev. C (median) oxide coeff. (temp.91 10537.) oxide coeff.00 Stress Multiplier = 1.031500 .60100 .724000 100.82200 .25Cr-0.343000E-07 2.47720 -8496.Larson Miller C-LMP (UTS > UTStr) median C-LMP (UTS < UTStr) median C-UTS C1sigma C2sigma C-LMP (UTS > UTStr) second par.00000 Std.) UTS transition . î – transition C1 (î>îtr) C2 (î>îtr) C3 (î>îtr) C1 (î<îtr) C2 (î<îtr) C3 (î<îtr) Dc . Multiplier = 1.second par.000 42951.000 .Damage diagram mid-point Df .100000 3. = .(stress) = . C (second par. C-LMP (UTS < UTStr) second par. Dev = .000000 1.272600 3.761500 1.825000 .5 15835.00000 Std.9 32035. q creep coeff.0 42951.1 26584. C (second par.000000 Current oxide crk depth = . q creep exp.033075 .00000 2 COMPONENT DATA: LIGAMENT SELECTION: b /b MATERIAL: 2 (1.100000 . n oxide exp.08300 . Dev. Evaluation Time (hrs)= 200000.108400 .00 -956. C (second par. strain-life curve .034729 . C (median) Q (oxide fatigue coeff.50 ..14700 -2.980000E-02 82432... Dev.9141 A-6 .0 .07260 3.000 511.9 21185.11800 4.000 511.040203 5289.000000 -5146.036465 . 00000 .00000 . 138472.042213 .4 77441.00000 .106670 .00000 . 183620. 196808.079599 .9 48689.00000 .7 65793.00000 .00000 . 119697.8 101312. 151191.00000 .9 > max 7 5289.101591 . 113524. 157608.092146 .00000 .9 > max 2 5289.9 > max 8 5289.096753 .00000 .062368 .3 60042. to fail 1 5289.182442 .075809 .065486 .00000 .191564 .117604 .00000 .9 > max 5 5289.092146 .560375 . 177068.9 > max 6 5289.142948 .9 > max 9 5289. .9 > max 10 5289. 132172. 164060.072199 . 144812.044324 .053876 .6 54341.046540 .083579 .068761 . 125913.588394 .136141 .9 > max 3 5289.00000 .00000 .157600 .00000 .3 89282.056569 .6 71593.129658 .00000 .123484 .00000 .EPRI Licensed Material Details of BLESS Example 43087.046540 .150096 ** Time exceeds user-specified maximum for analysis ** BLESS-Head Probabilistic Results ** first 10 simulations ** Simulation Time in Hours to init. 107395. 203441. 203441.048867 .00000 A-7 .051310 .051310 .00000 .150096 .9 > max a (in) .059398 .9 > max 4 5289. 190200.123484 .112004 .173754 Ox 1 1 1 1 1 1 1 1 1 1 Initiation C/F 0 0 0 0 0 0 0 0 0 0 Ai 0 0 0 0 0 0 0 0 0 0 Damage Creep Fatigue .8 83338. 170548.087758 .00000 .5 95273. 1.2000E-02 105000.000 .000 2.0 1.0000E-03 75000.0 1.000 .0000 .5000E-02 110000.0000 20000.0 sec 15775.000 7.000 9.0 1. 1. 1.0 1.000 6.3000E-02 Begin Analysis : Time .000 . 1.10:20:17a End Analysis : Time .000 .000 5.10:20:11a Begin Probabilistic : Time .0 1.000 3.3000E-02 145000.000 3.0500E-02 140000. 1.3500E-02 125000.0 1.0000E-03 60000.000 2. 1.0 1.0 1.4500E-02 175000.0500E-02 160000.0 1.2000E-02 170000.0 1.7500E-02 115000.0000 15000.000 6. 1.5500E-02 165000.0000 35000.00 .1000E-02 185000. 1.000 1.0000 25000.5000E-03 85000.0000 10000.9000E-02 135000.000 1. 1.0 1. 1.6500E-02 200000.3000E-02 205000.5000E-03 95000.0 1.0000 40000.0 1.000 5. 1. 1.000 3. 1.000 1.000 4.000 8.0 1.5000E-03 70000.3000E-02 195000. 1.000 8.0 1.000 3.7500E-02 180000.0 1.0000E-03 90000. 1.000 2.000 8.000 .EPRI Licensed Material Details of BLESS Example BLESS-Head Probabilistic Results Cumulative Probabilities of time to initiation & failure Total number of simulations = 2000 Time Cum. 1.000 4.000 6.5000E-03 65000.000 5.000 1.000 .8500E-02 190000.000 5.000 7.0000 55000. 1.0 sec 15769. Probability of (hours) Initiation Failure 5000.0000E-03 80000.0000 30000.9500E-02 120000.0000 50000.0000 45000. 1. 1.000 3. 1. 1. 1.2:43:06p Elapsed Time Elapsed Time Elapsed Time : : : (Begin Probabilistic) (End Analysis ) (End Analysis ) - Date Date Date - 01/03/2000 01/03/2000 01/03/2000 6.000 2.5000E-02 150000.000 2.9000E-02 155000.000 7.000 .000 .0500E-02 100000.000 .0 1.0 1.0 sec (Begin Analysis ) = (Begin Probabilistic) = (Begin Analysis ) = A-8 .000 1.7000E-02 130000. EPRI Licensed Material Details of BLESS Example Figure A-3 Recordings of Temperature and Pressure for a Typical Day A-9 . . Probability Concepts in Engineering Planning and Design.H. [ASME 98] Risk-Based Methods for Equipment Life Management: An Application Handbook. Fracture Mechanics: Fundamentals and Applications CRC Press. Part 1 – Light Water Reactor (LWR) Nuclear Power Plant Components. 20-1 1991.” Fracture Mechanics: Twenty-Second Symposium (Volume 1).C.II: Decision. Probability Concepts in Engineering Planning and Design. 1964. Handbook of Mathematical Functions.B REFERENCES M. Tang. 1995 pp.H.Development of Guidelines: Volume 1 . Anderson. Washington. [ASME 91] Risk-Based Inspection . Ang and W. [Ayyub 95] B. [Ang 75] A. 53-69. [Ang 84] Risk-Based Inspection . Ayyub and R. C.M. Florida 1995. National Bureau of Standards Applied Mathematics Series 55. Malito.A. & Reliability for Engineers.M. [ASME 94] Risk-Based Inspection . McCuen. Bloom and M. Tang. 1992 pp 393-411. [Anderson 95] A. John Wiley & Sons. Washington. [Abramowitz 64] T. Stegun. Statistics. 4 of Probabilistic Structural Mechanics Handbook. D. Chapman & Hall. H-S. 20-2 1992. Part 2 – Light Water Reactor (LWR) Nuclear Power Plant Components.Development of Guidelines:Volume 2. Florida 1997. Vol. 2000.” Ch. Boca Raton. Probability. New York 1984. under preparation by the ASME Center for Technology Development (CRTD).L. Boca Raton. ASME CRTD-Vol. ASME CRTD-Vol. Vol. [Bloom 92] B-1 . Pennsylvania. ed. 20-4 1998.H. New York. ASME CRTD-Vol. ASTM STP 1131. New York 1975. “Simulation-Based Reliability Methods. Abramowitz and I. [Ayyub 97] J. 20-3 1994. Sundarajan. [ASME 92] Risk-Based Inspection . McCuen.General Document. Ayyub and R.Development of Guidelines: Volume 2. Philadelphia.M.L. ASME CRTD-Vol. H-S. “Using Ct to Predict Component Life. Ang and W. [ASME 00] B. Risk and Reliability.C. John Wiley & Sons. 1: Basic Principles. CRC Press.Development of Guidelines: Volume 3 – Fossil Fuel-Fired Electric Power Generating Station Applications.H. D. Haldar and S. 251.” NUREG/CR–2189.C. Vol.S.O. “BLESS: Boiler Life Evaluation and Simulation System A Computer Code for Reliability Analysis of Headers and Piping. Wells. Inc.V. 251.O.O.M. [Harris 92b] D. ASME PVP-Vol.A. Harris. Vol. Harris. U. [Haldar 95] D. 3 of Probabilistic Structural Mechanics Handbook. ASME AD-Vol. 6. 1995. Dedhia. C. pp. project manager R. pp.S.” U. Chapman & Hall. 17-26. Harris. New York. ed. Harris. 1992. “First-Order and Second-Order Reliability Methods.0a User’s Manual. pp. Viswanathan. “Probabilistic Crack Growth and Modeling. “Probabilistic Fracture Mechanics. 1992 pp.J. Grunloh. Ch. [Harris 92a] D.Dedhia. C. “Simulation of Defects in Weld Construction. 265-331.H. Nuclear Regulatory Commission. “Setting Reinspection Intervals for Seam Welded Piping by Use of Fracture Mechanics and Target Reliability Values. 57-76. D. Chapman & Hall. 27-52. Marcel Dekker. ASME PVP-Vol. New York 1967. Schultz and R.D. New York. An Integrated Approach to Life Assessment of Boiler Pressure Parts. [EPRI 98] H. [Harris 81] D. Hahn and S. EPRI.EPRI Licensed Material References O. [Grunloh 92] G. edited by R. Chapman.J. CA: October 1998. Lim. Harris.” Reliability and Risk in Pressure Vessels and Piping.. Statistical Models in Engineering. ASME PVP-Vol. CM-110998.O. 1997.O. [Harris 95b] D. “Probabilistic Fracture Mechanics with Applications to Inspection Planning and Design. pp. “Probability of Pipe Fracture in the Primary Coolant Loop of a PWR Plant. report prepared for EPRI RP 3352-10. Cruse. [Chapman 93] R. and D.B. 304. Dedhia and S.” Ch. [Harris 97a] B-2 . “Theoretical and User’s Manual for pc-PRAISE.J. [D’Agostino 86] Turbine-Generator Maintenance Outage Interval Extension: Turbo-X Version 1. John Wiley & Sons. E. J. New York. Harris. Mahadevan. et al. Lu.H. Viswanathan 1992. 8 of Reliability-Based Mechanical Design. Washington. Bloom.. Harris and D. Ryder.Y. H. Volume 4: BLESS Code User’s manual and Life Assessment Guidelines. 1993 pp. C. 106-145. Marcel Dekker. Nuclear Regulatory Commission Report NUREG/CR-5864. New York 1986. 81-89. Washington. edited by T. [Harris 95a] D. R.O.” Probabilistic Structural Mechanics Handbook.” Fatigue and Fracture Mechanics in Pressure Vessels and Piping. D’Agostino and M. 5.” Reliability and Risk in Pressure Vessels and Piping. 1993 pp. Shapiro. Sundarajan. Grunloh. D. Stephens.” Ch.J.” Reliability Technology – 1992. 28. 1995. Goodness-of-fit Techniques. Palo Alto. Sundarajan. 1995. [Harris 93] D.O.C. DC 1981.C. 443-451.S. [Hahn 67] A. 5: Probabilistic Fracture Mechanics Analysis. 296/SERA-Vol. 288.O. [Harris 98] J.A. and Petrochemical Plants. Viswanathan. Khaleel. CA 1981.V. D. pp.” Fitness-for-Service Evaluations in Petroleum and Fossil Power Plants. and D. Cruse. 193-204. ASTM STP 1189. Lawrence. MathSoft. O. Advanced Fracture Mechanics. ASME-PVP Vol. New York 1985.F. [MATHCAD 91] D. Tilley.A. Mauney.” Fitness for Adverse Environments in Petroleum and Power Equipment.Dedhia. Material and Geometric Variables on the Creep/Fatigue Life of Headers in Fossil Power Service. ASME PVP-Vol. 1998 pp. New York. ed. 3. ASME PVP-Vol. 1999 pp.A. D.” Reliability and Risk in Pressure Vessels and Piping. Harris. 380.O. “Physics-Based Reliability Models. and D. Philadelphia.O. “The Effects of Initial Flaw Size and Inservice Inspection on Piping Reliability. Cruse. R. 4 of Reliability-Based Mechanical Design. ASME-PVP Vol. [Mauney 93] B-3 .A Simonen. and D. Simonen. Inc. “Flaw Size Distribution and Flaw Existence Frequencies in Nuclear Piping. “Comparisons of Predictions and Observations of Crack Growth in an Operating Header. Harris. [Keisler 95] M.1997 pp. and F. Marcel Dekker.. 474-497. Fossil. “The Impact of Inspection on Intergranular Stress Corrosion Cracking for Stainless Steel Piping. 1994 pp. T. 197-232 [Mahadevan 97b] MATHCAD User’s Guide. Harris. German. “Economic Optimization of Multiple Component Replacement/Inspection in the Power System Environment. [Henkener 93] M.” Ch. [Khaleel 94] M. “Statistical Analysis of Fatigue Strain-Life Data for Carbon and Low-Alloy Steels.J. MA 1991. [Mahadevan 97a] S. Khaleel. Mahadevan. Dedhia. Khaleel and F. 1997. and R. 359. 95-107. ASME PVPVol.G. [Harris 97b] D. 1993 pp. and C. T. Marcel Dekker. Shih. ASME PVP-Vol.A. 386. 123-146. Popelar. M. Chapman.A. 1997 pp.” Service Experience and Reliability Improvement: Nuclear.” Risk and Safety Assessment: Where is the Balance?. ASME PVP-Vol. [Kumar 81] S. 1993 pp. V. 1995 pp.A. Keisler and O. 355-366.A.” Fracture Mechanics: Twenty Third Symposium. 127-144. Harris. “Monte Carlo Simulation.. [Kanninen 85] J. Mahadevan.H. Dedhia. Oxford University Press. Palo Alto. 251. 129-133.M.A. 411-422. New York. “An Engineering Approach for Elastic-Plastic Fracture Mechanics.O.” EPRI Report NP-1931. Henkener.B. Forman. Simonen.A.3. F. [Khaleel 99] V. 1-16.” Risk and Safety Assessment: Where is the Balance?. Kanninen and C. Kumar..” Probabilistic and Environmental Aspects of Fracture and Fatigue. R.D.” Ch. Inc. Cambridge. [Khaleel 95] M.296/SERA-Vol. 1995 pp. Pennsylvania.F. Inc. “An Evaluation of Fracture Mechanics Properties of Various Aerospace Materials.EPRI Licensed Material References D. “The Effect of Operating Variables. Chopra. ed. 6 of Reliability-Based Mechanical Design. Metals Park.F. Tada. New York 2000. Rhode Island. Models. Viswanathan. Philadelphia. New York 1987 [Riedel 87] W.A. Cooper. Van Echo and W. Boca Raton. 1978 pp. Florida 1998. and G. 1984 pp. Struct.R. “Applications of NDE Reliability to Systems. [Shih 83] C. [Wolstenholme 99] B-4 . Chapman & Hall. 1983. 51. New York 1996. and S.. ASME Press.C. “Fully Plastic Crack Problems. H. 484-494. Inelastic Deformation of Metals. ASM International.L. Ninth Edition. Nondestructive Evaluation and Quality Control. 48-56 [Shih 84] D.C.D. “Life Assessment of Superheater/Reheater Tubes. John Wiley & Sons. Stress Analysis of Cracks Handbook. Dame. Ohio. Simmons. Chicago. [Tada 00] J.” ASME Piping and Pressure Vessel Conference. Hardy. Vol.” Metals Handbook. 674-688. Louisiana. American Society for Testing and Materials. New Orleans. Illinois 1992. Paterson. Fracture at High Temperatures. Vol. S. ASTM Data Series Publication No. [Saxena 98] C. [Rackwitz 78] H. DS6S1. Pennsylvania 1966.C. Viswanathan. Mechanical Properties and Metallurgy.” Materials Research Laboratory Report MRL E-147. [Viswanathan 92] L. Gehl. Grunloh. Stouffer and L.F. Ohio 1989. Nonlinear Fracture Mechanics for Engineers. and T. [Stouffer 96] H.F. “Tables of Hutchinson-Rice-Rosengren Singular Field Quantities. [Viswanathan 89] R. [Rummel 89] A. Wolstenholme. Part 1: Solution by a Penalty Method. Saxena. Accident Facts.” Journal of Applied Mechanics.EPRI Licensed Material References National Safety Council. [NSC 92] R. Riedel. Damage Mechanisms and Life Assessment of High-Temperature Components. Rackwitz and B. Third Edition. Reliability Modelling: A Statistical Approach. Paris. New York 1999. ASM International. 1992. 1989 pp. Irwin. Springer-Verlag. “Structural Reliability Under Combined Random Load Sequences.” Comput.D. Supplemental Report on the Elevated-Temperature Properties of Chromium-Molybdenum Steels. Rummel. P. Providence. Shih. Vol. Fiessler. Metals Park. Brown University. Shih and A. CRC Press. G. 17. [Van Echo 66] R.R.T. Needleman. 9. . PROCESS OR SIMILAR ITEM DISCLOSED IN THIS PACKAGE. in whole or in part. Palo Alto. CHOICE OF LAW This agreement will be governed by the laws of the State of California as applied to transactions taking place entirely in California between California residents.You agree to take all reasonable steps to prevent unauthorized disclosure or use of this package. IF YOU DO NOT AGREE TO THE TERMS OF THIS AGREEMENT. OR (III) THAT THIS PACKAGE IS SUITABLE TO ANY PARTICULAR USER’S CIRCUMSTANCE.EPRI member utilities and EPRI.S. or use the information contained in this package. 1. translate or modify this package. Electrify the World 2.epri. OR (B) ASSUMES RESPONSIBILITY FOR ANY DAMAGES OR OTHER LIABILITY WHATSOEVER (INCLUDING ANY CONSEQUENTIAL DAMAGES. INC. COPYRIGHT This package. EXPORT The laws and regulations of the United States restrict the export and re-export of any portion of this package. OR (II) THAT SUCH USE DOES NOT INFRINGE ON OR INTERFERE WITH PRIVATELY OWNED RIGHTS. if the consultant has entered into a contract agreeing not to disclose the package outside of its organization or to use the package for its own benefit or the benefit of any party other than your company. reproduce. PLEASE READ IT CAREFULLY BEFORE REMOVING THE WRAPPING MATERIAL.TERM AND TERMINATION This license and this agreement are effective until terminated. U. EPRI’s multidisciplinary team of scientists and engineers draws on a worldwide network of technical and business expertise to help solve today’s toughest energy and environmental problems.S. EPRI. EPRI has the right to terminate the license and this agreement immediately if you fail to comply with any term or condition of this agreement.ANY COSPONSOR. Printed on recycled paper in the United States of America 1000311 EPRI • 3412 Hillview Avenue. Inc. the company provides a wide range of innovative products and services to more than 1000 energyrelated organizations in 40 countries.Any EPRI member utility that does not have a Master Utility License Agreement may get one on request. No waiver. trademarks or any other intellectual property. trade names. All rights reserved.You may not. Now known simply as EPRI. GRANT OF LICENSE EPRI grants you the nonexclusive and nontransferable right during the term of this agreement to use this package only for your own benefit and the benefit of your organization.com • www. is either licensed to EPRI or owned by EPRI and is protected by United States and international copyright laws. Upon any termination you may destroy this package. RESTRICTIONS You may not rent. in any form. (EPRI). Inc. DISCLAIMER OF WARRANTIES AND LIMITATION OF LIABILITIES NEITHER EPRI. Electric Power Research Institute and EPRI are registered service marks of the Electric Power Research Institute. METHOD.ANY MEMBER OF EPRI. 6. trade secrets. lease.APPARATUS. rights or licenses in respect of this package. 7. (I) WITH RESPECT TO THE USE OF ANY INFORMATION. electric utilities established the Electric Power Research Institute in 1973 as a nonprofit research consortium for the benefit of utility members. Palo Alto. METHOD. INTEGRATION You have read and understand this agreement. and (III) a consultant to your company or related entities. superseding any prior related understanding or agreement. 8. About EPRI EPRI creates science and technology solutions for the global energy and energy services industry. copyrights. California 94303 • USA 800. PROCESS OR SIMILAR ITEM DISCLOSED IN THIS PACKAGE. this agreement does not grant you any right to patents. for the benefit of any third party or for any purpose other than as specified above unless such use is with the prior written permission of EPRI.Target: Boiler Life and Availability Improvement SINGLE USER LICENSE AGREEMENT THIS IS A LEGALLY BINDING AGREEMENT BETWEEN YOU AND THE ELECTRIC POWER RESEARCH INSTITUTE. © 2000 Electric Power Research Institute (EPRI). This shrink-wrap license agreement is subordinate to the terms of the Master Utility License Agreement between most U. Except as specified above.313. and you agree not to export or re-export this package or any related technical data in any form without the appropriate United States and foreign government approvals. license. EPRI. their customers. (II) its subsidiaries or other related entities.2121 • askepri@epri. APPARATUS. signed by an officer of EPRI. Inc.855. California 94304 • PO Box 10412. without the prior written permission of EPRI. and society. 4. but all obligations of nondisclosure will remain in effect. and acknowledge that it is the final. EXPRESS OR IMPLIED. 3.You may terminate them at any time by destroying this package. including the information contained in it.This means that the following may use this package: (I) your company (at any site owned or operated by your company). ELECTRIFY THE WORLD is a service mark of the Electric Power Research Institute. complete and exclusive agreement between you and EPRI concerning its subject matter. BY OPENING THIS SEALED PACKAGE YOU ARE AGREEING TO THE TERMS OF THIS AGREEMENT. INCLUDING ANY PARTY’S INTELLECTUAL PROPERTY. variation or different terms of this agreement will be enforceable against EPRI unless EPRI gives its prior written consent.com . 5. NOR ANY PERSON OR ORGANIZATION ACTING ON BEHALF OF ANY OF THEM: (A) MAKES ANY WARRANTY OR REPRESENTATION WHATSOEVER. disclose or give this package to any person or organization. INCLUDING MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE. EVEN IF EPRI OR ANY EPRI REPRESENTATIVE HAS BEEN ADVISED OF THE POSSIBILITY OF SUCH DAMAGES) RESULTING FROM YOUR SELECTION OR USE OF THIS PACKAGE OR ANY INFORMATION. or prepare any derivative work based on this package.PROMPTLY RETURN THE UNOPENED PACKAGE TO EPRI AND THE PURCHASE PRICE WILL BE REFUNDED.3774 • 650.
Copyright © 2024 DOKUMEN.SITE Inc.