ATheoreticalModelforHydraulicFracturingthroughTwo ...downloads.hindawi.com/journals/amse/2019/6094305.pdf · and gas wells [1]. During hydraulic fracturing, high water pressure is

$Page 1: ATheoreticalModelforHydraulicFracturingthroughTwo ...downloads.hindawi.com/journals/amse/2019/6094305.pdf · and gas wells [1]. During hydraulic fracturing, high water pressure is$
Research ArticleA Theoretical Model for Hydraulic Fracturing through TwoSymmetric Radial Perforations Emanating from A Borehole

Shibin Tang 1 Zhuo Dong1 Dong Duan2 and Yingchun Li 1

1School of Civil Engineering Dalian University of Technology Dalian 116024 China2College of Mining Engineering Taiyuan University of Technology Taiyuan 030024 China

Correspondence should be addressed to Shibin Tang tang_shibindluteducn

Received 15 January 2019 Revised 28 April 2019 Accepted 13 May 2019 Published 9 June 2019

Academic Editor Antonio Caggiano

Copyright copy 2019 Shibin Tang et al )is 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

)e production enhancement of oil gas or geothermal reservoirs through hydraulic fracturing requires an in-depth study onthe fracture initiation and propagation from the borehole According to the linear elastic fracture mechanics a theoreticalmodel is developed to calculate the stress intensity factors of two symmetric radial cracks emanating from a pressurizedborehole )e maximum tangential stress criterion under the mix-mode condition is developed to investigate the hydraulicfracture initiation )e critical water pressure and critical initiation angle predicted by the theoretical model match closely theexperimental results reported in the literature )e influence of the stress anisotropy coefficient the perforation angle andlength the borehole radius the ratio between the water pressures in the fracture and the borehole and Biotrsquos coefficient areinvestigated Moreover the effects of the injected high water pressure (ie larger than the critical water pressure) on thefracture initiation angle are studied to further understand the characteristics of hydraulic fracture initiation )e resultsindicate that the perforation angle and length the borehole radius and the stress anisotropy coefficient have a relatively stronginfluence on the critical water pressure and critical initiation angle During high-pressure water injection the fracture initiationangle decreases as the ratio between the water pressure in the fracture and the borehole and Biotrsquos ratio increase)e theoreticalmodel provides a comprehensive understanding of the fracture twist the mixed-mode fracture propagation feature and thehydraulic fracturing optimization

1 Introduction

Hydraulic fracturing has been widely used in the petroleumindustry since the 1930s for enhancing production from oiland gas wells [1] During hydraulic fracturing high waterpressure is pumped into boreholes to break down low-permeability oil and gas that flow into the well from thereservoirs One of the most important tasks in hydraulicfracturing design is to predict the fracturing path and theinjected water pressure Generally hydraulic fractures openin the direction of least resistance (in a plane perpendicularto the minimum horizontal stress) [2] An ideal horizontalwell is usually designed parallel to the orientation of theminimum in situ stress so that the fracture plane induced byhydraulic fracturing operation is perpendicular to the axis ofthe borehole However in practice the real trajectories of thehorizontal wells frequently deviate from the designed

direction due to the limitations of controlling technology indrilling )erefore the crack growth direction will changeunder different horizontal differential stress (the differencebetween the maximum and minimum horizontal stresses)[2] )e orientation of the newly formed fractures is criticalto the hydraulic fracturing operations and well production)e curved shape of a fracture strongly affects fluid flowbecause of additional frictions and changing fluid rheologyespecially for fracturing fluids with a high viscosity con-sisting of solid proppants )erefore a better understandingof how an induced fracture propagates under differentconditions such as different stress anisotropy coefficientfracturing pressure and azimuth is fundamental for de-signing well spacing well placement and expected ultimaterecovery [3ndash8]

)e hydraulic fractures initiated from the drillingborehole are broadly used in petroleum engineering in situ

HindawiAdvances in Materials Science and EngineeringVolume 2019 Article ID 6094305 21 pageshttpsdoiorg10115520196094305

stress measurement and geothermal development Donget al [6] have proposed a theoretical model for hydraulicfracturing through a single radial perforation emanatingfrom a borehole however two symmetrical radial cracksemanating from the borehole are more common in thehydraulic fracturing as shown in Figure 1(a) Mode-I stressintensity factor for symmetrical radial cracks emanatingfrom hollow cylinder in an infinite plane is first solved withthe weight function method by Jin et al [9] who showed thatthe borehole radius affects the stress intensity factor andstress distribution at the radial crack tips which conse-quently affects the crack initiation characteristics such as thefracture initiation angle and the critical water pressureHowever most studies neglected the borehole size andgenerally simplified the borehole and its associated sym-metrical radial precracks as a straight-line fracture [3 10]For instance Chen et al [2] characterized dynamic fracturepropagation path during hydraulic fracturing by simplifyingthe model as a straight line which was inconsistent with theactual experimental configuration Behnia et al [11] pre-sented a boundary element method based on the higherorder displacement discontinuity formulation to model thehydraulic fracture propagation in layered formations inwhich the fracture was shaped being a straight line )emechanical behavior of the line-shaped fracture under hy-draulic pressure has been extensively conducted by variousresearchers such as Sobhaniaragh et al [12] da Silva et al[13] and Cheng et al [14] To study the effect of the hy-draulic fracture initiated from a borehole Zhang et al [15]presented a 2Dmodel for the initiation and growth of one ormore hydraulic fractures from a well)ey suggested that forcomplex multiple fracture cases the misalignment angle andthe number of initiated fractures are important in near-wellbore fracture path selection Sesetty and Ghassemi [16]developed a 2D numerical model based on the displacementdiscontinuity method (DDM) and the fictitious stressmethod (FSM) to investigate the effect of rock anisotropy onwellbore stresses and hydraulic fracture propagation Ad-ditionally the pore pressure is also a critical factor affectingthe hydraulic fracture behavior [17 18] Gholami et al [19]investigated the effect of pore pressure on the hydraulicfracture propagation and indicated that a higher than av-erage pore pressure zone accelerates the fracture growthwhereas a lower pressure anomaly zone resists a growingfracture However few fracture criteria have been proposedto study the fracture initiation by considering the internalrock pore pressure

When a borehole with symmetric cracks is subjected toboth external compressive load and inner water pressure thestress intensity factors at the crack tips include those in bothopening mode (KI) and sliding mode (KII) Many studiesemphasized mathematical or numerical evolution of theopening mode stress intensity factor (KI) for two symmetriccracks emanating from a circular hole when the maximumand minimum horizontal stresses (denoted by σH σh) aretensile loads and the azimuth of the cracks is 0deg [20 21]However when the azimuth of the cracks in Figure 1(a) isnot zero the uniaxial or biaxial tensile loading results in aplane shear stress condition and both opening mode and the

sliding mode stress intensity factor (KI and KII) are nonzero[22] When the model shown in Figure 1(a) is subjected tocompressive loading (σH and σh are negative values) andboth the pore pressure (p) and the injected water pressure(P) are considered no theoretical work has been reported forcalculating the stress intensity factors (KI and KII) under thecombined loading conditions Since it is critical to predictthe accurate path of new fractures in the hydraulic fracturingoperations and well production the geometry and directionof the hydraulic fracture under the condition shown inFigure 1(a) are theoretically studied by an analytical methodin this work

2 Theoretical Model

Dong et al [6] have theoretically studied the feature ofhydraulic fracturing through a single radial perforationemanating from a borehole )e results indicated that pa-rameters such as the perforation length the borehole radiusand the stress anisotropy coefficient significantly affect thecritical water pressure and the critical initiation anglegreatly However the fracture growth behavior of twosymmetrical radial cracks (as shown in Figure 1(a)) is dif-ferent from that of a single radial crack which needs to befurther investigated Most of the existing studies assumedthat the perforation was horizontal (β 0deg) Consequentlythe far-field equivalent shear stress has no influence on themode-II stress intensity factors )e model shown inFigure 1(a) can be superposed by the models inFigures 1(b)ndash1(d) In addition Figures 1(b) and 1(c) are thehorizontal fractures subjected to far-field equivalent normaland shear stresses respectively whereas Figure 1(d) is thefracture subjected to injected water pressure and the porepressure in rock Since the perforation angle is zero there isno mode-II stress intensity factor under the loading con-ditions shown in Figures 1(b) and 1(d) and only the loadingconditions in Figure 1(c) lead to mode-II stress intensityfactor

For the model shown in Figure 1(b) the mode-I stressintensity factor caused by the far-field in situ stress isexpressed by the following equation [20 23]

KIminusσ σhprimeπa

radic1113896 1minus kprime( 1113857 05 3minus

a

R + a1113874 1113875 1 + 1243 1minus

a

R + a1113874 1113875

21113890 11138911113896 1113897

+ kprime 1 + 1minusa

R + a1113874 1113875 05 + 0743 1minus

a

R + a1113874 1113875

21113890 11138911113896 11138971113897

(1)

where σhprime is the far-field equivalent maximum normal kprimeis the equivalent stress anisotropy coefficient and a and Rare the perforation length and the borehole radiusrespectively

Following the convention in elastic mechanics thetensile stress and dilatant strain are referred as positivewhereas the compressive stress and contractive strain arereferred as negative in this study )erefore because the insitu stress shown in Figure 1(b) is compressive stress theparameter σhprime in equation (1) is negative According to thetheory of elastic mechanics for an inclined perforation

2 Advances in Materials Science and Engineering

angled β the far-eld equivalent maximum normal stress(σhprime) and far-eld equivalent minimum normal stress (σHprime )are written as follows

σHprime σhσ[(1 + k)minus(1minus k)cos 2β]

2

σhprime σhσ[(1 + k) +(1minus k)cos 2β]

2

(2)

where σh is the minimum horizontal stress and k is the ratioof the maximum horizontal stress to the minimum hori-zontal stress k σHσh

In addition the equivalent stress anisotropy coecient(kprime) in Figure 1(b) can be expressed by equation (3) whichrepresents the ratio of the far-eld equivalent maximumnormal stress to the far-eld equivalent minimum normalstress kprime σHprime σhprime

kprime (1 + k)minus(1minus k)cos 2β(1 + k) +(1minus k)cos 2β

(3)

Several methods have been proposed to estimate stressintensity factors through the nite element method (FEM)in which the interaction integral method yields higheraccuracy [24] Hence in this paper the stress intensityfactors are calculated by the interaction integral methodand for more details refer to the ANSYS software manual[24]

For the model shown in Figure 1(c) it is subjected to afar-eld equivalent shear stress which results in the mode-IIstress intensity factor For this loading condition Hsu [22]theoretically studied the mode-II stress intensity factor fortwo radial cracks emanating from a circular hole in aninnite sheet under uniform in-plane shear loading which islisted in Table 1 However for small cracks with aRlt 01the mode-II stress intensity factor was unavailable from Hsu[22] In this study such values are numerically calculated byusing the interaction integral method by ANSYS softwarewhich are marked by the star symbol (lowast) in Table 1 andplotted in Figure 2 Figure 2 shows the numerical resultsof stress intensity factor coecient under dierent values ofa(a+R) and the theoretical results are obtained from Hsu[22] e results indicate that our numerical results agreewell with the theoretical results

According to Figure 2 the mode-II stress intensity factorat the tip of the fracture shown in Figure 1(c) can be tted bythe following equations

YII 3127x3 minus 75216x2 + 53936x + 00089 (4)

KII τeffYIIπa

radic (5)

where YII is the normalized mode-II stress intensity factorx a(a+R) and τe is the far-eld equivalent shear stressFor an inclined perforation with the angle of β the far-eldequivalent shear stress can be written by as follows

σh σprime

kprimeσprimeσH = kσh

(a) (b)

(c) (d)

τeff

τeff

αp

αp

λp

λp

aP

P

R

Figure 1 Loading conditions on two inclined symmetrical radial perforations emanating from a borehole (a) can be created by superposing(b) (c) and (d)

Advances in Materials Science and Engineering 3

τeff σh(1minus k)sin 2β1113858 1113859

2 (6)

Newman [20] and Tada et al [23] theoretically de-rived the stress intensity factor formulae for the caseshown in Figure 1(d) without considering the porepressure )e mode-I stress intensity factor in the modelshown in Figure 1(d) can be given by the followingequation [20 23]

KIminusP Pπa

radic1113896(1minus λ) 1minus

a

R + a1113874 111387511138900637 + 0485 1minus

a

R + a1113874 1113875

2

+ 04a

R + a1113874 1113875

21minus

a

R + a1113874 11138751113891 + λ11138961 + 1minus

a

R + a1113874 1113875

middot 05 + 0743 1minusa

R + a1113874 1113875

21113890 111389111138971113897

(7)

where P is the water injection pressure and λ is the ratiobetween the water pressures in the fracture and theborehole

Based on equation (7) variation of the mode-I stressintensity factor with the perforation length (a) can bewritten by equation (8) and the results are illustrated inFigure 3

YIminusP( 1113857a KIminusP

P

πa


a

R + a1113874 1113875


R + a1113874 1113875

2+ 04

a

R + a1113874 1113875

21minus

a

R + a1113874 11138751113890 1113891

+ λ 1 + 1minusa

R + a1113874 1113875 05 + 0743 1minus

a

R + a1113874 1113875

21113890 11138911113896 11138971113897

(8)

According to Figure 3 in the case of λ 1 (the hydraulicpressure acts all over the fracture surface) the stress intensityfactor monotonically increases along the crack from theborehole to the perforation tip and the borehole radius isassumed constant (R 1 cm) )us the fracture propagatescontinuously once it is initiated On the other hand in thecase of λ 0 (the hydraulic pressure acts only inside theborehole) there is a critical point before which the stressintensity factor increases with the increment of the cracklength from the borehole to the perforation tip After thecritical point the stress intensity factor decreases along thecrack length from the borehole to the perforation tip)erefore when high-viscosity fluid is used as the fracturingfluid the stress intensity factors may not reach the fracturetoughness even if the perforation length increases Hence thefracture will not propagate unless the water injection pressureis increased to enlarge the value of stress intensity factor )isconclusion has also been demonstrated by the numerical

Table 1 Normalized mode-II stress intensity factor for uniform in-plane shear [22]ξ aR 001lowast 002lowast 003lowast 004lowast 005lowast 006lowast 007lowast 008lowast 009lowast 010FII 00347 01137 01604 02242 02570 02970 03337 03726 04177 04310ξ aR 015 020 030 040 050 060 080 10 150 200FII 05858 07114 08912 10086 10866 11380 11930 12134 12092 11854ξ aR 300 400 500 600 800 1000 2000 infinFII 11424 11130 10930 10788 10602 10486 10246 10000lowastNumerically calculated normalized mode-II stress intensity factors by ANSYS software using the interaction integral method

00

02

04

06

08

10

12

14

00 01 02 03 04 05 06 07 08 09 10

Y II

YII = 3127x3 ndash 75216x2 + 53936x + 00089x = a(R + a)

a(R + a)

R2 = 09997

FEM-YIIaR gt 01FEM-YIIaR lt 01

Hsu [20]Fitting formula

Figure 2 Normalized mode-II stress intensity factors

00

01

02

03

04

05

06

07

08

00

10

20

30

40

50

60

00 10 20 30 40 50 60 70 80 90 100

K I-P

P (λ

= 0

0)

K I-P

P (λ

= 1

0)

a (cm)

R = 10 cm

λ = 1λ = 0

Figure 3 Variation of the mode-I stress intensity factor with theperforation length


results in Zhang et al [25] who showed that a high value offluid viscosity would lead to high breakdown pressure

According to equation (7) the critical point when λ 0can be evaluated by the following equation

zKIminusPz(aR + a)

1113868111386811138681113868111386811138681113868λ0 0 (9)

)e solution of equation (9) shows that aR 0589is the critical point It means that in the case of λ 0 whenaRlt 0589 the stress intensity factor increases with theincrement of the crack length from the borehole to the cracktip but it decreases when aRgt 0589 )erefore for aninitially short perforation with aRlt 0589 if the input waterpressure is larger than the fracture initiation stress thefracture will extend without an additional pressure till thecrack length exceeds 0589R

Figure 4 shows the relationship between the normalizedmode-I stress intensity factor (KI-PP(πa)05) and thenormalized perforation length (a(R + a)) for differentvalues of the parameter λ )e results indicate that thevalue of KI-PP(πa)05 increases as λ increases for the samea(R + a) a(R + a)⟶ 0 or a≪R KI-PP(πa)05 1122 and2243 for λ 0 and 1 respectively which indicates theexistence of the borehole enlarges the stress intensityfactors many times no matter the water flows intothe fracture or not However a(R + a)⟶1 or a≫R KI-PP(πa)05 0 and 1 for λ 0 and 1 respectively which in-dicates the borehole can be ignored if the perforation lengthis much greater than the borehole radius For such case thestress intensity factor equals P

πa

radic

For the case with a borehole there is no theoreticalequation available to estimate the mode-I stress intensityfactor caused by the pore pressure (αp) shown in Figure 1(d))e widely used weight function which only depends on themodel geometry and crack type is developed to determinethe stress intensity factor If the weight function for themodel is known the associated stress intensity factors can beobtained by multiplying this function by the stress distri-bution in the intact body and integrating it along the cracklength [9 26] Hence in this paper the mode-I stress in-tensity factor caused by pore pressure was calculatedthrough the weight function method )e weight function isonly dependent on fracture geometry to an integrationexpression as follows

KI 1113946a

0σ(x)m(x a)dx (10)

m(a x) 2

2π(aminusx)

1113968 11138901 + M1 1minusx

a1113874 1113875

12+ M2 1minus

x

a1113874 1113875

+ M3 1minusx

a1113874 1113875

321113891

(11)

where σ(x) is the stress field resulting from the loads appliedon the uncracked body normal to the faces of the

prospective crackm(x a) is a weight function in which x is adummy variable and M1 M2 and M3 are weight functionparameters

To obtain the universal weight function m(x a) inequation (11) for a particular cracked body it is necessary todetermine the three parameters M1 M2 and M3 For aninfinite plate with two symmetric cracks emanating from aborehole Jin et al [9] determined these three parameterswhich can be written by

M1 0074x51 minus 01765x

41 minus 03516x

31 + 082x

21

+ 02964x1 minus 04255(12)

M2 minus0196x51 + 03508x

41 + 09246x

31 minus 15545x

21

minus 09476x1 + 07175(13)

M3 0098x51 minus 01754x

41 minus 04623x

31 + 07772x

21

+ 04738x1 + 01413(14)

where x1 log10(aR) and 0001le aRle 100For the loading condition with pore pressure σ(x) in

equation (10) becomes αp According to equations (10)ndash(14)the stress intensity factor caused by the pore pressure can begiven by the following equation

K(1)Iminusαp αp

πa

radicmiddot

2

radic

π2 + M1 +

23M2 +

12M31113874 1113875 (15)

Figure 5 depicts the relationship between the normalizedmode-I stress intensity factor caused by the pore pressureand the normalized crack length (a(a+R)) )en it can bewell predicted by the following equations

00

03

06

09

12

15

18

21

24

0 01 02 03 04 05 06 07 08 09 1

K I-P

P (π

a)0

5

a(R + a)

λ = 00λ = 05λ = 10

Figure 4 Normalized mode-I stress intensity factor


YIminusαp KIminusαp

αpπa

radic 11306minus 10552a

R + a( ) + 27211a

R + a( )2

minus 34525a

R + a( )3+ 1702

a

R + a( )4

(16)

KIminusαp αpπa

radicKIminusαp (17)

It can be seen from equations (16) and (17) that the valueof KI-αp increases as the pore pressure increases which has apositive eect on the hydraulic fracture initiation and re-duces the required pressure According to the above-mentioned discussion on the stress intensity factors undervarious cases the mode-I stress intensity factor caused by theloading conditions in Figure 1 can be evaluated by the sum ofequations (1) (7) and (17) as follows

KI KIminusσ +KIminusP +KIminusαp (18)

According to fracture mechanics the stress eld sur-rounding the fracture tip in the polar coordinate systemshown in Figure 6 can be given by the following equation

σr 1

22πr

radic KI cosθ2(3minus cos θ) +KII sin

θ2(3 cos θminus 1)[ ]

σθ 1

22πr

radic cosθ2KI(1 + cos θ)minus 3KII sin θ[ ]

τrθ 1

22πr

radic cosθ2KI sin θ +KII(3 cos θminus 1)[ ]

(19)

where r and θ are the polar coordinates originating at thefracture tip and the line of extension respectively e stressintensity factors KI and KII in equation (19) for the model

shown in Figure 1 can obtained by equations (18) and (5)respectively

To study the eects of combined loadings on the fractureinitiation several mixed-mode fracture criteria have beenproposed and veried under specic experimental condi-tions For tensile cracking the maximum tangential stresscriterion (MTS-criterion) proposed by Erdogan and Sih [27]is the simplest one and is widely used e MTS criterionstates that the fracture initiation direction coincides with thedirection of the maximum tangential stress along a constantradius around the crack tip Generally the MTS criterionprovides accurate results for tensile fracture in most brittlematerials including rock [28] e fracture initiation di-rection θc according to the MTS criterion can be found fromthe conditions specied by the following equation

σθ gt 0

zσθzθ

0

z2σθzθ2lt 0

(20)

By substituting the stress eld in equation (19) into theMTS criterion in equation (20) equation (21) is obtained

cosθc2KI 1 + cos θc( )minus 3KII sin θc[ ]gt 0

3 cosθ2KI sin θ +KII(3 cos θ minus 1)[ ] 0

3 sinθ2KI sin θ +KII(3 cos θminus 1)[ ]

minus 6 cosθ2KI cos θminus 3KII sin θ[ ]lt 0

(21)

Solving equation (21) to obtain the fracture initiationangle θc equation (22) is obtained

θc

0 KII 0

2 arctanKI minus

K2

I + 8K2II

radic( )

4KII

KII ne 0

(22)

090

095

100

105

110

115

00 02 04 06 08 10

Y I-α

p

a(R + a)

From equation (15)Fitting formula

YI-αp = 11306 ndash 10552x + 27211x2 ndash 34525x3 + 1702x4

x = a(R + a)R2 = 09989

Figure 5 Normalized mode-I stress intensity factor caused by porepressure

Fracture tipθ x

r

y σθ σrτrθ

Figure 6 Stresses at the fracture tip


where the fracture initiation angle (θc) is referred aspositive when it is measured along the counterclockwisedirection and negative when clockwise Equation (22)indicates that under the case of the pure mode-I fractureie KII 0 the fracture initiates along its original di-rection ie θc 0 However under the case of mix-modefracture ie KII ne 0 the fracture initiation angle dependson the positive or negative KII )e negative KII causes thepositive value of fracture initiation angle while the

positive KII results in the negative value of fracture ini-tiation angle

Moreover equation (23) can be obtained according tothe second formula in equation (18)

σθ( 1113857max

2πr

radic12cos

θc2

KI 1 + cos θc( 1113857minus 3KII sin θc1113858 1113859

(23)

By substituting equation (22) into equation (23)

σθ( 1113857max

2πr

radic

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(24)

According to fracture mechanics the mode-I fracturetoughness KIC can be obtained from equation (25) which isan important parameter of the mechanical properties of rocksand describes the materialrsquos resistance to crack propagation

KIC σt2πr

radic σθ( 1113857max

2πr

radic (25)

where σt is the critical tensile strengthWhen the water injection pressure reaches the critical

water pressure Pc the tangential stress σθ reaches its max-imum value which equals the tensile strength (σt) of therock Hence according to equations (24) and (25) thecritical water pressure can be obtained from equation (26)

KIC Keq

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(26)

where Keq is the equivalent stress intensity factor)e fracture criterion used in this study assumes that the

propagation direction is along a direction normal to themaximum tangential tensile stress and when the equivalentstress intensity factor Keq is greater than or equal to thefracture toughness of the rock KIC the fracture propagates)us equation (26) is an explicit solution for the criticalwater pressure Pc which can be easily applied in practiceConventional approaches commonly assume that the hy-draulic fracture initiates and propagates under the conditionthat is controlled by the mode-I (opening) stress intensityfactor ie KI reaches the value of the critical stress intensityfactor or the fracture toughness ie KIC Such a modelpredicts that no fracturing will occur until KI KIC Forexample Mogilevskaya et al [29] studied the hydraulicfracturing in which the critical water pressure was de-termined by adjusting the uniform pressure on the boundaryof a wellbore to meet the condition of KI KIC and theextension angle was determined by assuming KII 0 Suchmethod was also implemented in many numerical worksconducted by researchers such as Lecampion and Desroches[30] Zhang et al [15] Bao et al [31] and Garagash [32])us these studies have unanimously ignored the effect ofKII caused by the far-field in situ stress on the critical waterpressure which is only applicable given the condition thatthe perforation is perpendicular or parallel to the principalstress ie there is no effect of the equivalent shear stress inthe model However in the hydraulic fracturing engineering

practice the actual perforation angle is not always parallel tothe orientation of the maximum horizontal stress and theappearance of the shear stress is inevitable ie KIIne 0 Inaddition equation (26) indicates that KII is an importantfactor affecting the hydraulic fracture initiation )eequivalent stress intensity factor Keq conveniently addressesthe effect of both KI and KII on the critical injected waterpressure

3 Validations

To validate the proposed theoretical model the experi-mental data from Chen et al [2] and Jin [33] are used forvalidation In the experiments the samples were made ofcement and quartz sand at the volume ratio of 1 1 with ablock size of 300mm times 300mm times 300mm )e preexistingperforation length was a 30mm and the borehole di-ameter was R 15mm For numerical replication only a2D model with a size of 300mm times 300mm as illustrated inFigure 1(a) was used to investigate the hydraulic fractureinitiation )e minimum horizontal stress (ie σh shown inFigure 1(a)) was 10MPa Two types of maximum hori-zontal stresses (ie σH shown in Figure 1(a)) were used inthe experiments ie 4MPa and 6MPa respectively )erewas no pore pressure in the block specimens ie KIminusαp 0Based on the experiments Jin [33] suggested that thefracture initiation angle was zero when the perforationorientation was perpendicular to the plane of the minimum


horizontal stress and the newly created fracture propagatesto the direction perpendicular to the minimum horizontalprincipal stress

According to the parameters in the experiments thecritical water pressure and critical initiation angle are cal-culated theoretically by equations (26) and (22) respectivelywhich are also shown in Figure 7 together with the exper-imental results In Figure 7 the parameter k is the stressanisotropy coefficient defined in Figure 1(a) It can be seenfrom Figure 7(a) that when k 4 the theoretical results ofthe critical water pressure agree well with those obtainedfrom the experiments (the blue and red lines) As shown inFigure 7(b) the theoretical results of the critical waterpressure are consistent with the experimental results whenthe perforation angles equal 30deg and 60deg respectivelyHowever when β 15deg the difference between the theo-retical and the experimental result is roughly 18MPa Forthe critical initiation angle shown in Figure 7(c) the the-oretical and experimental data also fit well with each otherexcept for the azimuth of the fracture β 45deg in which thedifference is about 45deg When k 6 the theoretical results ofthe critical water pressure agree with the experimental dataat β 30deg and 45deg while at two other perforation angles(ie β 15deg and 60deg) the differences are significant for boththe critical water pressure and critical initiation angle )ediscrepancies between the theoretical and experimentalresults are possibly caused by following reasons (1) thetheoretical model is simplified from the practical experi-ments which may result in the experimental conditions thatcannot be fully realized by the theoretical model (2) Due tothe heterogeneity of the mortar blocks in the experiment(such as micropores pores cracks etc) the critical waterpressures obtained in the hydraulic fracturing experimentsmay have a certain range whereas the theoretical calculationassumes the material be homogeneous (3) Some inherenterrors in the experimental measurements are unavoidable Itis worth mentioning that an advantage of the presentedtheoretical model lies in its capability to investigate fractureinitiation in homogeneous rocks By considering the rockmaterial heterogeneity incorporating the proposed modelinto the numerical model benefits understanding the initi-ation and growth of hydraulic fracture in heterogeneousmaterial which potentially broadens the application of thepresent study

To further compare the theoretical and experimentalresults the experimental results of the critical water pressureare substituted into equation (22) to calculate the theoreticalresults of the fracture initiation angle for both k 4 and k 6which are plotted in Figure 7(c) and labelled ldquo)eoreticalwith EXP-Prdquo For both k 4 and k 6 the theoretical pre-dictions in the data series of ldquo)eoretical with EXP-Prdquo aremuch closer to the experimental results For example whenk 6 and β 15deg the differences between the theoretical andexperimental results are reduced from 36 to 12 if theexperimental results of the critical water pressure aresubstituted into equation (22) to calculate the critical ini-tiation angle Similarly when k 6 and β 60deg the theo-retical fracture initiation angle is in good agreement with theexperimental results

)e comparisons between the theoretical results ob-tained for k 4 and k 6 indicate that the horizontal stressanisotropy coefficient (k) has a great influence on thefracture initiation )e larger the horizontal stress anisot-ropy coefficient the shorter the distance of the reorientedfracture away from the maximum horizontal stress )eoverall good agreement between the theoretical and ex-perimental results shown in Figure 7 indicates that thetheoretical model proposed in this study can quantitativelypredict the critical water pressure and fracture initiationangle for the hydraulic fracturing model with the sym-metrical perforation emanating from a borehole )e effectof different parameters on the critical water pressure andfracture initiation angle is discussed in the following

4 Theoretical Results of the HydraulicFracture Initiation

Section 2 shows that the proposed theoretical model requiresseveral input parameters )ey are the far-field minimumand maximum horizontal stresses σh and σH whereσH kσh the perforation angle and length the boreholeradius Biotrsquos coefficient the pore pressure the ratio of thewater pressure in the fracture and borehole and the mode-Ifracture toughness All the parameters are directly mea-surable experimentally In the following sections the effectsof these parameters on the critical water pressure and criticalinitiation angle in hydraulic fracturing engineering practiceare discussed in which the mode-I fracture toughness KIC ofthe rock is assumed to be 21MPamiddotm12 [34]

41 Effect of Stress Anisotropy Coefficient )e ratio betweentwo horizontal stresses shown in Figure 1(a) affects signif-icantly the fracture dynamic reorientation )e larger thestress anisotropy coefficient the shorter the distance of thereoriented fracture away from the maximum horizontalstress [2] However there were no systematical analyticalstudies on the effect of the stress anisotropy coefficient onthe critical water pressure and critical initiation angle for themodel of a pressurized two symmetric radial perforationsemanating from a borehole In this study the stress an-isotropy coefficient is defined as the ratio of the maximum tominimum horizontal stresses and denoted by the parameterk which is illustrated in Figure 1(a) )e minimum hori-zontal stress is kept constant and hence the different values kof the stress anisotropy coefficient are obtained by changingthe maximum horizontal stress )e obtained results aredepicted in Figure 8 in which 8(a)ndash8(c) illustrate the effect ofstress anisotropy coefficient on the critical water pressureunder different perforation lengths ie a 10mm 100mmrespectively while 8(b)ndash8(d) show the corresponding criticalinitiation angle All other input parameters are presented inthese figures

It can be seen from Figure 8(a) that when k 1 ie in thehydrostatic stress state the value of critical water pressureremains constant for all the perforation angle ranging from0degsim90deg For all other stress ratios kne 1 when β 0deg the waterpressure reaches the critical water pressure which will result


in the fracture propagation in the direction parallel to themaximum horizontal stress As the perforation angle in-creases the critical water pressure increases gradually )istheoretical conclusion has also been demonstrated by thenumerical results obtained by Sepehri et al [4] who showedthat the breakdown pressure increases as the perforationazimuth increases When the perforating direction is per-pendicular to the maximum horizontal stress (β 90deg) thecritical water pressure reaches the maximum value

Figure 8(a) shows that there is an intersection point (β0)between the curves with different stress anisotropy co-efficients where the critical water pressure decreases withthe increment of stress anisotropy coefficient when0degle βle β0 while it shows the opposite trend whenβ0le βle 90deg What is the reason that there is an intersectionpoint β0 for different stress anisotropy coefficients whena 10mm Equations (18) and (5) indicate that KI-σ and KIIdepend on the perforation angle Both the equivalent normal

40

60

80

100

120

140

160

10 20 30 40 50 60

P c (M

Pa)

Theoretical k = 4Experimentalk = 4 [31]


β (deg)

(a)P c

(MPa

)



40

60

80

100

120

140

160

10 20 30 40 50 60β (deg)

(b)

00

50

100

150

200

250

300

350

10 20 30 40 50 60

ndashθc (

deg)

β (deg)

Theoretical k = 4Theoretical k = 6Experimentalk = 4 [31]

Experimental k = 6 [31]Theoretical with EXP-Pk = 4 [31]Theoretical with EXP-Pk = 6 [31]

(c)

Figure 7 Comparison of the critical water pressure and critical initiation angle obtained by using the theoretical model and the ex-perimental tests obtained from Chen et al [2] and Jin [33]


stresses (σprime) and KII shown in equations (2) and (5) aresymmetrical at β 45deg which is not the reason for the ex-istence of the transitional azimuth β0 However theequivalent stress anisotropy coefficient (krsquo) in equation (1)depends on both the stress anisotropy coefficient (k) and theperforation angle (β) which can be seen from equation (3)When the perforation length (a) and the wellbore radius (R)are constant the variation of stress anisotropy coefficientresults in the change of the equivalent stress anisotropycoefficient and consequently affects the value of KI-σ whichcan be calculated using equation (1) and is shown in Fig-ure 9 )e intersection angle is clearly observed inFigure 9(a) when the perforation length a 10mm |KI-σσh|increases with the increment of stress anisotropy coefficientwhen 0degle βle β0 while it shows the opposite pattern when

β0le βle 90deg which is same as the corresponding featureshown in Figure 8(a) In summary the reason for the ap-pearance of the intersection point in Figure 8(a) is that thestress anisotropy coefficient results in an intersection pointof KI-σ )e intersection point β0 will gradually approximatezero with the increment of perforation length as shown inFigure 9(b) )e value of KI-σσh in the model without thewellbore (ie there is no wellbore in Figure 1(a)) and theperforation length of (a+R) nearly equals that with wellborewhen a 300mm and k 3 which indicates that the effect ofthe borehole radius on KI-σ can be neglected when the valueof aR is larger than a critical value

Same as that in Figure 8(a) the critical water pressure fora 100mm is kept constant when the stress anisotropycoefficient k 10 as shown in Figure 8(c) Moreover the

60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25

a = 10 mmR = 50 mmσh = 30 MPaσH = kσhλ = 1α = 0

β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)

Figure 8 Effect of the stress anisotropy coefficient on the critical water pressure and critical initiation angle


critical water pressure increases with the increment of theperforation angle when the stress anisotropy coefficients areequal to 15 or 20 However for the higher stress anisotropycoefficient such as k 3 the critical water pressure slightlydecreases with the increment of the perforation anglepreceding the increase when the perforation angle exceeds acritical value

Figures 8(b)ndash8(d) show the effect of the stress anisotropycoefficient on the critical initiation angle With the increaseof the perforation angle the critical initiation angle deviatesfrom the perforation direction which increases at first thenmaximize at β 45deg and finally decreases )e curves of thecritical initiation angle versus the perforation angle aresymmetrical along β 45deg When the perforation angle isparallel to the maximum or minimum horizontal stressdirections the critical initiation angle equals zero )ecritical initiation angle increases as the stress anisotropycoefficient increases When the perforation length is smallerthan the wellbore radius the critical initiation angle is alsosmall even if the stress anisotropy coefficient is larger whichindicates that the fracture propagation direction is difficultto be changed away from the azimuth of the fracture Withthe perforation length increasing the critical initiation angleincreases remarkably Taking k 25 as an example themaximum fracture initiation angle is 152947deg whena 10mm which however increases to 627121deg whena 100mm )e critical initiation angle in the model withdifferent stress anisotropy coefficients can be addressed asfollows

(1) At a given perforation angle the critical initiationangle increases as the stress anisotropy coefficientincreases )e reason is that the mode-II stress in-tensity factorKII increases according to equations (4)and (5) According to equations (22) and (26) the

effect of stress intensity factors (KI and KII) on thecritical initiation angle (θc) and equivalent stressintensity factor (Keq) is shown in Figure 10 Both theequivalent stress intensity factor and critical initia-tion angle increase as the value of KIIKI increases)e critical initiation angle approaches 686deg whenthe ratio KIIKI is greater than 5 and finally reaches705deg if the ratio KIIKI is sufficiently large )ereforewhen the perforation angle is close to the direction ofthe maximum or minimum horizontal stresses (bothσh and σH) the value of KII is small which means thevalue ofKIIKI is small and the fracture opening is thedominant fracture growth mode the fracture doesnot twist much during its growth and the criticalinitiation angle is almost zero However whenperforation angle deviates much from the directionof the maximum or minimum horizontal stressesthe fracture propagation direction will turn and tendto align towards the orientation of the maximumhorizontal stresses because the absolute value of thecritical initiation angle increases gradually with theincrease of the value of KII as shown in Figure 10 Asthe value of KIIKI becomes large enough the shearsliding becomes the dominant mode of the fracturegrowth )us according to the proposed theoreticalmodel for the model shown in Figure 1(a) thefracture will twist more during its growth underhigher stress anisotropy coefficient conditionswhich is same as the experimental results obtained byIspas et al [35] )is is what hydraulic fracturingengineers have been trying to alleviate or avoid

(2) )e fracture does not twist during its growth whenthe maximum horizontal stress is equal to theminimum horizontal stress (ie k 1) because the

ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2

k = 3k = 3 no wellbore

a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)

Figure 9 Influence of stress anisotropy coefficient on the mode-I stress intensity factor KI-σ


mode-II stress intensity factor is zero which can beobserved from equations (4) and (5) and the shearsliding mode disappears under this condition Inother words the fracture plane will not be twisted)is theoretical conclusion has also been proven bythe experimental results [35]

(3) For the oil and gas reservoirs with higher stressanisotropy coefficient to mitigate the fracture twistnear the wellbore it is recommended to reduce theinclination angle through adjusting the wellboreazimuth andor perforation phasing )us theproposed model and corresponding theoreticalfindings are significant to further understand therefracturing process in the hydraulic fracturing

42 Effect of Borehole Radius It can be seen from the the-oretical model in Section 2 that the borehole radius affectsboth KI and KII and consequently affects the critical waterpressure and critical initiation angle which is shown inFigure 11 )e perforation length is a 20mm inFigures 11(a) and 11(b) while a 200mm in Figures 11(c)and 11(d) When a 20mm the critical water pressuredrops significantly when the borehole radius increases from50mm to 100mm Further increase of the borehole radiusresults in the continuous decrease of the critical waterpressure although the decreasing rate significantly declinesWhen the perforation length is a 200mm the critical waterpressure also decreases as the borehole radius increasesFigures 11(a) and 11(c) indicate that for a shorter perfo-ration length (such as smaller than the wellbore radius) theeffect of borehole radius on the critical water pressure isobvious when Ra is smaller than 5 otherwise it is un-remarkable )e effect of borehole radius on the criticalwater pressure should not be neglected when the perforationangle is close to the direction of the maximum horizontal

stress However if the perforation angle is close to the di-rection of the minimum horizontal stress the effect ofborehole radius on the critical water pressure can beneglected

)e effects of borehole radius on the critical initiationangle are illustrated in Figures 11(b) and 11(d) which de-pend on the perforation length When the perforation lengthis smaller than the borehole radius the critical initiationangle decreases as the borehole radius increases in which thedecreasing rate also slows down However when the per-foration length is greater than the borehole radius thecritical initiation angle was almost not affected by theborehole radius as shown in Figure 11(d)

43 Effect of Perforation Length According to fracturemechanics the increase of the perforation length increasesthe stress intensity factor which therefore decreases thefracture initiation stress As can be seen from Figure 11 forthe borehole with two symmetrical radial perforationsshown in Figure 1(a) the critical water pressure and criticalinitiation angle depend on the perforation length Figure 12shows the effect of the perforation length on the criticalwater pressure and critical initiation angle where theborehole radius is constant Figure 12(a) indicates theperforation length significantly affects the critical waterpressure especially when altR )e difference between thecritical water pressures obtained for a 20mm and 50mm is07982MPa at β 0deg and 16031MPa at β 90deg althoughthere is only 30mm difference in the perforation lengthHowever with the increase of the perforation length thedifference between the critical water pressures obtained forthe adjacent perforation lengths (such as a 150mm and200mm) is not significant which can also be observed inFigure 12(c) It shows that the critical water pressure sig-nificantly decreases at first as the perforation length in-creases and then linearly decreases as the perforation lengthincreases In other words when altR the critical waterpressure is very sensitive to the perforation length and nearlygrows exponentially with the perforation length decreasing)us the greater the perforation length the easier the hy-draulic fracture growth which can be used in practicalhydraulic fracture engineering However it is known thatthe preexisting fractures emanating from the horizontal wellin petroleum engineering are created by perforating gun)us the difficulties of the construction technology and thecosts are to be considered when the perforating gun is usedto produce the greater perforation length Practical hydraulicfracturing engineering shows that there is a reasonable andeconomical perforation length for a specific borehole radiusand a specific stress state

Figures 12(b) and 12(d) indicate that the critical initi-ation angle initially increases exponentially as the perfora-tion length increases when altR and then increases linearlywhen the perforation length is greater )is theoreticalconclusion has also been demonstrated by the numericalsimulation results of Sepehri et al [4] who indicated that thelonger the perforation length the easier the reorientationfrom the initial perforation plane In practical hydraulic

00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc

Figure 10 Effect of stress intensity factor on the critical initiationangle and equivalent stress intensity factor


fracture engineering the misalignment between the fractureinitiation direction and the direction of the in situ stressesusually occurs because the well may be drilled in a directionthat is not aligned with the principal stress directions or theperforations are not aligned with the preferred fracturedirection )erefore as shown in Figure 12 the fractureinitiation direction from a perforation or a natural fractureintersecting the wellbore may not align with the horizontalstress direction (except for the case of the perforation angleparallel to the orientation of the in situ principal stresses) Insuch cases the hydraulic fracture will reorient its propa-gation direction as it grows away from the wellbore until itspropagation direction is eventually parallel to the principalstress direction An optimal perforation length for fracturepropagation would require a minimal water injection

pressure and would generate a fracture at an achievablefracture initiation pressure with minimal tortuosity [2] Itcan be observed from Figure 12 that the greater perforationlength decreases the critical water pressure but increases thetwist of the fracture growth )us to accurately quantify thecritical water pressure in the hydraulic fracturing theknowledge of an accurate perforation length is requiredespecially when the perforation length is short

44 Effect of Ratio between theWater Pressures in the Fractureand the Borehole Generally the pressure in the fracture(ie so-called internal pressure) in impermeable rock causedby low viscous fluid is assumed to uniformly distribute alongthe fracture length and be equal to that in the borehole

40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

a = 20 mmσh = 30 MPaσH = 60 MPaλ = 1α = 0

P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)

Figure 11 Effect of the borehole radius on the critical water pressure and critical initiation angle


ie λ 1 However the viscous fluid and its permeation leadto that the internal pressure nonlinearly distributes along thefracture length in permeable rock Jeffrey [36] assumed thatthe internal pressure at the heel of the preexisting fracturewas equal to the borehole pressure while that at the tip of thepreexisting fracture was zero because of the existing fluid lagMany other studies also discussed the existence of the zonewithout fluid at the tip of the advancing fracture [37ndash39]

)e nonlinear distribution of the internal pressure inthe fracture caused by the fluid viscosity and rock per-meability can be considered in the proposed theoreticalmodel as proposed by Jeffrey [36] )e weight functionmethod formulated in equations (10)ndash(14) can then be

used to evaluate the mode-I stress intensity factor causedby the internal pressure and subsequently KI-P shown inequation (7) is replaced However in this study theobjective is not to derive an accurate decline function forthe internal pressure inside preexisting fracture but tohighlight the effect of the internal pressure on the criticalwater pressure and critical initiation angle )erefore theinternal pressure is assumed to uniformly distribute alongthe fracture but its value is lower than that in the boreholeby changing the parameter λ shown in Figure 1(a) )ecorresponding results are shown in Figure 13 For theinclined perforation with a high internal pressure (λ10)the critical water pressure is lowest as shown in

40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm

R = 50 mmσh = 30 MPaσH = 60 MPaλ = 1α = 0

P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)

β = 30degβ = 45deg


P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800

β = 15degβ = 30degβ = 45deg

ndashθc (

deg)

(d)

Figure 12 Effect of the perforation length on the critical water pressure and critical initiation angle


Figure 13(a) With the decrease of the ratio λ the criticalwater pressure increases rapidly according to a poly-nomial function of the parameter λ such as that shown inFigure 13(b) when β 15deg

However the critical initiation angle is not affected bythe ratio of the water pressures in the fracture and theborehole as shown in Figure 13(c) It is shown in Fig-ure 10 that the reason for the twist of the fracture growthis the influence of the mode-II stress intensity factor )echange of λ affects the mode-I stress intensity factor

caused by the water injection pressure (ie KI-P) inequation (7) but has no effect on KII in equation (5) Inother words KII is independent of λ hence the criticalinitiation angle does not change with the ratio λ changing)e water injection pressure distribution in the fracture isan important parameter for the evaluation of the criticalwater pressure )e effect of the other factors such as thefluid permeability and viscosity which may affect thepressure distribution in the fracture should be furtherdiscussed in future studies

30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03

a = 200 mmR = 50 mmσh = 30 MPaσH = 60 MPaα = 0

P c (M

Pa)

β (deg)

(a)


Pc = ndash28999λ3 + 72137λ2 ndash 6537λ + 27279R2 = 09991

50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)

Figure 13 Effect of the ratio of water pressure in the fracture and the borehole on the critical water pressure and critical initiation angle


45 Effect of Biotrsquos Coefficient )e elevated pore pressuredistribution generates a poroelastic stress whose magnitudeis proportional to the coefficient of effective stress during theelastic deformation of rocks [40] )is is usually referred toas Biotrsquos coefficient which has a value between 0 and 1Reported values of Biotrsquos coefficient for crystalline rocksinclude 027 for charcoal granite [41] Warpinski et al [42]reported the variation of α with changes in stress and porepressure for tight sandstone and chalk from 065 to 095 forwesterly granite and 063 for stone mountain granites )esevalues indicate that Biotrsquos coefficient is significantly greaterthan zero even for low porosity rocks

Figure 14 shows the effect of Biotrsquos coefficient (α) on thecritical water pressure and critical initiation angle It can beseen from Figure 14(a) that the critical water pressure de-creases as Biotrsquos coefficient increases which is beneficial tothe cost savings for the hydraulic fracturing engineeringAccording to equations (16) and (17) the value of KI-αpincreases as Biotrsquos coefficient increases Under a certain insitu stress situation the borehole radius the perforationlength and the mode-I and mode-II stress intensity factorsin equations (1) and (5) are constant )erefore for constantmode-I fracture toughness KIC KI in equation (18) is alsoconstant which indicates the increase of KI-αp decreases KI-Pand consequently reduces the critical water pressure Inother words the critical water pressure decreases as the porepressure increases )is result is very useful for the practicalhydraulic fracturing engineering For example the porepressure can be increased by allowing the fluid leak off intothe formation or decreasing the flow rate to allow thefracturing fluid more time to leak off into the formation)eincrease of the pore pressure then decreases the required thecritical water pressure according to the proposed theoreticalmodel and then helps cut the costs and enlarges the stim-ulated volume for the shale reservoirs)e theoretical resultsare consistent with the work of Gou et al [43] and Bruno andNakagawa [44] who showed that an increase of the porepressure decreases the critical pressure required to initiate afracture and reduce the width of the newly formed fractureFigure 14(b) also suggests that the critical initiation angle isnot affected by the pore pressure which is same as the effectof the parameter λ shown in Figure 13(c) )e reason is thatthe change of α affects KI but has no effect on KII

5 Effect of High-Pressure Water Injection

In hydraulic fracturing engineering such as oil and gasexploitation and geothermal utilization the stress state of theexisting fracture changes with the changing injected waterpressure )e high-pressure water injection may result inthree kinds of states (1) the fracture does not propagatebecause the injected water pressure is smaller than Pc (2) thefracture growth occurs because of the suitable injected waterpressure and (3) the injected water pressure is too largewhich leads to not only the fracture growth in the oil or gasreservoir but also the breakdown of other strata sub-sequently resulting in water leakage )e last state is usuallyobserved in the practical hydraulic fracturing engineeringwhich is bad for the oil or gas exploration and even results in

disasters For example in the deep geothermal developmentin Basel Switzerland due to the injected excessive waterpressure which induced earthquakes the entire project wasforced to be stopped )e injected different water pressuresresult in the different fracture initiation angles For examplewhen the parameters KIC 21MPamiddotm12 a 100mmR 50mm σh 3MPa σH 6MPa and λ 1 the porepressure is neglected as shown in Figure 8(a) )e minimumvalue and maximum value of critical water pressure are60726MPa and 92022MPa respectively )erefore if theinjected water pressure is greater than 92022MPa thefracture will propagate in the whole range of the azimuth ofthe fracture ie 0degsim90deg Four injected water pressures areconsidered ie P 12 15 20 and 25MPa and the corre-sponding results are shown in Figure 15)e results indicatethat fracture initiation angle under high-pressure waterinjection loading condition is smaller than that under criticalwater pressure At a given perforation angle the higher theinjected water pressure the lower the fracture initiationangle )is is because the high-pressure water injectionincreases KI while KII is constant which consequently de-creases fracture initiation angle according to equation (22)In other words the increase of KI indicates the openingmode is overcoming the sliding mode For β 30deg fractureinitiation angle is 420664deg however it quickly drops to169628deg when P 12MPa and 69006deg when P 25MPaWhen a horizontal well is fractured with a massive multi-stage hydraulic fracturing technology the results obtainedfrom this study indicate that the shape of the fracture nearhorizontal lateral is different at each stage in which thefracture plane twists less at stage I than that at stage II As aresult the increase of the injected water pressure will helpreduce the fracture tortuosity near the wellbore and will thendecrease the skin factor and increase the ultimate oil or gasrecovery However too large injected pressures will not onlyincrease the costs but also increase the risk of the pressurepenetration into other strata inducing earthquakes )us asuitable injected water pressure needs to be carefully esti-mated by the proposed theoretical model Figure 15 alsoindicates that fracture initiation angle is more sensitive to thechange of the pumping pressure for perforation angle near45deg which needs additional attentions since the fracture maytwist more in this region Different from the condition ofcritical water pressure the high-pressure water injectionloading condition leads to the fracture initiation angle be-coming nonsymmetric along the vertical line through β 45degin Figure 15 where the maximum value is about at β 50deg

)e fracture initiation angle may be different due to thevariation of the injected water pressure the in situ stressesand the geometries of the fracture and the borehole Fig-ure 16 shows the effect of these parameters on the fractureinitiation angle under high-pressure water injection loadingcondition (ie the injected water pressure is larger than thecritical water pressure) All the results are calculated byassuming the injected water pressure to be 20MPa

Figures 16(a) and 16(b) illustrate the effect of the stressanisotropy coefficient on the fracture initiation angle underhigh-pressure water pressure loading condition where theperforation lengths are different ie long and short


perforations )e results indicate that the increase of thestress anisotropy coefficient increases the fracture initiationangle for both short and long perforations )is is becausethe high stress anisotropy coefficient increasesKII indicatingthe sliding mode is enhanced )e results also suggest thatthe perforation length slightly affects the fracture initiationangle under the condition of the perforation length equal orlarger than the borehole radius

Both Figures 16(c) and 16(d) show the effect of theborehole radius on the fracture initiation angle under thehigh-pressure water injection loading condition with dif-ferent perforation lengths )e results suggest that under the

long perforation length (larger than the borehole radius) thefracture initiation angle is not significantly affected by theborehole radius However under the short perforationlength the fracture initiation angle decreases as the boreholeradius increases When the borehole radius is ten timeslarger than that of the perforation length the maximumfracture initiation angle is only about 246deg at β 50deg As theborehole radius increases further the fracture initiationangle reduces to about 0deg It should be noted that with thegrowth of the fracture the perforation length increasesresulting in the initially short fracture twists greatly duringits growth such as that in Figure 16(d) and finally propagatealong a constant angle such as that in Figure 16(c)

Figures 16(e) and 16(f) show the effect of perforationlength on the fracture initiation angle under differentborehole radius )e result shown in Figure 16(e) indicatesthat the perforation length does not affect the fractureinitiation angle when the perforation length is longer thanthe borehole radius ie ageR which can also be seen inFigures 16(a) and 16(b) However as shown in Figure 16(f )when the perforation length is shorter than the boreholeradius the increase of the perforation length increases thefracture initiation angle for the same perforation angle suchas that for a 50mm and 100mm When the perforationlength is larger than the wellbore radius the fracture ini-tiation angle does not change significantly with the increaseof the perforation length such as that for a 150mm and200mm which is same as the results shown in Figure 16(e))erefore if the perforation length is smaller than thewellbore radius the fracture initiation angle increases withthe growth of the fracture while finally tends to be a constantvalue if the perforation length is much larger than theborehole radius

Figure 16(g) shows that the fracture initiation angleincreases with the decrease of λ which suggests the higher

00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10

a = 100 mm R = 50 mmσh = 30 MPa σH = 60 MPaλ = 1 p = 50 MPa

P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00

a = 100 mmR = 50 mmσh = 30 MPaσH = 60 MPaλ = 1p = 50 MPa

ndashθc (

deg)

β (deg)

(b)

Figure 14 Effect of Biotrsquos coefficient on the critical water pressure and critical initiation angle

00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90

Critical FIAP = 12 MPaP = 15 MPa

P = 20 MPaP = 25 MPa

a = 100 mmR = 50 mmσh = 30 MPaσH = 60 MPaλ = 1α = 0

ndashθc (

deg)

β (deg)

Figure 15 Influence of the high-pressure water injection on thefracture initiation angle


00

50

100

150

200

250a = 200 mmR = 50 mmσh = 30 MPaPc = 20 MPaλ = 1α = 0

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)

a = 50 mmR = 50 mmσh = 30 MPaPc = 20 MPaλ = 1α = 0

00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105

120a = 200 mmσh = 30 MPaσH = 60 MPaPc = 20 MPaλ = 1α = 0

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

a = 20 mmσh = 30 MPaσH = 60 MPaPc = 20 MPaλ = 1α = 0

00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100

120 R = 50 mmσh = 30 MPaσH = 60 MPaPc = 20 MPaλ = 1α = 0

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

R = 150 mmσh = 30 MPaσH = 60 MPaPc = 20 MPaλ = 1α = 0

00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued


the fluid viscosity the larger the fracture initiation angleFigure 16(h) indicates that at a given perforation angle thehigher Biotrsquos coefficient the lower the fracture initiationangle since KI increases as Biotrsquos coefficient increasesaccording to equations (16) and (17) Biotrsquos coefficient is anintrinsic property of rock and is not homogeneous [45] thisinhomogeneity leads to different fracture initiation anglesunder different stages of the hydraulic fracturing Biotrsquoscoefficient is assumed to be a unit in the hydraulic fracturingsimulators reported in many studies which lead to anunderestimation of the fracture initiation angle An accuratelaboratory measurement of Biotrsquos coefficient is recom-mended to accurately predict the fracture geometry espe-cially for the hydraulic fracturing applied in shale gas and oilformations )e variation of Biotrsquos coefficient changes theeffective pore water pressure ie αp according to equations(16) and (17) In practical hydraulic fracturing engineeringBiotrsquos coefficient as well as the pore pressure surrounding thefracture is a time dependent parameter which suggests it isimportant to develop a fully coupled poroelastic fracturemodel to study the hydraulic fracture initiation behaviorsAs mentioned in Section 45 the mechanical effect of in-creasing Biotrsquos coefficient also can be regarded as keeping thevalue of Biotrsquos coefficient constant but increasing the porepressure )erefore for practical hydraulic fracturing en-gineering the pore pressure can be increased by allowingfluid leak off through injecting the high water pressure ordecreasing the flow rate to allow the fracturing fluid moretime to leak into the formation which avoids the fracture totwist during its growth and then benefit the design of theperforation angle

6 Conclusions

An analytical model is proposed for investigating the hy-draulic fracture initiation through a borehole with two

symmetric radial perforations )e proposed model isimplemented to predict the critical water pressure andfracture initiation angle for the hydraulic fracturing byconsidering the effect of the stress anisotropy coefficient theborehole radius the perforation angle and length the in-jected water pressure and the pore pressure in reservoirformation According to the theoretical results the followingconclusions can be drawn

(1) )e proposed model addresses the effect of both themode-I and mode-II stress intensity factors ie KIand KII on the critical water pressure and fractureinitiation angle which is critical for studying thehydraulic fracturing under complex loading condi-tions According to the theoretical model the exis-tence of the far-field equivalent shear stress is themain source of the twist of the fracture growth inhydraulic fracturing engineering )e critical waterpressure and critical initiation angle derived from thetheoretical model are compared against the resultsobtained from the experiments and good agreementis found

(2) )e critical initiation angle increases as the stressanisotropy coefficient and perforation length in-crease )e trend of the critical water pressure de-pends on the perforation angle and the ratioof perforation length to that of the borehole radius(aR) )e critical water pressure increases as theperforation angle increases when altR and kne 1while the critical water pressure slightly decreases asthe perforation angle increases at first and thenincreases when the perforation angle exceeds acritical value for agtR and k 3 In addition whenaltR the critical water pressure decreases with theincrement of stress anisotropy coefficient for theperforation angle smaller than β0 while just the

00

40

80

120

160

200a = 200 mmR = 50 mmσh = 30 MPaσH = 60 MPaPc = 20 MPaα = 0

λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10

a = 200 mmR = 50 mmσh = 30 MPaσH = 60 MPaPc = 20 MPaλ = 1p = 5 MPa

00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)

Figure 16 Effect of other factors on the fracture initiation angle under high-pressure water injection


opposite when the perforation angle is greater thanβ0

(3) )e critical water pressure decreases as the boreholeradius the ratio between the water pressure in thefracture and the borehole (λ) the perforation lengthand Biotrsquos ratio increase )e critical initiation angleis insensitive to the parameter λ and Biotrsquos co-efficient However under high water pressureloading condition the fracture initiation angle de-creases as the parameter λ and Biotrsquos coefficientincrease In addition the borehole radius has animportant influence on the critical initiation anglefor altR while the effect of the borehole radius onthe critical initiation angle can be ignored for agtR

)e theoretical model developed in this study can beused to investigate the hydraulic fracture initiation throughboth vertical and horizontal wells )e obtained results helpunderstand the characteristics of the hydraulic fracturingthrough the borehole with two symmetric radial cracks andprovide guidelines for the drilling and hydraulic fracturingpractice but direct applications without any critical analysismay lead to risks due to the assumptions of linear elasticityhomogeneous material symmetrical fracture two di-mension etc Moreover it is valuable to implement thistheoretical model in three-dimensional numerical simulatorby incorporating the material heterogeneity

Data Availability

)e figure and table data used to support the findings of thisstudy are included within the article

Conflicts of Interest

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

Acknowledgments

)is project was financially supported by the NationalNatural Science Foundation of China (51874065)

References

[1] A Vengosh N Warner R Jackson and T Darrah ldquo)eeffects of shale gas exploration and hydraulic fracturing on thequality of water resources in the United Statesrdquo ProcediaEarth and Planetary Science vol 7 pp 863ndash866 2013

[2] M Chen H Jiang G Q Zhang and Y Jin ldquo)e experimentalinvestigation of fracture propagation behavior and fracturegeometry in hydraulic fracturing through oriented perfora-tionsrdquo Petroleum Science and Technology vol 28 no 13pp 1297ndash1306 2010

[3] M Behnia K Goshtasbi G Zhang and S H M YazdildquoNumerical modeling of hydraulic fracture propagation andreorientationrdquo European Journal of Environmental and CivilEngineering vol 19 no 2 pp 152ndash167 2015

[4] J Sepehri M Y Soliman and S M Morse ldquoApplication ofextended finite element method to simulate hydraulic fracturepropagation from oriented perforationsrdquo in Proceedings of the

SPE Hydraulic Fracturing Technology Conference Society ofPetroleum Engineers )e Woodlands TX USA 2015

[5] H Zhu J Deng X Jin L Hu and B Luo ldquoHydraulic fractureinitiation and propagation from wellbore with orientedperforationrdquo Rock Mechanics and Rock Engineering vol 48no 2 pp 585ndash601 2014

[6] Z Dong S Tang P G Ranjith and Y Lang ldquoA theoreticalmodel for hydraulic fracturing through a single radial per-foration emanating from a boreholerdquo Engineering FractureMechanics vol 196 pp 28ndash42 2018

[7] S Tang ldquo)e effects of water on the strength of blacksandstone in a brittle regimerdquo Engineering Geology vol 239pp 167ndash178 2018

[8] S B Tang C Y Yu M J Heap P Z Chen and Y G Renldquo)e influence of water saturation on the short- and long-term mechanical behavior of red sandstonerdquo Rock Me-chanics and Rock Engineering vol 51 no 9 pp 2669ndash26872018

[9] X Jin S Shah J-C Roegiers and J McLennan ldquoWeightfunction of stress intensity factor for symmetrical radialcracks emanating from hollow cylinderrdquo Engineering FractureMechanics vol 159 pp 144ndash154 2016

[10] C Y Dong and C J De Pater ldquoNumerical implementation ofdisplacement discontinuity method and its application inhydraulic fracturingrdquo Computer Methods in Applied Me-chanics and Engineering vol 191 no 8ndash10 pp 745ndash760 2001

[11] M Behnia K Goshtasbi M F Marji and A GolshanildquoNumerical simulation of crack propagation in layered for-mationsrdquo Arabian Journal of Geosciences vol 7 no 7pp 2729ndash2737 2014

[12] B Sobhaniaragh V P Nguyen W J Mansur andF C Peters ldquoPore pressure and stress coupling in closely-spaced hydraulic fracturing designs on adjacent horizontalwellboresrdquo European Journal of MechanicsmdashASolids vol 67pp 18ndash33 2018

[13] B G da Silva and H Einstein ldquoPhysical processes involved inthe laboratory hydraulic fracturing of granite visual obser-vations and interpretationrdquo Engineering Fracture Mechanicsvol 191 pp 125ndash142 2018

[14] W Cheng G Jiang H Tian and Q Zhu ldquoNumerical in-vestigations of the fracture geometry and fluid distribution ofmultistage consecutive and alternative fracturing in a hori-zontal wellrdquo Computers and Geotechnics vol 92 pp 41ndash562017

[15] X Zhang R G Jeffrey A P Bunger and M )iercelinldquoInitiation and growth of a hydraulic fracture from a circularwellborerdquo International Journal of Rock Mechanics andMining Sciences vol 48 no 6 pp 984ndash995 2011

[16] V Sesetty and A Ghassemi ldquoEffect of rock anisotropy onwellbore stresses and hydraulic fracture propagationrdquo In-ternational Journal of Rock Mechanics and Mining Sciencesvol 112 pp 369ndash384 2018

[17] R Prabhakaran H De Pater and J Shaoul ldquoPore pressureeffects on fracture net pressure and hydraulic fracture con-tainment insights from an empirical and simulation ap-proachrdquo Journal of Petroleum Science and Engineeringvol 157 pp 724ndash736 2017

[18] H Shimizu S Murata and T Ishida ldquo)e distinct elementanalysis for hydraulic fracturing in hard rock consideringfluid viscosity and particle size distributionrdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 48 no 5pp 712ndash727 2011

[19] A Gholami M A Aghighi and S S Rahman ldquoEffect of non-uniform pore pressure fields on hydraulic fracture


propagationrdquo Journal of Petroleum Science and Engineeringvol 159 pp 889ndash902 2017

[20] J C Newman An Improved Method of Collocation for theStress Analysis of Cracked Plates with Various ShapedBoundaries National Aeronautics and Space AdministrationWashington DC USA 1971

[21] B M Liaw and M Kamel ldquoStress intensity factors for cracksemanating from a curvilinear hole in an anisotropic planeunder arbitrary loadingsrdquo Engineering Fracture Mechanicsvol 36 no 5 pp 669ndash681 1990

[22] Y C Hsu ldquo)e infinite sheet with two radial cracks fromcylindrical hole under inclined tension or in-plane shearrdquoInternational Journal of Fracture vol 13 no 6 pp 839ndash8451977

[23] H Tada P C Paris and G R Irwin De Stress Analysis ofCracks ASME Press New York NY USA 2000

[24] ANSYSMechanical)eory Reference Release 150 CanonsburgPA USA 2014

[25] B Zhang X Li Z Zhang Y Wu Y Wu and Y WangldquoNumerical investigation of influence of in-situ stress ratioinjection rate and fluid viscosity on hydraulic fracturepropagation using a distinct element approachrdquo Energiesvol 9 no 3 p 140 2016

[26] W Xu X R Wu Y Yu and Z H Li ldquoA weight functionmethod for mixed modes hole-edge cracksrdquo Fatigue ampFracture of Engineering Materials amp Structures vol 41 no 1pp 223ndash234 2018

[27] 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 519ndash525 1963

[28] S B Tang ldquo)e effect of T-stress on the fracture of brittle rockunder compressionrdquo International Journal of Rock Mechanicsand Mining Sciences vol 79 pp 86ndash98 2015

[29] S G Mogilevskaya L Rothenburg and M B DusseaultldquoGrowth of pressure-induced fractures in the vicinity of awellborerdquo International Journal of Fracture vol 104 no 4pp 23ndash30 2000

[30] B Lecampion and J Desroches ldquoSimultaneous initiation andgrowth of multiple radial hydraulic fractures from a hori-zontal wellborerdquo Journal of the Mechanics and Physics ofSolids vol 82 pp 235ndash258 2015

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

[32] D I Garagash ldquoPropagation of a plane-strain hydraulicfracture with a fluid lag early-time solutionrdquo InternationalJournal of Solids and Structures vol 43 no 18-19pp 5811ndash5835 2006

[33] X Jin An Integrated Geomechanics and Petrophysics Study ofHydraulic Fracturing in Naturally Fractured ReservoirsUniversity of Oklahoma Norman OK USA 2014

[34] Z Cui D Liu G An and M Zhou ldquoResearch progress inmode-I fracture toughness testing methods for rocksrdquo Journalof Test and Measurement Technology vol 23 no 3pp 189ndash196 2007

[35] I Ispas R Eve R Hickman R G Keck S M Willson andK E Olson ldquoLaboratory testing and numerical modelling offracture propagation from deviated wells in poorly consoli-dated formationsrdquo in Proceedings of the SPE Annual TechnicalConference and Exhibition Society of Petroleum EngineersSan Antonio TX USA October 2012

[36] R G Jeffrey ldquo)e combined effect of fluid lag and fracturetoughness on hydraulic fracture propagationrdquo in Proceedings

of the Joint Rocky Mountain RegionalLow PermeabilityReservoirs Symposium pp 269ndash276 Society of PetroleumEngineers Denver CO USA 1989

[37] A P Bunger and E Detournay ldquoAsymptotic solution for apenny-shaped near-surface hydraulic fracturerdquo EngineeringFracture Mechanics vol 72 no 16 pp 2468ndash2486 2005

[38] E Gordeliy and E Detournay ldquoA fixed grid algorithm forsimulating the propagation of a shallow hydraulic fracturewith a fluid lagrdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 35 no 5pp 602ndash629 2011

[39] B Lecampion and E Detournay ldquoAn implicit algorithm forthe propagation of a hydraulic fracture with a fluid lagrdquoComputer Methods in Applied Mechanics and Engineeringvol 196 no 49ndash52 pp 4863ndash4880 2007

[40] T Ito K Evans K Kawai and K Hayashi ldquoHydraulicfracture reopening pressure and the estimation of maximumhorizontal stressrdquo International Journal of Rock Mechanicsand Mining Sciences vol 36 no 6 pp 811ndash826 1999

[41] E Detournay and A H D Cheng ldquoFundamentals ofporoelasticityrdquo in Comprehensive Rock Engineering Perga-mon Press Oxford UK 1993

[42] N R Warpinski and L W Teufel ldquoDetermination of theeffective-stress law for permeability and deformation in low-permeability rocksrdquo SPE Formation Evaluation vol 7 no 2pp 123ndash131 1992

[43] F Guo N R Morgenstern and J D Scott ldquoInterpretation ofhydraulic fracturing breakdown pressurerdquo InternationalJournal of Rock Mechanics and Mining Sciences amp Geo-mechanics Abstracts vol 30 no 6 pp 617ndash626 1993

[44] M S Bruno and F M Nakagawa ldquoPore pressure influence ontensile fracture propagation in sedimentary rockrdquo In-ternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 28 no 4 pp 261ndash273 1991

[45] S Saurabh and S Harpalani ldquo)e effective stress law forstress-sensitive transversely isotropic rocksrdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 101pp 69ndash77 2018


CorrosionInternational Journal of

Hindawiwwwhindawicom Volume 2018

Advances in

Materials Science and EngineeringHindawiwwwhindawicom Volume 2018


Journal of

Chemistry

Analytical ChemistryInternational Journal of


ScienticaHindawiwwwhindawicom Volume 2018

Polymer ScienceInternational Journal of



Advances in Condensed Matter Physics


International Journal of

BiomaterialsHindawiwwwhindawicom

Journal ofEngineeringVolume 2018

Applied ChemistryJournal of


NanotechnologyHindawiwwwhindawicom Volume 2018

Journal of


High Energy PhysicsAdvances in

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

TribologyAdvances in



ChemistryAdvances in


Advances inPhysical Chemistry


BioMed Research InternationalMaterials

Journal of


Na

nom

ate

ria

ls


Journal ofNanomaterials

Submit your manuscripts atwwwhindawicom

stress measurement and geothermal development Donget al [6] have proposed a theoretical model for hydraulicfracturing through a single radial perforation emanatingfrom a borehole however two symmetrical radial cracksemanating from the borehole are more common in thehydraulic fracturing as shown in Figure 1(a) Mode-I stressintensity factor for symmetrical radial cracks emanatingfrom hollow cylinder in an infinite plane is first solved withthe weight function method by Jin et al [9] who showed thatthe borehole radius affects the stress intensity factor andstress distribution at the radial crack tips which conse-quently affects the crack initiation characteristics such as thefracture initiation angle and the critical water pressureHowever most studies neglected the borehole size andgenerally simplified the borehole and its associated sym-metrical radial precracks as a straight-line fracture [3 10]For instance Chen et al [2] characterized dynamic fracturepropagation path during hydraulic fracturing by simplifyingthe model as a straight line which was inconsistent with theactual experimental configuration Behnia et al [11] pre-sented a boundary element method based on the higherorder displacement discontinuity formulation to model thehydraulic fracture propagation in layered formations inwhich the fracture was shaped being a straight line )emechanical behavior of the line-shaped fracture under hy-draulic pressure has been extensively conducted by variousresearchers such as Sobhaniaragh et al [12] da Silva et al[13] and Cheng et al [14] To study the effect of the hy-draulic fracture initiated from a borehole Zhang et al [15]presented a 2Dmodel for the initiation and growth of one ormore hydraulic fractures from a well)ey suggested that forcomplex multiple fracture cases the misalignment angle andthe number of initiated fractures are important in near-wellbore fracture path selection Sesetty and Ghassemi [16]developed a 2D numerical model based on the displacementdiscontinuity method (DDM) and the fictitious stressmethod (FSM) to investigate the effect of rock anisotropy onwellbore stresses and hydraulic fracture propagation Ad-ditionally the pore pressure is also a critical factor affectingthe hydraulic fracture behavior [17 18] Gholami et al [19]investigated the effect of pore pressure on the hydraulicfracture propagation and indicated that a higher than av-erage pore pressure zone accelerates the fracture growthwhereas a lower pressure anomaly zone resists a growingfracture However few fracture criteria have been proposedto study the fracture initiation by considering the internalrock pore pressure

When a borehole with symmetric cracks is subjected toboth external compressive load and inner water pressure thestress intensity factors at the crack tips include those in bothopening mode (KI) and sliding mode (KII) Many studiesemphasized mathematical or numerical evolution of theopening mode stress intensity factor (KI) for two symmetriccracks emanating from a circular hole when the maximumand minimum horizontal stresses (denoted by σH σh) aretensile loads and the azimuth of the cracks is 0deg [20 21]However when the azimuth of the cracks in Figure 1(a) isnot zero the uniaxial or biaxial tensile loading results in aplane shear stress condition and both opening mode and the

sliding mode stress intensity factor (KI and KII) are nonzero[22] When the model shown in Figure 1(a) is subjected tocompressive loading (σH and σh are negative values) andboth the pore pressure (p) and the injected water pressure(P) are considered no theoretical work has been reported forcalculating the stress intensity factors (KI and KII) under thecombined loading conditions Since it is critical to predictthe accurate path of new fractures in the hydraulic fracturingoperations and well production the geometry and directionof the hydraulic fracture under the condition shown inFigure 1(a) are theoretically studied by an analytical methodin this work

2 Theoretical Model

Dong et al [6] have theoretically studied the feature ofhydraulic fracturing through a single radial perforationemanating from a borehole )e results indicated that pa-rameters such as the perforation length the borehole radiusand the stress anisotropy coefficient significantly affect thecritical water pressure and the critical initiation anglegreatly However the fracture growth behavior of twosymmetrical radial cracks (as shown in Figure 1(a)) is dif-ferent from that of a single radial crack which needs to befurther investigated Most of the existing studies assumedthat the perforation was horizontal (β 0deg) Consequentlythe far-field equivalent shear stress has no influence on themode-II stress intensity factors )e model shown inFigure 1(a) can be superposed by the models inFigures 1(b)ndash1(d) In addition Figures 1(b) and 1(c) are thehorizontal fractures subjected to far-field equivalent normaland shear stresses respectively whereas Figure 1(d) is thefracture subjected to injected water pressure and the porepressure in rock Since the perforation angle is zero there isno mode-II stress intensity factor under the loading con-ditions shown in Figures 1(b) and 1(d) and only the loadingconditions in Figure 1(c) lead to mode-II stress intensityfactor

For the model shown in Figure 1(b) the mode-I stressintensity factor caused by the far-field in situ stress isexpressed by the following equation [20 23]

KIminusσ σhprimeπa

radic1113896 1minus kprime( 1113857 05 3minus

a

R + a1113874 1113875 1 + 1243 1minus

a

R + a1113874 1113875

21113890 11138911113896 1113897

+ kprime 1 + 1minusa

R + a1113874 1113875 05 + 0743 1minus

a

R + a1113874 1113875

21113890 11138911113896 11138971113897

(1)

where σhprime is the far-field equivalent maximum normal kprimeis the equivalent stress anisotropy coefficient and a and Rare the perforation length and the borehole radiusrespectively

Following the convention in elastic mechanics thetensile stress and dilatant strain are referred as positivewhereas the compressive stress and contractive strain arereferred as negative in this study )erefore because the insitu stress shown in Figure 1(b) is compressive stress theparameter σhprime in equation (1) is negative According to thetheory of elastic mechanics for an inclined perforation




2


2

(2)




(3)




YII 3127x3 minus 75216x2 + 53936x + 00089 (4)

KII τeffYIIπa

radic (5)


σh σprime


(a) (b)

(c) (d)

τeff

τeff

αp

αp

λp

λp

aP

P

R




2 (6)


KIminusP Pπa


a

R + a1113874 111387511138900637 + 0485 1minus

a

R + a1113874 1113875

2

+ 04a

R + a1113874 1113875

21minus

a

R + a1113874 11138751113891 + λ11138961 + 1minus

a

R + a1113874 1113875


R + a1113874 1113875

21113890 111389111138971113897

(7)




P

πa


a

R + a1113874 1113875


R + a1113874 1113875

2+ 04

a

R + a1113874 1113875

21minus

a

R + a1113874 11138751113890 1113891

+ λ 1 + 1minusa

R + a1113874 1113875 05 + 0743 1minus

a

R + a1113874 1113875

21113890 11138911113896 11138971113897

(8)



00

02

04

06

08

10

12

14

00 01 02 03 04 05 06 07 08 09 10

Y II


a(R + a)

R2 = 09997




00

01

02

03

04

05

06

07

08

00

10

20

30

40

50

60

00 10 20 30 40 50 60 70 80 90 100

K I-P

P (λ

= 0

0)

K I-P

P (λ

= 1

0)

a (cm)

R = 10 cm

λ = 1λ = 0





zKIminusPz(aR + a)

1113868111386811138681113868111386811138681113868λ0 0 (9)



πa

radic


KI 1113946a

0σ(x)m(x a)dx (10)

m(a x) 2

2π(aminusx)

1113968 11138901 + M1 1minusx

a1113874 1113875

12+ M2 1minus

x

a1113874 1113875

+ M3 1minusx

a1113874 1113875

321113891

(11)




M1 0074x51 minus 01765x

41 minus 03516x

31 + 082x

21

+ 02964x1 minus 04255(12)

M2 minus0196x51 + 03508x

41 + 09246x

31 minus 15545x

21

minus 09476x1 + 07175(13)

M3 0098x51 minus 01754x

41 minus 04623x

31 + 07772x

21

+ 04738x1 + 01413(14)



K(1)Iminusαp αp

πa

radicmiddot

2

radic

π2 + M1 +

23M2 +

12M31113874 1113875 (15)


00

03

06

09

12

15

18

21

24

0 01 02 03 04 05 06 07 08 09 1

K I-P

P (π

a)0

5

a(R + a)

λ = 00λ = 05λ = 10




αpπa


R + a( ) + 27211a

R + a( )2

minus 34525a

R + a( )3+ 1702

a

R + a( )4

(16)

KIminusαp αpπa





σr 1

22πr



σθ 1

22πr


τrθ 1

22πr


(19)




σθ gt 0

zσθzθ

0

z2σθzθ2lt 0

(20)






(21)


θc

0 KII 0

2 arctanKI minus

K2

I + 8K2II

radic( )

4KII

KII ne 0

(22)

090

095

100

105

110

115

00 02 04 06 08 10

Y I-α

p

a(R + a)



x = a(R + a)R2 = 09989


Fracture tipθ x

r

y σθ σrτrθ






σθ( 1113857max

2πr

radic12cos

θc2


(23)


σθ( 1113857max

2πr

radic

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(24)


KIC σt2πr


2πr

radic (25)



KIC Keq

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(26)




3 Validations














40

60

80

100

120

140

160

10 20 30 40 50 60

P c (M

Pa)



β (deg)

(a)P c

(MPa

)



40

60

80

100

120

140

160

10 20 30 40 50 60β (deg)

(b)

00

50

100

150

200

250

300

350

10 20 30 40 50 60

ndashθc (

deg)

β (deg)



(c)






60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25


β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)








ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls






2


2

(2)




(3)




YII 3127x3 minus 75216x2 + 53936x + 00089 (4)

KII τeffYIIπa

radic (5)


σh σprime


(a) (b)

(c) (d)

τeff

τeff

αp

αp

λp

λp

aP

P

R




2 (6)


KIminusP Pπa


a

R + a1113874 111387511138900637 + 0485 1minus

a

R + a1113874 1113875

2

+ 04a

R + a1113874 1113875

21minus

a

R + a1113874 11138751113891 + λ11138961 + 1minus

a

R + a1113874 1113875


R + a1113874 1113875

21113890 111389111138971113897

(7)




P

πa


a

R + a1113874 1113875


R + a1113874 1113875

2+ 04

a

R + a1113874 1113875

21minus

a

R + a1113874 11138751113890 1113891

+ λ 1 + 1minusa

R + a1113874 1113875 05 + 0743 1minus

a

R + a1113874 1113875

21113890 11138911113896 11138971113897

(8)



00

02

04

06

08

10

12

14

00 01 02 03 04 05 06 07 08 09 10

Y II


a(R + a)

R2 = 09997




00

01

02

03

04

05

06

07

08

00

10

20

30

40

50

60

00 10 20 30 40 50 60 70 80 90 100

K I-P

P (λ

= 0

0)

K I-P

P (λ

= 1

0)

a (cm)

R = 10 cm

λ = 1λ = 0





zKIminusPz(aR + a)

1113868111386811138681113868111386811138681113868λ0 0 (9)



πa

radic


KI 1113946a

0σ(x)m(x a)dx (10)

m(a x) 2

2π(aminusx)

1113968 11138901 + M1 1minusx

a1113874 1113875

12+ M2 1minus

x

a1113874 1113875

+ M3 1minusx

a1113874 1113875

321113891

(11)




M1 0074x51 minus 01765x

41 minus 03516x

31 + 082x

21

+ 02964x1 minus 04255(12)

M2 minus0196x51 + 03508x

41 + 09246x

31 minus 15545x

21

minus 09476x1 + 07175(13)

M3 0098x51 minus 01754x

41 minus 04623x

31 + 07772x

21

+ 04738x1 + 01413(14)



K(1)Iminusαp αp

πa

radicmiddot

2

radic

π2 + M1 +

23M2 +

12M31113874 1113875 (15)


00

03

06

09

12

15

18

21

24

0 01 02 03 04 05 06 07 08 09 1

K I-P

P (π

a)0

5

a(R + a)

λ = 00λ = 05λ = 10




αpπa


R + a( ) + 27211a

R + a( )2

minus 34525a

R + a( )3+ 1702

a

R + a( )4

(16)

KIminusαp αpπa





σr 1

22πr



σθ 1

22πr


τrθ 1

22πr


(19)




σθ gt 0

zσθzθ

0

z2σθzθ2lt 0

(20)






(21)


θc

0 KII 0

2 arctanKI minus

K2

I + 8K2II

radic( )

4KII

KII ne 0

(22)

090

095

100

105

110

115

00 02 04 06 08 10

Y I-α

p

a(R + a)



x = a(R + a)R2 = 09989


Fracture tipθ x

r

y σθ σrτrθ






σθ( 1113857max

2πr

radic12cos

θc2


(23)


σθ( 1113857max

2πr

radic

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(24)


KIC σt2πr


2πr

radic (25)



KIC Keq

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(26)




3 Validations














40

60

80

100

120

140

160

10 20 30 40 50 60

P c (M

Pa)



β (deg)

(a)P c

(MPa

)



40

60

80

100

120

140

160

10 20 30 40 50 60β (deg)

(b)

00

50

100

150

200

250

300

350

10 20 30 40 50 60

ndashθc (

deg)

β (deg)



(c)






60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25


β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)








ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls





2 (6)


KIminusP Pπa


a

R + a1113874 111387511138900637 + 0485 1minus

a

R + a1113874 1113875

2

+ 04a

R + a1113874 1113875

21minus

a

R + a1113874 11138751113891 + λ11138961 + 1minus

a

R + a1113874 1113875


R + a1113874 1113875

21113890 111389111138971113897

(7)




P

πa


a

R + a1113874 1113875


R + a1113874 1113875

2+ 04

a

R + a1113874 1113875

21minus

a

R + a1113874 11138751113890 1113891

+ λ 1 + 1minusa

R + a1113874 1113875 05 + 0743 1minus

a

R + a1113874 1113875

21113890 11138911113896 11138971113897

(8)



00

02

04

06

08

10

12

14

00 01 02 03 04 05 06 07 08 09 10

Y II


a(R + a)

R2 = 09997




00

01

02

03

04

05

06

07

08

00

10

20

30

40

50

60

00 10 20 30 40 50 60 70 80 90 100

K I-P

P (λ

= 0

0)

K I-P

P (λ

= 1

0)

a (cm)

R = 10 cm

λ = 1λ = 0





zKIminusPz(aR + a)

1113868111386811138681113868111386811138681113868λ0 0 (9)



πa

radic


KI 1113946a

0σ(x)m(x a)dx (10)

m(a x) 2

2π(aminusx)

1113968 11138901 + M1 1minusx

a1113874 1113875

12+ M2 1minus

x

a1113874 1113875

+ M3 1minusx

a1113874 1113875

321113891

(11)




M1 0074x51 minus 01765x

41 minus 03516x

31 + 082x

21

+ 02964x1 minus 04255(12)

M2 minus0196x51 + 03508x

41 + 09246x

31 minus 15545x

21

minus 09476x1 + 07175(13)

M3 0098x51 minus 01754x

41 minus 04623x

31 + 07772x

21

+ 04738x1 + 01413(14)



K(1)Iminusαp αp

πa

radicmiddot

2

radic

π2 + M1 +

23M2 +

12M31113874 1113875 (15)


00

03

06

09

12

15

18

21

24

0 01 02 03 04 05 06 07 08 09 1

K I-P

P (π

a)0

5

a(R + a)

λ = 00λ = 05λ = 10




αpπa


R + a( ) + 27211a

R + a( )2

minus 34525a

R + a( )3+ 1702

a

R + a( )4

(16)

KIminusαp αpπa





σr 1

22πr



σθ 1

22πr


τrθ 1

22πr


(19)




σθ gt 0

zσθzθ

0

z2σθzθ2lt 0

(20)






(21)


θc

0 KII 0

2 arctanKI minus

K2

I + 8K2II

radic( )

4KII

KII ne 0

(22)

090

095

100

105

110

115

00 02 04 06 08 10

Y I-α

p

a(R + a)



x = a(R + a)R2 = 09989


Fracture tipθ x

r

y σθ σrτrθ






σθ( 1113857max

2πr

radic12cos

θc2


(23)


σθ( 1113857max

2πr

radic

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(24)


KIC σt2πr


2πr

radic (25)



KIC Keq

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(26)




3 Validations














40

60

80

100

120

140

160

10 20 30 40 50 60

P c (M

Pa)



β (deg)

(a)P c

(MPa

)



40

60

80

100

120

140

160

10 20 30 40 50 60β (deg)

(b)

00

50

100

150

200

250

300

350

10 20 30 40 50 60

ndashθc (

deg)

β (deg)



(c)






60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25


β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)








ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls






zKIminusPz(aR + a)

1113868111386811138681113868111386811138681113868λ0 0 (9)



πa

radic


KI 1113946a

0σ(x)m(x a)dx (10)

m(a x) 2

2π(aminusx)

1113968 11138901 + M1 1minusx

a1113874 1113875

12+ M2 1minus

x

a1113874 1113875

+ M3 1minusx

a1113874 1113875

321113891

(11)




M1 0074x51 minus 01765x

41 minus 03516x

31 + 082x

21

+ 02964x1 minus 04255(12)

M2 minus0196x51 + 03508x

41 + 09246x

31 minus 15545x

21

minus 09476x1 + 07175(13)

M3 0098x51 minus 01754x

41 minus 04623x

31 + 07772x

21

+ 04738x1 + 01413(14)



K(1)Iminusαp αp

πa

radicmiddot

2

radic

π2 + M1 +

23M2 +

12M31113874 1113875 (15)


00

03

06

09

12

15

18

21

24

0 01 02 03 04 05 06 07 08 09 1

K I-P

P (π

a)0

5

a(R + a)

λ = 00λ = 05λ = 10




αpπa


R + a( ) + 27211a

R + a( )2

minus 34525a

R + a( )3+ 1702

a

R + a( )4

(16)

KIminusαp αpπa





σr 1

22πr



σθ 1

22πr


τrθ 1

22πr


(19)




σθ gt 0

zσθzθ

0

z2σθzθ2lt 0

(20)






(21)


θc

0 KII 0

2 arctanKI minus

K2

I + 8K2II

radic( )

4KII

KII ne 0

(22)

090

095

100

105

110

115

00 02 04 06 08 10

Y I-α

p

a(R + a)



x = a(R + a)R2 = 09989


Fracture tipθ x

r

y σθ σrτrθ






σθ( 1113857max

2πr

radic12cos

θc2


(23)


σθ( 1113857max

2πr

radic

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(24)


KIC σt2πr


2πr

radic (25)



KIC Keq

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(26)




3 Validations














40

60

80

100

120

140

160

10 20 30 40 50 60

P c (M

Pa)



β (deg)

(a)P c

(MPa

)



40

60

80

100

120

140

160

10 20 30 40 50 60β (deg)

(b)

00

50

100

150

200

250

300

350

10 20 30 40 50 60

ndashθc (

deg)

β (deg)



(c)






60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25


β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)








ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls





αpπa


R + a( ) + 27211a

R + a( )2

minus 34525a

R + a( )3+ 1702

a

R + a( )4

(16)

KIminusαp αpπa





σr 1

22πr



σθ 1

22πr


τrθ 1

22πr


(19)




σθ gt 0

zσθzθ

0

z2σθzθ2lt 0

(20)






(21)


θc

0 KII 0

2 arctanKI minus

K2

I + 8K2II

radic( )

4KII

KII ne 0

(22)

090

095

100

105

110

115

00 02 04 06 08 10

Y I-α

p

a(R + a)



x = a(R + a)R2 = 09989


Fracture tipθ x

r

y σθ σrτrθ






σθ( 1113857max

2πr

radic12cos

θc2


(23)


σθ( 1113857max

2πr

radic

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(24)


KIC σt2πr


2πr

radic (25)



KIC Keq

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(26)




3 Validations














40

60

80

100

120

140

160

10 20 30 40 50 60

P c (M

Pa)



β (deg)

(a)P c

(MPa

)



40

60

80

100

120

140

160

10 20 30 40 50 60β (deg)

(b)

00

50

100

150

200

250

300

350

10 20 30 40 50 60

ndashθc (

deg)

β (deg)



(c)






60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25


β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)








ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls







σθ( 1113857max

2πr

radic12cos

θc2


(23)


σθ( 1113857max

2πr

radic

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(24)


KIC σt2πr


2πr

radic (25)



KIC Keq

KI KII 0

42

radicKII

111386811138681113868111386811138681113868111386811138683

KI + 3

8K2II + K2

I

1113969

1113874 1113875 K2I minusKI

8K2II + K2

I

1113969

+ 12K2II1113874 1113875minus32

KII ne 0

⎧⎪⎨

⎪⎩(26)




3 Validations














40

60

80

100

120

140

160

10 20 30 40 50 60

P c (M

Pa)



β (deg)

(a)P c

(MPa

)



40

60

80

100

120

140

160

10 20 30 40 50 60β (deg)

(b)

00

50

100

150

200

250

300

350

10 20 30 40 50 60

ndashθc (

deg)

β (deg)



(c)






60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25


β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)








ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls















40

60

80

100

120

140

160

10 20 30 40 50 60

P c (M

Pa)



β (deg)

(a)P c

(MPa

)



40

60

80

100

120

140

160

10 20 30 40 50 60β (deg)

(b)

00

50

100

150

200

250

300

350

10 20 30 40 50 60

ndashθc (

deg)

β (deg)



(c)






60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25


β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)








ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls






40

60

80

100

120

140

160

10 20 30 40 50 60

P c (M

Pa)



β (deg)

(a)P c

(MPa

)



40

60

80

100

120

140

160

10 20 30 40 50 60β (deg)

(b)

00

50

100

150

200

250

300

350

10 20 30 40 50 60

ndashθc (

deg)

β (deg)



(c)






60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25


β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)








ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls







60

80

100

120

140

160

0 15 30 45 60 75 90

k = 10k = 15

k = 20k = 25


β0

P c (M

Pa)

β (deg)

(a)

k = 10k = 15

k = 20k = 25


00

20

40

60

80

100

120

140

160

0 15 30 45 60 75 90

ndashθc (

deg)

β (deg)

(b)


50

60

70

80

90

100

110

120

P c (M

Pa)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(c)


00

100

200

300

400

500

600

700ndashθ

c (deg)

k = 10k = 15

k = 20k = 25

0 15 30 45 60 75 90β (deg)

(d)








ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls









ndash12

ndash10

ndash08

ndash06

ndash04

ndash02

00

0 10 20 30 40 50 60 70 80 90

K I-σ

σ

β0

k = 1k = 2k = 3

a = 10 mmR = 50 mmσh = 30 MPa

β (deg)

(a)K I

-σσ

ndash40

ndash35

ndash30

ndash25

ndash20

ndash15

ndash10

0 10 20 30 40 50 60 70 80 90

k = 1k = 2


a = 300 mmR = 50 mmσh = 30 MPa

β (deg)

(b)










00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls











00

150

300

450

600

750

00

50

100

150

200

250

300

350

400

0 5 10 15 20 25 30

ndashθc(

deg)

K eq

KIIKI

Keqθc






40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls







40

60

80

100

120

0 15 30 45 60 75 90

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


P c (M

Pa)

β (deg)

(a)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

50

100

150

200

250

0 15 30 45 60 75 90β (deg)

ndashθc (

deg)(b)


40

50

60

70

80

90

P c (M

Pa)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

(c)


00

100

200

300

400

500

600

700

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

β (deg)0 15 30 45 60 75 90

ndashθc (

deg)

(d)






40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls







40

60

80

100

120

0 15 30 45 60 75 90

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


P c (M

Pa)

β (deg)

(a)

a = 20 mma = 50 mma = 100 mm

a = 150 mma = 200 mm


00

100

200

300

400

500

600

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)


20

40

60

80

100

120

140

0 100 200 300 400 500 600 700 800a (mm)



P c (M

Pa)

(c)


a (mm)

00

150

300

450

600

750

900

0 100 200 300 400 500 600 700 800


ndashθc (

deg)

(d)






30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls







30

60

90

120

150

180

210

240

0 15 30 45 60 75 90

λ = 10λ = 07

λ = 05λ = 03


P c (M

Pa)

β (deg)

(a)



50

100

150

200

250

300

0204060810λ



P c (M

Pa)

(b)


00

100

200

300

400

500

600

0 15 30 45 60 75 90

λ = 10λ = 03

β (deg)

ndashθc (

deg)

(c)
















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls

















00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls









00

20

40

60

80

100

0 15 30 45 60 75 90

α = 00α = 03

α = 07α = 10


P c (M

Pa)

β (deg)

(a)

00

100

200

300

400

500

0 15 30 45 60 75 90

α = 10α = 00


ndashθc (

deg)

β (deg)

(b)


00

50

100

150

200

250

300

350

400

450

500

0 15 30 45 60 75 90




ndashθc (

deg)

β (deg)



00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls




00

50

100

150

200


k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(a)


00

50

100

150

200

250

k = 15k = 20

k = 25k = 30

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(b)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm

00

15

30

45

60

75

90

105


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(c)

R = 50 mmR = 100 mm

R = 150 mmR = 200 mm


00

15

30

45

60

75ndashθ

c (deg)

0 15 30 45 60 75 90β (deg)

(d)

a = 50 mma = 100 mm

a = 150 mma = 200 mm

00

20

40

60

80

100


ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(e)

a = 50 mma = 100 mm

a = 150 mma = 200 mm


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(f )

Figure 16 Continued



6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls





6 Conclusions





00

40

80

120

160


λ = 10λ = 08λ = 06

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(g)

α = 00α = 03

α = 07α = 10


00

20

40

60

80

100

120

ndashθc (

deg)

0 15 30 45 60 75 90β (deg)

(h)






Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls







Data Availability




Acknowledgments


References





















































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls



































Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls






Advances in



Journal of

Chemistry















Journal of





Volume 2018









Journal of


Na

nom

ate

ria

ls




Documents

ATheoreticalModelforHydraulicFracturingthroughTwo ...downloads.hindawi.com/journals/amse/2019/6094305.pdf · and gas wells [1]. During hydraulic fracturing, high water pressure is