research article determination of stress intensity factors...

Research ArticleDetermination of Stress Intensity Factors inLow Pressure Turbine Rotor Discs

Ivana Vasovic1 Stevan Maksimovic2 Katarina Maksimovic3

Slobodan Stupar4 Gordana Bakic4 and Mirko Maksimovic5

1 Institute Gosa Milana Rakica 35 Belgrade Serbia2Military Technical Institute Ratka Resanovica 1 Belgrade Serbia3 City of Belgrade Secretariat for Communal and Housing Affairs Office of Water Management Kraljice Marije 1XIII-XIV11000 Belgrade Serbia

4 Faculty of Mechanical Engineering University of Belgrade Kraljice Marije 16 Belgrade Serbia5 Belgrade Waterworks and Sewerage Kneza Milosa 27 Belgrade Serbia

Correspondence should be addressed to Ivana Vasovic ivanavvasovicgmailcom

Received 3 October 2013 Revised 13 January 2014 Accepted 16 January 2014 Published 1 June 2014

Academic Editor Marek Lefik

Copyright copy 2014 Ivana Vasovic et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

An attention in this paper is focused on the stress analysis and the determination of fracture mechanics parameters in low pressure(LP) turbine rotor discs and on developing analytic expressions for stress intensity factors at the critical location of LP steam turbinedisc Critical locations such as keyway and dovetail area experienced stress concentration leading to crack initiationMajor concernsfor the power industry are determining the critical locations with one side and fracture mechanics parameters with the other sideFor determination of the critical locations in LP turbine rotor disc conventional finite elements are used here For this initial cracklength and during crack growth it is necessary to determine SIFs In fatigue crack growth process it is necessary to have analyticformulas for the stress intensity factor To determine analytic formula for stress intensity factor (SIF) of cracked turbine rotor discspecial singular finite elements are used Using discrete values of SIFs which correspond to various crack lengths analytic formulaof SIF in polynomial forms is derived here For determination of SIF in this paper combined 119869-integral approach and singular finiteelements are used The interaction of mechanical and thermal effects was correlated in terms of the fracture toughness

1 Introduction

Stress intensity factors are very powerful parameters forpredicting the loads and crack length at which fracture canoccur The joint between the turbine blade and the discusually represents the most critical area from the point ofview of the static and fatigue approachesThe loads associatedwith these regions are mainly the centrifugal forces andthermal stresses Disks are the most strained and respon-sible steam turbine elements They contain technologicaland structural stress concentrators in which damages areeventually accumulated In low pressure (LP) steam turbinerotating discs are simultaneously subjected to mechanicaland moderate thermal load [1ndash3] A disc is loaded underinternal pressure due to shrink fit on a shaftThus additional

blade effectsmay be taken into consideration andmodeled byan external tensile load at the outer radius of the disc whenthe disc rotates with significant angular velocity Materialbehavior is temperature dependant and changes in mate-rial properties throughout the disc should be consideredIn order to attain a certain and reliable analysis solutionshould consider changes in material specification causedby temperature Therefore engineers have strong interestsin monitoring and analyzing of rotating components in jetengines to improve safety and to lower maintenance costTo prevent catastrophic failure of the engine they havedeveloped different techniques to analyze structures Theengineering field of fracture mechanics was established todevelop a basic understanding of such problems Todayfracture mechanics is not only of academic interest but also

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 304638 9 pageshttpdxdoiorg1011552014304638

2 Mathematical Problems in Engineering

plays an increasingly important role in structural design[4ndash6] Recently a more stringent safety criterion assuminga preexisting flaw in critical component has been adoptedto assess service life of aircraft [7ndash10] This emphasizes thesignificance of fracture mechanics as a tool for analysis Theconcepts of linear elastic fracture mechanics which lead tothe strain fracture toughness property 119870

119868119862 have already

been used in engineering applications Mechanical loadingis not the only factor considered in the design of structuresor structural components Other possible situations maybeoperational environments such as temperature have to beconsidered In the operation of gas turbine engines forexample thermal stresses can be as high as or higher than thecentrifugal stresses The worst condition of the combinationsof thermal centrifugal and gas bending stresses at elevatedtemperatures results in high local stresses which can leadto cracking of the LP turbine blades and rotor disks Thusthermal effects should not be ignored A special attentionin this paper is focused to develop analytic expressions forstress intensity factors at low pressure steam turbine discFinite element method (FEM) is used to determine criticallocation zone of the stress concentration at low pressuresteam turbine disc in which rotor blades are connectedwith rotation disc At critical location of rotation disc initialcrack lengths are assumed For various crack lengths at LPsteam turbine disc stress intensity factors are determinedunder thermomechanical loads Using these discrete valuesof SIFs here are derived analytic formulas for SIF 119870

119868 at

cracked LP disc Analytic formula of the stress intensity factorinformation is valuable in the prediction of service lives ofturbine discs Crack growth behavior is a major issue in avariety of rotating components for which analytic formula ofSIF is necessary

The primary objective of this research is to establishcomplete budgetary procedures related to the design of struc-tural elements of low pressure turbine disc in the presenceof initial crack This approach to design in the presence ofinitial cracks is known as the ldquodamage tolerance approachrdquoWith this approach where the initial damage is at the criticallocations of the turbines the risk of unforeseen catastrophicfailures during operation is reduced to a minimum

Low Pressure Turbine Large scale steam turbines employedfor the generation of electricity are expected to operate forworking lives of 30 years and beyond A harsh combination ofservice temperature and environment dictates that the mostseverely stressed components within the low pressure turbinewill need to be formed from what are generally classified ashigh performance materials

In light of such experiences the present study was com-missioned to investigate the fatigue response of a commonLP blade material the X12CrNiMoVNb steel An improvedunderstanding of the mechanical behaviour under variableloading conditions is deemed

Failures occur from time to time in power plants asthey do in other engineering structures However they arenot always examined closely to identify the causes Similarlywhen the failures are observed from time to time repairs areoften made without any careful analysis being undertaken

Most blade failure investigations end with a metallurgicalreport A metallurgical examination of the blade establisheswhether the failure was due to substandardmaterial or due tothe presence of flaws machine marks or corrosion pits thatcreated local stress raisers not accounted for in the allowablestress limits of the blade designA fracture surface of the bladecan suggest that fatigue is amore probable failuremechanismBut the correction of a blade problem requires more thanpositive identification of the mechanisms involved [3]

Finite element analysis needs to be performed becauseof complex geometry and boundary conditions such as theblade So finite element analysis is used to calculate thestresses and modal shapes of the models Previous researcheshave shown that the low pressure blades of a steam turbineare generally found to be more susceptible to failure thanintermediate pressure and high pressure blades [11]Themostcommon failure mechanisms which occur within the lowpressure blade are normally those associated with eithersympathetic or forced vibrations those caused by transientoperating conditions and those that occur as the result of thetransported and accumulated corrosive ions in working fluid[3 12ndash17] Blades and low pressures of steam turbine discs arecritical components in power plants that convert the availableenergy in steam into mechanical energy [1]

2 Strength Analysis with respect toFracture Mechanics

The objective of the analytical work is to provide the rela-tionship between the stress intensity factor and the variousmechanical and thermal load conditions Residual fatiguelife of cracked rotation components is generally based uponcrack growth consideration Fracture mechanics is the math-ematical tool that is employed It provides the concepts andequations used to determine how cracks grow and their effecton the strength of structure From an initial crack length 119886

119900

one must determine critical flow size 119886119888for the fast fracture

damage tolerance approach

119886119900997888rarr 119886119888

119870119868997888rarr 119870

119868119862

(1)

where 119870119868is the applied tensile mode 119868 is the stress intensity

factor and 119870119868119862

is the strain fracture toughness of materialIn fracture mechanics we try to correlate analytical

predictions of crack growth and failure with experimen-tal results The analytical predictions made by calculatingfracture parameters such as stress intensity factors in thecrack region can be used to estimate crack growth rateAlthough several stress intensity factor handbooks [5 18]have published the available solutions those solutions are notalways adequate for particular engineering applications Thisis especially true for cracks subjected to nonuniform stressfields near notch or thermal stresses [6 19] Stress intensityfactor calculation is an important issue for numerical analysisof fracture problems and there exist many approaches for

Mathematical Problems in Engineering 3

instance the extrapolation techniques 119869-integral approach[20 21] the virtual crack extension (VCE) technique andso forth Amongst these methods a displacement extrapola-tion method sometimes called crack opening displacementcorrelation technique used specially with the quarter-pointsingular finite elements (QPE) [21] is most widely used dueto its high accuracy and simplicity

Typical fracture parameters of interest are stress intensityfactors (119870

119868 119870119868119868 and 119870

119868119868119868) which correspond to three basic

modes of fracture 119869-integral [22] a path-independent lineintegral that measures the strength of the singular stressesand strains near a crack tip and energy release rate (119866)which represents the amount of work associated with a crackopening or closure

Firstly linear elastic or elastic-plastic static analysis has tobe performed and then we use postprocessing commands tocalculate fracture parameters

Finite element method (FEM) is the most widely usedtechnique for evaluating stress intensity factor (SIF) Themost important region in modeling the fracture region is theregion around the crack tip While the domain is meshedwe are using crack tip singular finite elements with nodalsingularity [6 20] Those elements exhibit the 119903minus12 singular-ity both on the boundary of the element and in the interiorDisplacement correlation was employed to determine stressintensity factorsMaksimovic et al [9] employed hybrid finiteelements with a square root surrounding the crack tip andregular elements in the rest of the domain to determine stressintensity factors

One main objective of this paper lies in developing acomplete computation procedure for the strength analysisof cracked structural components based on combining 119869-integral approach and singular finite elements

21 Determination of the Stress Intensity Factors Once a finiteelement solution has been obtained the values of the stressintensity factor can be extracted from it Determination ofthe stress intensity factors of cracked structural componentsbased on finite element analysis (FEA) and linear elasticfracture mechanics (LEFM) can be done In this paperthe stress analysis of LP turbine disc with initial crack isconsidered using the finite element programANSYS [23]Thecomputer program evaluates the stress intensity factors (SIFs)on the crack tip from the inner side of hole in turbine discTheprogram requires one global model of disc with or withoutmodeled crack and one detailed submodel of investigatedcrack The program makes it easy to build submodel froman already existing global model The global model canpreferably be meshed with tetrahedral elements which is oneof the strongest specified demands

The crack submodel is very easy to create in an alreadyexisting globally meshed model In this procedure the tra-ditional submodeling technique will be used where thedisplacement field of the global model is applied as boundaryconditions on the submodel The global model must containthe actual crack length otherwise the SIF will be incorrectdue to reduced stiffness of the submodel

x2nj

x1

Γ

Figure 1 Arbitrary contour and coordinate configuration

The global model has tetrahedral elements around thecrack tip making it impossible to achieve the SIF To solvethis problem a small local submodel with 20-node hexahedralelements with singular elements is generated around thecrack tip

SIFs are obtained from the FEM and 119869-integral methodsavailable in ANSYS software code [23] The stress intensityfactor was evaluated using the following equivalent relation-ship

119870 = radic119864119869

1 minus ]2 (2)

where 119864 is Youngrsquos modulus ] is Poissonrsquos ratio and 119869-integral is evaluated using ANSYS The path-independent 119869-line integral which was proposed by Rice [22] is defined as

119869 = intΓ

(1198821198891199092minus 120590119894119895

120597119906119894

1205971199091

119899119895119889119878) (3)

where119882 is the elastic strain energy density Γ is any contourabout the crack tip shown in Figure 1 119879

119894and 119906

119894are the

traction and displacement components along the contour 119878is arc length along the contour and 119909

1and 119909

2are the local

coordinates such that 1199091is along the crack

Generally the stress intensity factors are additive andprovide different loading conditions that induce the samemode of crack extension Hence the fracture condition of theinteraction of mechanical and thermal loads can be

119870119868= (119870119868)119872+ (119870119868)119879

(4)

when 119870119868= 119870119868119862 in which (119870

119868)119872is the stress intensity factor

due to mechanical load and (119870119868)119879is the stress intensity factor

due to temperature effectUnfortunately the 119869-integral is restricted to two-

dimensional bodies with external loading The 119869-integral ispath dependent for cases which include residual inertialor thermal stress terms or loadings along the crack faceIt cannot be used for three-dimensional structures ofnonhomogeneousmaterials in the direction of crack advance[24] Some efforts have been made to modify the expression


120596

304

R 215 R 350 R 460R 515

120601 15

2 Crack

10

Figure 2 Geometry properties of part of low pressure turbine discwith initial crack

of 119869-integral Wilson and Yursquos [25] integral is valid for use intwo-dimensional thermal cases Blackburn et alrsquos modifiedintegral [26] is applicable to three-dimensional cases In thisstudy the modified integral 119869lowast [26] was used as

119869lowast= intΓ

(119882lowast1198891199092minus 120590119894119895

120597119906119894

1205971199091

119899119895119889119878) +

119864120572

1 minus 2]int 120576119894119894

120597120579

1205971199091

119889119860 (5)

119882lowast= 119882 minus

119864120572120579

2 (1 minus 2])120576119894119894 (6)

in which

120590119894119895= 120582120576119894119894120575119894119895+ 2120583120576119894119895minus

119864120572

1 minus 2]120579120575119894119895

119882 =1

2120590119894119895120576119894119895

(7)

where 120583 and 120582 are Lamersquos constants 120579 is the temperatureand 120572 is the coefficient of thermal expansion If only mode119868 loading is considered then (5) reduces to

119869lowast=(1 minus ]2)1198702

119868

119864 (8)

Accordingly 119869-integral is modified to include the elastic-plastic and thermal effects whilst maintaining the indepen-dence of the trajectory along which integral is determined[27 28]

Combining FEM with modified 119869lowast-integral approach to

analyze thermomechanical problems with respect to fracturemechanics is considered in the articles [25ndash31]

3 Numerical Analysis and Validation

Attention in this investigation is focused on determinationof the stress intensity factors in the cracked turbine discs

1Nodal solution

Step = 1

Sub = 2

Time = 1

Temp (avg)

RSYS = 0

SMN =180

SMX = 193

MN

MX

ANSYS 90Al

Dec 30 2008

211825

180

181444

182889

184333

185778

187222

188667

190111

191556

193

Ring 5

Figure 3 Temperature distributions at part of a low pressure turbinedisc

In determining fracture mechanical parameters primarilystress intensity factors we used finite element method and119869-integral approach On basis of external load conditionsthe temperature on the upper surface of rotor disc is 193∘Cand on the lower surface is 180∘C is using SOLID70 typefinite element that is incorporated in ANSYS software code[23] Geometry properties of the part turbine disc includinglocation of initial crack length are shown in Figure 2 Turbinedisc is under combined mechanical load due to rotationspeed and due to temperature distributions Temperaturedistributions at part of a low pressure turbine disc are shownin Figure 3

In stress analysis disc was modeled with SOLID95 typeelements consisting of twenty-node solid elements (Figures4 and 5) These elements are suitable for modeling areaaround crack tip and correctly describe singular stress andstrain fields (Figure 6) Calculated stress values around crack


Crack front

Figure 4 Von Misses stresses from thermal and mechanical loads(119899 = 3000 omin) in elements crack length 119886 = 0001m

Crack tip top surface 00 m

Crack front

Crack tip bottom surface 001 m

Figure 5 Von Misses stresses in submodel from thermal andmechanical loads (119899 = 3300 omin) crack length 119886 = 00025m

Crack face Singular stresses at crack tip

Crack front

Figure 6 Von Misses stresses in submodel from thermal andmechanical loads (119899 = 3000 omin) in elements crack length 119886 =

0001m

Nodal solutionStep = 1

Sub = 1

Time = 1

EPELEQW (avg)DMX = 0001529

SMN = 0124Eminus4

SMX = 0006072

ANASYS 121

Jan 7 2014

132447

0124E minus 040686E minus 03

00013590002032

00027060003379

00040520004726

00053990006072

Turbine disc a = 20mm

MXMN

Figure 7 Elastic deformation in turbine disc (119899 = 3000 omin) inelements crack length 119886 = 2mm

tips were used to determine stress intensity factors using 119869-integral approach

In Figures 7 and 8 elastic deformations and displacementsin turbine discs are illustrated for the crack length 2mm VonMisses stress distributions in cracked turbine disc for initialcrack 2mm are shown in Figure 9

Values of stress intensity factors for different cracksizes and loads are shown in Table 1 and Figure 10 Diskswere made of 035C 065Mn 09Cr 03Mo steel grade steel(34HN3M steel according to GOST) Material properties forthe disk of the turbine are as follows density 120588 = 7820 kgm3Young modulus 119864 = 186GPa Poisson coefficient ] = 03coefficient of thermal expansion 120572 = 1332 120583mm∘C thermalconductivity 120582 = 36Wm∘C and119870

119868119862= 12000MPa(m)12

Previous finite element results can be used to determineanalytic expressions for stress intensity factorsThese analyticexpressions for stress intensity factors can be used for crackgrowth analyses and residual life estimations of cracked disc


Nodal solutionStep = 1Sub = 1Time = 1USUN (avg)RSYS = 0DMX = 0001529SMN = 0697E minus 03SMX = 0001529

ANASYS 121Jan 7 2014132659



00010670001159

00012520001344

00014370001529


MN

Figure 8 Displacement in turbine disc (119899 = 3000 omin) inelements crack length 119886 = 2mm

Figure 9 Von Misses stresses in turbine discs (119899 = 3000 omin) inelements crack length 119886 = 2mm

Stress intensity factors

KI(M

Pa(m)12)

0 0002 0004 0006 0008 0010

Distance from the top surface (m)

0

2

4

6

8

10

12

14

SIF (n = 3000 omin a = 00010m)

SIF (n = 3300 omin a = 00010m)

SIF (n = 3000 omin a = 00015m)

SIF (n = 3300 omin a = 00015m)

SIF (n = 3000 omin a = 00020m)

SIF (n = 3300 omin a = 00020m)

SIF (n = 3000 omin a = 00025m)

SIF (n = 3300 omin a = 00025m)

SIF (n = 3000 omin a = 00030m)

SIF (n = 3300 omin a = 00030m)

Figure 10 Stress intensity factors for different crack lengths androtational speed

By using discrete values of stress intensity factors for differentcrack lengths as shown in Table 1 analytic expressions wereestablished for different load conditions

The values of stress intensity factors can be found bysolving (9) and (10) for different crack lengths 119886 on the topsurface of disc and rotational speed (119899 = 3000 omin and119899 = 3300 omin Figure 11) respectively

1198701198681= 23746666667119886

3minus 178131429119886

2+ 517289119886 + 400

(9)

1198701198682= 24046666667119886

3minus 185431429119886

2+ 561644119886 + 502

(10)

in which 119870119868[MPa(m)12] a [m] is the crack length

Using previously derived analytic formulas for the stressintensity factors 119870

119868we can to control situation in which

119870119868will be equal to its critical value 119870

119868119862 In accordance

to computation results under thermomechanical loads themaximum value of the SIF is 119870

119868= 1165MPa(m)12 Table 1

and its fracture toughness is 119870119868119862

= 1200MPa(m)12 Inthis investigation derived analytic relations (9) and (10) forthe stress intensity factors are used for determination of thereserve factor (RF) with respect to ldquostaticrdquo fracturemechanicsthat is defined as

RF =119870119868119862

119870119868

(11)


Table 1 Stress intensity factors for different crack lengths and rotational speed

119870119868[MPa(m)12]

Crack lengths [m]

119886 = 00010m

Rotational speed [omin]

119899 = 3000 73373 76519 76059 75651 74727 72176119886 = 00010m 119899 = 3300 86310 89799 89979 89631 88407 85053119886 = 00015m 119899 = 3000 82728 86502 86411 86006 85149 81400119886 = 00015m 119899 = 3300 97197 101864 101855 101434 100165 95799119886 = 00020m 119899 = 3000 85604 90005 89857 89428 88593 84211119886 = 00020m 119899 = 3300 100724 106150 106085 105644 104798 99285119886 = 00025m 119899 = 3000 90887 95932 95977 95547 94731 89392119886 = 00025m 119899 = 3300 106963 113151 113304 112683 112036 105392119886 = 00030m 119899 = 3000 93531 95389 98753 98345 97341 91821119886 = 00030m 119899 = 3300 108985 116519 116484 116067 115023 108108

Position of crack tip from the top surface [m] 0 0002 0004 0006 0008 0010

775

885

99510

10511

11512

00005 0001 00015 0002 00025 0003 00035 0004 00045 0005

Poly (KI2-3300omin)Poly (KI1-3000 omin)

KI2-3300ominKI1-3000 omin

KI2 = 24046666667a3minus 185431429a

2+ 561644a + 502

KI1 = 23746666667a3minus 178131429a2 + 517289a + 400

Figure 11 Polynomial form of derived stress intensity factors basedon discrete values of stress intensity factors

The relations (9) and (10) can be used too for the crack growthanalyses of damaged structural components and residualfatigue lives estimations

The crack growth of the disc due to fatigue or stresscorrosion can be predicted in accordance with the theory ofthe conventional fracture mechanics The dependence of thefatigue crack growth rate on the stress intensity factor (9) or(10) can be conveniently represented as follows

119889119886

119889119873= 119862(Δ119870)

119898[1 minus (Δ119870

119905ℎΔ119870)1198991

1 minus (Δ119870max119870119888)]

1198993

(12)

where 119889119886119889119873 is the fatigue crack growth the rate 119870119905ℎis the

threshold value of stress intensity factor 119870119888is the critical

value and 119862 and 119899 are constants depending on the materialand environmental conditions

31 Estimating Remaining Life The remaining life of LPturbine discs for keyway cracking is calculated by the initial

crack size (119886119894) the critical size (119886cr) and the crack growth rate

(119889119886119889119905) and is expressed by the following relationship

119905rem =119886cr minus 119886

119894

119889119886119889119905 (13)

Previous relations (12) and (13) can be used for residual lifeestimation of low pressure turbine discs

4 Conclusions

This work established the complete computation procedurefor derivation of analytic expressions for the stress intensityfactors of the complex cracked structural element suchas steam low pressure turbine disc In order to get highoperation availability and to establish a maintenance andspares management strategy capable of developing the lifeextension of steam turbine suitable assessment of agingdamage and residual life estimation for most relevant parts ofturbine is essential This investigation focused on developinganalytic expressions for stress intensity factors at the criticallocation of LP turbine disc FEM is used for defining thecritical location with respect to fracture mechanics and forthe determination of the stress intensity factors The stressintensity factor information is valuable in the prediction ofthe service lives of turbine discs The method developedin this paper was applied to compute the stress intensityfactors in domain of linear elastic facturemechanics (LEFM)in plain stress problem The finite element method hasbeen used to calculate thermal and mechanical stresses ina low pressure turbine disc too These results were usedto determinate the stress intensity factors using 119869-integralapproach In many cases the 119869-integral can be the easiestmeans of calculating stress intensity factors This method iseasy to use when the software supports determination ofthe contour integral The results are fairly accurate even forthe coarse meshes because the contours can be chosen to beremote from the near crack tip regionThe computed discretevalues of stress intensity factors by finite elements are usedfor derivation equations (9) and (10) which present a newanalytical expression for the stress intensity factors where


value 119886 is the crack length The derived analytic expressionsfor the stress intensity factors can be used for the ldquostaticrdquofracture mechanics analysis (11) or residual life estimation ofdamaged low pressure turbine components using relations(12) and (13)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was financially supported by the Ministry ofScience and Technological Developments of Serbia underProjects OI-174001 and TR-35045

References

[1] C Liu and D D Macdonald ldquoPrediction of failures of low-pressure steam turbine disksrdquo Journal of Pressure Vessel Tech-nology vol 119 no 4 pp 393ndash400 1997

[2] G R Jovicic V K Grabulov S M Maksimovic et al ldquoResiduallife estimation of a thermal power plant componentmdashthe high-pressure turbine housing caserdquo Thermal Science vol 13 no 4pp 99ndash106 2009

[3] H-J Kim ldquoFatigue failure analysis of laststage blade in a lowpressure steam turbinerdquo Engineering Failure Analysis vol 6 no2 pp 93ndash100 1999

[4] M Shankar K Kumar and S L Ajit Prasad ldquoT-root blades in asteam turbine rotor A case studyrdquo Engineering Failure Analysisvol 17 no 5 pp 1205ndash1212 2010

[5] A G Evans ldquoPerspectives on the development of high-toughness ceramicsrdquo Journal of the American Society vol 73 pp187ndash206 1990

[6] R D Henshell and K G Shaw ldquoCrack tip finite elements areunnecessaryrdquo International Journal for Numerical Methods inEngineering vol 9 no 3 pp 495ndash507 1975

[7] M Blazic K Maksimovic and Y Assoul ldquoDetermination ofstress intensity factors of of structural elements by surfacecracksrdquo in Proceedings of the 3rd Serbian Congress Theoreti-cal and Applied Mechanics pp 374ndash383 Serbian Society ofMechanics Vlasina Lake Serbia July 2011

[8] S Boljanovic and S Maksimovic ldquoFatigue crack growthmodel-ing of attachment lugsrdquo International Journal of Fatigue vol 58pp 66ndash74 2013

[9] S Maksimovic S Posavljak K Maksimovic V Nikolic and VDjurkovic ldquoTotal fatigue life estimation of notched structuralcomponents using low-cycle fatigue propertiesrdquo Strain vol 47no 2 pp 341ndash349 2011

[10] S Maksimovic I Vasovic M Maksimovic and M ETHuricldquoResidual life estimation of damaged structural componentsusing low-cycle fatigue propertiesrdquo in Proceedings of the 3rdInternational Congress of Serbian Society of Mechanics VlasinaLake Serbia July 2011

[11] G Das S Ghosh Chowdhury A Kumar Ray S Kumar DasandD Kumar Bhattacharya ldquoTurbine blade failure in a thermalpower plantrdquo Engineering Failure Analysis vol 10 no 1 pp 85ndash91 2003

[12] WZWang F-Z XuanK-L Zhu and S-T Tu ldquoFailure analysisof the final stage blade in steam turbinerdquo Engineering FailureAnalysis vol 14 no 4 pp 632ndash641 2007

[13] L C White Modern Power Station Practice British ElectricityInternational Pergamon Press 1992

[14] R Viswanathan Damage Mechanisms and Life Assessmentof High Temperature Components ASM International MetalsPark Ohio USA 1989

[15] Y Zhang M Urquidi-MacDonald G R Engelhardt and D DMacDonald ldquoDevelopment of localized corrosion damage onlow pressure turbine disks and blades I Passivityrdquo Electrochim-ica Acta vol 69 pp 1ndash11 2012

[16] Y Zhang M Urquidi-MacDonald G R Engelhardt and D DMacDonald ldquoDevelopment of localized corrosion damage onlow pressure turbine disks and blades II Passivity breakdownrdquoElectrochimica Acta vol 69 pp 12ndash18 2012

[17] Y Zhang M Urquidi-MacDonald G R Engelhardt and D DMacDonald ldquoDevelopment of localized corrosion damage onlowpressure turbine disks and blades III application of damagefunction analysis to the prediction of damagerdquo ElectrochimicaActa vol 69 pp 19ndash29 2012

[18] J W Hutchinson and Z Suo ldquoMixed mode cracking in layeredmaterialsrdquo in Advances in Applied Mechanics J W Hutchinsonand T YWu Eds vol 29 pp 63ndash191 Academic Press OrlandoFla USA 1992

[19] S T Lin and R E Rowlands ldquoThermoelastic stress analysis oforthotropic compositesrdquo Experimental Mechanics vol 35 no 3pp 257ndash265 1995

[20] R S Barsoum ldquoTriangular quarter-point elements as elastic andperfectly-plastic crack tip elementsrdquo International Journal forNumerical Methods in Engineering vol 11 pp 85ndash98 1977

[21] R S Barsoum ldquoOn the use of isoparametric finite elements inlinear fracture mechanicsrdquo International Journal for NumericalMethods in Engineering vol 10 no 1 pp 25ndash37 1976

[22] J R Rice ldquoA path independent integral and approximateanalysis of strain concentration by notches and cracksrdquo Journalof Applied Mechanics vol 35 pp 379ndash386 1968

[23] ANSYS Finite Element Software Code[24] A R Zak and M L Williams ldquoCrack point singularities at a

bimaterial interfacerdquo Journal of Applied Mechanics vol 30 pp142ndash143 1963

[25] W K Wilson and I-W Yu ldquoThe use of the J-integral in thermalstress crack problemsrdquo International Journal of Fracture vol 15no 4 pp 377ndash387 1979

[26] W S Blackburn A D Jackson and T K Hellen ldquoAn integralassociated with the state of a crack tip in a non-elastic materialrdquoInternational Journal of Fracture vol 13 no 2 pp 183ndash199 1977

[27] S Maksimovic ldquoFinite elements in thermoelastic and elasto-plastic fracture mechanicsrdquo in Proceedings of the 3rd Interna-tional Conference Held University Held at College Swansea pp495ndash504 March 1984

[28] J R Rice ldquoElastic fracture mechanics concepts for interfacialcracksrdquo Journal of Applied Mechanics vol 55 no 1 pp 98ndash1031988

[29] S Maksimovic ldquoAn investigation of the effect of thermalgradients on fracturerdquo in Proceedings of the 6th InternationalConference on Fracture (ICF rsquo6) vol 2 pp 4ndash10 PergamonPress New Delhi India December 1984


[30] D Stamenkovic Evaluation Fracture Mechanics Parameters ofThermally Loaded Structures Scientific Technical Review no 22008

[31] D Stamenkovic ldquoDetermination of fracture mechanics param-eters using FEM and J-integral approach finite element sim-ulation of the high risk constructionsrdquo in Proceedings of the2ndWSEAS International Conference onApplied andTheoreticalMechanics (MECHANICS rsquo06) D Mijuca and S MaksimovicEds Venice Italy 2006

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

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Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


plays an increasingly important role in structural design[4ndash6] Recently a more stringent safety criterion assuminga preexisting flaw in critical component has been adoptedto assess service life of aircraft [7ndash10] This emphasizes thesignificance of fracture mechanics as a tool for analysis Theconcepts of linear elastic fracture mechanics which lead tothe strain fracture toughness property 119870

119868119862 have already

been used in engineering applications Mechanical loadingis not the only factor considered in the design of structuresor structural components Other possible situations maybeoperational environments such as temperature have to beconsidered In the operation of gas turbine engines forexample thermal stresses can be as high as or higher than thecentrifugal stresses The worst condition of the combinationsof thermal centrifugal and gas bending stresses at elevatedtemperatures results in high local stresses which can leadto cracking of the LP turbine blades and rotor disks Thusthermal effects should not be ignored A special attentionin this paper is focused to develop analytic expressions forstress intensity factors at low pressure steam turbine discFinite element method (FEM) is used to determine criticallocation zone of the stress concentration at low pressuresteam turbine disc in which rotor blades are connectedwith rotation disc At critical location of rotation disc initialcrack lengths are assumed For various crack lengths at LPsteam turbine disc stress intensity factors are determinedunder thermomechanical loads Using these discrete valuesof SIFs here are derived analytic formulas for SIF 119870

119868 at

cracked LP disc Analytic formula of the stress intensity factorinformation is valuable in the prediction of service lives ofturbine discs Crack growth behavior is a major issue in avariety of rotating components for which analytic formula ofSIF is necessary

The primary objective of this research is to establishcomplete budgetary procedures related to the design of struc-tural elements of low pressure turbine disc in the presenceof initial crack This approach to design in the presence ofinitial cracks is known as the ldquodamage tolerance approachrdquoWith this approach where the initial damage is at the criticallocations of the turbines the risk of unforeseen catastrophicfailures during operation is reduced to a minimum

Low Pressure Turbine Large scale steam turbines employedfor the generation of electricity are expected to operate forworking lives of 30 years and beyond A harsh combination ofservice temperature and environment dictates that the mostseverely stressed components within the low pressure turbinewill need to be formed from what are generally classified ashigh performance materials

In light of such experiences the present study was com-missioned to investigate the fatigue response of a commonLP blade material the X12CrNiMoVNb steel An improvedunderstanding of the mechanical behaviour under variableloading conditions is deemed

Failures occur from time to time in power plants asthey do in other engineering structures However they arenot always examined closely to identify the causes Similarlywhen the failures are observed from time to time repairs areoften made without any careful analysis being undertaken

Most blade failure investigations end with a metallurgicalreport A metallurgical examination of the blade establisheswhether the failure was due to substandardmaterial or due tothe presence of flaws machine marks or corrosion pits thatcreated local stress raisers not accounted for in the allowablestress limits of the blade designA fracture surface of the bladecan suggest that fatigue is amore probable failuremechanismBut the correction of a blade problem requires more thanpositive identification of the mechanisms involved [3]

Finite element analysis needs to be performed becauseof complex geometry and boundary conditions such as theblade So finite element analysis is used to calculate thestresses and modal shapes of the models Previous researcheshave shown that the low pressure blades of a steam turbineare generally found to be more susceptible to failure thanintermediate pressure and high pressure blades [11]Themostcommon failure mechanisms which occur within the lowpressure blade are normally those associated with eithersympathetic or forced vibrations those caused by transientoperating conditions and those that occur as the result of thetransported and accumulated corrosive ions in working fluid[3 12ndash17] Blades and low pressures of steam turbine discs arecritical components in power plants that convert the availableenergy in steam into mechanical energy [1]

2 Strength Analysis with respect toFracture Mechanics

The objective of the analytical work is to provide the rela-tionship between the stress intensity factor and the variousmechanical and thermal load conditions Residual fatiguelife of cracked rotation components is generally based uponcrack growth consideration Fracture mechanics is the math-ematical tool that is employed It provides the concepts andequations used to determine how cracks grow and their effecton the strength of structure From an initial crack length 119886

119900

one must determine critical flow size 119886119888for the fast fracture

damage tolerance approach

119886119900997888rarr 119886119888

119870119868997888rarr 119870

119868119862

(1)

where 119870119868is the applied tensile mode 119868 is the stress intensity

factor and 119870119868119862

is the strain fracture toughness of materialIn fracture mechanics we try to correlate analytical

predictions of crack growth and failure with experimen-tal results The analytical predictions made by calculatingfracture parameters such as stress intensity factors in thecrack region can be used to estimate crack growth rateAlthough several stress intensity factor handbooks [5 18]have published the available solutions those solutions are notalways adequate for particular engineering applications Thisis especially true for cracks subjected to nonuniform stressfields near notch or thermal stresses [6 19] Stress intensityfactor calculation is an important issue for numerical analysisof fracture problems and there exist many approaches for




119868 119870119868119868 and 119870








x2nj

x1

Γ




119870 = radic119864119869

1 minus ]2 (2)


119869 = intΓ

(1198821198891199092minus 120590119894119895

120597119906119894

1205971199091

119899119895119889119878) (3)


119894and 119906

119894are the


1and 119909

2are the local



119870119868= (119870119868)119872+ (119870119868)119879

(4)

when 119870119868= 119870119868119862 in which (119870






120596

304

R 215 R 350 R 460R 515

120601 15

2 Crack

10



119869lowast= intΓ

(119882lowast1198891199092minus 120590119894119895

120597119906119894

1205971199091

119899119895119889119878) +

119864120572

1 minus 2]int 120576119894119894

120597120579

1205971199091

119889119860 (5)


119864120572120579

2 (1 minus 2])120576119894119894 (6)

in which

120590119894119895= 120582120576119894119894120575119894119895+ 2120583120576119894119895minus

119864120572

1 minus 2]120579120575119894119895

119882 =1

2120590119894119895120576119894119895

(7)


119869lowast=(1 minus ]2)1198702

119868

119864 (8)






1Nodal solution

Step = 1

Sub = 2

Time = 1

Temp (avg)

RSYS = 0

SMN =180

SMX = 193

MN

MX

ANSYS 90Al

Dec 30 2008

211825

180

181444

182889

184333

185778

187222

188667

190111

191556

193

Ring 5





Crack front



Crack front




Crack front


0001m


Sub = 1

Time = 1


SMN = 0124Eminus4

SMX = 0006072

ANASYS 121

Jan 7 2014

132447


00013590002032

00027060003379

00040520004726

00053990006072


MXMN





119868119862= 12000MPa(m)12




ANASYS 121Jan 7 2014132659



00010670001159

00012520001344

00014370001529


MN




KI(M

Pa(m)12)

0 0002 0004 0006 0008 0010


0

2

4

6

8

10

12

14

SIF (n = 3000 omin a = 00010m)

SIF (n = 3300 omin a = 00010m)

SIF (n = 3000 omin a = 00015m)

SIF (n = 3300 omin a = 00015m)

SIF (n = 3000 omin a = 00020m)

SIF (n = 3300 omin a = 00020m)

SIF (n = 3000 omin a = 00025m)

SIF (n = 3300 omin a = 00025m)

SIF (n = 3000 omin a = 00030m)

SIF (n = 3300 omin a = 00030m)




1198701198681= 23746666667119886

3minus 178131429119886

2+ 517289119886 + 400

(9)

1198701198682= 24046666667119886

3minus 185431429119886

2+ 561644119886 + 502

(10)







119868= 1165MPa(m)12 Table 1



RF =119870119868119862

119870119868

(11)



119870119868[MPa(m)12]

Crack lengths [m]

119886 = 00010m


119899 = 3000 73373 76519 76059 75651 74727 72176119886 = 00010m 119899 = 3300 86310 89799 89979 89631 88407 85053119886 = 00015m 119899 = 3000 82728 86502 86411 86006 85149 81400119886 = 00015m 119899 = 3300 97197 101864 101855 101434 100165 95799119886 = 00020m 119899 = 3000 85604 90005 89857 89428 88593 84211119886 = 00020m 119899 = 3300 100724 106150 106085 105644 104798 99285119886 = 00025m 119899 = 3000 90887 95932 95977 95547 94731 89392119886 = 00025m 119899 = 3300 106963 113151 113304 112683 112036 105392119886 = 00030m 119899 = 3000 93531 95389 98753 98345 97341 91821119886 = 00030m 119899 = 3300 108985 116519 116484 116067 115023 108108


775

885

99510

10511

11512

00005 0001 00015 0002 00025 0003 00035 0004 00045 0005



KI2 = 24046666667a3minus 185431429a

2+ 561644a + 502

KI1 = 23746666667a3minus 178131429a2 + 517289a + 400




119889119886

119889119873= 119862(Δ119870)

119898[1 minus (Δ119870

119905ℎΔ119870)1198991

1 minus (Δ119870max119870119888)]

1198993

(12)







119905rem =119886cr minus 119886

119894

119889119886119889119905 (13)


4 Conclusions






Acknowledgments


References








































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119868 119870119868119868 and 119870








x2nj

x1

Γ




119870 = radic119864119869

1 minus ]2 (2)


119869 = intΓ

(1198821198891199092minus 120590119894119895

120597119906119894

1205971199091

119899119895119889119878) (3)


119894and 119906

119894are the


1and 119909

2are the local



119870119868= (119870119868)119872+ (119870119868)119879

(4)

when 119870119868= 119870119868119862 in which (119870






120596

304

R 215 R 350 R 460R 515

120601 15

2 Crack

10



119869lowast= intΓ

(119882lowast1198891199092minus 120590119894119895

120597119906119894

1205971199091

119899119895119889119878) +

119864120572

1 minus 2]int 120576119894119894

120597120579

1205971199091

119889119860 (5)


119864120572120579

2 (1 minus 2])120576119894119894 (6)

in which

120590119894119895= 120582120576119894119894120575119894119895+ 2120583120576119894119895minus

119864120572

1 minus 2]120579120575119894119895

119882 =1

2120590119894119895120576119894119895

(7)


119869lowast=(1 minus ]2)1198702

119868

119864 (8)






1Nodal solution

Step = 1

Sub = 2

Time = 1

Temp (avg)

RSYS = 0

SMN =180

SMX = 193

MN

MX

ANSYS 90Al

Dec 30 2008

211825

180

181444

182889

184333

185778

187222

188667

190111

191556

193

Ring 5





Crack front



Crack front




Crack front


0001m


Sub = 1

Time = 1


SMN = 0124Eminus4

SMX = 0006072

ANASYS 121

Jan 7 2014

132447


00013590002032

00027060003379

00040520004726

00053990006072


MXMN





119868119862= 12000MPa(m)12




ANASYS 121Jan 7 2014132659



00010670001159

00012520001344

00014370001529


MN




KI(M

Pa(m)12)

0 0002 0004 0006 0008 0010


0

2

4

6

8

10

12

14

SIF (n = 3000 omin a = 00010m)

SIF (n = 3300 omin a = 00010m)

SIF (n = 3000 omin a = 00015m)

SIF (n = 3300 omin a = 00015m)

SIF (n = 3000 omin a = 00020m)

SIF (n = 3300 omin a = 00020m)

SIF (n = 3000 omin a = 00025m)

SIF (n = 3300 omin a = 00025m)

SIF (n = 3000 omin a = 00030m)

SIF (n = 3300 omin a = 00030m)




1198701198681= 23746666667119886

3minus 178131429119886

2+ 517289119886 + 400

(9)

1198701198682= 24046666667119886

3minus 185431429119886

2+ 561644119886 + 502

(10)







119868= 1165MPa(m)12 Table 1



RF =119870119868119862

119870119868

(11)



119870119868[MPa(m)12]

Crack lengths [m]

119886 = 00010m


119899 = 3000 73373 76519 76059 75651 74727 72176119886 = 00010m 119899 = 3300 86310 89799 89979 89631 88407 85053119886 = 00015m 119899 = 3000 82728 86502 86411 86006 85149 81400119886 = 00015m 119899 = 3300 97197 101864 101855 101434 100165 95799119886 = 00020m 119899 = 3000 85604 90005 89857 89428 88593 84211119886 = 00020m 119899 = 3300 100724 106150 106085 105644 104798 99285119886 = 00025m 119899 = 3000 90887 95932 95977 95547 94731 89392119886 = 00025m 119899 = 3300 106963 113151 113304 112683 112036 105392119886 = 00030m 119899 = 3000 93531 95389 98753 98345 97341 91821119886 = 00030m 119899 = 3300 108985 116519 116484 116067 115023 108108


775

885

99510

10511

11512

00005 0001 00015 0002 00025 0003 00035 0004 00045 0005



KI2 = 24046666667a3minus 185431429a

2+ 561644a + 502

KI1 = 23746666667a3minus 178131429a2 + 517289a + 400




119889119886

119889119873= 119862(Δ119870)

119898[1 minus (Δ119870

119905ℎΔ119870)1198991

1 minus (Δ119870max119870119888)]

1198993

(12)







119905rem =119886cr minus 119886

119894

119889119886119889119905 (13)


4 Conclusions






Acknowledgments


References








































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










120596

304

R 215 R 350 R 460R 515

120601 15

2 Crack

10



119869lowast= intΓ

(119882lowast1198891199092minus 120590119894119895

120597119906119894

1205971199091

119899119895119889119878) +

119864120572

1 minus 2]int 120576119894119894

120597120579

1205971199091

119889119860 (5)


119864120572120579

2 (1 minus 2])120576119894119894 (6)

in which

120590119894119895= 120582120576119894119894120575119894119895+ 2120583120576119894119895minus

119864120572

1 minus 2]120579120575119894119895

119882 =1

2120590119894119895120576119894119895

(7)


119869lowast=(1 minus ]2)1198702

119868

119864 (8)






1Nodal solution

Step = 1

Sub = 2

Time = 1

Temp (avg)

RSYS = 0

SMN =180

SMX = 193

MN

MX

ANSYS 90Al

Dec 30 2008

211825

180

181444

182889

184333

185778

187222

188667

190111

191556

193

Ring 5





Crack front



Crack front




Crack front


0001m


Sub = 1

Time = 1


SMN = 0124Eminus4

SMX = 0006072

ANASYS 121

Jan 7 2014

132447


00013590002032

00027060003379

00040520004726

00053990006072


MXMN





119868119862= 12000MPa(m)12




ANASYS 121Jan 7 2014132659



00010670001159

00012520001344

00014370001529


MN




KI(M

Pa(m)12)

0 0002 0004 0006 0008 0010


0

2

4

6

8

10

12

14

SIF (n = 3000 omin a = 00010m)

SIF (n = 3300 omin a = 00010m)

SIF (n = 3000 omin a = 00015m)

SIF (n = 3300 omin a = 00015m)

SIF (n = 3000 omin a = 00020m)

SIF (n = 3300 omin a = 00020m)

SIF (n = 3000 omin a = 00025m)

SIF (n = 3300 omin a = 00025m)

SIF (n = 3000 omin a = 00030m)

SIF (n = 3300 omin a = 00030m)




1198701198681= 23746666667119886

3minus 178131429119886

2+ 517289119886 + 400

(9)

1198701198682= 24046666667119886

3minus 185431429119886

2+ 561644119886 + 502

(10)







119868= 1165MPa(m)12 Table 1



RF =119870119868119862

119870119868

(11)



119870119868[MPa(m)12]

Crack lengths [m]

119886 = 00010m


119899 = 3000 73373 76519 76059 75651 74727 72176119886 = 00010m 119899 = 3300 86310 89799 89979 89631 88407 85053119886 = 00015m 119899 = 3000 82728 86502 86411 86006 85149 81400119886 = 00015m 119899 = 3300 97197 101864 101855 101434 100165 95799119886 = 00020m 119899 = 3000 85604 90005 89857 89428 88593 84211119886 = 00020m 119899 = 3300 100724 106150 106085 105644 104798 99285119886 = 00025m 119899 = 3000 90887 95932 95977 95547 94731 89392119886 = 00025m 119899 = 3300 106963 113151 113304 112683 112036 105392119886 = 00030m 119899 = 3000 93531 95389 98753 98345 97341 91821119886 = 00030m 119899 = 3300 108985 116519 116484 116067 115023 108108


775

885

99510

10511

11512

00005 0001 00015 0002 00025 0003 00035 0004 00045 0005



KI2 = 24046666667a3minus 185431429a

2+ 561644a + 502

KI1 = 23746666667a3minus 178131429a2 + 517289a + 400




119889119886

119889119873= 119862(Δ119870)

119898[1 minus (Δ119870

119905ℎΔ119870)1198991

1 minus (Δ119870max119870119888)]

1198993

(12)







119905rem =119886cr minus 119886

119894

119889119886119889119905 (13)


4 Conclusions






Acknowledgments


References








































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Crack front



Crack front




Crack front


0001m


Sub = 1

Time = 1


SMN = 0124Eminus4

SMX = 0006072

ANASYS 121

Jan 7 2014

132447


00013590002032

00027060003379

00040520004726

00053990006072


MXMN





119868119862= 12000MPa(m)12




ANASYS 121Jan 7 2014132659



00010670001159

00012520001344

00014370001529


MN




KI(M

Pa(m)12)

0 0002 0004 0006 0008 0010


0

2

4

6

8

10

12

14

SIF (n = 3000 omin a = 00010m)

SIF (n = 3300 omin a = 00010m)

SIF (n = 3000 omin a = 00015m)

SIF (n = 3300 omin a = 00015m)

SIF (n = 3000 omin a = 00020m)

SIF (n = 3300 omin a = 00020m)

SIF (n = 3000 omin a = 00025m)

SIF (n = 3300 omin a = 00025m)

SIF (n = 3000 omin a = 00030m)

SIF (n = 3300 omin a = 00030m)




1198701198681= 23746666667119886

3minus 178131429119886

2+ 517289119886 + 400

(9)

1198701198682= 24046666667119886

3minus 185431429119886

2+ 561644119886 + 502

(10)







119868= 1165MPa(m)12 Table 1



RF =119870119868119862

119870119868

(11)



119870119868[MPa(m)12]

Crack lengths [m]

119886 = 00010m


119899 = 3000 73373 76519 76059 75651 74727 72176119886 = 00010m 119899 = 3300 86310 89799 89979 89631 88407 85053119886 = 00015m 119899 = 3000 82728 86502 86411 86006 85149 81400119886 = 00015m 119899 = 3300 97197 101864 101855 101434 100165 95799119886 = 00020m 119899 = 3000 85604 90005 89857 89428 88593 84211119886 = 00020m 119899 = 3300 100724 106150 106085 105644 104798 99285119886 = 00025m 119899 = 3000 90887 95932 95977 95547 94731 89392119886 = 00025m 119899 = 3300 106963 113151 113304 112683 112036 105392119886 = 00030m 119899 = 3000 93531 95389 98753 98345 97341 91821119886 = 00030m 119899 = 3300 108985 116519 116484 116067 115023 108108


775

885

99510

10511

11512

00005 0001 00015 0002 00025 0003 00035 0004 00045 0005



KI2 = 24046666667a3minus 185431429a

2+ 561644a + 502

KI1 = 23746666667a3minus 178131429a2 + 517289a + 400




119889119886

119889119873= 119862(Δ119870)

119898[1 minus (Δ119870

119905ℎΔ119870)1198991

1 minus (Δ119870max119870119888)]

1198993

(12)







119905rem =119886cr minus 119886

119894

119889119886119889119905 (13)


4 Conclusions






Acknowledgments


References








































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











ANASYS 121Jan 7 2014132659



00010670001159

00012520001344

00014370001529


MN




KI(M

Pa(m)12)

0 0002 0004 0006 0008 0010


0

2

4

6

8

10

12

14

SIF (n = 3000 omin a = 00010m)

SIF (n = 3300 omin a = 00010m)

SIF (n = 3000 omin a = 00015m)

SIF (n = 3300 omin a = 00015m)

SIF (n = 3000 omin a = 00020m)

SIF (n = 3300 omin a = 00020m)

SIF (n = 3000 omin a = 00025m)

SIF (n = 3300 omin a = 00025m)

SIF (n = 3000 omin a = 00030m)

SIF (n = 3300 omin a = 00030m)




1198701198681= 23746666667119886

3minus 178131429119886

2+ 517289119886 + 400

(9)

1198701198682= 24046666667119886

3minus 185431429119886

2+ 561644119886 + 502

(10)







119868= 1165MPa(m)12 Table 1



RF =119870119868119862

119870119868

(11)



119870119868[MPa(m)12]

Crack lengths [m]

119886 = 00010m


119899 = 3000 73373 76519 76059 75651 74727 72176119886 = 00010m 119899 = 3300 86310 89799 89979 89631 88407 85053119886 = 00015m 119899 = 3000 82728 86502 86411 86006 85149 81400119886 = 00015m 119899 = 3300 97197 101864 101855 101434 100165 95799119886 = 00020m 119899 = 3000 85604 90005 89857 89428 88593 84211119886 = 00020m 119899 = 3300 100724 106150 106085 105644 104798 99285119886 = 00025m 119899 = 3000 90887 95932 95977 95547 94731 89392119886 = 00025m 119899 = 3300 106963 113151 113304 112683 112036 105392119886 = 00030m 119899 = 3000 93531 95389 98753 98345 97341 91821119886 = 00030m 119899 = 3300 108985 116519 116484 116067 115023 108108


775

885

99510

10511

11512

00005 0001 00015 0002 00025 0003 00035 0004 00045 0005



KI2 = 24046666667a3minus 185431429a

2+ 561644a + 502

KI1 = 23746666667a3minus 178131429a2 + 517289a + 400




119889119886

119889119873= 119862(Δ119870)

119898[1 minus (Δ119870

119905ℎΔ119870)1198991

1 minus (Δ119870max119870119888)]

1198993

(12)







119905rem =119886cr minus 119886

119894

119889119886119889119905 (13)


4 Conclusions






Acknowledgments


References








































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119870119868[MPa(m)12]

Crack lengths [m]

119886 = 00010m


119899 = 3000 73373 76519 76059 75651 74727 72176119886 = 00010m 119899 = 3300 86310 89799 89979 89631 88407 85053119886 = 00015m 119899 = 3000 82728 86502 86411 86006 85149 81400119886 = 00015m 119899 = 3300 97197 101864 101855 101434 100165 95799119886 = 00020m 119899 = 3000 85604 90005 89857 89428 88593 84211119886 = 00020m 119899 = 3300 100724 106150 106085 105644 104798 99285119886 = 00025m 119899 = 3000 90887 95932 95977 95547 94731 89392119886 = 00025m 119899 = 3300 106963 113151 113304 112683 112036 105392119886 = 00030m 119899 = 3000 93531 95389 98753 98345 97341 91821119886 = 00030m 119899 = 3300 108985 116519 116484 116067 115023 108108


775

885

99510

10511

11512

00005 0001 00015 0002 00025 0003 00035 0004 00045 0005



KI2 = 24046666667a3minus 185431429a

2+ 561644a + 502

KI1 = 23746666667a3minus 178131429a2 + 517289a + 400




119889119886

119889119873= 119862(Δ119870)

119898[1 minus (Δ119870

119905ℎΔ119870)1198991

1 minus (Δ119870max119870119888)]

1198993

(12)







119905rem =119886cr minus 119886

119894

119889119886119889119905 (13)


4 Conclusions






Acknowledgments


References








































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













Acknowledgments


References








































Volume 2014




Journal of











Journal of


Function Spaces






Algebra



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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