Coupled Finite Volume Methods and Extended Finite Element ... · ResearchArticle Coupled Finite...

Research ArticleCoupled Finite Volume Methods and ExtendedFinite Element Methods for the Dynamic Crack PropagationModelling with the Pressurized Crack Surfaces

Shouyan Jiang and Chengbin Du

Department of Engineering Mechanics Hohai University Nanjing 210098 China

Correspondence should be addressed to Shouyan Jiang syjianghhueducn and Chengbin Du cbduhhueducn

Received 25 July 2017 Revised 7 October 2017 Accepted 11 October 2017 Published 28 November 2017

Academic Editor Mickael Lallart

Copyright copy 2017 Shouyan Jiang and Chengbin Du This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

Wemodel the fluid flowwithin the crack as one-dimensional flowand assume that the flow is laminar the fluid is incompressible andaccounts for the time-dependent rate of crack openingHerewe discretise the flowequation byfinite volumemethodsThe extendedfinite element methods are used for solving solid medium with crack under dynamic loads Having constructed the approximationof dynamic extended finite element methods the derivation of governing equation for dynamic extended finite element methods ispresented The implicit time algorithm is elaborated for the time descritisation of dominant equation In addition the interactionintegral method is given for evaluating stress intensity factors Then the coupling model for modelling hydraulic fracture can beestablished by the extended finite element methods and the finite volume methods We compare our present numerical resultswith our experimental results for verifying the proposed model Finally we investigate the water pressure distribution along cracksurface and the effect of water pressure distribution on the fracture property

1 Introduction

For hydraulic concrete structures the external dynamicloads such as strong earthquake may cause cracking ofthese structures The cracking of the structure causes thefluid injecting into the solid medium The injecting fluidproduces fluid pressures along crack surface and affectsthe deformation of the solid medium and the fracturepropagation again Many researchers [1ndash4] contributed tothe study of hydraulic fracturing problem and these effortsled to a progressive recognition of the multiscale nature ofthe hydraulic fracturing problem In simulation of hydraulicfracturing these important aspects need to be speciallyconcerned namely the flow of viscous fracturing fluid thecreation of fracture surfaces in the solid the formation ofa lag between the crack edge and the fluid front the elasticdeformation of the solid and the leak-off of fluid from thefracture

Some researches (2005) [5] showed that in hydraulicstructures the water could penetrate an initiated crack and

the crack opening velocity the magnitude of the openingand crack mouth pressure had an important effect on thewater pressure distribution Boone and Ingraffea (1990) [6]proposed a finite difference approximation for modellingfluid flow along the fracture Wu and Wong (2014) [7] incor-porated the cubic law into the numerical manifold methodfor modelling fluid flow through fractures Lisjak et al (2017)[8] assumed that the fluid flow in discontinuous porousrock masses was a viscous compressible fluid and the flowwas explicitly solved based on a cubic law approximationThe fluid flow along a propagating crack surface satisfiessome natural flow law Some hypotheses [9ndash12] such aslinear distribution of the water pressure along crack case fullreservoir pressure case for evaluating water pressure along apropagating crack cannot reflect accurately the variation ofthe water pressure along new developing cracks in structures

In recent years many numerical methods have beendeveloped for hydraulic fracturing modeling such as finiteelement methods [13 14] generalized finite element methods[15] finite-discrete element methods [8] numerical manifold

HindawiShock and VibrationVolume 2017 Article ID 3751340 16 pageshttpsdoiorg10115520173751340

2 Shock and Vibration

methods [7] boundary element methods [16] discontinuousdeformation analysis methods [17] and extended finite ele-ment methods (XFEM) The XFEM shows huge advantagefor dealing with discontinuous problems [18ndash20] and alsohydraulic fracture problems The XFEMmesh does not needto align with a discontinuity For moving discontinuitiessuch as crack propagation problem it does not need to carryon remeshing Mesh refinement is also unnecessary arounda discontinuous feature The first simulation of hydraulicfracture in XFEM was due to Rethore et al [21] and theydeveloped a two-scale numerical model for fluid flow infractured deforming porous media In 2009 Lecampion [22]adopted the XFEM for investigating the solution of hydraulicfracture problems Gordeliy and Peirce (2013) [23] proposedcoupled algorithms that used the XFEM to solve the elasticcrack component of the elastohydrodynamic equations thatgoverned the propagation of hydraulic fractures in an elasticmedium Subsequently they (2013) [24] proposed two novelXFEM schemes for modeling fluid driven fractures both ofwhich exploited an implicit level set algorithm for locatingthe singular free boundary that occurred when the fluidand fracture fronts coalesce Their excellent works providemathematical proof for using XFEM to solve hydraulicfracturing problems Khoei et al [25] simulated the crackgrowth in saturated porous media using XFEM Taleghani[26] also developed an XFEM code to simulate fracturepropagation initiation and intersection and the presentedcoupled fluid flow-fracture mechanics simulations extendedavailable modeling efforts and provided a unified frameworkfor evaluating fracture design parameters and their con-sequences Salimzadeh and Khalili (2015) [27] proposed athree-phase hydromechanical model for hydraulic fracturingand they handled discontinuity by using XFEM while cohe-sive crack model was used as fracturing criterion Wang etal (2015) [28] proposed a hybrid approach combining theXFEM and the finite volume method to simulate hydraulicfracturing in concrete dams Our current study concernsdeveloping a model for cracking modeling of structureunder water pressure along a propagating crack surface anddynamic loads Additionally we will compare our presentnumerical results with our experimental results for verifyingthe proposed model

This paper is organized as follows Section 2 introducessome governing equations for elastic dynamic responses ofthe solid medium and fluid flow pressure within the crackSection 3 discusses numerical approximation of extendedfinite element methods and finite volume methods of theflow along a crack Section 4 gives a numerical examplefor investigating the water pressure distribution along cracksurface and the effect of water pressure distribution on thefracture property We also compare our present numericalresults with our experimental results Section 5 summarisesthe major conclusions that can be drawn from this study

2 Governing Equations

21 Elastic Dynamic Responses of the Solid Medium Theboundary of a bounded domain Ω isin 1198772 is partitionedinto three parts the displacement boundary (Γu) the traction

boundary (Γt) and the crack boundary (Γc) that is traction-free The elastodynamic basic equation is expressed asnabla sdot 120590 + b = 120588u in Ω

120576 = nablasu in Ω120590 = D 120576 in Ω (1)

with the following boundary and initial conditions

u (x 119905) = u (x 119905) on Γt120590 sdot n = t on Γu120590 sdot n = 0 on Γc

u (x 119905 = 0) = u (0) u (x 119905 = 0) = u (0)

(2)

where 120590 is the Cauchy stress tensor b is the body forcevector 120576 is the strain tensor 120588 is the material density u isthe acceleration field vector nablas is the symmetric part of thegradient operator u is the displacement field vector D isthe constitutive matrix n is the unit outward normal vectorto the crack surface u is the prescribed displacement t isthe external traction vector u(0) is the initial displacementvector and u(0) is the initial velocity vector22 Fluid Flow Pressure within the Crack In this paper wemodel the fluid flow within the crack as one-dimensionalflow Assume that the flow is laminar and the fluid isincompressible But here we account for the time-dependentrate of crack opening

The conservation of the incompressible fluid in thefracture can be expressed as [6]nabla sdot q + + 119892 = 0 (3)

wherenabla is the divergence operator defined in x direction q isthe fluid flux is the time-dependent rate of crack openingand = 120597119908120597119905 and 119892 is the fluid loss into the solid mediaand here we ignore the fluid loss that is 119892 = 0

Additionally Poiseuillersquos law [29] gives the followingexpression

q = minus 119908312120583nabla119901 (4)

where119908 is the crack opening 120583 is the fluid viscosity and 119901 isthe fluid pressure

The pressure boundary conditions at the fluid injectionpoint in the crack are 119901 (119909 = 0 119905) = 1199010 (5)

where 1199010 is the pressure of the fluid injection pointIn the fluid lag zone [30]119901 (119909 119905) lt 0 (119897t le 119909 le 119897) (6)

According to the fluid pressure continuity lag condition (6)provides the net-pressure boundary condition at the fluidfront 119909 = 119897t for the fluid flow equations (3)

Shock and Vibration 3

Crack

Heaviside enriched element

Crack tip enriched element

Ilowast＜Ｍ

Ilowast＜Ｌ

Figure 1 A strategy for enriched elements and nodes

3 Numerical Approximation

31 XFEM for Dynamic Problems

311 XFEM Approximation The XFEM approximation for2D cracked domains can be written as

uh (x) = sum119894isin119868

119873119894 (x) u119894 + sum119894isin119868lowastabs

119873lowast119894 (x) [119867 (x) minus 119867 (x119894)] a119894+ sum119894isin119868lowastbr

119873lowast119894 (x) 4sum119895=1

[119865119895 (x) minus 119865119895 (x119894)] b119895119894 (7)

where 119873119894(x) is the standard finite element shape function ofnode 119894 u119894 is the unknown of the standard finite element partat node 119894 119868 is the set of all nodes in the domain and119873lowast119894 (x) isthe partition of unity functions and the function can hold thesame form with the standard finite element shape functionbut is not necessary a119894 and b119895119894 are the nodal enriched degreeof freedom 119868lowastabs and 119868lowastbr are the set of enrichment nodes shownin Figure 1 and 119868lowastabs 119868lowastbr sub 119868

For these elements that are cut completely by a crackthe nodes of these elements that are the nodal subset 119868lowastabsare enriched by Heaviside function 119867(x) The definition ofHeaviside function119867(x) follows

119867(x) = +1 (x minus xlowast) sdot n gt 0minus1 (x minus xlowast) sdot n lt 0 (8)

where xlowast is the projection of a point x on the crack surface nis the unit outward normal to the crack surface

For these elements that are cut partially by a crack thenodes of these elements that are the nodal subset 119868lowastbr are

enriched by the crack tip enrichment function 119865119895(x) Thedefinition of the crack tip enrichment function 119865119895(x) follows119865119895=1234 (119903 120579)= radic119903 sin 1205792 radic119903 cos 1205792 radic119903 sin 120579 sin 1205792 radic119903 sin 120579 cos 1205792 (9)

where 119903 and 120579 are the local crack tip coordinate system

312 Discrete Equations By the principle of virtual work thefollowing discrete equations can be obtained

Muh + Kuh = f (10)

where K (M) is the global stiffness (mass) matrix assembledby the element stiffness (mass) matrix f is the global externalforce vector uh and uh denote the vector of nodal parameters(which include the classic degrees of freedom u and theenrichment degrees of freedom a b) and its second timederivative respectively and

uh = u a bT uh = u a bT (11)

The element stiffness matrix is expressed by

ke = [[[[kuu kua kub

kau kaa kab

kbu kba kbb

]]]] (12)


where

k119903119904 = intΩe(B119903)T DB119904dΩ

= intΩe[B1199031 B1199032 B1199033 B1199034]T D [B1199041 B1199042 B1199043 B1199044] dΩ

(119903 119904 = u a b) Bu119894 = [[[[

120597119873119894120597119909 0 1205971198731198941205971199100 120597119873119894120597119910 120597119873119894120597119909]]]]T

119894 = 1 2 3 4

Ba119894 = [[[[

120597 (119873lowast119894 1006704119867)120597119909 0 120597 (119873lowast119894 1006704119867)1205971199100 120597 (119873lowast119894 1006704119867)120597119910 120597 (119873lowast119894 1006704119867)120597119909]]]]T

1006704119867 = 119867 (x) minus 119867 (x119894) 119894 = 1 2 3 4

Bb119894 = [Bb1

119894 Bb2119894 Bb3119894 Bb4119894 ]

Bb119895119894 = [[[[[

120597 (119873lowast119894 1006704119865119895)120597119909 0 120597 (119873lowast119894 1006704119865119895)1205971199100 120597 (119873lowast119894 1006704119865119895)120597119910 120597 (119873lowast119894 1006704119865119895)120597119909

]]]]]

T

1006704119865119895 = 119865119895 (x) minus 119865119895 (x119894) 119894 119895 = 1 2 3 4

(13)

The element mass matrix is expressed by

me = [[[[muu mua mub

mau maa mab

mbu mba mbb

]]]] (14)

where

muu = intΩe120588 (N)T N 119889Ω

mua = [mau]T = intΩe120588 (N)T (N1006704119867)119889Ω

mub119895 = [mbu

119895 ]T = intΩe120588 (N)T (N 1006704119865119895) 119889Ω

maa = intΩe120588 (N1006704119867)T (N1006704119867)119889Ω

mab119895 = [mba

119895 ]T = intΩe120588 (N1006704119867)T (N 1006704119865119895) 119889Ω

mbb119895119896 = int

Ωe120588 (N 1006704119865119895)T (N 1006704119865119896) 119889Ω

119895 119896 = 1 2 3 4

N = [N1 N2 N3 N4] N119894 = [119873119894 00 119873119894] 119894 = 1 2 3 4

(15)

The element external force vector is

fe = [fu fa fb]T (16)

where

fu = intΩe

NTb dΩ + intΓet

NTt dΓfa = int

Ωe(Nlowast1006704H)T b dΩ + int

Γet

(Nlowast1006704H)T t dΓ+ intΓec

(Nlowast1006704H)T (p sdot n) dΓfb119895 = int

Ωe(Nlowast1006704F119895)T b dΩ + int

Γet

(Nlowast1006704F119895)T t dΓ+ intΓec

(Nlowast1006704F119895)T (p sdot n) dΓ 119895 = 1 2 3 4

(17)

313 Time Integration Schemes The following time-integra-tion scheme is used in dynamic analysis Equation (10) for aspecific time 119905 + Δ119905 is expressed as

Mu119905+Δ119905 + Ku119905+Δ119905 = f119905+Δ119905u119905+Δ119905 = u119905 + (1 minus 120574) u119905Δ119905 + 120574u119905+Δ119905Δ119905u119905+Δ119905 = u119905 + u119905Δ119905 + (12 minus 120573) u119905Δ1199052+ 120573u119905+Δ119905Δ1199052

(18)

where u119905 u119905 and u119905 are the displacement velocity andacceleration vectors at time 119905 respectivelyΔ119905 is the time step120574 and 120573 are parameters that can be determined to obtainintegration accuracy and stability with

120573 = 14 (1 minus 120572)2 120574 = 12 minus 120572

minus13 le 120572 le 0(19)

Here referring to the software ABAQUS (ABAQUSTheory Manual Version 69) we set parameter 120572 = minus005to remove the slight high frequency noise in the solutionwithout having any significant effect on themeaningful lowerfrequency response

The following steps describe the prescribe integrationmethod procedure while neglecting the damping effects


Crack CrackCrack

Crack

(a) Elements partitioned completely by a straight crack

Crack Crack

Crack

(b) Elements partitioned completely by a broken crack

Crack

Crack

(c) Elements partitioned partially by astraight crack

Figure 2 Element partitioning method for these elements containing a discontinuous interface

(I) Initial Calculations

(i) Form stiffness matrix K and mass matrixM(ii) Give the initial displacement vector u0 and the initial

velocity vector u0 Then calculate the initial accelera-tion vector u0 by the equilibrium equation

Mu0 + Ku0 = f0 (20)

(iii) Select a time step Δ119905 and the parameters 120573 and 120574Here 120573 = 0275625 and 120574 = 055 are used Calculateintegration constants

1198880 = 1120573Δ1199052 1198881 = 120574120573Δ119905 1198882 = 1120573Δ119905 1198883 = 12120573 minus 11198884 = 120574120573 minus 11198885 = Δ119905 ( 1205742120573 minus 1) 1198886 = Δ119905 (1 minus 120574) 1198887 = 120574Δ119905

(21)

(iv) Form the effective stiffness matrix K

K = K + 1198880M (22)

(II) For Each Time Step

(i) Calculate effective loads f119905+Δ119905 at time 119905 + Δ119905f119905+Δ119905 = f119905+Δ119905 +M (1198880u119905 + 1198882u119905 + 1198883u119905) (23)

(ii) Solve for the displacement vector u119905+Δ119905 at time 119905 + Δ119905Ku119905+Δ119905 = f119905+Δ119905 (24)

(iii) Calculate the acceleration vector u119905+Δ119905 and the veloc-ity vector u119905+Δ119905 at time 119905 + Δ119905

u119905+Δ119905 = 1198880 (u119905+Δ119905 minus u119905) minus 1198882u119905 minus 1198883u119905u119905+Δ119905 = u119905 + 1198886u119905 + 1198887u119905+Δ119905 (25)

314 Integration Schemes at theDiscontinuities For these ele-ments partitioned by a crack the ordinary Gauss quadraturerules cannot accurately calculate the integration of enrich-ment function An alternative method that is dividing theenrichment element into a set of subpolygons usually needsto be used [31] Additionally some simplified numericalintegration methods also had been proposed in literaturesVentura [32] conducted an important first attempt to sim-plify numerical integration His work is based on replac-ing nonpolynomial functions by ldquoequivalentrdquo polynomialsHowever the proposed method is exact for triangular andtetrahedral elements but for quadrilateral elements whenthe opposite sides are not parallel additional approximationis introduced Another method is strain smoothing [33]In strain smoothing the surface integration is transformedinto equivalent boundary integration by use of the Green-Ostrogradsky theorem Natarajan et al [34] used the newnumerical integration proposed for arbitrary polygons [35] tointegrate the discontinuous and singular integrands appear-ing in the XFEM stiffness matrix In this paper the methodsubdividing the element into subquads is used For theseelements partitioned completely or partially by a crack themethod subdividing these elements into subquads is shownin Figure 2

To solve the element stiffness or mass matrix of theseenrichment elements each subquad element is respectivelytransferred into the standard element (minus1 1) times (minus1 1) by themethod of the coordinate transformationTheGauss integra-tion points are distributed into each subquad To improve theaccuracy of crack tip integration 15 Gauss integration points


1

Crack

0 00 0

00 11 0 0

10 1A

1 1 0

10 1 1 0

00 11 1 0

0 00 0 0

Γd+

Γdminus

Figure 3 Elements selection for the interaction integral near thecrack tip

are distributed into each subquad for elements containingcrack tip The numerical integration is firstly performed ineach subquad element domain and then the element stiffnessor mass matrix of the enrichment element can be obtained byassembling the numerical integration results of each subquadelement It is worthwhile pointing out that these subquads areonly necessary for integration purposes They do not provideadditional degrees of freedom for the global stiffness andmass matrix

315 Interaction Integral for Computing Stress Intensity Fac-tors Take field 1 (120590(1)119894119895 120576(1)119894119895 119906(1)119894 ) for the actual field andfield 2 (120590(2)119894119895 120576(2)119894119895 119906(2)119894 ) for the auxiliary field The actual fieldis obtained from numerical solutions computed by usingXFEM and the auxiliary field refers to the asymptotic resultsof linear elastic fracture dynamics [36] The interactionintegral equation which is used to evaluate the stress intensityfactors is as follows

119868(12) = intA(120590(1)119894119895 119906(2)1198941 + 120590(2)119894119895 119906(1)1198941 minus 12120590(1)119894119896 120576(2)119894119896 1205751119895) 119902119895119889Ω

+ intΓd+cupΓdminus

(12120590(1)119894119896 120576(2)119894119896 1205751119895 minus 120590(1)119894119895 119906(2)1198941 minus 120590(2)119894119895 119906(1)1198941 )sdot 119902119899119895dΓ

(26)

The second term in (26) denotes the contribution oftraction along crack interface As shown in Figure 3 Adenotes the circle domain with centre at the crack tip and theradius 119877 Γd+ cup Γdminus consists of a interface starting from theexternal integration radius to crack tip in a two-way manner119877 is defined as 119877 = 119903kℎe (27)

where ℎe is the crack-tip element size 119903k is a user-specifiedscalar multiple 119902 is the weight function 119902 = 1 if the node liesin A and 119902 = 0 if the node lies outside of A or lies on the

boundary of A The weight function 119902 in the interior of anelement is obtained by the interpolation of the nodal value

119902 = 4sum119894=1

119873119894119902119894 (28)

Additionally the interaction integral relates to the stressintensity factors through the relation

119872(12) = 2119864lowast [119870I119870auxI + 119870II119870aux

II ] (29)

where 119870auxI and 119870aux

II are the local auxiliary stress intensityfactors for the auxiliary fields respectively and the definitionof 119864lowast is

119864lowast = 119864 (plane stress)1198641 minus ]2

(plane strain) (30)

By setting 119870auxI = 1 and 119870aux

II = 0 as well as 119872(12) =119872(12)1 we obtain the expression of119870I as follows

119870I = 119864lowast119872(12)12 (31)

Similarly we obtain the equality

119870II = 119864lowast119872(12)22 (32)

by setting119870auxI = 0 and119870aux

II = 1 and119872(12) = 119872(12)2

316 Crack Propagation Criteria The maximum circumfer-ential stress criterion [37] is used to determine the crackgrowth direction Once 119870I and 119870II are calculated the crite-rion gives the following crack growth direction

120579c = 2 tanminus1 14 ( 119870I119870IIplusmn radic( 119870I119870II

)2 + 8) (33)

where 120579c is the crack growth angle in the local crack-tipcoordinate system If 119870II = 0 then 120579c = 0 It should alsobe noted that if 119870II gt 0 the crack growth angle 120579c lt 0 andif 119870II lt 0 then 120579c gt 0 Sukumar and Prevost (2003) [38]proposed a computationally more amenable expression for120579c

120579c = 2 tanminus1( minus2119870II119870I1 + radic1 + 8 (119870II119870I)2) (34)

The equivalent stress intensity factor then follows

119870e = cos120579c2 (119870Icos

2 120579c2 minus 15119870II sin 120579c) (35)

The double-K criterion [39] is used for determiningcrack propagation The crack propagation processes can beexpressed as follows


Control volume

2 3

Nodes

Q

L1 L2

ewP

p0 pt1

Figure 4 Dividing fluid domain into discrete control volumes

(i) If 119870e lt 119870iniIc the crack does not propagate

(ii) When119870e = 119870iniIc the performed crack begins to crack

initially(iii) When 119870ini

Ic lt 119870e lt 119870unIc the propagating crack

develops steadily(iv) When119870e ge 119870un

Ic the crack propagates unsteadily119870iniIc is the initiation toughness 119870un

Ic is the unstable fracturetoughness

32 Finite Volume Methods of the Flow along a Crack Con-sidering (3) and (4) and ignoring the fluid loss into the solidmedia the following expression can be obtained120597120597119909 (119896120597119901120597119909) + 120597119908120597119905 = 0 (36)

where

119896 = minus 119908312120583 (37)

321 Discretised Equation Equation (36) is one-dimensionalsteady state diffusion equation Here we discretise the equa-tion by finite volume methods The first step in the finitevolume method is to divide the domain into discrete controlvolumes As shown in Figure 4 we place a number of nodesin the fluid space between119875 and119876The boundaries of controlvolumes are positioned midway between adjacent nodesThus each node is surrounded by a control volumeThe westside face of the control volume is referred to by ldquowrdquo and theeast side face of the control volume is referred to by ldquoerdquo

Integrating the governing equation in (36) over a controlvolume the following discretised equation can be obtained

intΔ119881

120597120597119909 (119896120597119901120597119909) d119881 + intΔ119881

120597119908120597119905 d119881= ∮119860119899(119896120597119901120597119909) d119860 + int

Δ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ intΔ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ Δ119908Δ119905 sdot Δ119881 = 0

(38)

where119860 is the cross-sectional area of the control volume faceΔ119881 is the volume 119908 is the average crack opening width overa control volume

According to (36) and observing a typical finite volumeelement with node 2 we can obtain

(119896119860120597119901120597119909)eminus (119896119860120597119901120597119909)w

= minusΔ1199082Δ119905 sdot Δ1198812 (39)

To calculate gradients at the control volume faces a linearapproximate distribution of pressure is considered here So(39) yields

119896e119860e1199013 minus 11990121198712 minus 119896w119860w

1199012 minus 11990111198711 = minusΔ1199082Δ119905 sdot Δ1198812 (40)

119896w 119860w 119896e and 119860e can be obtained by the way of linearlyinterpolated values and

119896w = 120575w1198961 + (1 minus 120575w) 1198962119860w = 120575w1198601 + (1 minus 120575w) 1198602120575w = 119871w21198711 119896e = 120575e1198962 + (1 minus 120575e) 1198963119860e = 120575e1198602 + (1 minus 120575e) 1198603120575e = 119871e31198712

(41)

Equation (40) can be rewritten as

119886211199011 + 119886221199012 + 119886231199013 = 1198872 (42)

where

11988621 = 119896w119860w1198711 11988622 = minus119896w119860w1198711 minus 119896e119860e1198712 11988623 = 119896e119860e1198712 1198872 = minusΔ1199082Δ119905 sdot Δ1198812

(43)

322 Boundary Conditions

(I) Boundary Conditions for Fluid Injection Point As shownin Figure 5 integration of (36) over the control volumesurrounding node 1 gives

119896e119860e1199012 minus 11990111198711 minus 119896w119860w

1199011 minus 119901w1198710 = minusΔ1199081Δ119905 sdot Δ1198811 (44)


119886111199011 + 119886121199012 = 1198871 (45)


2

L1L0

ew 1

Figure 5 Boundary conditions for fluid injection point

Lnminus1 Ln

n minus 1 w n e

Figure 6 Boundary conditions for fluid front

where

11988611 = minus119896e119860e1198711 minus 119896w119860w1198710 11988612 = 119896e119860e1198711 1198871 = minusΔ1199081Δ119905 sdot Δ1198811 minus 119896w119860w1198710 119901w119896e = 120575e1198961 + (1 minus 120575e) 1198962 119860e = 120575e1198601 + (1 minus 120575e) 1198602

(46)

119896w and 119860w can be calculated directly according to crackopening displacement and 119901w is the pressure of the fluidinjection point

(II) Boundary Conditions for Fluid Front Similarly as shownin Figure 6 integration of (36) over the control volumesurrounding node n gives

119896e119860e119901t minus 119901119899119871119899 minus 119896w119860w

119901119899 minus 119901119899minus1119871119899minus1 = minusΔ119908119899Δ119905 sdot Δ119881119899 (47)


119886119899(119899minus1)119901119899minus1 + 119886119899119899119901119899 = 119887119899 (48)

Dynamic loads

Solid medium Crack opening

Fluid pressureFluid injection

Figure 7 The coupling of the elasticity fluid flow and fracturegrowth

where

119886119899(119899minus1) = 119896w119860w119871119899minus1 119886119899119899 = minus119896w119860w119871119899minus1 minus 119896e119860e119871119899 119887119899 = minusΔ119908119899Δ119905 sdot Δ119881119899 minus 119896e119860e

119901t119871119899 119896w = 120575w119896119899minus1 + (1 minus 120575w) 119896119899 119860w = 120575w119860119899minus1 + (1 minus 120575w) 119860119899(49)

119896e and119860e can be calculated directly according to crack widthat the fluid front and 119901t is the pressure of the fluid front and119901t = 0

By setting up discretised equations of the forms (42) (45)and (48) at each node the fluid pressure distribution along acrack can be obtained

33 Coupling Scheme To correctly solve the system of equa-tions the elasticity fluid flow and fracture growth shouldbe coupled together see Figure 7 The crack opening causesthe fluid injecting into the solid medium The injecting fluidproduces fluid pressures along crack surface and affects thedeformation of the solidmedium and crack propagation Dueto the deformation of the solid medium the crack openingchanges again

According to (36) the relations between the fluid pressureand crack width are nonlinear It is necessary to use analgorithm that can solve iteratively the fluid pressure andcrack width The staggered Newton algorithm [40] is usedhereThe iteration processes for solving fluid pressure at time119905 are stated as follows

(i) Assume that the fluid pressure 119901119896 has been computedat the kth step iteration Compute dynamic responsesof solid medium under dynamic loads and fluidpressure by using XFEM According to dynamicresponses we can obtain the crack width119908119896 at the 119896thstep iteration

(ii) According to the known crackwidth119908119896 at the kth stepiteration compute the fluid pressure 119901119896+12 by usingFVM

(iii) Compute the fluid pressure 119901119896+1 at the (119896 + 1)th stepiteration by the flowing expression119901119896+1 = (1 minus 120572) 119901119896 + 120572119901119896+12 (50)


Iron

F F

Crack

Crack

80

30

Concrete

y

O

x

200

200

nminus n+

p tt

summinussum+

Figure 8 Schematic of a notched cubic concrete specimen subjectedto a splitting force (unit mm)

(iv) Compute dynamic responses of solid medium underdynamic loads and fluid pressure 119901119896+1 by usingXFEM According to dynamic responses we canobtain the crack width 119908119896+12

(v) Compute the crack width 119908119896+1 at the (119896 + 1)th stepiteration by the flowing expression

119908119896+1 = (1 minus 120572)119908119896 + 120572119908119896+12 (51)

(vi) Given tolerance 120576 = 001 compute the followingexpression

120575119901 = sum119899119894=1 1003816100381610038161003816119901119896+1 minus 1199011198961003816100381610038161003816sum119899119894=1 1003816100381610038161003816119901119896+11003816100381610038161003816 (52)

If 120575119901 le 120576 finish iteration or else go to (ii)

4 Numerical Example

As shown in Figure 8 a notched cubic concrete specimenwiththe dimension of 200mm times 200mm times 200mm subjectedto a splitting force was considered in this example Herethe splitting force was applied on the iron The iron wasfastened on the concrete specimen The material propertiesof concrete were Young modulus 119864 = 36GPa Poisson ratioV = 0167 and mass density 120588 = 2400 kgm3 The materialproperties of iron were Youngmodulus 119864 = 200GPa Poissonratio V = 03 and density 120588 = 7800 kgm3 In numericalmodel the specimen was discretised into 10154 elements and10370 nodes see Figure 9 The concrete materialrsquos initiationtoughness119870ini

Ic equalled 0888MPasdotm12 for slow loading caseand 0909MPasdotm12 for fast loading case and its unstablefracture toughness 119870un

Ic equalled 2914MPasdotm12 for slowloading case and 3028MPasdotm12 for fast loading case Theparameters were tested by experiment For the slow loadingcase we set the time step Δ119905 = 10 s while Δ119905 = 01 s for the fastloading case

Figure 9 XFEMmesh

In this section we mainly investigated the water pressuredistribution along crack surface and the effect of waterpressure distribution on the fracture property

41 Splitting Force versus CMOD Curve Comparison withOur Experiments [41] As shown in Figures 10 and 11 weplotted splitting force versus CMOD curve for slow loadingcase and fast loading case respectively with three differentwater pressure cases (i) without water pressure (ii) waterpressure 119901 = 02MPa at the crack mouth (iii) water pressure119901 = 04MPa at the crack mouth We also compared ourpresent numerical results with our experimental results

Both numerical and experimental results show that(i) with the increase of the applied water pressure at thecrack mouth the peak value of the splitting force decreaseddramatically (ii) under the same case compared with theslow loading case the maximum splitting force from the fastloading case had an obvious increase Additionally a fairlysatisfactory agreement can still be observed for the curvein ascending section This verification indicated that theproposed model was quite effective for simulating hydraulicfracture problems

We also listed the maximum splitting force with threedifferent water pressure cases in Table 1 The maximum errorreached to 1698

42 Effect ofWater Pressure at the CrackMouth on the FractureProperty For experiment the maximum water pressure 119901119899= 04MPa at the crack mouth was applied Due to thelimitation of experimental conditions it was quite difficultin the experiment to increase the water pressure at the crackmouth for investigating its effect on the fracture propertyThe numerical simulation showed its huge advantages inextending test conditions Here we observed numerically


Table 1 The maximum splitting force 119865maxkNSlow loading case Fast loading case119901119899 = 00 119901119899 = 02MPa 119901119899 = 04MPa 119901119899 = 00MPa 119901119899 = 02MPa 119901119899 = 04MPa

Numerical results 1560 1331 1075 1951 1702 1400Experimental results 1879 1428 1052 1982 1616 1295Error 1698 679 219 156 532 811

Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10

Present numerical resultsExperimental results (Chen et al 2017)

(a) Without water pressure

Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10


(b) Water pressure 119901119899 = 02MPa at the crack mouth

Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10


(c) Water pressure 119901119899 = 04MPa at the crack mouth

Figure 10 Splitting force versus CMOD curve for slow loading case with different water pressures

six cases 119901119899 = 00 02MPa 04MPa 06MPa 08MPa and10MPa

Figure 12(a) showed the effect of water pressure distribu-tion on the fracture property for a slow loading case with sixdifferent water pressure cases and Figure 12(b) for fast loadingcase We also plotted maximum splitting force versus appliedwater pressure at the crack mouth curve see Figure 13 Asdepicted in Section 41 the results shown in these figures were

still straightforward With the increase of the applied waterpressure at the crack mouth the peak value of the splittingforce decreased dramatically Under the same case comparedwith the slow loading case the maximum splitting force fromthe fast loading case had an obvious increase It was obviousthat if the applied water pressure increased the mechanicalsplitting force was designed to decrease to maintain thesame CMOD In Figure 13 it also could be found that if


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


Figure 11 Splitting force vesus CMOD curve for fast loading case with different water pressures

119901119899 ge 06MPa the decrease velocity of the maximum splittingforce slowed down

The F-CMOD curve contained two phases the pre-peakpeak response and the postpeak response The CMODvalue responding to peak response was always less than02mmWith the increase of the applied water pressure at thecrack mouth the CMOD value responding to peak responsedecreased significantly

43 Water Pressure Distribution along a Propagating CrackSurface Figure 14 showed the water pressure distributionalong a propagating crack surface for slow and fast loadingcases To limit the length of this paper only the case 119901119899 =04MPa was shown here It could be seen that the waterpressure distribution along crack surface followed parabolic

distribution But at the beginning the water pressure distri-bution closely approximated linear distribution At last thewater pressure for most crack segments approximated theapplied water pressure at the crack mouth Adjacent to tipthe water pressure dropped rapidly to zero

44 Failure Patterns As shown in Figures 15 and 16 weshowed the failure mode for slow and fast loading cases with119901119899 = 02MPa In numerical model we assumed that the con-crete specimen was ideal homogeneous material Thereforethe numerical results showed that the crack was pure mode-Icrack and propagated in a straight mannerThe experimentalresults also showed that the crack propagated approximatelyin a straight manner but a slight deflection could still beobserved This phenomenon could be interpreted as theheterogeneity of the actual concrete specimen


CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0

pn = 02 MPapn = 04 MPa

pn = 06 MPapn = 08 MPapn = 10 MPa

(a) Slow loading case

CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0



(b) Fast loading case

Figure 12 Effect of water pressure at the crack mouth on the fracture property

Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12

Fast loading caseSlow loading case

Figure 13Maximum splitting force versus applied water pressure atthe crack mouth curve

5 Summary and Conclusions

In this paper wemodel the fluid flowwithin the crack as one-dimensional flow and assume that the flow is laminar thefluid is incompressible and accounts for the time-dependentrate of crack openingHere we discretise the flow equation byfinite volume methodsThe extended finite element methodsare used for solving solid medium with crack under dynamicloads Having constructed the approximation of dynamicextended finite element methods the derivation of governingequation for dynamic extended finite elementmethods is pre-sentedThe implicit time algorithm is elaborated for the timedescritisation of dominant equation In addition the interac-tion integral method is given for evaluating stress intensity

factors Then the coupling model for modelling hydraulicfracture can be established by the extended finite elementmethods and the finite volume methods We compare ourpresent numerical results with our experimental results forverifying the proposed model Finally we investigate thewater pressure distribution along crack surface and the effectof water pressure distribution on the fracture property Somevaluable conclusions can be drawn from this study

(i) Some conclusions between numerical results andexperimental results were quite identical A fairlysatisfactory agreement could also be observed for F-CMOD curve in ascending section Therefore theproposed model was quite effective for simulatinghydraulic fracture problems

(ii) The F-CMOD curve contained two phases the pre-peakpeak response and the postpeak response TheCMODvalue responding to peak responsewas alwaysless than 02mm With the increase of the appliedwater pressure at the crack mouth the CMOD valueresponding to peak response decreased significantly

(iii) With the increase of the applied water pressure atthe crack mouth the peak value of the splittingforce decreased dramatically Under the same casecompared with the slow loading case the maximumsplitting force from the fast loading case had anobvious increase

(iv) The water pressure distribution along crack surfacefollowed parabolic distribution But at the beginningthe water pressure distribution closely approximatedlinear distribution At last the water pressure formost crack segments approximated the applied waterpressure at the crackmouth Adjacent to tip the waterpressure dropped rapidly to zero


0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)

Distance from notched tip y (mm)

t = 150 st = 400 st = 200 s

t = 500 st = 300 s


Distance from notched tip y (mm) 0 10 20 30 40 50 60 70 80

Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


Figure 14 Water pressure distribution along a propagating crack surface

minus2 minus15 minus1 minus05 0 05 1 15

(a) x-directional displacement contoursmmminus04 minus02 0 02

(b) y-directional displacement contoursmm

(c) Deformed mesh (d) Failure mode of experiment

Figure 15 Failure mode for slow loading case with 119901119899 = 02MPa






Figure 16 Failure mode for fast loading case with 119901119899 = 02MPa

(v) In numerical model we assumed that the concretespecimen was ideal homogeneous material There-fore the numerical results showed that the crackwas pure mode-I crack and propagated in a straightmanner The experimental results also showed thatthe crack propagated approximately in a straightmanner but a slight deflection could still be observedThis phenomenon could be interpreted as the hetero-geneity of the actual concrete specimen

Our future work will further investigate the water pres-sure distribution along a propagating crack surface and theeffect of water pressure distribution on the fracture propertywith considering crack opening-closing

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledge support for this researchfrom the National Natural Science Foundation of China(Grants nos 51309088 11372098 51579084 and 51308188)the Fundamental Research Funds for the Central Universities(Grant no 2015B01714) the National Sci-Tech Support Planof China (Grant no 2015BAB07B10) and China PostdoctoralScience Foundation funded project (Grant no 2014T70466)The authors also would like to thank Professor CharlesAugarde for the discussion during their visit to DurhamUniversity who stayed in Hohai University supported by theRoyal Society

References

[1] E Detournay ldquoMechanics of Hydraulic Fracturesrdquo AnnualReview of Fluid Mechanics vol 48 pp 311ndash339 2016

[2] J Adachi E Siebrits A Peirce and J Desroches ldquoComputersimulation of hydraulic fracturesrdquo International Journal of RockMechanics andMining Sciences vol 44 no 5 pp 739ndash757 2007


[3] A Linkov ldquoBench-mark solution for a penny-shaped hydraulicfracture driven by a thinning fluidrdquo Physics vol 77 no 77 pp14ndash23 2015

[4] A P Bunger and E Detournay ldquoEarly-time solution for a radialhydraulic fracturerdquo Journal of Engineering Mechanics vol 133no 5 pp 534ndash540 2007

[5] F Javanmardi P Leger and R Tinawi ldquoSeismic structuralstability of concrete gravity dams considering transient upliftpressures in cracksrdquo Engineering Structures vol 27 no 4 pp616ndash628 2005

[6] T J Boone and A R Ingraffea ldquoA numerical procedure forsimulation of hydraulically-driven fracture propagation inporoelastic mediardquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 14 no 1 pp 27ndash471990

[7] Z Wu and L N Y Wong ldquoExtension of numerical manifoldmethod for coupled fluid flow and fracturing problemsrdquoInternational Journal for Numerical and Analytical Methods inGeomechanics vol 38 no 18 pp 1990ndash2008 2014

[8] A Lisjak P Kaifosh L He B S A Tatone O KMahabadi andG Grasselli ldquoA 2D fully-coupled hydro-mechanical FDEMformulation for modelling fracturing processes in discontinu-ous porous rock massesrdquo Computers amp Geosciences vol 81 pp1ndash18 2017

[9] U Ohlsson M Nystrom T Olofsson and K WaagaardldquoInfluence of hydraulic pressure in fracture mechanics mod-elling of crack propagation in concreterdquo Materials and Struc-turesMateriaux et Constructions vol 31 no 207 pp 203ndash2081996

[10] F Javanmardi P Leger and R Tinawi ldquoSeismic water pressurein cracked concrete gravity dams Experimental study andtheoretical modelingrdquo Journal of Structural Engineering vol 131no 1 pp 139ndash150 2005

[11] G Bolzon and G Cocchetti ldquoDirect assessment of structuralresistance against pressurized fracturerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 27 no5 pp 353ndash378 2003

[12] J M Segura and I Carol ldquoNumerical modelling of pressurizedfracture evolution in concrete using zero-thickness interfaceelementsrdquo Engineering Fracture Mechanics vol 77 no 9 pp1386ndash1399 2010

[13] J Q Bao E Fathi and S Ameri ldquoA coupled finite elementmethod for the numerical simulation of hydraulic fracturingwith a condensation techniquerdquo Engineering Fracture Mechan-ics vol 131 pp 269ndash281 2014

[14] A Shojaei A Dahi Taleghani and G Li ldquoA continuumdamage failure model for hydraulic fracturing of porous rocksrdquoInternational Journal of Plasticity vol 59 pp 199ndash212 2014

[15] P Gupta and C A Duarte ldquoSimulation of non-planar three-dimensional hydraulic fracture propagationrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 38 no 13 pp 1397ndash1430 2014

[16] S Cherny V Lapin D Esipov et al ldquoSimulating fully 3D non-planar evolution of hydraulic fracturesrdquo International Journal ofFracture vol 201 no 2 pp 181ndash211 2016

[17] L Q Choo Z Zhao H Chen and Q Tian ldquoHydraulic frac-turing modeling using the discontinuous deformation analysis(DDA) methodrdquo Computers amp Geosciences vol 76 pp 12ndash222016

[18] X Zhuang J Chun and H Zhu ldquoA comparative study onunfilled and filled crack propagation for rock-like brittle mate-rialrdquo Theoretical and Applied Fracture Mechanics vol 72 pp110ndash120 2014

[19] G Yi T Yu T Q Bui C Ma and S Hirose ldquoSIFs evaluationof sharp V-notched fracture by XFEM and strain energyapproachrdquo Theoretical and Applied Fracture Mechanics vol 89pp 35ndash44 2017

[20] J Shouyan D Chengbin G Chongshi and C Xiaocui ldquoXFEManalysis of the effects of voids inclusions and other cracks onthe dynamic stress intensity factor of a major crackrdquo Fatigue ampFracture of Engineering Materials Structures vol 37 no 8 pp866ndash882 2014

[21] J Rethore R de Borst and M-A Abellan ldquoA two-scaleapproach for fluid flow in fractured porous mediardquo Interna-tional Journal for Numerical Methods in Engineering vol 71 no7 pp 780ndash800 2007

[22] B Lecampion ldquoAn extended finite element method forhydraulic fracture problemsrdquo Communications in NumericalMethods in Engineering vol 25 no 2 pp 121ndash133 2009

[23] E Gordeliy and A Peirce ldquoCoupling schemes for modelinghydraulic fracture propagation using the XFEMrdquo ComputerMethods Applied Mechanics and Engineering vol 253 pp 305ndash322 2013

[24] E Gordeliy and A Peirce ldquoImplicit level set schemes for mod-eling hydraulic fractures using the XFEMrdquo Computer MethodsApplied Mechanics and Engineering vol 266 pp 125ndash143 2013

[25] A R KhoeiMVahab EHaghighat and SMoallemi ldquoAmesh-independent finite element formulation for modeling crackgrowth in saturated porous media based on an enriched-FEMtechniquerdquo International Journal of Fracture vol 188 no 1 pp79ndash108 2014

[26] A D Taleghani Analysis of hydraulic fracture propagationin fractured reservoirs an improved model for the interactionbetween induced and natural fractures The University of Texasat Austin 2009

[27] S Salimzadeh and N Khalili ldquoA three-phase XFEM modelfor hydraulic fracturing with cohesive crack propagationrdquoComputers amp Geosciences vol 69 pp 82ndash92 2015

[28] K Wang Q Zhang X Xia L Wang and X Liu ldquoAnalysisof hydraulic fracturing in concrete dam considering fluid-structure interaction using XFEM-FVM modelrdquo EngineeringFailure Analysis vol 57 pp 399ndash412 2015

[29] G K Batchelor Introduction to Fluid Dynamics CambridgeUniversity Press Cambridge UK 1967

[30] D I Garagash ldquoPropagation of a plane-strain hydraulic fracturewith a fluid lag Early-time solutionrdquo International Journal ofSolids and Structures vol 43 no 18-19 pp 5811ndash5835 2006

[31] NMoes J Dolbow andT Belytschko ldquoA finite elementmethodfor crack growth without remeshingrdquo International Journal forNumerical Methods in Engineering vol 46 no 1 pp 131ndash1501999

[32] G Ventura ldquoOn the elimination of quadrature subcells for dis-continuous functions in the eXtended Finite-Element MethodrdquoInternational Journal for Numerical Methods in Engineering vol66 no 5 pp 761ndash795 2006

[33] S P A Bordas T Rabczuk N-XHung et al ldquoStrain smoothingin FEM and XFEMrdquo Computers amp Structures vol 88 no 23-24pp 1419ndash1443 2010

[34] S Natarajan D R Mahapatra and S P Bordas ldquoIntegratingstrong and weak discontinuities without integration subcells


and example applications in an XFEMGFEM frameworkrdquoInternational Journal for Numerical Methods in Engineering vol83 no 3 pp 269ndash294 2010

[35] S Natarajan S Bordas and D Mahapatra ldquoNumerical inte-gration over arbitrary polygonal domains based on Schwarz-Christoffel conformal mappingrdquo International Journal forNumerical Methods in Engineering vol 80 no 1 pp 103ndash1342009

[36] M Attigui and C Petit ldquoMixed-mode separation in dynamicfracture mechanics New path independent integralsrdquo Interna-tional Journal of Fracture vol 84 no 1 pp 19ndash36 1997

[37] F Erdogan and G C Sih ldquoOn the crack extension in platesunder plane loading and transverse shearrdquo Journal of BasicEngineering vol 85 no 4 pp 519ndash527 1963

[38] N Sukumar and J-H Prevost ldquoModeling quasi-static crackgrowth with the extended finite element method Part IComputer implementationrdquo International Journal of Solids andStructures vol 40 no 26 pp 7513ndash7537 2003

[39] S Xu and H W Reinhardt ldquoDetermination of double-Kcriterion for crack propagation in quasi-brittle fracture part Iexperimental investigation of crack propagationrdquo InternationalJournal of Fracture vol 98 no 2 pp 111ndash149 1999

[40] A R Khoei M Hirmand M Vahab and M Bazargan ldquoAnenriched FEM technique for modeling hydraulically drivencohesive fracture propagation in impermeable media withfrictional natural faults Numerical and experimental investiga-tionsrdquo International Journal for Numerical Methods in Engineer-ing vol 104 no 6 pp 439ndash468 2015

[41] X Chen C Du M You S Jiang and L Sun ldquoExperimentalstudy on water fracture interactions in concreterdquo EngineeringFracture Mechanics vol 179 pp 314ndash327 2017

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Modelling amp Simulation in EngineeringHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Chemical EngineeringInternational Journal of Antennas and

Propagation




Navigation and Observation



DistributedSensor Networks



methods [7] boundary element methods [16] discontinuousdeformation analysis methods [17] and extended finite ele-ment methods (XFEM) The XFEM shows huge advantagefor dealing with discontinuous problems [18ndash20] and alsohydraulic fracture problems The XFEMmesh does not needto align with a discontinuity For moving discontinuitiessuch as crack propagation problem it does not need to carryon remeshing Mesh refinement is also unnecessary arounda discontinuous feature The first simulation of hydraulicfracture in XFEM was due to Rethore et al [21] and theydeveloped a two-scale numerical model for fluid flow infractured deforming porous media In 2009 Lecampion [22]adopted the XFEM for investigating the solution of hydraulicfracture problems Gordeliy and Peirce (2013) [23] proposedcoupled algorithms that used the XFEM to solve the elasticcrack component of the elastohydrodynamic equations thatgoverned the propagation of hydraulic fractures in an elasticmedium Subsequently they (2013) [24] proposed two novelXFEM schemes for modeling fluid driven fractures both ofwhich exploited an implicit level set algorithm for locatingthe singular free boundary that occurred when the fluidand fracture fronts coalesce Their excellent works providemathematical proof for using XFEM to solve hydraulicfracturing problems Khoei et al [25] simulated the crackgrowth in saturated porous media using XFEM Taleghani[26] also developed an XFEM code to simulate fracturepropagation initiation and intersection and the presentedcoupled fluid flow-fracture mechanics simulations extendedavailable modeling efforts and provided a unified frameworkfor evaluating fracture design parameters and their con-sequences Salimzadeh and Khalili (2015) [27] proposed athree-phase hydromechanical model for hydraulic fracturingand they handled discontinuity by using XFEM while cohe-sive crack model was used as fracturing criterion Wang etal (2015) [28] proposed a hybrid approach combining theXFEM and the finite volume method to simulate hydraulicfracturing in concrete dams Our current study concernsdeveloping a model for cracking modeling of structureunder water pressure along a propagating crack surface anddynamic loads Additionally we will compare our presentnumerical results with our experimental results for verifyingthe proposed model

This paper is organized as follows Section 2 introducessome governing equations for elastic dynamic responses ofthe solid medium and fluid flow pressure within the crackSection 3 discusses numerical approximation of extendedfinite element methods and finite volume methods of theflow along a crack Section 4 gives a numerical examplefor investigating the water pressure distribution along cracksurface and the effect of water pressure distribution on thefracture property We also compare our present numericalresults with our experimental results Section 5 summarisesthe major conclusions that can be drawn from this study

2 Governing Equations

21 Elastic Dynamic Responses of the Solid Medium Theboundary of a bounded domain Ω isin 1198772 is partitionedinto three parts the displacement boundary (Γu) the traction

boundary (Γt) and the crack boundary (Γc) that is traction-free The elastodynamic basic equation is expressed asnabla sdot 120590 + b = 120588u in Ω

120576 = nablasu in Ω120590 = D 120576 in Ω (1)

with the following boundary and initial conditions

u (x 119905) = u (x 119905) on Γt120590 sdot n = t on Γu120590 sdot n = 0 on Γc

u (x 119905 = 0) = u (0) u (x 119905 = 0) = u (0)

(2)

where 120590 is the Cauchy stress tensor b is the body forcevector 120576 is the strain tensor 120588 is the material density u isthe acceleration field vector nablas is the symmetric part of thegradient operator u is the displacement field vector D isthe constitutive matrix n is the unit outward normal vectorto the crack surface u is the prescribed displacement t isthe external traction vector u(0) is the initial displacementvector and u(0) is the initial velocity vector22 Fluid Flow Pressure within the Crack In this paper wemodel the fluid flow within the crack as one-dimensionalflow Assume that the flow is laminar and the fluid isincompressible But here we account for the time-dependentrate of crack opening

The conservation of the incompressible fluid in thefracture can be expressed as [6]nabla sdot q + + 119892 = 0 (3)

wherenabla is the divergence operator defined in x direction q isthe fluid flux is the time-dependent rate of crack openingand = 120597119908120597119905 and 119892 is the fluid loss into the solid mediaand here we ignore the fluid loss that is 119892 = 0

Additionally Poiseuillersquos law [29] gives the followingexpression

q = minus 119908312120583nabla119901 (4)

where119908 is the crack opening 120583 is the fluid viscosity and 119901 isthe fluid pressure

The pressure boundary conditions at the fluid injectionpoint in the crack are 119901 (119909 = 0 119905) = 1199010 (5)

where 1199010 is the pressure of the fluid injection pointIn the fluid lag zone [30]119901 (119909 119905) lt 0 (119897t le 119909 le 119897) (6)

According to the fluid pressure continuity lag condition (6)provides the net-pressure boundary condition at the fluidfront 119909 = 119897t for the fluid flow equations (3)


Crack



Ilowast＜Ｍ

Ilowast＜Ｌ





uh (x) = sum119894isin119868



119873lowast119894 (x) 4sum119895=1

[119865119895 (x) minus 119865119895 (x119894)] b119895119894 (7)









Muh + Kuh = f (10)





kau kaa kab

kbu kba kbb

]]]] (12)


where

k119903119904 = intΩe(B119903)T DB119904dΩ


(119903 119904 = u a b) Bu119894 = [[[[

120597119873119894120597119909 0 1205971198731198941205971199100 120597119873119894120597119910 120597119873119894120597119909]]]]T

119894 = 1 2 3 4

Ba119894 = [[[[


1006704119867 = 119867 (x) minus 119867 (x119894) 119894 = 1 2 3 4

Bb119894 = [Bb1

119894 Bb2119894 Bb3119894 Bb4119894 ]

Bb119895119894 = [[[[[


]]]]]

T

1006704119865119895 = 119865119895 (x) minus 119865119895 (x119894) 119894 119895 = 1 2 3 4

(13)



mau maa mab

mbu mba mbb

]]]] (14)

where

muu = intΩe120588 (N)T N 119889Ω


mub119895 = [mbu

119895 ]T = intΩe120588 (N)T (N 1006704119865119895) 119889Ω

maa = intΩe120588 (N1006704119867)T (N1006704119867)119889Ω

mab119895 = [mba

119895 ]T = intΩe120588 (N1006704119867)T (N 1006704119865119895) 119889Ω

mbb119895119896 = int

Ωe120588 (N 1006704119865119895)T (N 1006704119865119896) 119889Ω

119895 119896 = 1 2 3 4

N = [N1 N2 N3 N4] N119894 = [119873119894 00 119873119894] 119894 = 1 2 3 4

(15)



where

fu = intΩe

NTb dΩ + intΓet

NTt dΓfa = int


Γet




Γet



(17)



(18)


120573 = 14 (1 minus 120572)2 120574 = 12 minus 120572

minus13 le 120572 le 0(19)




Crack CrackCrack

Crack


Crack Crack

Crack


Crack

Crack






Mu0 + Ku0 = f0 (20)



(21)


K = K + 1198880M (22)









1

Crack

0 00 0

00 11 0 0

10 1A

1 1 0

10 1 1 0

00 11 1 0

0 00 0 0

Γd+

Γdminus




119868(12) = intA(120590(1)119894119895 119906(2)1198941 + 120590(2)119894119895 119906(1)1198941 minus 12120590(1)119894119896 120576(2)119894119896 1205751119895) 119902119895119889Ω



(26)




119902 = 4sum119894=1

119873119894119902119894 (28)



II ] (29)




(plane strain) (30)



119870I = 119864lowast119872(12)12 (31)


119870II = 119864lowast119872(12)22 (32)


II = 1 and119872(12) = 119872(12)2



)2 + 8) (33)




119870e = cos120579c2 (119870Icos

2 120579c2 minus 15119870II sin 120579c) (35)



Control volume

2 3

Nodes

Q

L1 L2

ewP

p0 pt1










where

119896 = minus 119908312120583 (37)



intΔ119881

120597120597119909 (119896120597119901120597119909) d119881 + intΔ119881

120597119908120597119905 d119881= ∮119860119899(119896120597119901120597119909) d119860 + int

Δ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ intΔ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ Δ119908Δ119905 sdot Δ119881 = 0

(38)



(119896119860120597119901120597119909)eminus (119896119860120597119901120597119909)w

= minusΔ1199082Δ119905 sdot Δ1198812 (39)


119896e119860e1199013 minus 11990121198712 minus 119896w119860w




(41)


119886211199011 + 119886221199012 + 119886231199013 = 1198872 (42)

where


(43)



119896e119860e1199012 minus 11990111198711 minus 119896w119860w



119886111199011 + 119886121199012 = 1198871 (45)


2

L1L0

ew 1


Lnminus1 Ln

n minus 1 w n e


where


(46)






119886119899(119899minus1)119901119899minus1 + 119886119899119899119901119899 = 119887119899 (48)

Dynamic loads




where











Iron

F F

Crack

Crack

80

30

Concrete

y

O

x

200

200

nminus n+

p tt

summinussum+




119908119896+1 = (1 minus 120572)119908119896 + 120572119908119896+12 (51)


120575119901 = sum119899119894=1 1003816100381610038161003816119901119896+1 minus 1199011198961003816100381610038161003816sum119899119894=1 1003816100381610038161003816119901119896+11003816100381610038161003816 (52)


4 Numerical Example




Figure 9 XFEMmesh









Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10








CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










Crack



Ilowast＜Ｍ

Ilowast＜Ｌ





uh (x) = sum119894isin119868



119873lowast119894 (x) 4sum119895=1

[119865119895 (x) minus 119865119895 (x119894)] b119895119894 (7)









Muh + Kuh = f (10)





kau kaa kab

kbu kba kbb

]]]] (12)


where

k119903119904 = intΩe(B119903)T DB119904dΩ


(119903 119904 = u a b) Bu119894 = [[[[

120597119873119894120597119909 0 1205971198731198941205971199100 120597119873119894120597119910 120597119873119894120597119909]]]]T

119894 = 1 2 3 4

Ba119894 = [[[[


1006704119867 = 119867 (x) minus 119867 (x119894) 119894 = 1 2 3 4

Bb119894 = [Bb1

119894 Bb2119894 Bb3119894 Bb4119894 ]

Bb119895119894 = [[[[[


]]]]]

T

1006704119865119895 = 119865119895 (x) minus 119865119895 (x119894) 119894 119895 = 1 2 3 4

(13)



mau maa mab

mbu mba mbb

]]]] (14)

where

muu = intΩe120588 (N)T N 119889Ω


mub119895 = [mbu

119895 ]T = intΩe120588 (N)T (N 1006704119865119895) 119889Ω

maa = intΩe120588 (N1006704119867)T (N1006704119867)119889Ω

mab119895 = [mba

119895 ]T = intΩe120588 (N1006704119867)T (N 1006704119865119895) 119889Ω

mbb119895119896 = int

Ωe120588 (N 1006704119865119895)T (N 1006704119865119896) 119889Ω

119895 119896 = 1 2 3 4

N = [N1 N2 N3 N4] N119894 = [119873119894 00 119873119894] 119894 = 1 2 3 4

(15)



where

fu = intΩe

NTb dΩ + intΓet

NTt dΓfa = int


Γet




Γet



(17)



(18)


120573 = 14 (1 minus 120572)2 120574 = 12 minus 120572

minus13 le 120572 le 0(19)




Crack CrackCrack

Crack


Crack Crack

Crack


Crack

Crack






Mu0 + Ku0 = f0 (20)



(21)


K = K + 1198880M (22)









1

Crack

0 00 0

00 11 0 0

10 1A

1 1 0

10 1 1 0

00 11 1 0

0 00 0 0

Γd+

Γdminus




119868(12) = intA(120590(1)119894119895 119906(2)1198941 + 120590(2)119894119895 119906(1)1198941 minus 12120590(1)119894119896 120576(2)119894119896 1205751119895) 119902119895119889Ω



(26)




119902 = 4sum119894=1

119873119894119902119894 (28)



II ] (29)




(plane strain) (30)



119870I = 119864lowast119872(12)12 (31)


119870II = 119864lowast119872(12)22 (32)


II = 1 and119872(12) = 119872(12)2



)2 + 8) (33)




119870e = cos120579c2 (119870Icos

2 120579c2 minus 15119870II sin 120579c) (35)



Control volume

2 3

Nodes

Q

L1 L2

ewP

p0 pt1










where

119896 = minus 119908312120583 (37)



intΔ119881

120597120597119909 (119896120597119901120597119909) d119881 + intΔ119881

120597119908120597119905 d119881= ∮119860119899(119896120597119901120597119909) d119860 + int

Δ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ intΔ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ Δ119908Δ119905 sdot Δ119881 = 0

(38)



(119896119860120597119901120597119909)eminus (119896119860120597119901120597119909)w

= minusΔ1199082Δ119905 sdot Δ1198812 (39)


119896e119860e1199013 minus 11990121198712 minus 119896w119860w




(41)


119886211199011 + 119886221199012 + 119886231199013 = 1198872 (42)

where


(43)



119896e119860e1199012 minus 11990111198711 minus 119896w119860w



119886111199011 + 119886121199012 = 1198871 (45)


2

L1L0

ew 1


Lnminus1 Ln

n minus 1 w n e


where


(46)






119886119899(119899minus1)119901119899minus1 + 119886119899119899119901119899 = 119887119899 (48)

Dynamic loads




where











Iron

F F

Crack

Crack

80

30

Concrete

y

O

x

200

200

nminus n+

p tt

summinussum+




119908119896+1 = (1 minus 120572)119908119896 + 120572119908119896+12 (51)


120575119901 = sum119899119894=1 1003816100381610038161003816119901119896+1 minus 1199011198961003816100381610038161003816sum119899119894=1 1003816100381610038161003816119901119896+11003816100381610038161003816 (52)


4 Numerical Example




Figure 9 XFEMmesh









Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10








CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










where

k119903119904 = intΩe(B119903)T DB119904dΩ


(119903 119904 = u a b) Bu119894 = [[[[

120597119873119894120597119909 0 1205971198731198941205971199100 120597119873119894120597119910 120597119873119894120597119909]]]]T

119894 = 1 2 3 4

Ba119894 = [[[[


1006704119867 = 119867 (x) minus 119867 (x119894) 119894 = 1 2 3 4

Bb119894 = [Bb1

119894 Bb2119894 Bb3119894 Bb4119894 ]

Bb119895119894 = [[[[[


]]]]]

T

1006704119865119895 = 119865119895 (x) minus 119865119895 (x119894) 119894 119895 = 1 2 3 4

(13)



mau maa mab

mbu mba mbb

]]]] (14)

where

muu = intΩe120588 (N)T N 119889Ω


mub119895 = [mbu

119895 ]T = intΩe120588 (N)T (N 1006704119865119895) 119889Ω

maa = intΩe120588 (N1006704119867)T (N1006704119867)119889Ω

mab119895 = [mba

119895 ]T = intΩe120588 (N1006704119867)T (N 1006704119865119895) 119889Ω

mbb119895119896 = int

Ωe120588 (N 1006704119865119895)T (N 1006704119865119896) 119889Ω

119895 119896 = 1 2 3 4

N = [N1 N2 N3 N4] N119894 = [119873119894 00 119873119894] 119894 = 1 2 3 4

(15)



where

fu = intΩe

NTb dΩ + intΓet

NTt dΓfa = int


Γet




Γet



(17)



(18)


120573 = 14 (1 minus 120572)2 120574 = 12 minus 120572

minus13 le 120572 le 0(19)




Crack CrackCrack

Crack


Crack Crack

Crack


Crack

Crack






Mu0 + Ku0 = f0 (20)



(21)


K = K + 1198880M (22)









1

Crack

0 00 0

00 11 0 0

10 1A

1 1 0

10 1 1 0

00 11 1 0

0 00 0 0

Γd+

Γdminus




119868(12) = intA(120590(1)119894119895 119906(2)1198941 + 120590(2)119894119895 119906(1)1198941 minus 12120590(1)119894119896 120576(2)119894119896 1205751119895) 119902119895119889Ω



(26)




119902 = 4sum119894=1

119873119894119902119894 (28)



II ] (29)




(plane strain) (30)



119870I = 119864lowast119872(12)12 (31)


119870II = 119864lowast119872(12)22 (32)


II = 1 and119872(12) = 119872(12)2



)2 + 8) (33)




119870e = cos120579c2 (119870Icos

2 120579c2 minus 15119870II sin 120579c) (35)



Control volume

2 3

Nodes

Q

L1 L2

ewP

p0 pt1










where

119896 = minus 119908312120583 (37)



intΔ119881

120597120597119909 (119896120597119901120597119909) d119881 + intΔ119881

120597119908120597119905 d119881= ∮119860119899(119896120597119901120597119909) d119860 + int

Δ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ intΔ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ Δ119908Δ119905 sdot Δ119881 = 0

(38)



(119896119860120597119901120597119909)eminus (119896119860120597119901120597119909)w

= minusΔ1199082Δ119905 sdot Δ1198812 (39)


119896e119860e1199013 minus 11990121198712 minus 119896w119860w




(41)


119886211199011 + 119886221199012 + 119886231199013 = 1198872 (42)

where


(43)



119896e119860e1199012 minus 11990111198711 minus 119896w119860w



119886111199011 + 119886121199012 = 1198871 (45)


2

L1L0

ew 1


Lnminus1 Ln

n minus 1 w n e


where


(46)






119886119899(119899minus1)119901119899minus1 + 119886119899119899119901119899 = 119887119899 (48)

Dynamic loads




where











Iron

F F

Crack

Crack

80

30

Concrete

y

O

x

200

200

nminus n+

p tt

summinussum+




119908119896+1 = (1 minus 120572)119908119896 + 120572119908119896+12 (51)


120575119901 = sum119899119894=1 1003816100381610038161003816119901119896+1 minus 1199011198961003816100381610038161003816sum119899119894=1 1003816100381610038161003816119901119896+11003816100381610038161003816 (52)


4 Numerical Example




Figure 9 XFEMmesh









Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10








CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










Crack CrackCrack

Crack


Crack Crack

Crack


Crack

Crack






Mu0 + Ku0 = f0 (20)



(21)


K = K + 1198880M (22)









1

Crack

0 00 0

00 11 0 0

10 1A

1 1 0

10 1 1 0

00 11 1 0

0 00 0 0

Γd+

Γdminus




119868(12) = intA(120590(1)119894119895 119906(2)1198941 + 120590(2)119894119895 119906(1)1198941 minus 12120590(1)119894119896 120576(2)119894119896 1205751119895) 119902119895119889Ω



(26)




119902 = 4sum119894=1

119873119894119902119894 (28)



II ] (29)




(plane strain) (30)



119870I = 119864lowast119872(12)12 (31)


119870II = 119864lowast119872(12)22 (32)


II = 1 and119872(12) = 119872(12)2



)2 + 8) (33)




119870e = cos120579c2 (119870Icos

2 120579c2 minus 15119870II sin 120579c) (35)



Control volume

2 3

Nodes

Q

L1 L2

ewP

p0 pt1










where

119896 = minus 119908312120583 (37)



intΔ119881

120597120597119909 (119896120597119901120597119909) d119881 + intΔ119881

120597119908120597119905 d119881= ∮119860119899(119896120597119901120597119909) d119860 + int

Δ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ intΔ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ Δ119908Δ119905 sdot Δ119881 = 0

(38)



(119896119860120597119901120597119909)eminus (119896119860120597119901120597119909)w

= minusΔ1199082Δ119905 sdot Δ1198812 (39)


119896e119860e1199013 minus 11990121198712 minus 119896w119860w




(41)


119886211199011 + 119886221199012 + 119886231199013 = 1198872 (42)

where


(43)



119896e119860e1199012 minus 11990111198711 minus 119896w119860w



119886111199011 + 119886121199012 = 1198871 (45)


2

L1L0

ew 1


Lnminus1 Ln

n minus 1 w n e


where


(46)






119886119899(119899minus1)119901119899minus1 + 119886119899119899119901119899 = 119887119899 (48)

Dynamic loads




where











Iron

F F

Crack

Crack

80

30

Concrete

y

O

x

200

200

nminus n+

p tt

summinussum+




119908119896+1 = (1 minus 120572)119908119896 + 120572119908119896+12 (51)


120575119901 = sum119899119894=1 1003816100381610038161003816119901119896+1 minus 1199011198961003816100381610038161003816sum119899119894=1 1003816100381610038161003816119901119896+11003816100381610038161003816 (52)


4 Numerical Example




Figure 9 XFEMmesh









Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10








CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










1

Crack

0 00 0

00 11 0 0

10 1A

1 1 0

10 1 1 0

00 11 1 0

0 00 0 0

Γd+

Γdminus




119868(12) = intA(120590(1)119894119895 119906(2)1198941 + 120590(2)119894119895 119906(1)1198941 minus 12120590(1)119894119896 120576(2)119894119896 1205751119895) 119902119895119889Ω



(26)




119902 = 4sum119894=1

119873119894119902119894 (28)



II ] (29)




(plane strain) (30)



119870I = 119864lowast119872(12)12 (31)


119870II = 119864lowast119872(12)22 (32)


II = 1 and119872(12) = 119872(12)2



)2 + 8) (33)




119870e = cos120579c2 (119870Icos

2 120579c2 minus 15119870II sin 120579c) (35)



Control volume

2 3

Nodes

Q

L1 L2

ewP

p0 pt1










where

119896 = minus 119908312120583 (37)



intΔ119881

120597120597119909 (119896120597119901120597119909) d119881 + intΔ119881

120597119908120597119905 d119881= ∮119860119899(119896120597119901120597119909) d119860 + int

Δ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ intΔ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ Δ119908Δ119905 sdot Δ119881 = 0

(38)



(119896119860120597119901120597119909)eminus (119896119860120597119901120597119909)w

= minusΔ1199082Δ119905 sdot Δ1198812 (39)


119896e119860e1199013 minus 11990121198712 minus 119896w119860w




(41)


119886211199011 + 119886221199012 + 119886231199013 = 1198872 (42)

where


(43)



119896e119860e1199012 minus 11990111198711 minus 119896w119860w



119886111199011 + 119886121199012 = 1198871 (45)


2

L1L0

ew 1


Lnminus1 Ln

n minus 1 w n e


where


(46)






119886119899(119899minus1)119901119899minus1 + 119886119899119899119901119899 = 119887119899 (48)

Dynamic loads




where











Iron

F F

Crack

Crack

80

30

Concrete

y

O

x

200

200

nminus n+

p tt

summinussum+




119908119896+1 = (1 minus 120572)119908119896 + 120572119908119896+12 (51)


120575119901 = sum119899119894=1 1003816100381610038161003816119901119896+1 minus 1199011198961003816100381610038161003816sum119899119894=1 1003816100381610038161003816119901119896+11003816100381610038161003816 (52)


4 Numerical Example




Figure 9 XFEMmesh









Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10








CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










Control volume

2 3

Nodes

Q

L1 L2

ewP

p0 pt1










where

119896 = minus 119908312120583 (37)



intΔ119881

120597120597119909 (119896120597119901120597119909) d119881 + intΔ119881

120597119908120597119905 d119881= ∮119860119899(119896120597119901120597119909) d119860 + int

Δ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ intΔ119881

120597119908120597119905 d119881= (119896119860120597119901120597119909)e

minus (119896119860120597119901120597119909)w+ Δ119908Δ119905 sdot Δ119881 = 0

(38)



(119896119860120597119901120597119909)eminus (119896119860120597119901120597119909)w

= minusΔ1199082Δ119905 sdot Δ1198812 (39)


119896e119860e1199013 minus 11990121198712 minus 119896w119860w




(41)


119886211199011 + 119886221199012 + 119886231199013 = 1198872 (42)

where


(43)



119896e119860e1199012 minus 11990111198711 minus 119896w119860w



119886111199011 + 119886121199012 = 1198871 (45)


2

L1L0

ew 1


Lnminus1 Ln

n minus 1 w n e


where


(46)






119886119899(119899minus1)119901119899minus1 + 119886119899119899119901119899 = 119887119899 (48)

Dynamic loads




where











Iron

F F

Crack

Crack

80

30

Concrete

y

O

x

200

200

nminus n+

p tt

summinussum+




119908119896+1 = (1 minus 120572)119908119896 + 120572119908119896+12 (51)


120575119901 = sum119899119894=1 1003816100381610038161003816119901119896+1 minus 1199011198961003816100381610038161003816sum119899119894=1 1003816100381610038161003816119901119896+11003816100381610038161003816 (52)


4 Numerical Example




Figure 9 XFEMmesh









Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10








CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










2

L1L0

ew 1


Lnminus1 Ln

n minus 1 w n e


where


(46)






119886119899(119899minus1)119901119899minus1 + 119886119899119899119901119899 = 119887119899 (48)

Dynamic loads




where











Iron

F F

Crack

Crack

80

30

Concrete

y

O

x

200

200

nminus n+

p tt

summinussum+




119908119896+1 = (1 minus 120572)119908119896 + 120572119908119896+12 (51)


120575119901 = sum119899119894=1 1003816100381610038161003816119901119896+1 minus 1199011198961003816100381610038161003816sum119899119894=1 1003816100381610038161003816119901119896+11003816100381610038161003816 (52)


4 Numerical Example




Figure 9 XFEMmesh









Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10








CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










Iron

F F

Crack

Crack

80

30

Concrete

y

O

x

200

200

nminus n+

p tt

summinussum+




119908119896+1 = (1 minus 120572)119908119896 + 120572119908119896+12 (51)


120575119901 = sum119899119894=1 1003816100381610038161003816119901119896+1 minus 1199011198961003816100381610038161003816sum119899119894=1 1003816100381610038161003816119901119896+11003816100381610038161003816 (52)


4 Numerical Example




Figure 9 XFEMmesh









Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10








CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation












Split

ting

forc

eF(k

N)

24

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

20

16

12

8

4

0

CMOD (mm)00 02 04 06 08 10



Split

ting

forc

eF(k

N)

14

12

10

8

6

4

2

0

CMOD (mm)00 02 04 06 08 10








CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

24

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0


CMOD (mm)00 02 04 06 08 10


Split

ting

forc

eF(k

N)

20

16

12

8

4

0









CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










CMOD (mm)00 02 04 06 08 10

Split

ting

forc

eF(k

N)

16

12

8

4

0

pn = 0




CMOD (mm)

Split

ting

forc

eF(k

N)

16

20

12

8

4

000 02 04 06 08 10

pn = 0





Max

imum

split

ting

forc

eFＧＲ

(kN

)

20

18

16

14

12

10

8

6

4

Pressure (MPa)00 02 04 06 08 10 12











0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










0 10 20 30 40 50 60 70 80000

005

010

015

020

025

030

035

040

045

050Pr

essu

re p

(MPa

)


t = 150 st = 400 st = 200 s

t = 500 st = 300 s



Pres

sure

p (M

Pa)

t = 20 st = 40 st = 25 s

t = 50 st = 30 s

000

005

010

015

020

025

030

035

040

045

050


















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation



















Acknowledgments


References













































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation



















































RoboticsJournal of





Journal of



RotatingMachinery



Journal of

Volume 201


VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation









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Transcript of Coupled Finite Volume Methods and Extended Finite Element ... · ResearchArticle Coupled Finite...