A Unified Constitutive Model for Both Clay and Sand With

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Computers and Geotechnics 35 (2008) 210–222

A unified constitutive model for both clay and sand withhardening parameter independent on stress path

Y.P. Yao a,*, D.A. Sun b, H. Matsuoka c

a Department of Civil Engineering, Beihang University, Beijing 100083, Chinab Department of Civil Engineering, Shanghai University, Shanghai 200072, China

c Department of Civil Engineering, Nagoya Institute of Technology, Nagoya 466-8555, Japan

Received 31 August 2006; received in revised form 30 March 2007; accepted 30 March 2007Available online 21 May 2007

Abstract

A unified constitutive model for both clay and sand under three-dimensional stress conditions is derived from the modified Cam-claymodel, by taking the following two points into consideration. First, a transformed stress tensor based on the SMP (spatially mobilizedplane) criterion is applied to the Cam-clay model. The proposed model consistently describes shear yielding and shear failure and com-bines critical state theory with the SMP criterion for clay. Secondly, a new hardening parameter, which is independent of the stress path,is derived in order to develop a unified constitutive model for both clay and sand. It not only describes the dilatancy for lightly to heavilydilatant sand, but also reduces to the plastic volumetric strain for clay. The validity of the hardening parameter is confirmed by the testresults of triaxial compression and extension tests on sand under various stress paths. Only five conventional soil parameters are neededin the proposed model.� 2007 Elsevier Ltd. All rights reserved.

Keywords: Clay; Dilatancy; Elastoplastic model; Sand; Stress path; Three-dimensional stress

1. Introduction

The modified Cam-clay model, called the Cam-claymodel for short in this paper, which is suitable for clay(in this paper, the term clay is to be interpreted as normallyconsolidated clay), was proposed by Roscoe and Burland[18]. The Cam-clay model and many other models havebeen generalized by assuming a section of the yield surfaceto be circular in the p-plane. The mean stress pð¼ rii=3Þand deviator stress qð¼

ffiffiffiffiffiffiffiffi3=2

p ffiffiffiffiffiffiffiffiffisijsijp Þ are used as the stress

parameters in the models, where sijð¼ rij � pdijÞ is a devia-toric stress tensor and dij is Kronecker’s delta. That is tosay, the criterion of the Extended Mises type (g = q/p = const.) is adopted for the shear yielding and the shearfailure of clay in the Cam-clay model. Shear yielding iscaused by an increase in stress ratio g = q/p, while com-

0266-352X/$ - see front matter � 2007 Elsevier Ltd. All rights reserved.

doi:10.1016/j.compgeo.2007.04.003

* Corresponding author. Tel.: +86 10 82316111.E-mail address: [email protected] (Y.P. Yao).

pressive yielding is caused by an increase in the mean stressp. However, as experimental evidence shows, the ExtendedMises criterion grossly overestimates the strength in triaxialextension, and also results in incorrect intermediate stressratios in plane strain [22]. In contrast to the ExtendedMises failure, the SMP failure criterion [8], which is consid-ered to be a three-dimensional extension of the Mohr–Cou-lomb criterion, is a failure criterion that explains the highquality test results for soils. A transformed stress tensorhas been proposed by Matsuoka et al. [9] which makesthe SMP criterion become circular in the transformed p-plane. A revised transformed stress tensor ~rij for theSMP criterion is developed in this paper based on thetransformed stress method [9] and the work of Yao andSun [24,25] in order to take the plastic strain incrementdirection into account additionally. The new transformedstress tensor ~rij is applied to the Cam-clay model. The pro-posed model consistently describes the shear yielding andshear failure of soils under three-dimensional stresses, both

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Circle

SMP

( )22~σσ ( )33

~σσ

( )11~σσ

0~ rr =

AA′

θ~

θ

Fig. 1. SMP criterion in p-plane (solid curve) and transformed p-plane(broken circle).

Y.P. Yao et al. / Computers and Geotechnics 35 (2008) 210–222 211

of which obey the SMP criterion, and combines criticalstate theory with the SMP criterion for clay.

Traditionally, the plastic volumetric strain epv is taken as

the hardening parameter in the Cam-clay model, which isnot appropriate for dilatant sand. To date, a lot of harden-ing parameters which can describe the dilatancy of soilshave been assumed (e.g., [5,15,2,11,1,6]), and several dilat-ant plasticity models have been proposed (e.g., [4,16,17,27,3,26]). In this paper, a new hardening parameter, thatis independent of stress path, is derived by consideringunified yield and plastic potential functions which are thesame those for Cam-clay. As will be described later, thephysical meaning of this hardening parameter is clear.The proposed hardening parameter not only describes thedilatancy from lightly to heavily dilatant sand, but alsoreduces to the plastic volumetric strain ep

v for clay. Thevalidity of the hardening parameter H is confirmed by tri-axial compression and extension test results on sand undervarious stress paths.

The ability of the proposed model to predict the drainedbehavior of normally consolidated clay and saturated sandis examined along various stress paths under triaxial com-pression and extension conditions. The results predicted bythe proposed model agree well with the test results. Onlyfive soil parameters are needed in the proposed model.The values of these parameters can be determined by aloading and unloading isotropic compression test and aconventional triaxial compression test. In this paper, theterm stress is to be interpreted as effective stress. The imple-mentation of the model into the finite element method fol-lows the methods described in Sheng et al. [20] and Sloanet al. [21].

2. A transformed stress tensor based on the SMP criterion

A transformed stress tensor, by which the SMP criterionis made circular in the transformed p-plane, has been pro-posed by Matsuoka et al. [9]. A revised transformed stresstensor ~rij is developed in this paper based on the trans-formed stress method [9] and the work of Yao and Sun[24,25] in order to take the plastic strain increment direc-tion into account additionally. The outline of the newtransformed stress tensor ~rij is introduced as follows:

The SMP criterion can be written as

I1I2=I3 ¼ const ð1Þwhere I1, I2 and I3 are the first, second and third stressinvariants, respectively. The solid curve in Fig. 1 is theshape of the SMP criterion in the p-plane. The transformedstress is deduced from what makes the SMP curve in the p-plane become a circle, the dot line of Fig. 1, with the centerbeing the origin in the transformed p-plane (~p-plane).

When p = const, we assume that ~r is equal to r0, where ~ris the stress radius in the p-plane of the transformed stressspace and r0 is the stress radius in the p-plane of the ordin-ary stress space when h = 0. From Eq. (1), r0 can beexpressed by Matsuoka et al. [9]

~r ¼ r0 ¼ffiffiffi2

3

rq� ¼

ffiffiffi2

3

r2I1

3ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðI1I2 � I3Þ=ðI1I2 � 9I3Þ

p� 1

ð2Þ

The generalized deviatoric stress in the transformed stressspace ~qð¼


p~rÞ can be written as

~q ¼ q� ¼ 2I1

3ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðI1I2 � I3Þ=ðI1I2 � 9I3Þ

p� 1

ð3Þ

To realize the transformation from rij to ~rij, the follow-ing equations should be made:

~p ¼ p

~q ¼ q�

~h ¼ f1ðhÞ

8><>: ð4Þ

where ~p is the mean transformed stress; ~h and h are Lode’sstress angles in the transformed stress space and the ordin-ary stress space respectively, which are one-to-one corre-spondence with intermediate principal stress parameters ~band b (=(r2 � r3)/(r1 � r3)) in the transformed stress spaceand the ordinary stress space respectively; f1(h) is a func-tion of h which also can be expressed by b. So, Eq. (4)can be rewritten as

~p ¼ p

~q ¼ q�

~b ¼ f2ðbÞ

8><>: ð5Þ

where

~b ¼ ~r2�~r3

~r1�~r3

~p ¼ 13ð~r1 þ ~r2 þ ~r3Þ ¼ 1

3~rii

~q ¼ 1ffiffi2p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið~r1 � ~r2Þ2 þ ð~r2 � ~r3Þ2 þ ð~r3 � ~r1Þ2

q

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi32ð~rij � ~pdijÞð~rij � ~pdijÞ

q

8>>>>>><>>>>>>:

ð6Þ

f1(h) or f2(b) is a function to transform the flow directionof the plastic strain increment in the ordinary stress spaceinto the flow direction in the transformed stress space. By

212 Y.P. Yao et al. / Computers and Geotechnics 35 (2008) 210–222

f1(h) or f2(b), the deviation of the plastic strain incrementdirection from the stress direction in the ordinary p-planecan be taken into consideration, although the directionsof stress and plastic strain increments in transformed p-plane are coincident.

A reasonable plastic potential surface for frictionalmaterials should be between the curved-triangle shape(I1I2/I3 = const) and the circular shape (q/p = const) inthe p-plane. So, the flow direction of the plastic strainincrement is also between the normal to the triangularshape of SMP criterion and that to the circular shape, asshown Fig. 1. The following f2(b) [24] has the abovecharacteristics

f2ðbÞ ¼ffiffiffiffiffir1p þ ffiffiffiffiffi

r3p

ffiffiffiffiffir2p þ ffiffiffiffiffi

r3p b ¼

ffiffiffiffiffir2p � ffiffiffiffiffi

r3p

ffiffiffiffiffir1p � ffiffiffiffiffi

r3p ð7Þ

So, the transformed stress tensors under the general stressstate can be obtained as

~rij ¼ pdij þq�

qsðrs

ij � psdijÞ ð8Þ

and

ps ¼ rsii=3

qs ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið3=2Þðrs

ij � psdijÞðrsij � psdijÞ

q

rsij ¼ ðI s

1rik þ I s3dikÞðrkj þ I s

3dkjÞ�1

I s1 ¼

ffiffiffiffiffir1p þ ffiffiffiffiffi

r2p þ ffiffiffiffiffi

r3p

I s2 ¼

ffiffiffiffiffiffiffiffiffir1r2p þ ffiffiffiffiffiffiffiffiffi

r2r3p þ ffiffiffiffiffiffiffiffiffi

r3r1p

I s3 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffir1r2r3p

9>>>>>>>>>>=>>>>>>>>>>;

ð9Þ

When the stress tensor rij is given, the transformed stresstensor ~rij can be calculated from Eq. (8). From the abovederivation, it can be shown that the shape of the SMP cri-terion becomes a cone with the axis being the space diago-nal ~r1 ¼ ~r2 ¼ ~r3 in the transformed principal stress space.Its cross-section is a circle with the center being the originO in the ~p-plane (see Fig. 1). Noting the similarity in theshapes of the Extended Mises criterion in the principalstress space and the SMP criterion in the transformed prin-cipal stress space, we can revise existing elastoplastic mod-els such as the Cam-clay model by using the transformedstress tensor ~rij based on the SMP criterion.

3. Basic method for constructing a unified hardeningparameter independent of stress path

Hardening of materials is due to the occurrence of plasticstrain. Therefore, a parameter representing the degree ofhardening must depend on the plastic strain. Hardeningparameters are internal variables that are used to indicatethe rate of plastic deformation, and should be a functionof plastic volumetric strain and plastic shear strain. For asingle yield surface model, the hardening parameter shouldhave the following properties: (1) its increments should bethe same from one point on a yield surface to differentpoints on another yield surface, and (2) its increment should

also be same from one point to another point along differentstress paths. Here, the hardening parameter is different withthe hardening modulus which obviously changes from pointto point. However, the plastic strain is usually dependent onthe stress path. For example, the plastic volumetric strainand plastic shear strain for sand are all dependent on stresspath (the test results in the following section can confirmthis characteristic). So, the plastic strain increments insteadof the total plastic strains should be used to construct a uni-fied hardening parameter for various kinds of soils. That isto say, the hardening parameter should be a function of theplastic volumetric strain increment dep

v and the plastic shearstrain increment dep

d. It is assumed that the plastic volumet-ric strain increment and plastic shear strain increment canbe related through the stress-dilatancy equationdep

v=depd ¼ f1ðgÞ. For example, the stress-dilatancy equation

in the original Cam-clay model is depv=dep

d ¼ M � g (stressratio g = q/p; M is the value of g at the critical state), so thatf1(g) = M � g. Therefore, we can choose either the plasticvolumetric strain increment dep

v or the plastic shear strainincrement dep

d as a basic parameter to construct the harden-ing parameter. However, plastic shear strain does not occurduring isotropic compression, which means that it is impos-sible to describe hardening under isotropic compression byuse of the plastic shear strain increment. Therefore, we haveto choose the plastic volumetric strain increment dep

v to con-struct the hardening parameter. Realizing that the incre-ment of the hardening parameter should be independentof the stress path, we assume there exists a stress functionR(g) such that the integral

R depv

RðgÞ is independent of stresspath. Thus, this integral can be used as a general hardeningparameter

H ¼Z

depv

RðgÞ ð10Þ

In the Cam-clay model, R(g) = 1. A suitable R(g) for bothclay and sand will be introduced in the next section.

4. A unified hardening parameter for both clay and sand

A basic method for constructing a unified hardeningparameter independent of stress path has been introducedin the preceding section. We will now derive a new harden-ing parameter for both clay and sand in this section. Themodified Cam-clay model is considered to be one of thebest basic elastoplastic models for clay, which is better thanthe original one [19]. In the modified Cam-clay model, theyield and plastic potential functions are assumed to be ofthe same form as follows:

f ¼ g ¼ lnpp0

þ ln 1þ q2

M2p2

� ��Z

depv

cp¼ 0 ð11Þ

where p0 is the initial mean stress, depvð¼ dep

iiÞ is the plasticvolumetric strain increment and cp is written as

cp ¼k� j1þ e0

ð12Þ


with k being the compression index, j being the swelling in-dex, and e0 the initial void ratio at p = p0. It is reasonablethat Eq. (11) is also chosen as the plastic potential functionfor dilatant sand, because this equation predicts that theplastic volumetric strain increment is positive before thecharacteristic state (g = M) [7] and negative after the char-acteristic state (see Fig. 2). To adopt an associated flow rulein new model, a yield function similar to Eq. (11) is also as-sumed for sand. However, the plastic volumetric straincannot be used as the hardening parameter for sand be-cause it is dependent on the stress path and does not in-crease monotonically with loading. In this paper, a newhardening parameter H is derived to describe the hardeningbehavior of clay and sand. The yield and plastic potentialfunctions for sand are written as

f ¼ g ¼ lnpp0

þ ln 1þ q2

M2p2

� �� H ¼ 0 ð13Þ

Nova and Wood [14,15] have chosen a combination ofthe plastic volumetric strain and plastic deviator strain asa hardening parameter. In the present model, the harden-ing parameter is considered to be a combination of thestress tensor rij and plastic strain increment tensor dep

ij,i.e., a plastic work type of hardening parameter. Thus,the following hardening parameter is assumed by usingthe plastic strain invariants

H ¼Z

dH ¼Z

c1ðrijÞdepv þ c2ðrijÞdep

d

� �ð14Þ

where c1(rij) and c2(rij) are the functions of the stress

tensor, respectively, and depd ¼


p�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðdep

ij � depvdij=3Þðdep

ij � depvdij=3Þ

qÞ is the plastic deviator

strain increment. In the modified Cam-clay model, the fol-lowing stress-dilatancy equation is adopted:

depv

depd

¼ M2 � g2

2gð15Þ

Substituting Eq. (15) into (14) gives

H ¼Z

c1ðrijÞdepvþc2ðrijÞ

2g

M2�g2dep

v

� ¼Z

cðrijÞdepv ð16Þ

where c(rij) is a function of the stress tensor. Note thatc(rij) in Eq. (16) is also equal to 1/R(g) in Eq. (10). Fig. 3

q , d

ε

p , d ε0

η=Mη=M f

vp

dp

d ε

d εd ε

d ε

vp

dp

vp

dp

Fig. 2. Direction of plastic strain increment vectors.

shows the constant mean stress path (path AB) and the iso-tropic compression stress path (path AC) along which thehardening parameter changes from H0 to H. The processfor finding the hardening parameter H is (1) to derive ageneral equation of H along path AB and (2) to determinean explicit equation of H along path AC.

4.1. Along the constant mean stress path AB

After substituting Eq. (16) into Eq. (13), the total differ-ential form of the yield function is expressed as

df ¼ ofop

dp þ ofoq

dqþ ofoH

dH

¼ ofop

dp þ ofoq

dq� cðrijÞKofop¼ 0 ð17Þ

So the proportionality constant K can be written as

K ¼ 1

cðrijÞ

ofop dp þ of

oq dqofop

ð18Þ

Based on Eq. (13), the following two differential equa-tions can be obtained:

ofop¼ 1

pM2 � g2

M2 þ g2ð19Þ

ofoq¼ 1

p2g

M2 þ g2ð20Þ

By substituting Eqs. (19) and (20) into Eq. (18), the plas-tic deviator strain increment along a constant mean stresspath is as follows:

depd ¼ K

ofoq¼ 1

cðrijÞ1

p4g2

M4 � g4dq ð21Þ

Fig. 4 shows the results of triaxial compression tests onclay and sand (data from [12]). It can be seen from Fig. 4athat the shapes of the curves g � ed for clay and sand arealike. The stress ratios (q/p) at failure for clay and sandare M and Mf respectively. In fact, Mf is not constant dur-ing shearing. If the variation of Mf is considered duringshearing, more complex behavior (e.g., the softening) ofsoils can be described [23]. In this paper, Mf is assumedto be constant for the sake of simplicity. The parametersM and Mf are similar to those of Nova and Wood [15].

q

C (px,0)0

η= Mη=M f

A (p0,0)

B (p0,q)

p

HH

0

Fig. 3. Stress paths for deriving hardening parameter H.

0 100 200 3000

2

4

6

8

Clay

Sand

Comp.p=196 kPa

d /d d

(Mf-

)/4

44

2

0 2 4 6 80

1

2

d(%)ε

Clay M

Comp.p=196 kPa

Sand Mf

ηη

η

η ε

Fig. 4. Triaxial compression test results for clay and sand arranged in (a)g � ed; (b) ðM4

f � g4Þ=4g2 � dg=ded.


Therefore, compared with the equation of the plastic devi-ator strain increment for clay in the Cam-clay model whenthe mean stress is constant (Eq. (22)), the equation of theplastic deviator strain increment for sand is assumed tobe Eq. (23) when the mean stress is constant

depd ¼ cp

1

p4g2

M4 � g4dq ðfor clayÞ ð22Þ

depd ¼ q

1

p4g2

M4f � g4

dq ðfor sandÞ ð23Þ

where q is a constant. Eqs. (22) and (23) can also be writtenin the linear forms

M4 � g4

4g2¼ cp

dgdep

d

ðfor clayÞ ð24Þ

M4f � g4

4g2¼ q

dgdep

d

ðfor sandÞ ð25Þ

The validity of Eqs. (24) and (25) is confirmed by the tri-axial compression results in Fig. 4b. So, the assumed formof Eq. (23) is appropriate. It is worth noting that Mf = Mfor clay and the elastic deviator strain is very small under aconstant mean stress in Fig. 4b. By combining Eqs. (21)and (23), c(rij) in Eq. (21) is expressed as

cðrijÞ ¼1

qM4

f � g4

M4 � g4ð26Þ

Substituting Eq. (26) into Eq. (16) gives

H ¼Z

dH ¼Z

1

qM4

f � g4

M4 � g4dep

v ð27Þ

4.2. Along the isotropic compression stress path AC

When g = 0 (path AC), Eq. (27) becomes

H ¼Z

dH ¼Z

1

qM4

f

M4dep

v ð28Þ

Moreover, under the isotropic compression condition(g = q/p = 0) and ep

v ¼ cp lnðp=p0Þ. In addition, Eq. (13)becomes H = ln(p/p0) when g = 0. So, the following equa-tion can be obtained from the above two equations:

H ¼Z

dH ¼Z

depv

cpð29Þ

Letting Eq. (28) equal Eq. (29) gives

q ¼ cpM4

f

M4ð30Þ

Finally, we can obtain the following equation of the newhardening parameter for sand by substituting Eq. (30) intoEq. (27):

H ¼Z

dH ¼Z

1

cp

M4

M4f

M4f � g4

M4 � g4dep

v ð31Þ

Comparing Eq. (10) with Eq. (31) gives

RðgÞ ¼ cpM4

f

M4

M4 � g4

M4f � g4

ð32Þ

If M = Mf, Eq. (31) becomes H ¼R

dH ¼R

depv=cp, which

is the same as the hardening parameter for clay in theCam-clay model. H in Eq. (31) is a unified one for bothclay and sand.

5. A unified elastoplastic model for both clay and sand

In the proposed model, the equations of the yield locusand plastic potential remain the same as in Cam-clay, butthe transformed stress tensor ~rij (based on the SMP crite-rion) and the new hardening parameter are adopted tomodel the mechanical behavior of clay and sand underthree-dimensional stresses.

The total strain increment is given by the summation ofthe elastic component and the plastic component as usual:

deij ¼ deeij þ dep

ij ð33Þ

Here, the elastic component is given by the followingequation

deeij ¼

1þ mE

drij �mE

drmmdij ð34Þ

where m is Poisson’s ratio, and the elastic modulus E is ex-pressed as


E ¼ 3ð1� 2mÞð1þ e0Þj

p ð35Þ

The plastic component is given by assuming the flow rulenot in rij space but in ~rij space.

depij ¼ K

ogo~rij

ð36Þ

where the plastic potential function g (or the yield functionf), the hardening parameter H, the proportionality con-stant K and the differential equation og=o~rij are given,respectively, as follows:

f ¼ g ¼ ln~p~p0

þ ln 1þ ~q2

M2~p2

� �� H ¼ 0 ð37Þ

H ¼Z

dH ¼Z

1

cp

M4

M4f

M4f � ~g4

M4 � ~g4dep

v ð38Þ

K ¼ cpM4

f

M4

M4 � ~g4

M4f � ~g4

d~p þ 2~p~q

M2~p2 � ~q2d~q

� �ð39Þ

ogo~rij¼ 1

M2~p2 þ ~q2

M2~p2 � ~q2

3~pdij þ 3ð~rij � ~pdijÞ

� ð40Þ

In the above equations, the deviator stress ~q and the trans-formed stress ratio ~g in ~rij space, Mð~g at critical state) andM fð~g at peak) are written respectively as follows:

~g ¼ ~q=~p ð41Þ

~q ¼ffiffiffiffiffiffiffiffiffiffiffiffi3

2~sij~sij

rð42Þ

M ¼ 6 sin /c

3� sin /c

ð43Þ

M f ¼6 sin /

3� sin /ð44Þ

where ~p0ð¼ p0Þ is the initial mean stress, and (/c,/) are theinternal friction angles at the characteristic state and shearfailure, respectively. The main features of the proposedmodel for both clay and sand are presented as follows.

5.1. For sand

The stress-dilatancy equation of the proposed model isexpressed as follows:

depv

depd

¼ M2~p2 � ~q2

2~p~qð45Þ

Eq. (45) can be drawn as Fig. 5a in the ~q=~p � �depv=dep

d

plane and as Fig. 5b in the q=p � �depv=dep

d plane undertriaxial compression and extension conditions. It is seenfrom these two figures that the unique relationship between~q=~p and �dep

v=depd can explain the difference in the q=p �

�depv=dep

d relations between triaxial compression andextension.

Under triaxial compression and extension stress condi-tions, the yield function in Eq. (37) is plotted in Fig. 6,where CL and FL show the characteristic state line andthe shear failure line, respectively. ~ra and ~rr are the trans-

formed stresses corresponding to ra and rr respectively, inwhich ra and rr are the axial and radial stresses in triaxialstress conditions. It is seen from Fig. 6 that although theyield curves in triaxial compression and extension are sym-metrical with respect to the ~p-axis in the ~p � ð~ra � ~rrÞ plane(see Fig. 6a), the yield curves are not symmetrical withrespect to p-axis, and the value of q in triaxial extensionis smaller than that in triaxial compression in thep � (ra � rr) plane at the same p (see Fig. 6b). This trendis similar to the test results from various kinds of soils(e.g., [10]).

How the proposed model describes the dilatancy of soilis explained as follows. From Eq. (38), we obtain

depv ¼ cp

M4f

M4

M4 � ~g4

M4f � ~g4

dH ð46Þ

Since dH is always larger than or equal to 0, the follow-ing conclusions can be obtained from Eq. (46):

(1) ~g ¼ 0 (isotropic compression condition): depv ¼ cpdH .

(2) 0 6 ~g < M (negative dilatancy condition): depv > 0.

(3) ~g ¼ M (characteristic state condition): depv ¼ 0.

(4) M < ~g 6 M f (positive dilatancy condition): depv < 0.

As indicated above, the dilatancy of clay and sand arereasonably described by the new hardening parameter H.The validity of H and the other quantities used usually ashardening parameters will be checked in the various stresspaths as follows.

Fig. 7 shows the stress paths in triaxial tests on Toyourasand (data from [11]) in terms of the relation between meanstress p and deviator stress q. The values of the principalstress ratios are the same (r1/r3 = 4) at points F and F 0

in Fig. 7. We will check the stress path dependency of fourquantities, the plastic volumetric strain ep

v, the plastic devi-ator strain ep

d, the plastic work Wp and the proposed hard-ening parameter H, in four kinds of triaxial compressiontests (paths: ADEF, ABCF, AGF and ABEF) and threekinds of triaxial extension tests (paths: AD 0F 0, ACF 0 andAF 0).

Figs. 8–11 show the variations of those quantities alongthe seven kinds of stress paths under triaxial compressionand extension conditions. It is obvious from Figs. 8 and9 that the plastic volumetric strain ep

v and the plastic devi-ator strain ep

d are unsuitable for the hardening parameterfor sand because these quantities depend on the stress pathsat the same stress state. From Fig. 10 the plastic work Wp isalmost independent of the stress paths only in triaxial com-pression or triaxial extension, but its value at the stressstate F is different from that at F 0 although the stress invar-iants (~p and ~q) of the stress states F and F 0 are the same.The other problem induced by the plastic work hardeningparameter is that sand will always harden even though itis at a peak or failure state. Hence, the plastic work is alsonot good for the hardening parameter in general stressstates. However, Fig. 11 shows that the values of the

0 200 400 600 800

-500

0

500

1000

p(kPa)

Comp.

Ext.

A B C

E

D

F

D'

F'

q(kP

a)

G

Fig. 7. Stress paths of triaxial tests for examining new hardeningparameter.

0 200 400 600 800

-1

-0.5

0

Comp.

v(%

)

p(kPa)

p

AB

C

F

F

F

ED

G

0 200 400 600 800

-0.4

-0.2

0

0.2

0.4

Ext.

v(%

)ε

ε

p(kPa)

AB

CF'

F'D'

F'

p

Fig. 8. Relation between plastic volumetric strain epv and mean principal

stress p.

qp

-d v /d dp p

Mf

M

Comp. Ext.

M

Mf

q

p

-d ε v /d dp p

~

~

Comp. and Ext.

ε ε ε

Fig. 5. Stress-dilatancy relationships of proposed model under triaxial compression and extension conditions expressed in (a) ~q=~p � �depv=dep

d and (b)q=p � �dep

v=depd.

aσσ

σσ

-r

p0

CL

CL

~~

~

FL

FL

a-r

p0

CL

CL

FL

FL

Fig. 6. Yield curves of proposed model under triaxial compression andextension conditions expressed in (a) ~p � ð~ra � ~rrÞ plane and (b)p � (ra � rr) plane.


proposed hardening parameter H are uniquely determinedat the same stress state, regardless of the stress path in tri-axial compression and extension and the previous stress

history. So, the proposed hardening parameter H is a statequantity, and we employ it as a new hardening parameterfor sand.

5.2. For clay

As mentioned before, if M = Mf, the new hardeningparameter H becomes the plastic volumetric strain, whichis the same as the hardening parameter for normally con-solidated clay in the Cam-clay model. Therefore, in thiscase the difference between the proposed model and theCam-clay model is only the stress tensor used. Let us dis-cuss the critical state conditions in three-dimensional stres-ses in detail. The critical state conditions of the proposedmodel in three-dimensional stresses can be expressed asfollows:

0 200

HH

400 600 8000

1

2

3

p(kPa)

Comp.

A

B C

D

EF

G

0 200 400 600 8000

1

2

3

p(kPa)

Ext.

A

B C

D'F'

Fig. 11. Relation between new hardening parameter H and meanprincipal stress p.

0 200 400 600 8000

1

2

3

4

5

d(%

)

p(kPa)

Comp.

p

A B C

DE

E

F

F

F

0 200 400 600 8000

1

2

3

4

Ext.

d(%

)ε

ε

p(kPa)

p

A B C

D'

F'

F'

G

Fig. 9. Relation between plastic deviator strain epd and mean principal

stress p.

0 200 400 600 8000

5

10

15

20

p(kPa)

Wp (k

Pa)

Comp.

A B C

D

E

F

G

0 200 400 600 8000

5

10

p(kPa)

Wp (k

Pa) Ext.

A B C

D'

F'

Fig. 10. Relation between plastic work Wp and mean principal stress p.


~gcs ¼ ~qcs=~pcs ¼ M

depv=dep

d ¼ 0

epv ¼ cp½lnð~pcs=~p0Þ þ ln 2�

epd !1

9>>>=>>>;

ð47Þ

where the suffix cs means the critical state. When Eq. (47) issatisfied, soil will be continuously distorted. Eq. (47) is the

same as the critical state conditions of the Cam-clay modelin triaxial compression, so we might say Eq. (47) is theextension form for the critical state conditions of theCam-clay model under three-dimensional stresses.

6. Predictions versus experimental results

A series of triaxial compression and triaxial extensiontests on normally consolidated Fujinomori clay andsaturated Toyoura sand have been completed by Nakaiand Matsuoka [12] and Nakai [11]. Fig. 12 shows the initialyield surfaces from Eq. (37) and the test stress paths con-ducted in triaxial compression and extension for clay andsand. The test data are used to examine the capability ofthe proposed model in predicting drained behavior of clayand sand. The values of soil parameters used in the modelare M = Mf = 1.45, k/(1 + e0) = 0.0508, j/(1 + e0) =0.0112 and m = 0.3 for Fujinomori clay, and M = 0.95,Mf = 1.66, k/(1 + e0) = 0.00403, j/(1 + e0) = 0.00251 andm = 0.3 for Toyoura sand, respectively. The above soilparameters are determined from isotropic compressiontests and conventional triaxial compression tests.

6.1. Path p = const

Fig. 13 compares the predicted and observed results forthe drained behavior of Fujinomori clay and Toyoura sandunder triaxial compression and extension conditions when

a-σσ r

p0

η=M

η=M

Clay

p=c

p=c

1=c

1=c

3=c

3=c

~

~

p0

a-r

p0

=M

η=M

Sand

p=c

p=c

1=c

1=c

3=c

3=c

=Mf

η=Mf

~

~

~~

p0

σ σ

σ σ

σσ

σ σ

σ σ

η

ηFig. 12. Stress paths in triaxial compression and extension tests for (a)clay and (b) sand.

(a) For clay

1/ 3

V(%)

i(%)

ClayCompressionp=196 kPa

-52

01

3

5

5-10 10 15

4

5

4

2

1/ 3

V(%)

i(%)

ClayExtensionp=196 kPa

-52

01

3

5

5-10 10 15

4

5

4

σ σ

εε

ε

σ σ

ε

Fig. 13. Comparison between predicted and test results under triaxial comp

(a) For clay

1/ 3

V(%)

i(%)

ClayCompression

3=196 kPa

-54

01

3

5

5-10 10 15

8

5

4

2

1/ 3

V(%)

i(%)

ClayExtension

3=196 kPa

-54

01

3

5

5-10 10 15

8

5

4

2

σ

σ

σ

σ

σ

σ

ε

ε

ε

ε

Fig. 14. Comparison between predicted and test results under triaxial comp


p = 196 kPa. It can be seen from this figure that the predic-tions (solid lines) of the proposed model agree well with theobserved test results (marked s) for clay and sand at con-stant mean stress under triaxial compression and extensionconditions.

6.2. Path r3 = const

Fig. 14 compares the predicted and observed test resultsfor the drained behavior of Fujinomori clay and Toyourasand under triaxial compression and extension conditionswhen r3 = 196 kPa. It can be seen from this figure thatthe predictions (solid lines) of the proposed model agreewell with the observed test results (marked s) for clay

(b) For sand

-3

-2

-1

0

1

SandExtensionp=196 kPa

1/ 3

v(%

)1(%)

1

3

5

-4 -2 0 2 43(%)

-3

-2

-1

0

1

SandCompressionp=196 kPa

1/ 3

v(%

)

1(%)

1

3

5

-4 -2 0 2 43(%)

σ σ

ε ε

εε ε

ε

σ σ

ression and extension conditions when p = constant for clay and sand.

(b) For sand

-3

-2

-1

0

1

SandCompression

3=196 kPa

1/ 3

v(%

)

1(%)

1

3

5

-4 -2 0 2 43(%)

-3

-2

-1

0

1

SandExtension

3=196 kPa

1/ 3

v(%

)

1(%)

1

3

5

-4 -2 0 2 43(%)

σ

σ

σ

σ

σ σ

ε ε

εε

εε

ression and extension conditions when r3 = constant for clay and sand.

(a) For clay (b) For sand

-3

-2

-1

0

1

SandCompression

1=196 kPa

1/ 3

v(%

)

1(%)

1

3

5

-4 -2 0 2 43(%)

-3

-2

-1

0

1

SandExtension

1=196 kPa

1/ 3

v(%

)

1(%)

1

3

5

-4 -2 0 2 43(%)

1/ 3

V(%)

1(%)

ClayCompression

1=196 kPa

-52

01

3

5

5-10 10 15

4

5

3(%)

1/ 3

V(%)

1(%)

ClayExtension

1=196 kPa

-52

01

3

5

5-10 10 15

4

5

3(%)

σ σ

σ σ σ

σ

σ

σ

σ σ

ε

σ

ε

ε

ε

ε

ε ε

ε ε

εε

σ

ε

Fig. 15. Comparison between predicted and test results under triaxial compression and extension conditions when r1 = constant for clay and sand.

0 200 400 600 8000

100

200

300

400

p(kPa)

q(kP

a)

B(240, 280) C

E(240,180) F

A

D

Fig. 16. Stress paths including constant and decrease in stress ratio intriaxial compression for sand. 0 200 400 600 800

0

200

400

600

800

x(kPa)

y(k

Pa)

'C

'B

B

A

C

1R =2R =3R =

σ

σ

Fig. 18. Stress paths under plane strain condition.


and sand with increasing mean stress under triaxial com-pression and extension conditions.

6.3. Path r1 = const

Fig. 15 compares the predicted and observed test resultsfor the drained behavior of Fujinomori clay and Toyoura

-4 -2 0 2 4

2

0

-2

-4

1

2

3

σ1/σ

3

ε3(%) ε

1(%)

compression

-1

ABC

ε v(%)

-4

c

Fig. 17. Comparison between predicted and test results along st

sand under triaxial compression and extension conditionswhen r1 = 196 kPa. Since the most of the stress path in tri-axial compression for clay is within the initial yield surface(Fig. 12a), the predicted strain is a little smaller than the

-2 0 2 4

2

0

-2

-4

1

2

3

σ1/σ

3

ε3(%) ε

1(%)

ompression

-1

DEF

ε v(%)

ress paths ABC and DEF in triaxial compression for sand.

(a) Stress-strain behavior

(b) Intermediate principal stress

0 1 2 3 4 50

1

2

3

Path AB

z/

x

y / x

-20 -10 0 10 20

8

4

0

4

6

2Path AB

x = 98kPa

( )%y( )%x

( )%v

xyσ

σ

σ σ

σσ

ε

ε

ε

σ

Fig. 19. Comparison between predicted and measured (a) stress–strainbehavior and (b) intermediate principal stress under plane strain path ABwithout change in principal stress directions (data after [13]).



0 1 2 3 4 50

1

2

3

y / x

z/

x

Path AC

-20 -10 0 10 20

-4

-2

0

2

4

6

Path AC

( ) 2yx + =147kPa

( )%x ( )%y

( )%v

xyσ

σ σ

ε

ε

ε

σ

σ σ

σσ

Fig. 20. Comparison between predicted and measured (a) stress–strainbehavior and (b) intermediate principal stress under plane strain path ACwithout change in principal stress directions (data after [13]).



0 1 2 3 4 50

1

2

3

x / y

z/

y

Path AB’

-30 -20 -10 0 10 20 30 40

6

6

2

12

0

4

Path AB’

y =196kPa

( )%y ( )%x

( )%v

yxσ

σ

σ σ

σσ

σ

ε

ε

ε

Fig. 21. Comparison between predicted and measured (a) stress–strainbehavior and (b) intermediate principal stress under plane strain path AB 0

with change in principal stress directions (data after [13]).


test results (Fig. 15a). But, it can be seen from Fig. 15a andb that the predicted results (solid lines) agree well with thetest results (marked �) for clay under triaxial extension con-dition and for sand under triaxial compression and exten-sion conditions.

6.4. Stress paths including constant and decrease in stress

ratio in triaxial compression for sand

Fig. 16 shows two special stress paths, which contain aconstant and a decrease in stress ratio under triaxialcompression, for tests on a sand. The values of the sandparameters used in the model are M = 1.08, Mf = 1.68, k/(1 + e0) = 0.00462, j/(1 + e0) = 0.00259 and m = 0.3,respectively. The above soil parameters were determinedby an isotropic compression test and a conventional triax-ial compression test.

Fig. 17 shows the predicted and observed test results forthe drained stress–strain behavior along the stress pathsABC and DEF, respectively. It can be seen from this figurethat the results (solid lines) predicted by the proposedmodel agree well with the test results (marked s) in thosespecial stress paths.

6.5. Stress paths in plane strain for clay

In order to validate the present model under plane straincondition, the model is used to predict the results of theplane strain tests on Fujinomori clay [13].

Fig. 18 shows the stress paths tested in the plane straintests. The tests are conducted along four kinds of stresspaths (AB, AC, AB 0 and AC 0) under plane strain condition

from K0-consolidation state (point A: ry = 196 kPa,rx = rz = 98 kPa). The model parameters used in the pre-dictions are the same as the above for Fujinomori clay.

Figs. 19 and 20 show the comparison between predictedand measured stress–strain behavior and intermediate prin-cipal stress along paths AB and path AC, respectively,without change in principal stress directions.

Figs. 21 and 22 show the comparison between predictedand measured stress–strain behavior and intermediate prin-



0 1 2 3 4 50

1

2

3

x/ y

z/

y

Path AC’

-20 -10 0 10 20

8

4

0

2

4

6

Path AC’

( ) 2yx + =147kPa

( )%v

( )%y ( )%x

yxσ

σ σ

σ

σσ

σ

σ

ε

ε

ε

Fig. 22. Comparison between predicted and measured (a) stress–strainbehavior and (b) intermediate principal stress under plane strain path AC 0

with change in principal stress directions (data after [13]).


cipal stress along paths AB 0 and path AC 0, respectively,with change in principal stress directions.

It is seen from Figs. 19–22 that the present model givesrelatively good predictions of the measured stress–strainbehavior and the measured intermediate principal stressvalues.

Therefore, it can be seen from the above comparisonsthat the proposed model can reasonably describe thestress–strain characteristics of clay and sand under three-dimensional stresses, as well as the dilatancy of sand alongvarious stress paths.

7. Conclusions

(1) A new hardening parameter is derived by consideringunified yield and plastic potential functions for bothclay and sand. It not only describes the dilatancyfor lightly to heavily dilatant sand, but also reducesto the plastic volumetric strain for clay. The validityof the hardening parameter is confirmed by triaxialcompression and extension tests on sand along vari-ous stress paths.

(2) An elastoplastic model is proposed by applying thetransformed stress tensor ~rij (based on the SMP crite-rion) and the new hardening parameter H to theCam-clay model. The proposed model can reason-ably describe the stress–strain behavior of clay andsand under three-dimensional stress states.

(3) The five soil parameters (k, j, M, Mf and m) in theproposed model can be determined by a loadingand unloading isotropic compression test and a con-ventional triaxial compression test.

Acknowledgements

This study was supported by the National Natural Sci-ence Foundation of China, NSFC (No. 10672010 andNo. 50479001). The authors thank Prof. S.W. Sloan andDr. D.C. Sheng at the University of Newcastle, Australia,for their helps in improving the quality of the paper. Theauthors are also appreciative of the contributions madeby D.C. Lu and W. Hou at Beihang University, China,who made part of the predictions in this paper.

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A Unified Constitutive Model for Both Clay and Sand With

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