formulationofanisotropicstrengthcriterionfor...

Research ArticleFormulation of Anisotropic Strength Criterion forGeotechnical Materials

Liping Chen12 Sui Wang12 Bin Chen 123 Xiaokai Niu4 Guogang Ying5 and BoWu126

1School of Civil and Transportation Engineering Ningbo University of Technology Ningbo 315211 China2Engineering Research Center of Industrial Construction in Civil Engineering of Zhejiang Ningbo University of TechnologyNingbo 315016 China3Ningbo Rail Transit Group Co Ltd Ningbo Zhejiang 315211 China4Beijing Municipal Engineering Research Institute Beijing 100037 China5NingboTech University Ningbo 315100 China6Guangxi Key Laboratory of Disaster Prevention and Engineering Safety Guangxi University Nanning China

Correspondence should be addressed to Bin Chen chenbinnbuteducn

Received 3 March 2020 Revised 23 July 2020 Accepted 3 August 2020 Published 7 September 2020

Academic Editor Fan Gu

Copyright copy 2020 Liping Chen et al ampis 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

A new nonlinear unified strength (NUS) criterion is obtained based on the spatially mobilized plane (SMP) criterion and Misescriterion New criterion is a series of smooth curves between SMP curved triangle and Mises circle in the π plane and therebyunifies the strength criteria ampe new criterion can reflect the effect of the intermediate principal stress and consider the strengthnonlinearity of a material Based on the fabric tensor the anisotropic parameter A is defined and the anisotropic equation isproposed and introduced into the NUS criterion to form a nonlinear unified anisotropic strength criterion ampe new criterion canbe used to predict the strength variation of granular materials and cohesive materials under three-dimensional stress and canpresent the strength anisotropy of the geomaterials ampe validity of the new criterion was checked using rock and soil materials Itis shown that the prediction results for the criterion agree well with the test data

1 Introduction

ampe research of strength criterion theory [1ndash11] is an im-portant topic in geotechnical engineering ampere are usuallythree types of methods for establishing strength criterianamely theoretical methods empirical methods andcombined theoretical-empirical methods ampeoreticalmethods are usually based on a hypothesis and there is amodel to interpret the failure mechanism of geomaterialsAll the model parameters in the strength criteria establishedby this method have clear physical meanings ampis type ofstrength criterion uses the law of friction to explain thefailure of geomaterials including the MohrndashCoulombstrength theory the D-P criterion [11] the spatially mobi-lized plane (SMP) criterion [2] and the twin shear unifiedstrength theory [9 12 13] Strength criteria established byempirical methods are usually based on the fitting of test

data and include the famous WillamndashWarnke criterion [14]and the HoekndashBrown criterion [6 15] Compared to thecriteria that are based on theoretical methods the physicalmeanings of some of the parameters of the strength criteriaestablished by empirical methods are often not very clearFinally combined theoretical-empirical methods are usuallyused to establish special strength criteria for specific materialpropertiesampis type of criterion applies the MohrndashCoulombstrength theory to jointed rock materials [16] by introducingadditional parameters [17ndash19]

However the above strength criteria are associatedwith a single shape in the π plane and are thus unable toreflect the factors of material strength that may vary withthe change of internal factors To address this problemresearchers [10 11 20ndash26] have proposed to unify thestrength criteria ampe common method is to introducesome parameters to change the size of the shape function

HindawiAdvances in Civil EngineeringVolume 2020 Article ID 8825816 10 pageshttpsdoiorg10115520208825816

ampe above criteria are all isotropic strength criteria andcannot depict the strength anisotropy of geomaterialsHowever numerous geomaterials exhibit transverseisotropy due to deposition that is the strength of thematerial is the same in the deposition surface but differentin the depositional surfaces of different directions ampedirectional shear test of most soils shows that the changein the direction of the principal stress also leads tostrength anisotropy reflecting the initial anisotropy of thematerial amperefore it is necessary to establish a uniformstrength criterion that considers the depositional char-acteristics or initial anisotropy of geomaterials

ampe above previous studies show that the strength cri-terion is researched through a simple single form to a unifiedstrength theory system with a relatively wide scope of ap-plication Consequently it is important to explore a widelyapplicable strength theory system In the present paper anew anisotropic nonlinear unified strength criterion isestablished based on Mises criterion and SMP criterion ampenew criterion is approximated as a series of smooth curvesbetween the SMP curved triangle and the von Mises circle inthe π plane to unify the strength criteria Based on the fabrictensor a nonlinear anisotropic strength criterion is pro-posed ampe correctness of the criterion is verified using sandclay and rock materials

2 Form of the NUS Criterion

21 Linear Unified Strength (LUS) Criterion SMP criterion(Matsuoka and Nakai [2]) has been widely used to check thestrength of geomaterials because of its simple form and clearphysical meaning In the present study the SMP criterion ismodified to make it more widely applicable ampe SMP cri-terion is shown in the following equation

I1I2

I3 C (1)

where I1 I2 and I3 are stress invariants

I1 σ1 + σ2 + σ3I2 σ1σ2 + σ2σ3 + σ1σ3I3 σ1σ2σ3

⎫⎪⎪⎬

⎪⎪⎭ (2)

C is a model parameter Under conventional triaxialcompression (σ2 σ3) the sine of the friction angle φ0 of agranular material is

sinφ0 σ1 minus σ3σ1 + σ3

(3)

Substituting equations (2) and (3) into equation (1) gives

I31

I3

3 minus sinφ0( 11138573

1 minus sinφ0 minus sin2φ0 + sin3φ0 (4)

Using the trigonometric identity the expression of q isobtained as

3p3

minus L2q2p + L3 cos(3θ)q

3 0 (5)

where

L2 3 + sin2φ0

12 sin2φ0 (6)

L3 9 minus sin2φ0

108 sin2φ0 (7)

p σ1 + σ2 + σ3

3 (8)

q

12

σ1 minus σ2( 11138572

+ σ2 minus σ3( 11138572

+ σ1 minus σ3( 11138572

1113960 1113961

1113970

(9)

where p is the mean stress and q is generalized shear stressampus the shape function of the SMP criterion is

g0(θ) q

q0

cos (13)arccos minus33

radicL32L

3221113872 11138731113960 1113961


radicL3 cos(3θ)2L

3221113872 11138731113960 1113961

(10)

where q0 is the data for q when θ 0 amperefore the forms ofp and q of the SMP criterion are

q Mg0(θ)p (11)

where g0(θ) is obtained by equation (10) and M is thecoefficient of friction

M 6 sinφ0

3 minus sinφ0 (12)

ampe shape function of the SMP criterion cannot reflectthe influence of internal factors on material therefore theapplication of the criterion is limited In this report theshape function is rewritten as the following expression

q Mg(θ) p + σ0( 1113857 (13)

where M is obtained from equation (12) σ0 which is thetriaxial tensile strength of the material is given by

σ0 c cotφ0 (14)

and p and q are the average stress and the generalized shearstress as defined by equations (9) and (10) respectively

amperefore

g(θ) rθ

r0 (15)

where

rθ

3

radic(1 minus α)

2L2

1113968cos δθ

+6α sinφ03 minus sinφ0

r0

3

radic(1 minus α)

2L2

1113968cos δ0

+6α sinφ0

3 minus sinφ0

⎫⎪⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎪⎭

(16)

and α is the model parameter determined by the strengthratio under the triaxial compression and extension condi-tions ampe value of α ranges from 0 to 1 One has

2 Advances in Civil Engineering

δθ 13arccos minus

33

radicL3 cos(3θ)

2L322

1113888 1113889

δ0 13arccos minus

33

radicL3

2L322

1113888 1113889

⎫⎪⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎪⎭

(17)

where L2 and L3 are obtained from equations (6) and (7)respectively

Figure 1 illustrates the influences of parameters φ0 and αon linear criterion in the deviatoric plane As Figure 1(b)shows the shape of the criterion is shifted from the SMPcurved triangle to the von Mises circle when α graduallyincreases from 0 to 1

22 Nonlinear Isotropic Strength (NIS) Criterion In thetriaxial test of the rockfill material tests [27] the largeconfining pressure causes the granules to break [28ndash30] andthe material exhibits remarkable strength nonlinearitiesamperefore to obtain the nonlinear of the rockfill material theabove new criterion equation (13) need be further modifiedas expressed in the following equation

q Mfg(θ)pr

p + σ0pr

1113888 1113889

n

(18a)

ampat is

lnq

pr

1113888 1113889 n lnp + σ0

pr

1113888 1113889 + ln Mfg(θ)1113960 1113961 (18b)

where pr is reference stress usually taken as pr 101 kPa Mfis usually obtained from the intercept of equation (18b) andthe shape function and n is a curvature parameter

In the space of ln(qpr) minus ln(p + σ0pr) ln[Mfg(θ)] isthe intercept and n is the slope

When n 1

Mf M 6 sinφ0

3 minus sinφ0 (19)

where n is the curvature parameter and the shape functionin the deviatoric plane is

g(θ)

3

radic(1 minus α)2

L2

1113968cos δθ1113872 1113873 + 6α sinφ03 minus sinφ0( 1113857

3

radic(1 minus α)2

L2

1113968cos δ01113872 1113873 + 6α sinφ03 minus sinφ0( 1113857

(20)

23 Nonlinear Unified Strength (NUS) CriterionNumerous geomaterials exhibit transverse isotropy due todeposition as shown in Figure 2 that is the strength of thematerials is the same within the depositional surface butdifferent in the depositional surfaces of different directionsreflecting the initial anisotropy of the material amperefore itis necessary to establish a strength criterion that considersthe depositional characteristics or initial anisotropy of thegeomaterial ampe premise of establishing the anisotropicstrength criterion is to quantify the degree and direction of

the initial anisotropy of the geomaterial ampe fabric tensorproposed by Brewer is a good choice ampe anisotropicfunction in the present study refers to the form in [31ndash33]

f(A) exp η1 A minus A0( 1113857 + η2 A minus A0( 11138572

1113960 1113961 (21)

where A is the anisotropic parameter and A0 is the value thatA takes from b 0 and α 0 η1 and η2 in the anisotropicfunction are selected to depict the anisotropy of the materialampe anisotropic function and the nonlinear isotropicstrength criterion are combined to obtain an anisotropicstrength criterion as shown in the following equation

q f(A)Mfg(θ)pr

p + σ0pr

1113888 1113889

n

(22)

24 Effects of η1 and η2 on ANUS ampe effect of isotropicparameters on the shape of the strength criterion in thedeviatoric plane has been discussed above ampis sectionfocuses on the effect of anisotropic parameters on the shapeof the strength criterion According to the characteristics ofthe test data the parameters η1 and η2 in the anisotropicfunction are selected to depict the anisotropy of the materialDifferent formulas can be selected for the anisotropic pa-rameter A in equation (22) according to the true triaxial testor the hollow torsional shear test In the present study theanisotropic strength in the true triaxial test is mainlydiscussed

In the true triaxial test when the stress tensor and thefabric tensor are coaxial (β 0) as shown in Figure 2 theshape of the strength criterion in the π plane varies with theparameters η1 and η2 as shown in Figures 3(a) 3(c) and3(e) In this section with η2 fixed at minus025 minus0333 and 0 theeffect of varying η1 on the strength curve in the π plane isdiscussed As shown in Figures 3(a) and 3(b) withη2 minus025 the strength values of the anisotropic criterionand the isotropic criterion at θ 180deg are equal with differentη1 values and the criterion is symmetric about the σz axisampe criterion expands in the π plane when η1gt 0 and shrinksin the π plane when η1lt 0 and the degrees of expansion orshrinkage are related to the value of η1 ampe anisotropicfunction variation in the π plane is consistent with that of thestrength criterion as shown in Figure 3(b)With η2 minus0333as shown in Figure 3(c) the strength values of the aniso-tropic criterion and the isotropic criterion at θ 120deg and240deg respectively are equal with different η1 values and thecriterion is symmetric about the σz axis Relative to theisotropic criterion the anisotropic criterion expands inregions I and II and shrinks in region III of the π plane whenη1gt 0 and it shrinks in regions I and II and expands inregion III of the π plane when η1lt 0 with the degrees ofexpansion or shrinkage being related to the value of η1 ampeanisotropic function variation in the π plane is consistentwith that of the strength criterion as shown in Figure 3(d)When η2 0 as shown in Figure 3(e) the criterion at dif-ferent η1 values is symmetric about the σz axis when η1gt 0the anisotropic criterion expands in the π plane relative tothe isotropic criterion and when η1lt 0 the anisotropic

Advances in Civil Engineering 3

criterion shrinks in the π plane relative to the isotropiccriterion ampe degrees of expansion or shrinkage are relatedto the value of η1 ampe anisotropic function variation in the πplane is consistent with that of the strength criterion asshown in Figure 3(f)

When the stress tensor and the fabric tensor are notcoaxial (βne 0) the changes in the shape of the strengthcriterion in the π plane at different included angles areshown in Figures 4(a) and 4(c) With fixed η1 and η2 values(η1 minus008 η2 0) the effect of the change in the includedangle between the deposition surface and the vertical stresson the strength curve in the π plane is discussedWhen β 0as shown in Figure 4 the strength curve is symmetric aboutthe σz axis in the π plane When the deposition surface isrotated in the x minus z plane as shown in Figure 4(a) thestrength curve is symmetric with respect to the σy axis in theπ plane when β 90deg ampe changes in the strength curvewhen β 30deg and 60deg are shown in Figure 4(a) As the figure

shows when the deposition surface is rotated in the x minus z

plane the strength values of the anisotropic strength cri-terion are equal at θ 60deg and 240deg ampe included angle has alarge effect on the position and size of the strength curveampemethod proposed in the present study can predict the trendand position of the strength curve at different includedangles When the deposition surface is rotated in the y minus z

plane as shown in Figure 4(c) the strength curve is sym-metric about the σx axis in the π plane when β 90deg ampechanges in the strength curve when β 30deg and 60deg are shownin Figure 4(c) As the figure indicates when the depositionsurface is rotated in the y minus z plane the strength values ofthe anisotropic strength criterion are equal at θ 120deg and300degampe variations of the anisotropic function in the π planein both cases are shown in Figure 4(b) and 4(d) where asimilar situation with the change in the strength criterion isobserved

3 Test Verification

31 Parameter Determination To facilitate the applicationof the criteria the model parameters should be determinedas much as possible using conventional triaxial compressionor extension tests In this section the proposed ANUScriterion is applied to various geomaterials ampe geomaterialmodel parameters need to be determined only by conven-tional triaxial compression or extension tests In this sectionthe strength parameters and anisotropy parameters in theanisotropic criterion are mainly determined Callisto et al[34] performed a large number of conventional triaxial testsand true triaxial tests on Pietrafitta clay to determine all theparameters in the anisotropic criterion In the present studythe parameters are determined by conventional triaxial testdata and other test data are used for model verification ampespecific determination steps are as follows

(1) Determination of Mf n and σ0 ampe p minus q curve isdrawn using the failure data of the conventional

10deg20deg30deg40deg

50deg60deg90deg

σxσy

σz

(a)

σxσy

σz

α = 10 (Mises)α = 08α = 04α = 0 (SMP)

(b)

Figure 1 Variations of criterion with φ0 and α on the octahedral plane

σy

σ1

σ2σ3

β

σ3

σ1σ2

β

σ3

σ1σ2

βσx

θ = 60deg

θ = 120deg

θ = 180deg

θ = 240deg

θ = 300deg

IIIIII

II II

II

θ

θ = 0deg σzz

y

x

Figure 2 Application of true triaxial tests on cross-anisotropicsoils in the octahedral plane


triaxial test as shown in Figure 5(a) As shown in thefigure p minus q exhibits a linear relationship ampereforen 1 the slope of the straight line isM(M 6 sinφ03 minus sinφ0) and the intercept is C(C 6c cosφ03 minus sinφ0) where c is the cohesionand φ0 is the friction angle From the fitted linearrelation in the figure it can be seen that for thePietrafitta clay M 05472 and C 10559 kPaamperefore it can be determined that c 4998 kPaand φ0 14478deg and σ0 ccotφ0 19358 kPa can beobtained from the values of c and φ0 taking ξ 0

(2) Determination of η1 and η2 η1 and η2 can be de-termined by the q value of other stress paths withbne 0 Usually the strength of triaxial extension istaken as the reference value for calibration In mosttriaxial tests only one of the terms in equation (21) isneeded to satisfy the prediction accuracy require-ment For this reason only the η1 term is taken inthis test In the triaxial extension test when b 1 andθ 180degAndashA0 4 Using qe f(B)Mfg(θ)(p + σ0)qe 25441 kPa (θ180deg p 250 kPa) can be obtained

using the same method amperefore g(θ) 08462f(A) exp(4η1) and η1 00417

Using the above steps all parameters of the anisotropiccriterion for Pietrafitta clay are determined ampe test pa-rameter determination method is shown in Figure 5(a) ampecomparison of the prediction results by the isotropic cri-terion and the anisotropic criterion in Figure 5(b) shows thatthe anisotropic criterion proposed in the present study betterpredicts the test data

32 Strength Nonlinearity A series of large triaxial tests forrockfill materials were conducted to research strengthnonlinearity of this material [27] In the meridional plane(p minus q) the test results show the strength nonlinearitiesamperefore it is necessary to use nonlinear strength criteria todepict the strength development pattern ampe two strengthcriteria are compared by using them to predict the test dataas shown in Figure 6 As the figure shows the nonlinearstrength criterion proposed this paper can depict thestrength nonlinearity of this material

Isotropicη1 = 01

σx σy

σzM = 142σ0 = 0n = 1α = 02

η2 = ndash025 η1 = ndash01

(a)

η1 = 01

η1 = ndash01η2 = ndash025

Isotropic

σyσx

σz

(b)

Isotropic

M = 142σ0 = 0n = 1α = 02

η2 = ndash0333

σz

σyσx

η1 = ndash01

η1 = 01

(c)

η1 = 01

η1 = ndash01

Isotropicσx σy

σzη2 = ndash0333

(d)

Isotropic

M = 142σ0 = 0n = 1α = 02η2 = 0

σz

σyσx

η1 = ndash005

η1 = 005

(e)

Isotropic

σz

σyσx

η2 = 0η1 = ndash005

η1 = 005

(f )

Figure 3 Effect of η1 on the failure loci and f(A) in the deviatoric plane with fixed η2 in the octahedral plane (a c e) the failure curves and(b d f ) the f(A)


33 Anisotropy

331 True Triaxial Test of Soil In addition to the above truetriaxial test of Pietrafitta clay fine glass-bead sand is thenused to verify the anisotropic criterion Haruyama [35]conducted a large number of conventional triaxial tests andtrue triaxial tests to study the initial anisotropy of fine glass-bead sand ampe effective stress of the test is p 294 kPa ampeconventional triaxial compression test and the triaxial tensiletest are used to determine the parameters ampe parametersdetermined by the test data are as follows M 1142 σ0 0

n 1 η1 minus0038 and η2 0 ampe prediction results and testdata in the deviatoric plane and b-φ plane are shown inFigure 7 It can be seen from the figure that the anisotropiccriterion of the present study can reasonably predict thestrength of the fine glass sand in the π plane However theisotropic criterion fails to predict the strength of regions IIand III in the deviatoric plane and overestimates thematerialstrength In addition in the b-φ plane the isotropic criterioncan only predict one variation curve of the friction anglewhereas the anisotropic criterion results in a continuousprediction curve in each region reflecting the anisotropy of

β = 90degβ = 60deg

β = 30deg

β = 0degIsotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy

σz

(a)

Isotropic

σx σy

σzβ = 90deg

β = 60deg

β = 30deg

β = 0deg

η1 = ndash008η2 = 0

(b)

Isotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy

σz β = 90degβ = 60deg

β = 0deg

β = 30deg

(c)

Isotropic

σx σy


β = 0deg

β = 30deg

η1 = ndash008η2 = 0

(d)

Figure 4 ampe variations of the failure surface and f(A) in the octahedral plane (a b) the y minus z plane and (c d) the x minus z plane

0 250 500 750 10000

250

500

750

M = 05472n = 0

σ0 = 19358kPa

q (k

Pa)

p (kPa)

y = 0547x + 10559R2 = 09784

(a)

Isotropic Anisotropic Test data

M = 05472σ0 = 192845kPa

n = 1ξ = 0

η1 = 00417η2 = 0

σx σy

σz

(b)

Figure 5 Illustration on calibrating the parameters Mf c σ0 (b) Comparison of the two criteria for natural Pietrafitta clay (test data from[34])


the material amperefore the anisotropic criterion can rea-sonably predict the peak strength of the fine glass-bead sandcompared to the isotropic criterion

332 Torsional Shear Tests for Soils ampe failure frictionangle measured from torsional shear tests on LeightonBuzzard sand of different densities and on spherical glassbeads conducted by Yang et al [37] was analyzed using theproposed ANUS criterion ampe effective stress of the test isp 200 kPa ampe parameters determined by the test data areas follows for glass ballotini M 143 σ0 0 n 1η1 minus0368 and η2 01187 for medium dense LB sandM 1805 σ0 0 n 1 η1 minus02463 and η2 00593 for

dense LB sand M 191 σ0 0 n 1 η1 minus02668 andη2 00655 ampe prediction results (ANUS criterion andALD criterion [38]) and test data in the α-φ plane are shownin Figure 8 As shown in Figure 8 under torsional shear teststhe sand shows obvious anisotropy the new anisotropiccriterion can predict the failure friction angle well and theprediction curve is smooth

333 Rock Triaxial Test considering Deposition Surfaceampis test is mainly used to verify the peak strength of the rockconsidering the different angles between the rock depositionsurface and the principal stress ampe isotropic strengthcriterion and the anisotropic strength criterion were used for

10 15 20 25 30 35

20

25

30

35

40

ln (q

pa)

ln (ppa)

y = 0885x + 0922R2 = 0999

n = 0885σ0 = 0

Mf = 2514

(a)

0 1000 2000 3000 40000

2000

4000

6000

8000

q (k

Pa)

p (kPa)

LinearNonlinearTest data

(b)

Figure 6 Verifications of the two criteria (a) Parameter determination (b) Test data in the p minus q diagram


σx σy

σzM = 1142σ0 = 0n = 1

ξ = ndash002η1 = ndash0038

η2 = 0

(a)

00 02 04 06 08 1015

20

25

30

35Fr

ictio

n an

gle

b

Isotropic Sector 1Anisotropicsector 2 Anisotropicsector 3

Test data sector 1 Test data sector 2Test data sector 3

(b)

Figure 7 Verifications of the two criteria for glass beads in (a) the octahedral plane (b) the bndashφ diagram (test data from [35] this figure isreproduced from [36])


prediction and a comparison of the prediction results withthe test data shows the superiority of the anisotropic strengthcriterion Here based on the schist triaxial tests with dif-ferent deposition surface angles in the literature the pa-rameters of the anisotropic strength criterion are M 236σ0 8 Mpa n 074 η1 minus247 and η2 minus033 It can beseen from the test results that the two types of rock exhibitstrong strength anisotropy in different deposition directionsIn comparison the use of the isotropic strength criterion topredict strength variations can lead to large errors making itdifficult to promote the isotropic criterion in engineeringapplications

As shown in Figure 9 the anisotropic criterion canreasonably predict the strength failure line with different

deposition surface angles in the p minus q plane In contrastthe isotropic strength criterion can only predict thefailure line when β 0deg cannot reflect the strength an-isotropy and overestimates the strength values withdifferent deposition surface angles especially for β 45degMeanwhile the anisotropic criterion can reasonablypredict the trend of strength variation under differentconfining pressures at different deposition surfaces asshown in Figure 9(a) It can be seen from Figure 9(b) thatthe strengths of rocks with different deposition surfaceangles differ considerably ampe peak strength of the rock isthe largest when β 0deg or 90deg and the smallest when thedeposition surface angle is approximately 45deg In contrastthe isotropic criterion is a straight line in the β minus q plane

0 30 60 9010

20

30

40

50

ϕ (deg

)

α (deg)

Glass ballotini Medium dense LB sandDense LB sand ALD criterionANUS criterion

Figure 8 Verifications of ANUS criterion for different granular materials (data from [37])

0 50 100 1500

100

200

300

400

500

600

Mf = 236 n = 0737 σ0 = 8MPaη1 = ndash2471 η2 = ndash0333

q (k

Pa)

p (kPa)

Anisotropic (β = 0) Anisotropic (β = 45deg)Test data (β = 15deg)

Anisotropic (β = 15deg)Test data (β = 0)Test data (β = 45deg)

(a)

0 30 60 90ndash100

0

100

200

300

q (k

Pa)

β (degree)

Isotropic Anisotropic Test data (σc = 40MPa)

Test data (σc = 5MPa)


(b)

Figure 9 Comparison of the two criteria for Tournemire shale in (a) the p minus q diagram and (b) the β minus q diagram (test data from [39] thisfigure is reproduced from [40])


and hence cannot reflect the strength anisotropy causedby the direction of the deposition surface

4 Conclusion

(1) In this paper a new strength criterion is proposed onthe basis of the SMP criterion In the π plane the newcriterion is a series of smooth curves and can unifythe strength criteria

(2) ampe new criterion can depict the 3D strength vari-ation of the material and reflect the effect of theintermediate principal stress and the nonlinearcharacteristics of the material strength

(3) ampe unified anisotropic strength criterion based onthe pattern of the fabric evolution of granular ma-terials can reflect the strength anisotropy caused bythe deposition characteristics of the material ampiscriterion can be applied to the true triaxial test toconsider the direction angle of the deposition sur-face ampe correctness of the criterion was verifiedusing sand clay and rock materials

Data Availability

ampe underlying data used in the presented study were ob-tained from the literature

Conflicts of Interest

ampe authors declare that they have no conflicts of interest

Acknowledgments

ampis work was sponsored by Ningbo Natural ScienceFoundation Project (2019A610394) and the Initial ScientificResearch Fund of Young Teachers in Ningbo University ofTechnology (2140011540012) and supported by the Sys-tematic Project of Guangxi Key Laboratory of DisasterPrevention and Structural Safety (no 2019ZDK005) and theNingbo Public Welfare Science and Technology PlanningProject (no 2019C50012) ampese financial supports aregratefully acknowledged

References

[1] D C Drucker W Prager and H J Greenberg ldquoExtendedlimit design theorems for continuous mediardquo Quarterly ofApplied Mathematics vol 9 no 4 pp 381ndash389 1974

[2] H Matsuoka and T Nakai ldquoStress-deformation and strengthcharacteristics of soil under three different principal stressesrdquoProceedings of the Japan Society of Civil Engineers vol 1974no 232 pp 59ndash70 1974

[3] H Matsuoka ldquoOn the significance of the ldquospatial mobilizedplanerdquordquo Soils and Foundations vol 16 no 1 pp 91ndash100 1976

[4] P V Lade and J M Duncan ldquoElastoplastic stress-straintheory for cohesionless soilrdquo Journal of the GeotechnicalEngineering Division vol 101 no GT10 pp 1037ndash1053 1975

[5] P V Lade ldquoElasto-plastic stress-strain theory for cohesionlesssoil with curved yield surfacesrdquo International Journal of Solidsand Structures vol 13 no 11 pp 1019ndash1035 1977

[6] E Hoek and E Brown ldquoEmpirical strength criterion for rockmassesrdquo Journal of the Geotechnical Engineering Divisionvol 106 no 9 pp 1013ndash1035 1980

[7] W F Chen ldquoStrength theory applications developments andprospects for 21st centuryrdquo Concrete Plasticiry Past Presentand Future pp 7ndash48 Science Press Beijing New York 1998

[8] M D Liu and J P Carter ldquoGeneral strength criterion forgeomaterialsrdquo International Journal of Geomechanics vol 3no 2 pp 253ndash259 2003

[9] M-H Yu Y-W Zan and J Zhao ldquoA unified strength cri-terion for rock materialrdquo International Journal of Rock Me-chanics and Mining Sciences vol 39 no 8 pp 975ndash989 2002

[10] Y Yoshimine D Lu A Zhou and B Zou ldquoGeneralized non-linear strength theory and transformed stress spacerdquo Sciencein China Series E vol 47 no 6 pp 691ndash709 2004

[11] Y Xiao H Liu and R Y Liang ldquoModified Cam-Clay modelincorporating unified nonlinear strength criterionrdquo ScienceChina Technological Sciences vol 54 no 4 pp 805ndash810 2011

[12] M Yu L He and L Song ldquoTwin shear stress theory and itsgeneralizeationrdquo Scientia Sinica Series A vol 28 no 11pp 1174ndash1183 1985

[13] M-h Yu ldquoAdvances in strength theories for materials undercomplex stress state in the 20th Centuryrdquo Applied MechanicsReviews vol 55 no 3 pp 169ndash218 2002

[14] K J Willam and E P Warnke ldquoConstitutive model for thetriaxialbehavior of concreterdquo in Proceedings of the Interna-tional Association for Bridge and Structural EngineeringSeminar on Concrete Structure Subjected to Triaxial StressesISMES vol 19 pp 1ndash30 Bergamo Italy May 1975

[15] E Hoek and E T Brown ldquoPractical estimates of rock massstrengthrdquo International Journal of Rock Mechanics andMining Sciences vol 34 no 8 pp 1165ndash1186 1997

[16] S Mahendra and S Bhawani ldquoModified Mohr-Coulombcriterion for non-linear triaxial and polyaxial strength ofjointed rocksrdquo International Journal of Rock Mechanics andMining Sciences vol 51 pp 43ndash52 2012

[17] Z W Gao J D Zhao and Y P Yao ldquoA generalized an-isotropic failure criterion for geomaterialsrdquo InternationalJournal of Solids and Structures vol 47 no 22ndash23pp 3166ndash3185 2012

[18] Z Tong P Fu S Zhou and Y F Dafalias ldquoExperimentalinvestigation of shear strength of sands with inherent fabricanisotropyrdquoActa Geotechnica vol 9 no 2 pp 257ndash275 2014

[19] JWang Z Song B Zhao X Liu J Liu and J Lai ldquoA study onthe mechanical behavior and statistical damage constitutivemodel of sandstonerdquo Arabian Journal for Science and Engi-neering vol 43 no 10 pp 5179ndash5192 2018

[20] L Li M Aubertin R Simon and B Bussiere ldquoFormulationand application of a general inelastic locus for geomaterialswith variable porosityrdquo Canadian Geotechnical Journalvol 42 no 2 pp 601ndash623

[21] G BussiereSimon ldquoA new yield and failure criterion forgeomaterialsrdquo Geotechnique vol 58 no 2 pp 125ndash132 2008

[22] G Mortara ldquoA hierarchical single yield surface for frictionalmaterialsrdquo Computers And Geotechnics vol 36 no 6pp 960ndash967 2009

[23] D Su Z-L Wang and F Xing ldquoA two-parameter expressionfor failure surfacesrdquo Computers And Geotechnics vol 36no 3 pp 517ndash524 2009

[24] M Liu Y Gao and H Liu ldquoA nonlinear Drucker-Prager andMatsuoka-Nakai unified failure criterion for geomaterialswith separated stress invariantsrdquo International Journal of RockMechanics And Mining Sciences vol 50 pp 1ndash10 2012


[25] D C Lu C Ma X L Du L Jin and Q M Gong ldquoDe-velopment of a new nonlinear unified strength theory forgeomaterials based on the characteristic stress conceptrdquo In-ternational Journal of Geomechanics vol 17 no 2 Article ID04016058 2016

[26] P V Osswald and T A Osswald ldquoA strength tensor basedfailure criterion with stress interactionsrdquo Polymer Compositesvol 39 no 8 pp 2826ndash2834 2017

[27] M C Liu Y F Gao and X M Huang ldquoStudy on elasto-plastic constitutive model of rockfills with nonlinear strengthcharacteristicsrdquo Chinese Journal of Geotechnical Engineeringvol 27 no 3 pp 294ndash298 2005 in Chinese

[28] A Varadarajan K G Sharma S M Abbas andA K Dhawan ldquoConstitutive model for rockfill materials anddetermination of material constantsrdquo International Journal ofGeomechanics vol 6 no 4 pp 226ndash237 2006

[29] A K Gupta Constitutive Modeling of Rockfill MaterialsIndian Institute of Technology Delhi India 2000

[30] X R Liu Y L Tu P Wang Z L Zhong W L Tang andL B Du ldquoParticle breakage of soil-rock aggregate based onlarge-scale direct shear testsrdquo Chinese Journal of GeotechnicalEngineering vol 39 no 8 pp 1425ndash1434 2017 in Chinese

[31] Y F Dafalias A G Papadimitriou and X S Li ldquoSandplasticity model accounting for inherent fabric anisotropyrdquoJournal of Engineering Mechanics vol 130 no 11 pp 1319ndash1333 2004

[32] S Pietruszczak and Z Mroz ldquoFormulation of anisotropicfailure criteria incorporating a microstructure tensorrdquoComputers and Geotechnics vol 26 no 2 pp 105ndash112 2000

[33] S Pietruszczak and Z Mroz ldquoOn failure criteria for aniso-tropic cohesive-frictional materialsrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 25no 5 pp 509ndash524 2001

[34] L Callisto and S Rampello ldquoShear strength and small-strainstiffness of a natural clay under general stress conditionsrdquoGeotechnique vol 52 no 8 pp 547ndash560 2002

[35] M Haruyama ldquoAnisotropic deformation-strength charac-teristics of an assembly of spherical particles under threedimensional stressesrdquo Soils and Foundations vol 21 no 4pp 41ndash55 1981

[36] S Wang Z Zhong and X Liu ldquoDevelopment of an aniso-tropic nonlinear strength criterion for geomaterials based onSMP criterionrdquo International Journal of Geomechanicsvol 20 no 3 Article ID 04019183 2020

[37] L-T Yang D Wanatowski X Li H-S Yu and Y CaildquoLaboratory and micromechanical investigation of soil an-isotropyrdquo Japanese Geotechnical Society Special Publicationvol 2 no 9 pp 411ndash416 2016

[38] W Cao R Wang and J M Zhang ldquoFormulation of aniso-tropic strength criteria for cohesionless granular materialsrdquoInternational Journal of Geomechanics ASCE vol 17 no 7pp 1ndash10 Article ID 04016151 2017

[39] H Niandou J F Shao J P Henry and D FourmaintrauxldquoLaboratory investigation of the mechanical behaviour ofTournemire shalerdquo International Journal of Rock Mechanicsand Mining Sciences vol 34 no 1 pp 3ndash16 1997

[40] R Wang P Fu and Z X Tong ldquoDiscussion of ldquodescription ofinherent and induced anisotropy in granular media withparticles of high sphericityrdquordquo International Journal of Geo-mechanics ASCE vol 18 no 1 Article ID 07017014 2018


ampe above criteria are all isotropic strength criteria andcannot depict the strength anisotropy of geomaterialsHowever numerous geomaterials exhibit transverseisotropy due to deposition that is the strength of thematerial is the same in the deposition surface but differentin the depositional surfaces of different directions ampedirectional shear test of most soils shows that the changein the direction of the principal stress also leads tostrength anisotropy reflecting the initial anisotropy of thematerial amperefore it is necessary to establish a uniformstrength criterion that considers the depositional char-acteristics or initial anisotropy of geomaterials

ampe above previous studies show that the strength cri-terion is researched through a simple single form to a unifiedstrength theory system with a relatively wide scope of ap-plication Consequently it is important to explore a widelyapplicable strength theory system In the present paper anew anisotropic nonlinear unified strength criterion isestablished based on Mises criterion and SMP criterion ampenew criterion is approximated as a series of smooth curvesbetween the SMP curved triangle and the von Mises circle inthe π plane to unify the strength criteria Based on the fabrictensor a nonlinear anisotropic strength criterion is pro-posed ampe correctness of the criterion is verified using sandclay and rock materials

2 Form of the NUS Criterion

21 Linear Unified Strength (LUS) Criterion SMP criterion(Matsuoka and Nakai [2]) has been widely used to check thestrength of geomaterials because of its simple form and clearphysical meaning In the present study the SMP criterion ismodified to make it more widely applicable ampe SMP cri-terion is shown in the following equation

I1I2

I3 C (1)

where I1 I2 and I3 are stress invariants

I1 σ1 + σ2 + σ3I2 σ1σ2 + σ2σ3 + σ1σ3I3 σ1σ2σ3

⎫⎪⎪⎬

⎪⎪⎭ (2)

C is a model parameter Under conventional triaxialcompression (σ2 σ3) the sine of the friction angle φ0 of agranular material is

sinφ0 σ1 minus σ3σ1 + σ3

(3)

Substituting equations (2) and (3) into equation (1) gives

I31

I3

3 minus sinφ0( 11138573

1 minus sinφ0 minus sin2φ0 + sin3φ0 (4)

Using the trigonometric identity the expression of q isobtained as

3p3

minus L2q2p + L3 cos(3θ)q

3 0 (5)

where

L2 3 + sin2φ0

12 sin2φ0 (6)

L3 9 minus sin2φ0

108 sin2φ0 (7)

p σ1 + σ2 + σ3

3 (8)

q

12

σ1 minus σ2( 11138572

+ σ2 minus σ3( 11138572

+ σ1 minus σ3( 11138572

1113960 1113961

1113970

(9)

where p is the mean stress and q is generalized shear stressampus the shape function of the SMP criterion is

g0(θ) q

q0


radicL32L

3221113872 11138731113960 1113961


radicL3 cos(3θ)2L

3221113872 11138731113960 1113961

(10)

where q0 is the data for q when θ 0 amperefore the forms ofp and q of the SMP criterion are

q Mg0(θ)p (11)

where g0(θ) is obtained by equation (10) and M is thecoefficient of friction

M 6 sinφ0

3 minus sinφ0 (12)

ampe shape function of the SMP criterion cannot reflectthe influence of internal factors on material therefore theapplication of the criterion is limited In this report theshape function is rewritten as the following expression

q Mg(θ) p + σ0( 1113857 (13)

where M is obtained from equation (12) σ0 which is thetriaxial tensile strength of the material is given by

σ0 c cotφ0 (14)

and p and q are the average stress and the generalized shearstress as defined by equations (9) and (10) respectively

amperefore

g(θ) rθ

r0 (15)

where

rθ

3

radic(1 minus α)

2L2

1113968cos δθ

+6α sinφ03 minus sinφ0

r0

3

radic(1 minus α)

2L2

1113968cos δ0

+6α sinφ0

3 minus sinφ0

⎫⎪⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎪⎭

(16)

and α is the model parameter determined by the strengthratio under the triaxial compression and extension condi-tions ampe value of α ranges from 0 to 1 One has


δθ 13arccos minus

33

radicL3 cos(3θ)

2L322

1113888 1113889

δ0 13arccos minus

33

radicL3

2L322

1113888 1113889

⎫⎪⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎪⎭

(17)




q Mfg(θ)pr

p + σ0pr

1113888 1113889

n

(18a)

ampat is

lnq

pr

1113888 1113889 n lnp + σ0

pr

1113888 1113889 + ln Mfg(θ)1113960 1113961 (18b)



When n 1

Mf M 6 sinφ0

3 minus sinφ0 (19)


g(θ)

3

radic(1 minus α)2

L2


3

radic(1 minus α)2

L2


(20)




1113960 1113961 (21)


q f(A)Mfg(θ)pr

p + σ0pr

1113888 1113889

n

(22)









3 Test Verification




50deg60deg90deg

σxσy

σz

(a)

σxσy

σz

α = 10 (Mises)α = 08α = 04α = 0 (SMP)

(b)


σy

σ1

σ2σ3

β

σ3

σ1σ2

β

σ3

σ1σ2

βσx

θ = 60deg

θ = 120deg

θ = 180deg

θ = 240deg

θ = 300deg

IIIIII

II II

II

θ

θ = 0deg σzz

y

x








Isotropicη1 = 01

σx σy

σzM = 142σ0 = 0n = 1α = 02


(a)

η1 = 01


Isotropic

σyσx

σz

(b)

Isotropic

M = 142σ0 = 0n = 1α = 02

η2 = ndash0333

σz

σyσx

η1 = ndash01

η1 = 01

(c)

η1 = 01

η1 = ndash01

Isotropicσx σy

σzη2 = ndash0333

(d)

Isotropic

M = 142σ0 = 0n = 1α = 02η2 = 0

σz

σyσx

η1 = ndash005

η1 = 005

(e)

Isotropic

σz

σyσx

η2 = 0η1 = ndash005

η1 = 005

(f )



33 Anisotropy




β = 30deg

β = 0degIsotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy

σz

(a)

Isotropic

σx σy

σzβ = 90deg

β = 60deg

β = 30deg

β = 0deg

η1 = ndash008η2 = 0

(b)

Isotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy


β = 0deg

β = 30deg

(c)

Isotropic

σx σy


β = 0deg

β = 30deg

η1 = ndash008η2 = 0

(d)


0 250 500 750 10000

250

500

750

M = 05472n = 0

σ0 = 19358kPa

q (k

Pa)

p (kPa)

y = 0547x + 10559R2 = 09784

(a)


M = 05472σ0 = 192845kPa

n = 1ξ = 0

η1 = 00417η2 = 0

σx σy

σz

(b)







10 15 20 25 30 35

20

25

30

35

40

ln (q

pa)

ln (ppa)

y = 0885x + 0922R2 = 0999

n = 0885σ0 = 0

Mf = 2514

(a)

0 1000 2000 3000 40000

2000

4000

6000

8000

q (k

Pa)

p (kPa)


(b)



σx σy

σzM = 1142σ0 = 0n = 1


η2 = 0

(a)

00 02 04 06 08 1015

20

25

30

35Fr

ictio

n an

gle

b



(b)






0 30 60 9010

20

30

40

50

ϕ (deg

)

α (deg)



0 50 100 1500

100

200

300

400

500

600


q (k

Pa)

p (kPa)



(a)

0 30 60 90ndash100

0

100

200

300

q (k

Pa)

β (degree)




(b)




4 Conclusion




Data Availability




Acknowledgments


References











































δθ 13arccos minus

33

radicL3 cos(3θ)

2L322

1113888 1113889

δ0 13arccos minus

33

radicL3

2L322

1113888 1113889

⎫⎪⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎪⎭

(17)




q Mfg(θ)pr

p + σ0pr

1113888 1113889

n

(18a)

ampat is

lnq

pr

1113888 1113889 n lnp + σ0

pr

1113888 1113889 + ln Mfg(θ)1113960 1113961 (18b)



When n 1

Mf M 6 sinφ0

3 minus sinφ0 (19)


g(θ)

3

radic(1 minus α)2

L2


3

radic(1 minus α)2

L2


(20)




1113960 1113961 (21)


q f(A)Mfg(θ)pr

p + σ0pr

1113888 1113889

n

(22)









3 Test Verification




50deg60deg90deg

σxσy

σz

(a)

σxσy

σz

α = 10 (Mises)α = 08α = 04α = 0 (SMP)

(b)


σy

σ1

σ2σ3

β

σ3

σ1σ2

β

σ3

σ1σ2

βσx

θ = 60deg

θ = 120deg

θ = 180deg

θ = 240deg

θ = 300deg

IIIIII

II II

II

θ

θ = 0deg σzz

y

x








Isotropicη1 = 01

σx σy

σzM = 142σ0 = 0n = 1α = 02


(a)

η1 = 01


Isotropic

σyσx

σz

(b)

Isotropic

M = 142σ0 = 0n = 1α = 02

η2 = ndash0333

σz

σyσx

η1 = ndash01

η1 = 01

(c)

η1 = 01

η1 = ndash01

Isotropicσx σy

σzη2 = ndash0333

(d)

Isotropic

M = 142σ0 = 0n = 1α = 02η2 = 0

σz

σyσx

η1 = ndash005

η1 = 005

(e)

Isotropic

σz

σyσx

η2 = 0η1 = ndash005

η1 = 005

(f )



33 Anisotropy




β = 30deg

β = 0degIsotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy

σz

(a)

Isotropic

σx σy

σzβ = 90deg

β = 60deg

β = 30deg

β = 0deg

η1 = ndash008η2 = 0

(b)

Isotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy


β = 0deg

β = 30deg

(c)

Isotropic

σx σy


β = 0deg

β = 30deg

η1 = ndash008η2 = 0

(d)


0 250 500 750 10000

250

500

750

M = 05472n = 0

σ0 = 19358kPa

q (k

Pa)

p (kPa)

y = 0547x + 10559R2 = 09784

(a)


M = 05472σ0 = 192845kPa

n = 1ξ = 0

η1 = 00417η2 = 0

σx σy

σz

(b)







10 15 20 25 30 35

20

25

30

35

40

ln (q

pa)

ln (ppa)

y = 0885x + 0922R2 = 0999

n = 0885σ0 = 0

Mf = 2514

(a)

0 1000 2000 3000 40000

2000

4000

6000

8000

q (k

Pa)

p (kPa)


(b)



σx σy

σzM = 1142σ0 = 0n = 1


η2 = 0

(a)

00 02 04 06 08 1015

20

25

30

35Fr

ictio

n an

gle

b



(b)






0 30 60 9010

20

30

40

50

ϕ (deg

)

α (deg)



0 50 100 1500

100

200

300

400

500

600


q (k

Pa)

p (kPa)



(a)

0 30 60 90ndash100

0

100

200

300

q (k

Pa)

β (degree)




(b)




4 Conclusion




Data Availability




Acknowledgments


References
















































3 Test Verification




50deg60deg90deg

σxσy

σz

(a)

σxσy

σz

α = 10 (Mises)α = 08α = 04α = 0 (SMP)

(b)


σy

σ1

σ2σ3

β

σ3

σ1σ2

β

σ3

σ1σ2

βσx

θ = 60deg

θ = 120deg

θ = 180deg

θ = 240deg

θ = 300deg

IIIIII

II II

II

θ

θ = 0deg σzz

y

x








Isotropicη1 = 01

σx σy

σzM = 142σ0 = 0n = 1α = 02


(a)

η1 = 01


Isotropic

σyσx

σz

(b)

Isotropic

M = 142σ0 = 0n = 1α = 02

η2 = ndash0333

σz

σyσx

η1 = ndash01

η1 = 01

(c)

η1 = 01

η1 = ndash01

Isotropicσx σy

σzη2 = ndash0333

(d)

Isotropic

M = 142σ0 = 0n = 1α = 02η2 = 0

σz

σyσx

η1 = ndash005

η1 = 005

(e)

Isotropic

σz

σyσx

η2 = 0η1 = ndash005

η1 = 005

(f )



33 Anisotropy




β = 30deg

β = 0degIsotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy

σz

(a)

Isotropic

σx σy

σzβ = 90deg

β = 60deg

β = 30deg

β = 0deg

η1 = ndash008η2 = 0

(b)

Isotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy


β = 0deg

β = 30deg

(c)

Isotropic

σx σy


β = 0deg

β = 30deg

η1 = ndash008η2 = 0

(d)


0 250 500 750 10000

250

500

750

M = 05472n = 0

σ0 = 19358kPa

q (k

Pa)

p (kPa)

y = 0547x + 10559R2 = 09784

(a)


M = 05472σ0 = 192845kPa

n = 1ξ = 0

η1 = 00417η2 = 0

σx σy

σz

(b)







10 15 20 25 30 35

20

25

30

35

40

ln (q

pa)

ln (ppa)

y = 0885x + 0922R2 = 0999

n = 0885σ0 = 0

Mf = 2514

(a)

0 1000 2000 3000 40000

2000

4000

6000

8000

q (k

Pa)

p (kPa)


(b)



σx σy

σzM = 1142σ0 = 0n = 1


η2 = 0

(a)

00 02 04 06 08 1015

20

25

30

35Fr

ictio

n an

gle

b



(b)






0 30 60 9010

20

30

40

50

ϕ (deg

)

α (deg)



0 50 100 1500

100

200

300

400

500

600


q (k

Pa)

p (kPa)



(a)

0 30 60 90ndash100

0

100

200

300

q (k

Pa)

β (degree)




(b)




4 Conclusion




Data Availability




Acknowledgments


References
















































Isotropicη1 = 01

σx σy

σzM = 142σ0 = 0n = 1α = 02


(a)

η1 = 01


Isotropic

σyσx

σz

(b)

Isotropic

M = 142σ0 = 0n = 1α = 02

η2 = ndash0333

σz

σyσx

η1 = ndash01

η1 = 01

(c)

η1 = 01

η1 = ndash01

Isotropicσx σy

σzη2 = ndash0333

(d)

Isotropic

M = 142σ0 = 0n = 1α = 02η2 = 0

σz

σyσx

η1 = ndash005

η1 = 005

(e)

Isotropic

σz

σyσx

η2 = 0η1 = ndash005

η1 = 005

(f )



33 Anisotropy




β = 30deg

β = 0degIsotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy

σz

(a)

Isotropic

σx σy

σzβ = 90deg

β = 60deg

β = 30deg

β = 0deg

η1 = ndash008η2 = 0

(b)

Isotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy


β = 0deg

β = 30deg

(c)

Isotropic

σx σy


β = 0deg

β = 30deg

η1 = ndash008η2 = 0

(d)


0 250 500 750 10000

250

500

750

M = 05472n = 0

σ0 = 19358kPa

q (k

Pa)

p (kPa)

y = 0547x + 10559R2 = 09784

(a)


M = 05472σ0 = 192845kPa

n = 1ξ = 0

η1 = 00417η2 = 0

σx σy

σz

(b)







10 15 20 25 30 35

20

25

30

35

40

ln (q

pa)

ln (ppa)

y = 0885x + 0922R2 = 0999

n = 0885σ0 = 0

Mf = 2514

(a)

0 1000 2000 3000 40000

2000

4000

6000

8000

q (k

Pa)

p (kPa)


(b)



σx σy

σzM = 1142σ0 = 0n = 1


η2 = 0

(a)

00 02 04 06 08 1015

20

25

30

35Fr

ictio

n an

gle

b



(b)






0 30 60 9010

20

30

40

50

ϕ (deg

)

α (deg)



0 50 100 1500

100

200

300

400

500

600


q (k

Pa)

p (kPa)



(a)

0 30 60 90ndash100

0

100

200

300

q (k

Pa)

β (degree)




(b)




4 Conclusion




Data Availability




Acknowledgments


References











































33 Anisotropy




β = 30deg

β = 0degIsotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy

σz

(a)

Isotropic

σx σy

σzβ = 90deg

β = 60deg

β = 30deg

β = 0deg

η1 = ndash008η2 = 0

(b)

Isotropic

M = 142σ0 = 0n = 1ξ = 02

η1 = ndash008η2 = 0

σx σy


β = 0deg

β = 30deg

(c)

Isotropic

σx σy


β = 0deg

β = 30deg

η1 = ndash008η2 = 0

(d)


0 250 500 750 10000

250

500

750

M = 05472n = 0

σ0 = 19358kPa

q (k

Pa)

p (kPa)

y = 0547x + 10559R2 = 09784

(a)


M = 05472σ0 = 192845kPa

n = 1ξ = 0

η1 = 00417η2 = 0

σx σy

σz

(b)







10 15 20 25 30 35

20

25

30

35

40

ln (q

pa)

ln (ppa)

y = 0885x + 0922R2 = 0999

n = 0885σ0 = 0

Mf = 2514

(a)

0 1000 2000 3000 40000

2000

4000

6000

8000

q (k

Pa)

p (kPa)


(b)



σx σy

σzM = 1142σ0 = 0n = 1


η2 = 0

(a)

00 02 04 06 08 1015

20

25

30

35Fr

ictio

n an

gle

b



(b)






0 30 60 9010

20

30

40

50

ϕ (deg

)

α (deg)



0 50 100 1500

100

200

300

400

500

600


q (k

Pa)

p (kPa)



(a)

0 30 60 90ndash100

0

100

200

300

q (k

Pa)

β (degree)




(b)




4 Conclusion




Data Availability




Acknowledgments


References















































10 15 20 25 30 35

20

25

30

35

40

ln (q

pa)

ln (ppa)

y = 0885x + 0922R2 = 0999

n = 0885σ0 = 0

Mf = 2514

(a)

0 1000 2000 3000 40000

2000

4000

6000

8000

q (k

Pa)

p (kPa)


(b)



σx σy

σzM = 1142σ0 = 0n = 1


η2 = 0

(a)

00 02 04 06 08 1015

20

25

30

35Fr

ictio

n an

gle

b



(b)






0 30 60 9010

20

30

40

50

ϕ (deg

)

α (deg)



0 50 100 1500

100

200

300

400

500

600


q (k

Pa)

p (kPa)



(a)

0 30 60 90ndash100

0

100

200

300

q (k

Pa)

β (degree)




(b)




4 Conclusion




Data Availability




Acknowledgments


References














































0 30 60 9010

20

30

40

50

ϕ (deg

)

α (deg)



0 50 100 1500

100

200

300

400

500

600


q (k

Pa)

p (kPa)



(a)

0 30 60 90ndash100

0

100

200

300

q (k

Pa)

β (degree)




(b)




4 Conclusion




Data Availability




Acknowledgments


References












































4 Conclusion




Data Availability




Acknowledgments


References











































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