Numerical analysis of strain localization in Rocks with ... · The full stress-strain response of...

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Numerical analysis of strain localization in Rocks withThermo-Hydro-Mechanical couplings using Cosserat

continuumHadrien Rattez Ioannis Stefanou Jean Sulem Manolis Veveakis Thomas

Poulet

To cite this versionHadrien Rattez Ioannis Stefanou Jean Sulem Manolis Veveakis Thomas Poulet Numerical analysisof strain localization in Rocks with Thermo-Hydro-Mechanical couplings using Cosserat continuumRock Mechanics and Rock Engineering Springer Verlag 2018 101007s00603-018-1529-7 hal-01821909

See discussions stats and author profiles for this publication at httpswwwresearchgatenetpublication325905084

Numerical analysis of strain localization in Rocks with Thermo-Hydro-

Mechanical couplings using Cosserat continuum

Article in Rock Mechanics and Rock Engineering middot June 2018

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Rock Mechanics and Rock Engineering manuscript No(will be inserted by the editor)

Numerical analysis of strain localization in Rocks with

Thermo-Hydro-Mechanical couplings using Cosserat

continuum

Hadrien Rattez middot Ioannis Stefanou middot Jean

Sulem middot Manolis Veveakis middot Thomas Poulet

Received 2018

Abstract A numerical model for Thermo-Hydro-Mechanical (THM) strong cou-plings in an elasto-plastic Cosserat continuum is developed in order to explorethe influence of frictional heating and thermal pore fluid pressurization on thestrain localization phenomenon This model allows specifically to study the com-plete stress-strain response of a rock specimen as well as the size of the strainlocalization zone for a rock taking into account its microstructure The numer-ical implementation in a finite element code is presented matching adequatelyanalytical solutions or results from other simulations found in the literature Twodifferent applications of the numerical model are also presented to highlight itscapabilities The first one is a biaxial test on a saturated weak sandstone forwhich the influence on the stress-strain response of the characteristic size of themicrostructure and of thermal pressurization is investigated The second one isthe rapid shearing of a mature fault zone in the brittle part of the lithosphere Inthis example the evolution of the thickness of the localized zone and the influenceof the permeability change on the stress-strain response are studied

Keywords Thermo-Hydro-Mechanical couplings middot Cosserat continuum middot strainlocalization middot Finite Element Model

H Rattez I Stefanou JSulemLaboratoire NavierCERMES UMR 8205 Ecole des Ponts ParisTech IFSTTAR CNRS68 avenue Blaise Pascal 77455 Marne-la-Valle FranceTel +331 64 15 34 56E-mail hadrienrattezenpcfr

T PouletCSIRO Mineral ResourcesKensington WA 6151 Australia

M Veveakis T PouletSchool of Petroleum Engineering University of New South Wales Sydney Australia

2 Hadrien Rattez et al

List of Symbols

c CohesionC Heat capacity of the porous materialchy Hydraulic diffusivityCe

ijkl Elastic stiffness tensorchy Thermal diffusivityeijk Levi-Civita symbolF Yield surfaceG Shear modulusGc Cosserat shear modulusgi Coefficients in the generalized plastic strain invarianthi Coefficients in the generalized stress invariantK Bulk modulusL First modulus of the elastic flexural bending rigidity tensorM Second modulus of the elastic flexural bending rigidity tensorMc Third modulus of the elastic flexural bending rigidity tensorMe

ijkl Elastic flexural bending rigidity tensorn Eulerian porosityp Pore pressureQ Plastic potentialq Hardening variableR Cosserat internal lengthS External surface of the specimenT Temperatureui Displacementsα Coefficient of thermal expansion of the porous mediumβ Dilatancyβf Compressibility of the fluid phase per unit volumeβs Compressibility of the solid phase per unit volumeβlowast Mixture compressibility per unit volumeχ Permeability of the porous mediumδij Kronecker symbolηf Viscosity of the liquid phaseγij Strain tensorγ[ij] Antisymmetric part of the strain tensorγeij Elastic strain tensorγp Generalized plastic strain invariantγpij Plastic strain tensorκij Curvature tensorκ[ij] Antisymmetric part of the curvature tensorκeij Elastic part of the curvature tensorκpij Plastic part of the curvature tensorκ(ij) Symmetric part of the curvature tensor

λf Coefficient of thermal expansion of the fluid phase per unit volumeλs Coefficient of thermal expansion of the solid phase per unit volumeλlowast Coefficient of thermal expansion of the soil-water mixture per unit volumemicro Friction coefficientmicroij Couple-stress tensor

Title Suppressed Due to Excessive Length 3

Ω Volume of the specimenωci Cosserat rotationsωcij Tensor of rotations of the microstructureΩij Macroscopic rotation tensorψi Test functionsρ Density of the porous materialσprime Effective mean stressσij Symmetric part of the total stress tensorτij Total stress tensorτ[ij] Antisymmetric part of the total stress tensorτ Generalized deviatoric stress invariant for Cosserat continuaτ primeij Effective stress tensorεij Symmetric part of the strain tensorξ Parameter that enables to switch between different hardening

1 Introduction

Thermo-Hydro-Mechanical (THM) couplings play a major role in various applica-tions in rock mechanics like for instance reservoir mechanics (Longuemare et al2002) fault mechanics (Sulem et al 2011) geothermal energy (Tung et al 2017) ornuclear waste disposals (Menaceur et al 2016) The reason is that the conditions of5

temperature and the presence of pore fluids greatly affect the mechanical responseand vice versa These couplings are particularly relevant in the context of strainlocalization which is one of the most important feature encountered in the rocks ofthe lithosphere and is encountered at different scales (Regenauer-Lieb et al 2017)Beyond some level of deformation the strength of the material decreases and this10

softening of the rock resistance leads to a localization of the deformation into nar-row zones This softening behavior of the material is not limited to mechanicalprocesses and can be enhanced by multi-physical couplings

A first approach to study strain localization consists in looking at the possi-ble critical conditions for which the constitutive equations of the material allow15

the formation of deformation bands (Rice 1975) In many cases though it isinteresting to track the evolution of the system after the onset of localizationThe full stress-strain response of the material is essential in many rock mechanicsapplications (Goodman 1989) even in the absence of THM couplings For thatpurpose we need to simulate numerically the solution of a nonlinear boundary20

value problem which is a challenging task due to the complications that arise dueto softening Indeed the classical continuum theory cannot be used because thegoverning system of equations is ill-posed (Vardoulakis 1985) and a regulariza-tion of the problem is needed to alleviate mesh dependency (de Borst 1991) ielocalization in only one element25

This pathology leads to nonphysical results and can be remediated by resortingto appropriate theories (Muhlhaus and Vardoulakis 1987 Pijaudier and Bazant1987 Needleman 1988) Among these theories Cosserat continuum is particularlysuitable for modeling the microstructure of granular rocks (Papanastasiou andVardoulakis 1992 Papanastasiou and Zervos 2016 Stefanou et al 2017)30

In this paper we present a numerical model that can address strain localizationphenomena in rocks modeled as a Cosserat continuum and considering THM cou-


plings Special attention is devoted to the validation of the numerical tool againstanalytical solutions and numerical results from the literature Two specific exam-ples of application are also developed biaxial loading on sandstone specimens and35

seismic slip in fault gouges

2 Constitutive model formulation

In this section a brief summary of the governing equations is given A more detailedpresentation can be found in Stefanou et al (2017) Rattez et al (2018a)

21 Cosserat continuum40

As compared to the classical (Cauchy) continuum in a 3D-Cosserat continuumeach material point has three additional rotational degrees of freedom ωc

k relatedto the tensor of rotation of the microstructure ωc

ij by

ωcij = minuseijk ωc

k (1)

where eijk is the Levi-Civita symbol and the indices are equal to 123Several kinematic fields are introduced the deformation tensor γij - which is45

split into its symmetric γ(ij)] = εij and antisymmetric part γ[ij] - and the curvaturetensor κij - also split into split its symmetric κ(ij) and antisymmetric part κ[ij]

εij =1

2(uij + uji) Ωij =

1

2(uij minus uji)

γ[ij] = Ωij minus ωcij κij = ωc

ij

γij = εij + γ[ij] = uij minus ωcij = uij + eijk ω

ck (2)

The macroscopic strain and rotation tensors (εij and Ωij) are obtained as thesymmetric and antisymmetric part of the displacement gradient as in the classicalCauchy continuum γ[ij] is the difference between the global rotation Ωij and the50

rotation of the microstructure ωcij The curvature κij is defined as the gradient

of the vector of Cosserat rotations and represents the gradient of microscopicdeformations

Correspondingly the stress tensor τij is also divided into its symmetric σijand antisymmetric part τ[ij] The symmetric part corresponds to the macroscopic55

stresses (the ones that are considered in Cauchy continuum) Besides the stresstensor a couple stress tensor microij is introduced which is energy conjugate to thecurvature

Neglecting the volumetric forces acting on the medium and inertia terms themomentum balance equations are written as follows (Stefanou et al 2017)60

τijj = 0 (3)

microijj minus eijk τjk = 0 (4)


22 Constitutive equations for a Cosserat continuum

The general constitutive equations for a linear isotropic elastic Cosserat continuumare defined by 6 coefficients the two classical deformation moduli K and G andfour new coefficients LM Mc and Gc (Mindlin 1964) The effective stress tensorτ primeij and couple stress tensor microij can be expressed as65

τ primeij = Ceijkl γ

ekl and microij =Me

ijkl κekl (5)

The effective stress tensor is linked to the total stress tensor by τ primeij = τij + pδijCe

ijkl is the elastic stiffness tensor and Meijkl the elastic flexural bending rigidity

tensor

Meijkl = (Lminus 2

3M)δijδkl + (M +Mc)δikδjl

+(M minusMc)δilδjk (6)

Ceijkl = (K minus 2

3G)δijδkl + (G+Gc)δikδjl

+(GminusGc)δilδjk (7)

For an isotropic Cosserat continuum we introduce 6 coefficients 3 of them havethe dimensions of stress (KG and Gc) and the other 3 the dimensions of length70

squared times stress (L M and Mc) Thus any ratio of L M or Mc to K G orGc will introduce a material parameter with dimension of length squared (Cowin

1970) For geomaterials we usually consider R =radic

MG

as a characteristic length

related to the microstructure of the material (Unterreiner 1994) Moreover weassume that M =Mc to ensure that no out-of-plane couple stresses are produced75

during the shearing of a layer (invariance in x1 and x3 directions)A flow theory of plasticity for granular media with Cosserat microstructure

can be derived by keeping the same definitions for the yield surface and the plasticpotential as in the classical theory but by generalizing the stress and strain in-variants to Cosserat continua (Muhlhaus and Vardoulakis 1987) We decomposethe deformation and curvature rate tensors into elastic plastic and thermal parts(small strains theory)

γij = γeij + γpij + γthij and κij = κeij + κpij (8)

The thermal strain rate is γthij = α T δij with α the coefficient of thermalexpansion The effect of temperature on the curvature tensor is neglected herein

The generalized deviatoric strain and stress second invariants are generalizedas

τ =

radic

h1 sij sij + h2 sij sji +1

R2(h3mij mij + h4mij mji)

γp =radic

g1 epij e

pij + g2 e

pij e

pji +R2(g3 k

pij k

pij + g4 k

pij k

pji) (9)

where sij mij eij and kij are the deviatoric parts of the effective stress couple-stress strain and curvature respectively hi and gi are coefficients determined80


by micro-mechanical considerations (Vardoulakis and Sulem 1995 Rattez et al2018a) With the above definition of the second deviatoric stress invariants aDrucker Prager plasticity model can be derived as in Muhlhaus and Vardoulakis(1987) or a von Mises plasticity model as in de Borst (1991)

23 Mass balance equation85

We consider a porous medium consisting of two phases ie the solid skeleton andthe pore fluid

The mass balance of the fluid phase leads to the following form of the equationfor the pore pressure p

partp

partt= chy pii +

λlowast

βlowast

partT

parttminus 1

βlowast

partεvpartt

(10)

where chy = χ(ηfβlowast) is the hydraulic diffusivity βlowast = nβf + (1 minus n)βs is the90

mixture compressibility λlowast = nλf+(1minusn)λs is the coefficient of thermal expansionof the soil-water mixture (Vardoulakis 1986) χ is the intrinsic permeability ofthe porous medium and ηf is the viscosity of the pore fluid βf and βs are thecompressibilities per unit volume of the pore fluid and the solid phase respectivelyand λf λs are the thermal expansivities per unit volume n is the eulerian porosity95

of the medium

24 Energy balance equation

The energy balance equation yields the following form of the evolution equationfor the temperature T

ρC(partT

parttminus cthTii) = τ primeij γ

pij + microijκ

pij (11)

where cth = kT

ρC is the thermal diffusivity kT is the thermal conductivity of the100

material ρC is the specific heat per unit volume of the material in its referencestate

3 Finite element implementation and validation tests

In this section the numerical implementation of a Cosserat continuum with THMcouplings is presented and tested through a series of simulations based on various105

benchmarks for the different physical couplings The tests are summarized in thissection


31 Numerical implementation

A displacement-rotation-temperature-pore pressure incremental finite element for-mulation is used to numerically integrate the model presented in the previous sec-110

tion The integration in time is implicit using the backward Euler method whichis implemented in REDBACK (Poulet et al 2016)

The weak form of the linear and angular momentum balance equations Eqs 3-4is written as (Godio et al 2015)

minusint

Ω

τijψijdΩ +

int

partΩΣ

τijnjψidS = 0 (12)

minusint

Ω

microijψijdΩ +

int

partΩΣ

microijnjψidS minusint

Ω

εijkτjkψidΩ = 0 (13)

The energy and fluid mass balance equations are also written in their weak115

forms

int

Ω

pψdΩ + chy(

int

Ω

piψidΩ minusint

partΩ

piniψdS)minusλlowast

βlowast

int

Ω

T ψdΩ +1

βlowast

int

Ω

εvψdΩ = 0

(14)int

Ω

T ψdΩ + cth(

int

Ω

TiψidΩ minusint

partΩ

TiniψdS)minus1

ρC

int

Ω

(τ primeij γpij + microijκ

pij)ψidΩ = 0

(15)

where ψ and ψi are tests functions chy cth ρC βlowast and λlowast are considered constant

hereinLinear Lagrange test functions are chosen for all the fields and full integration

is performed Quadratic interpolation functions could be used as well for the dis-120

placement field linear for the Cosserat rotations and linear or constant for thepressure and temperature fields in order to assure compatible interpolation orderand lead to a more efficient formulation Moreover reduced integration could bealso used to improve performance (see (Godio et al 2015) for a detailed discus-sion) However the mesh convergence analyses performed herein show acceptable125

performance and convergence despite the use of linear shape functions for all thefields

32 Mechanical tests

As a first step the implementation of the equations for a Cosserat continuumand of the constitutive laws are checked For elasticity the numerical results are130

compared to the analytical solution of shearing of an elastic layer and for plasticitywith results found in the literature

321 Elasticity tests

The numerical results are compared with an analytical solution called the bound-ary layer effect in Cosserat continua described in Vardoulakis and Sulem (1995)135


We consider the shearing of an elastic infinite layer of thickness h in plane strainconditions The geometry is described in Fig 2

We use the stress-strain relationships of a 2D-linear isotropic elastic Cosseratmedium in plane strain conditions

All mechanical properties are assumed to be independent from the x1 coordi-140

nates which results in γ11 = 0 and κ31 = 0 We also assume that τ22 = 0 at thetop and u2 = 0 at the bottom which results in τ11 = τ22 = γ22 = 0 everywhere inthe layer

In that case the equilibrium equations can be reduced to only two equations

partτ12partx2

= 0 or τ12 = constant = τc (16)

partmicro32

partx2+ τ21 minus τ12 = 0 (17)

Then we insert the constitutive equations to obtain a coupled system of partialdifferential equations in u1 and ωc

3

(G+Gc)part2u1partx22

+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)

The general solution of this system is

ωc3 = K1 e

δx2 +K2 eminusδx2 minus τc

2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M

The four constants (K1 K2 K3 and τc) are determined from the boundary145

conditions in u1 and ωc3

At the bottom of the layer (x2 = 0) the displacement is equal to 0 and theCosserat rotation is constrained to be equal to the rigid body rotation

u1(x2 = 0) = 0 and ωc3(x2 = 0) = Ω21(x2 = 0) = minus1

2

partu1partx2

(x2 = 0) (22)

At the top of the layer (x2 = h) the displacement is imposed equal to 001mmand the Cosserat rotation is equal to -01 It corresponds to an interface with aparticular roughness (Vardoulakis et al 1992 Vardoulakis and Unterreiner 1995)

u1(x2 = h) = 001mm and ωc3(x2 = 0) = minus01 (23)

The set of parameters chosen as an example are given in Table 1We choose a characteristic lengthR (or bending length) one order of magnitude

less than the thickness of the band and we observe that the Cosserat effects areconfined in a layer of size about three times R adjacent to the upper boundary150

where the rotation and displacement are imposedFor the geometry of the numerical simulation we consider a cube of dimensions

111 mm3 with 50 elements along x2-direction 10 along x1-direction and 1 along


Parameters values unitsG 10 GPaGc 20 GPah 10minus3 mR 10minus4 mM G R2 kN

Table 1 Values of the parameters used for the boundary layer elastic test

x3-direction We prescribe periodic boundaries in the x1- and x3-directions Wehave 6 variables in our model (u1 u2 u3 ω

c1 ω

c2 ω

c3) thus their values or the155

values of their derivatives have to be imposed at the top and bottom edgesBottom edge (x2=0)

u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)

In Fig 3(a) the results for the displacements are plotted on the deformedmesh In Figs 3(b) we compare the values obtained for ωc

3 with the analyticalsolution presented above The difference between the numerical and the analyticalsolutions is negligible Moreover a mesh convergence analysis is conducted for160

the value of the Cosserat rotation at x2 = 09 (see Fig 4) which shows that therelative error is less than 12 for 40 elements and decreases to 001 for 400elements (see also Godio et al (2015) for a mesh convergence in terms of modalfrequencies with Cosserat continuum)

322 Plasticity tests165

The Drucker-Prager plastic model for three dimensional Cosserat continua is devel-oped in Rattez et al (2018a) assuming a hardening law arising from the evolutionof the friction coefficient In order to compare with results in the literature thatconsider a von Mises yield criterion we assume the following expression for theyield criterion170

F = τ + microσ(1 + hsqξ)minus c(1 + hsq(1minus ξ)) (26)

Q = τ + βσ(1 + hsqξ)minus c(1 + hsq(1minus ξ)) (27)

where σ = τkk

3 and τ is the generalized second invariant of the deviatoricstresses ξ is a parameter that allows us either to have a hardening on the frictioncoefficient (ξ = 1) as for the Drucker-Prager model derived in Rattez et al (2018a)or on the cohesion (ξ = 0) as for the von Mises yield criterion hs is a parametercontrolling the magnitude of the hardeningsoftening rate and q is the hardening175

variable (chosen as q = γp )As an example we consider an infinite layer of height h = 60mm subjected to

pure shear at constant velocity V (see Fig 5) The same material parameters withGodio et al (2016) are chosen in order to have a comparison


The yield criterion is based on a von Mises model adapted for Cosserat con-180

tinua as in de Borst et al (2012) In equation 26 we consider ξ = 0 micro = 0 inorder to have the same expression

F = τ minus c(γp) (28)

where c is the equivalent plastic stress (or cohesion) that depends linearly on γp theequivalent plastic strain The coefficients in the stress invariant are h1 = h2 = 3

4 h3 = 3

2 and h4 = 0 In the plastic strain invariant they are g1 = g2 = 13 g3 = 2

3185

and g4 = 0 These values for the coefficients differ from Muhlhaus and Vardoulakis(1987) Muhlhaus (1986) Vardoulakis and Sulem (1995) because of the use of avon Mises criterion for which yielding occurs when τ reaches the yield strengthof the material in simple tension leading to the conditions h1 + h2 = 3

2 andg1 + g2 = 2

3 In the other papers yielding is chosen to occur when τ reaches the190

yield strength of the material in pure shear leading to the conditions h1 +h2 = 12

and g1 + g2 = 2In this purely mechanical example a linear softening rule is chosen to exhibit

a localization of the deformations The initial cohesion is c = 100 MPa and thehardening modulus is hs = minus4195

The elastic properties of the material are K = 4000 MPa G = 4000 MPa andGc = 2000 MPa The internal length of the microstructured continuum is chosento be R = 25 mm

We investigate first the mesh-convergence of the model which is the one of themost interesting features of the Cosserat continua A three-dimensional geometry200

is considered with periodic boundary conditions for the right left front and backside of the cube A regular mesh is chosen with one element in directions x1 andx3 and a range of 32 to 300 in the direction x2 In Fig 6(b) the shear stress τ12 atthe top of the layer is plotted versus the normalized horizontal displacement at thetop The plastic regime is reached for τ12 = cradic

3as expected and shows a softening205

behavior exactly like Godio et al (2016) and de Borst (1991) The results for 200and 300 elements in the vertical direction exhibit no clear difference showing amesh-convergence

In Fig 6(a) the Cosserat rotation around direction x3 is plotted on the de-formed mesh for τ12 = 26MPa after softening The magnitude of the rotations is210

higher inside the zone of localized deformations as observed experimentally (Hallet al 2010)

In Fig 7 the shear strain γ12 is plotted along the height of the layer for differentdiscretizations and for τ12 = 26MPa after softening The deformation profile ispractically identical for 80 and 200 elements showing also a mesh convergence215

In Fig 8(a) we present the stress-strain evolution of the shear layer discretizedwith only one element This allows to test the implementation of the harden-ingsoftening constitutive behavior Both the static Muhlhaus-Vardoulakis and deBorst coefficients for the shear stress and shear strain generalized invariants areused and a hardeningsoftening coefficient of 4-4 MPa is used We observe that220

the yield criterion is reached for τ12 = cradic3considering de Borst coefficients and

τ12 = c considering Muhlhaus-Vardoulakis choice of coefficientsIn Fig 8(b) we plot the generalized stress invariant τ as a function of the

generalized plastic strain invariant γp The slopes in this diagram enable us toverify that the value of the hardening modulus is the one expected225


33 Thermo-Mechanical tests

These tests concern the verification of the coupling between the energy balanceequation and the mechanical equations (Eqs 12 13 and 15)

First we look at the heating of a rod of length 10mm with a section of 1times1mm2

(Fig 9)230

For the first test the evolution of temperature at the middle of the rod iscompared to the solution obtained analytically using series (Fig 10(a)) (Kreyszig2006) A good agreement is found In Fig 10(b) we plot the evolution of thedisplacement at the free end of the rod for cth = 1 mm2s It tends to a value of25times 10minus2mm as expected (thermal expansion coefficient 25times 10minus2C)235

Additional tests are performed for the same geometry of the rod but withdifferent boundary conditions In these tests the normal displacements of all facesare fixed except on one side as shown in Fig 11 The rod is no longer free toexpand in all directions and the precluded deformations cause the development ofstresses that can be calculated analytically in elasticity240

The deformation along the x3-axis of the rod is given by

γ33 =3K

K + 43G

α∆T (29)

and the stresses perpendicular to the axis of the rod are

τ11 = τ22 = (K minus 2

3G)γ33 minus 3Kα∆T (30)

Therefore when the temperature reaches a steady state ∆T = 1C we obtainγ33 = 375times 10minus2 and τ11 = τ22 = minus1125times 10minus2MPa (with K = 3MPa andG = 15MPa) These values are in agreement with the results of the numerical245

simulation shown in Fig 12The two last tests are dedicated to Thermo-Mechanical couplings considering

an elasto-plastic behavior in order to explore the influence of a temperature riseon the plastic behavior of the material The configuration for the first test is theone shown in Fig 11 but this time a yield stress (cohesion) is set to 001MPa250

Perfect plasticity is considered in this example but with no plastic dissipation inthe energy balance equation In Fig 13 the results for the stress and displacementare compared We observe that τ11 remains constant after reaching the value of001MPa The development of plastic deformations also has a consequence on thedisplacement in the rod axis direction (x3) which is higher than that for the elastic255

rod Nevertheless no difference is observed for the evolution of the temperature asthe mechanical dissipation term is neglected in this example Thus no feedbackbetween the mechanical behavior and the energy balance is possible

In order to have an insight into the feedback of the mechanics on the energybalance equation (due to the plastic dissipation term) a shear test is modeled The260

same parameters as the plastic tests with a height of 60mm and only one elementis considered for the mesh in the vertical direction For simplicity we neglect heatdiffusion and only consider the effect of plastic dissipation on the temperaturechange

ρCpartT

partt= σij ε

pij + τ[ij]γ

p

[ij] + microijκpij (31)


From this test we verify the value of the mechanical dissipation τ12∆γ

p

12

∆tat each265

time step Moreover from the energy balance equation (Eq 31) the temperature

increase between two time steps in the plastic regime is equal toτ12∆γ

p

12∆t

ρC (with

ρC taken equal to 1 MPaC here) The comparison of this expression and of theactual temperature increase is plotted in Fig 14 showing no difference

34 Hydro-Mechanical tests270

Three tests have been designed in order to check the Hydro-Mechanical (HM) cou-plings (Eqs 12 13 and 14) An undrained oedometer configuration is consideredThe geometry and boundary conditions are shown in Fig 15

No mass diffusion is considered in these tests and therefore the mass conser-vation equation becomes

partp

partt= minus 1

βlowast

partεvpartt

(32)

A cubic single element of dimensions 1times 1times 1 mm3 representing a fully satu-rated sample with its lateral displacements prevented is considered Moreover all275

boundaries are impervious A constant downward velocity v3 is applied on the topboundary

The deformation of the cube in the direction of the applied velocity is

γ33 = minusv3tL

(33)

The effective normal stresses are linked to the elastic deformation by

τ prime11 = τ prime22 = (K minus 2G

3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)

No shear stresses are developed and the rise in pore pressure is

p = minusγ33βlowast

(36)

We consider first two tests with an elastic behavior for the material The compari-280

son of the pore pressure rise compared to the analytical solution is plotted in Fig 16and the two evolution match perfectly (v3 = 10minus2mmsminus1 and βlowast=2MPaminus1)

The solution in terms of total stresses is validated against analytical results isshown in Fig 17(a)

A last test is designed for HM couplings considering an elasto-plastic behavior285

and allows us to test the numerical implementation of the plastic constitutivebehavior and its coupling with the pore pressure In this test a cohesion of 01MPais considered with zero friction

In oedometric conditions the generalized shear stress invariant is equal to

τ =

radic

2

3(h1 + h2)(τ prime11 minus τ prime33)

2 (37)


Combining Eqs 37 34 and 34 the yield criterion is reached at a time290

t =

radic3cL

2Gv3radic

2(h1 + h2)(38)

which is equal to 333s for the parameters considered here (G = 15MPa v3 =10minus2mmsminus1 and de Borst coefficients in the invariant) This value is compatiblewith the numerical results presented in Fig 17(b)

35 Thermo-Hydro-Mechanical tests

In order to test the implementation of THM couplings we consider a shear test295

taking into account the full set of equations of our model (Eqs 14 15 12 and 13)Similar to the shearing test for Thermo-Mechanical couplings a height of 60mmwith only one element in the vertical direction is considered Assuming no heatand fluid flux (undrained adiabatic case) we obtain

partp

partt=λlowast

βlowast

partT

partt(39)

In this test we focus on the thermal pressurization term in the mass balance300

equation The coefficient λlowast

βlowastis taken equal to 05 MPaC The results for the

pressure and temperature are plotted in Fig 18 We can see that there is a ratioof 05 MPaC between the two graphs which corresponds to the value of thethermal pressurization coefficient We observe here that the plastic dissipation in-duces an increase of temperature which leads to a pore pressure increase (thermal305

pressurization)

4 Examples of numerical simulations

In order to illustrate the capabilities of the proposed model in geomechanicalapplications two different problems are considered in this section The first oneis a standard biaxial test for a weak sandstone The second is an infinite sheared310

layer modeling a mature fault zone in the brittle part of the lithosphere

41 Biaxial test for sandstones

Sandstones are sedimentary rocks formed by cemented sand grains These grainspresent diameters within the range of 00625 mm to 2 mm according to Udden-Wentworth particle size classification (Wentworth 1922) Due to this microstruc-315

ture one of the main deformation mechanisms observed in sandstones is a granularflow characterized by grain sliding and rolling (Fossen et al 2007) Therefore aCosserat continuum appears to be an appropriate framework to describe this classof materials Moreover because of their generally high porosity and permeabil-ity many of the oil gas and groundwater reservoirs in the world are found in320

sandstones (Bjorlykke 2010) The study of deformation bands is thus of primary


importance for hydrogeologists due to hydraulic properties modifications whichusually accompany their development (eg permeability reduction) and can act asbarriers or channels to fluid flow

The conditions of temperature and pore pressure at the depth of the reser-325

voirs influence their mechanical behavior A Thermo-Hydro-Mechanical model isnecessary to capture correctly the behavior of the medium at depth

In this study we consider a Drucker-Prager yield surface that can describe thebehavior of geomaterials that exhibit dilatancy when sheared like sandstones atlow confining pressures It should be noted that this yield surface is not suitable to330

describe compactional shear bands formed at higher confining pressures (Stefanouand Sulem 2014a) In this case a yield surface presenting a cap at high meanstresses is necessary (Issen and Rudnicki 2000)

We consider a biaxial geometry and the mechanical parameters are derivedfrom triaxial compression test carried in Castlegate sandstone and described in335

Zervos et al (2001) The geometry and the boundary conditions are shown inFig 19 The specimen has a height of 10cm and a width of 5cm A displacement isprogressively imposed at the top boundary with a constant velocity The normaldisplacement at the bottom boundary are precluded as well as the horizontaldisplacement of the point at the bottom right corner340

A Drucker-Prager yield surface is chosen as it adequately captures the behaviorof the sandstone at shearing

F = τ + microσprime minus c (40)

The coefficients in the generalized stress invariant τ are h1 = h3 = 23 and

h2 = h4 = minus16 The coefficients the generalized plastic strain invariant γp are

g1 = g3 = 85 and g2 = g4 = 2

5 The cohesion c evolves with the equivalent plastic345

strain following a hyperbolic law (Zervos et al 2001)

c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)

whereK2 andK3 are coefficients calibrated against experimental resultsK2 =1323times10minus5 and K3 = 61271times10minus2 c0 is a threshold value for the cohesion thatgoverns the initiation of the plastic behavior taken here equal to 625 MPa K1 isan open parameter that defines the hardening (or softening) rate of the material350

Its influence on the stress-strain response is shown in Fig 21Based on micromechanical considerations that enable to link the internal lengths

of Cosserat continua to the mean grain size of a granular medium (eg (Rattezet al 2018a)) it is possible to investigate qualitatively the influence of the charac-teristic size of the microstructure on the overall stress-strain response of the system355

and the shear band thickness This is accomplished by varying the representativemean grain size as shown in Fig 22

The softening behavior of the stress-strain diagram is more pronounced forsmaller grains as the shear band tends to be thinner The elastic part and thehardening part of the diagrams are not affected by the grain size The difference360

in the shear band size is illustrated in Fig 23 where the generalized plastic straininvariant is plotted on the deformed mesh for the same value of the shear stress


(τ12=40MPa after softening) In Rattez et al (2018b) the shear band size is evalu-ated using the plastic strain rate γp as a proxy because it allows an instantaneousevaluation of the localization process (γp depends on the deformation history)365

Therefore a thinner plastic zone as obtained in the case of the smaller grain sizecorresponds to the specimen in which a thinner zone of localized deformation hasformed

In order to trigger the localization an imperfection has to be introduced Inthis paper a notch is introduced at the bottom left of the specimen This enables370

to have the same deformation pattern in all the tests and to obtain stress-straindiagrams that can be compared (Besuelle et al 2006)

The effect of the THM couplings on the response of the material is shown inFig 24 The test taking into account only the mechanical equations corresponds toan isothermal and drained material In the THM model we consider the feedback375

of the evolution of the temperature and pore pressure on the mechanical behaviorWe observe softening due to the thermal pressurization phenomenon but it isnot very pronounced as the plastic strain remains small and the magnitude oftemperature and pore pressure fields are not high enough to strongly affect themechanical response However in the case of the shearing of an infinite layer380

modeling a fault zone the effect of thermal pressurization is much more significantas described in the next section

42 Shearing of a fluid saturated mature fault zone

Observations of faults from drilling or outcrops show the formation of a slip zoneof finite but very small thickness composed of cataclastic material due to exces-385

sive shearing called gouge material (Sibson 2003) These zones are of primaryimportance as they accommodate most of the slip during a seismic event Fur-ther investigations indicate that an even thinner zone of ultracataclastic materialcalled the Principal Slip Zone (PSZ) exists inside the gouge (Rice 2006a) Thethickness of these shear bands is a key parameter for understanding fault behavior390

(Kanamori and Brodsky 2004) as it is related to the triggering and evolution ofvarious multiphysical couplings and to the energy dissipation during seismic slip

A substantial grain size reduction (grain cataclasis) is usually used as a markerto identify the principal slip zone (Sammis and Ben-Zion 2008) Moreover manyphysical processes are directly affected by grain size evolution First of all the395

production of fine particles can significantly weaken the material Previous studieshave shown that the production of nanograins during slip reduces the frictioncoefficient due the lubrication of the slip surface (Han et al 2010 Di Toro et al2011) The grain size affects the frictional properties (Anthony and Marone 2005)and the strength of the gouge (Dieterich 1981) Furthermore grain size also has400

an effect on the dissolution or dehydration reactions occuring in faults as chemicalreactions kinetics depend on the specific surface (Hu and Hueckel 2007 Stefanouand Sulem 2014b) Therefore it is necessary to include information about the sizeof the microstructure and its evolution in the constitutive model of a gouge

In order to describe the influence of weakening processes and of the size of the405

microstructure in a gouge during a seismic event a THM model of a fault gougeis developed in this section The fault zone is modeled here as a saturated infinite


parameters values units parameters values unitsK 20times 103 MPa micro 05G 10times 103 MPa β 0Gc 5times 103 MPa λlowast 74times 10minus5 oCR 10 microm ρC 28 MPaoCρ 2500 kgm3 cth 1 mm2sβlowast 82times 10minus5 MPaminus1 chy 12 mm2sn 004 α 25times 10minus5 oC

Table 2 Numerical values for the parameters of a deep rock gouge from (Sulem et al 2011Rice 2006b)

layer sheared at its top with a constant velocity of 1ms a typical rate reachedduring a seismic slip

The values of the different parameters are retrieved from Rattez et al (2017)410

and Rice (2006a) for a gouge material lying at 7km a centroidal depth for crustalfaults These values are summarized in Table 2 Considering the great depth ofthe fault zero dilatancy is assumed here (cf Sleep (1999))

In order to emphasize the role of the softening due to thermal pressurizationperfect plasticity is considered (Fig 25) For the shearing of a gouge considering415

only the mechanical equations no softening is observed as expected Whereas withTHM couplings a significant decrease of the shear stress appears in the plasticregime However it requires a sufficient accumulated slip before a notable THMsoftening takes place With this set of parameters for a decrease of 10 of theshear stress a shear deformation of 30 of the layer is necessary420

The hydraulic parameters of fault gouges are poorly constrained (Segall andRice 2006) and are central in the behavior of a fault For instance the thermalpressurization weakening is magnified by the capacity of the porous space to trapa pressurized heated pore fluid If the fluid can diffuse in the surrounding mediumthe pressurization is precluded and cannot operate efficiently causing the insta-425

bility propagation to terminate (Wibberley et al 2008) Thus we investigate theinfluence of the permeability on the response of the gouge material in Fig 26

The permeability of fault zones depends on many parameters such as the rocktype the stress state the fault architecture among others (Wibberley and Shi-mamoto 2003) Moreover its value is not constant along the fault and evolve430

during shearing (Sulem and Famin 2009) A parametric study on this parameteris consequently conducted for typical values ranging from 10minus17m2 to 10minus19m2 inFig 26 For higher permeabilities the softening is less severe as the shear bandis thicker Indeed the diffusion of the pore fluid tends to enlarge the band Thisoutcome is illustrated in Fig 27 The deformed mesh for two different perme-435

abilities are plotted for the same level of deformation and exhibit different shearband thicknesses For lower permeabilities the pressurization inside the shear bandis stronger enhancing softening (see also Scuderi and Collettini (2016) for recentexperimental evidence)

5 Concluding Remarks440

A numerical model for Thermo-Hydro-Mechanical (THM) couplings in Cosseratcontinua is presented implemented and validated in this paper The fully coupled


nonlinear set of equations is integrated using an incremental finite element solverNotice that staggered (sequential) schemes may lead to peculiar numerical artifactsand instabilities445

The implementation is thoroughly tested in order to certify the results obtainedafterwards The proposed tests can be used also as guidelines for implementationsof other THM codes A particular attention is given to the coupling terms relatedto temperature and pore pressure evolution and their impact on the mechanicalbehavior450

Thereafter the numerical tool is used to study biaxial tests for sandstonesusing a hyperbolic law to describe their softening The influence of the size of themicrostructure on the shear band thickness and as a result on the stress-straindiagram is shown However as the range of deformation of these tests remainslimited the thermal pressurization weakening effect is rather small455

Finally the problem of a sheared saturated mature fault zone is studied Aperfect plastic law is introduced to exhibit clearly the role of the THM couplingson the softening behavior of the material which are significant A parametric studyon the role of the permeability is carried out and highlights the major role playedby the diffusion processes on the strain localization phenomenon460

Acknowledgements The second author IS would like to acknowledge support of the Eu-ropean Research Council (ERC) under the European Unions Horizon 2020 research and inno-vation program (grant agreement No 757848 CoQuake)

References

Anthony JL Marone CJ (2005) Influence of particle characteristics on granular465

friction Journal of Geophysical Research B Solid Earth 110(8)1ndash14 DOI1010292004JB003399

Besuelle P Chambon R Collin F (2006) Switching deformations modes in post-localization solutions with a quasi-brittle material journal of materials andstructures 1(7)1115ndash1134470

Bjorlykke K (2010) PetroleumGeoscience from Sedimentary Enviroments to RockPhysics DOI 101007978-3-642-02332-3 arXiv10111669v3

de Borst R (1991) Simulation of Strain Localization a Reappraisal of the CosseratContinuum Engineering Computations 8(4)317ndash332 DOI 101108eb023842

de Borst R Crisfield MA Remmers JJC Verhoosel CV (2012) Nonlinear Finite475

Element Analysis of Solids and Structures John Wiley amp SonsCowin SC (1970) Stress Functions for Cosserat Elasticity International Journal

of Solids and Structures 6(4)389ndash398Di Toro G Han R Hirose T De Paola N Nielsen S Mizoguchi

K Ferri F Cocco M Shimamoto T (2011) Fault lubrication during480

earthquakes Nature 471(7339)494ndash498 DOI 101038nature09838 URLhttpwwwnaturecomdoifinder101038nature09838

Dieterich JH (1981) Constitutive properties of faults with sim-ulated gouge Geophysical Monograph Series 24103ndash120 URLhttpwwwaguorgbooksgmv024GM024p0103GM024p0103shtml485

Fossen H Schultz RA Shipton ZK Mair K (2007) Deforma-tion bands in sandstone a review Journal of the Geologi-


cal Society 164(4)755ndash769 DOI 1011440016-76492006-036 URLhttpjgslyellcollectionorgcgidoi1011440016-76492006-036

Godio M Stefanou I Sab K Sulem J (2015) Dynamic finite element for-490

mulation for Cosserat elastic plates International Journal for NumericalMethods in Engineering 101(13)992ndash1018 DOI 101002nme4833〉 URLhttpshalarchives-ouvertesfrhal-01114343

Godio M Stefanou I Sab K Sulem J (2016) Multisurface plasticity for Cosseratmaterials plate element implementation and validation International Journal495

for Numerical Methods in Engineering DOI 101002nme5219Goodman RE (1989) Introduction to rock mechanics John Wi-

ley amp Sons Inc DOI 1010160148-9062(81)90521-0 URLhttplinkinghubelseviercomretrievepii01489062819052109809069v1500

Hall SA Bornert M Desrues J Lenoir N Pannier Y Viggiani G Besuelle P(2010) Discrete and continuum analysis of localised deformation in sand usingX-ray microCT and volumetric digital image correlation Geotechnique 60(5)315ndash322 DOI 101680geot2010605315

Han R Hirose T Shimamoto T (2010) Strong velocity weakening and powder505

lubrication of simulated carbonate faults at seismic slip rates Journal of Geo-physical Research Solid Earth 115(3) DOI 1010292008JB006136

Hu LB Hueckel T (2007) Coupled chemo-mechanics of intergranular contactToward a three-scale model Computers and Geotechnics 34(4)306ndash327 DOI101016jcompgeo200702009510

Issen KA Rudnicki JW (2000) Conditions for compaction bands in porous rockJournal of Geophysical Research Solid Earth 105(B9)21529 DOI 1010292000JB900185 URL httpdoiwileycom1010292000JB900185

Kanamori H Brodsky EE (2004) The physics of earthquakes Reports on Progressin Physics 67(8)1429ndash1496 DOI 1010880034-4885678R03515

Kreyszig E (2006) Advanced Engineering Mathematics 9th ednWiley New York DOI 101017CBO9781107415324004 URLhttpsbooksgooglecombooksid=diBo7dXRLwoCamppgis=1arXiv10111669v3

Longuemare P Mainguy M Lemonnier P Onaisi A Gerard C Koutsabeloulis520

N (2002) Geomechanics in reservoir simulation Overview of coupling methodsand field case study Oil and Gas Science and Technology 57(5)471ndash483 DOI102516ogst2002031

Menaceur H Delage P Tang AM Conil N (2016) On the Thermo-Hydro-Mechanical Behaviour of a Sheared Callovo-Oxfordian Claystone525

Sample with Respect to the EDZ Behaviour Rock Mechanics andRock Engineering 49(5)1875ndash1888 DOI 101007s00603-015-0897-5 URLhttpdxdoiorg101007s00603-015-0897-5

Mindlin R (1964) Micro-structure in linear elasticity Archive for Rational Me-chanics and Analysis 16(1) DOI 101007BF00248490530

Muhlhaus HB (1986) Scherfugenanalyse bei granularem Material im Rahmen derCosserat-Theorie Ingenieur-Archiv 56389ndash399 DOI 101007BF02570619

Muhlhaus HB Vardoulakis I (1987) The thickness of shear bands in granularmaterials Geotechnique 37(3)271ndash283 DOI 101680geot1987373271 URLhttpwwwicevirtuallibrarycomcontentarticle101680geot1987373271535


Needleman A (1988) Material rate dependence and mesh sensitivity in localizationproblems Computer Methods in Applied Mechanics and Engineering 67(1)69ndash85 DOI 1010160045-7825(88)90069-2

Papanastasiou PC Vardoulakis I (1992) Numerical treatment of progressive lo-calization in relation to borehole stability International Journal for Numeri-540

cal and Analytical Methods in Geomechanics 16(6)389ndash424 DOI 101002nag1610160602 URL httpdoiwileycom101002nag1610160602

Papanastasiou PC Zervos A (2016) Numerical modelling of strain localizationIn ALERT Doctoral School 2016 Modelling of instabilities and bifurcation inGeomechanics pp 133ndash164545

Pijaudier CG Bazant ZP (1987) Non local Damage Theory Jour-nal of engineering mechanics 113(10)1512ndash1533 DOI httpdxdoiorg101061(ASCE)0733-9399(1987)11310(1512) URLhttpdxdoiorg101061(ASCE)0733-9399(1987)11310(1512)

Poulet T Paesold M Veveakis M (2016) Multi-Physics Modelling of Fault550

Mechanics Using REDBACK A Parallel Open-Source Simulator for TightlyCoupled Problems Rock Mechanics and Rock Engineering pp 1ndash17 DOI101007s00603-016-0927-y

Rattez H Stefanou I Veveakis M Sulem J Poulet T (2017) Localisation of defor-mation for shearing of a rock layer with Cosserat microstructure and different555

couplings In 11th International Workshop on Bifurcation and Degradation inGeomaterials

Rattez H Stefanou I Sulem J (2018a) The importance of Thermo-Hydro-Mechanical couplings and microstructure to strain localization in 3D con-tinua with application to seismic faults Part I Theory and linear stability560

analysis Journal of the Mechanics and Physics of Solids 11554ndash76 DOI101016jjmps201803004

Rattez H Stefanou I Sulem J Veveakis M Poulet T (2018b) The importanceof Thermo-Hydro-Mechanical couplings and strain localisation in 3D continuawith microstructure Part II Numerical implementation and post-bifurcation565

analysis Journal of the Mechanics and Physics of Solids 1151ndash29 DOI 101016jjmps201803003

Regenauer-Lieb K Poulet T Veveakis M Alevizos S (2017) The dynamics of mul-tiscale multiphysics faults Part I - The long-term behaviour of the lithosphereTectonophysics (October) DOI 101016jtecto201706011570

Rice JR (1975) On the Stability of Dilatant Hardeningfor Saturated Rock Masses Journal of Geophysical Re-search 80(11)1531ndash1536 DOI 101029JB080i011p01531 URLhttpwwwaguorgpubscrossref1975JB080i011p01531shtml

Rice JR (2006a) Heating and weakening of faults during earthquake slip575

Journal of Geophysical Research 111(B5) DOI 1010292005JB004006 URLhttpwwwaguorgpubscrossref20062005JB004006shtml

Rice JR (2006b) Heating and weakening of faults during earthquake slip Journalof Geophysical Research Solid Earth 111(5) DOI 1010292005JB004006

Sammis CG Ben-Zion Y (2008) Mechanics of grain-size reduction in fault580

zones Journal of Geophysical Research Solid Earth 113(2)1ndash12 DOI 1010292006JB004892

Scuderi MM Collettini C (2016) The role of fluid pressure in inducedvs triggered seismicity Insights from rock deformation experiments on


carbonates Scientific Reports 6(April)1ndash9 DOI 101038srep24852 URL585

httpdxdoiorg101038srep24852

Segall P Rice JR (2006) Does shear heating of pore fluid contribute to earthquakenucleation Journal of Geophysical Research Solid Earth 111(9)1ndash17 DOI1010292005JB004129

Sibson RH (2003) Thickness of the seismic slip zone Bulletin of the Seismological590

Society of America 93(3)1169ndash1178 DOI 1017850120020061Sleep NH (1999) Rate- and state-dependent friction of intact rock and

gouge Journal Of Geophysical Research 104(B8)17847ndash17855 DOI 1010291999JB900185

Stefanou I Sulem J (2014a) Chemically induced compaction bands Trig-595

gering conditions and band thickness Journal of Geophysical Re-search Solid Earth 119(2)880ndash899 DOI 1010022013JB010342 URLhttpdoiwileycom1010022013JB010342

Stefanou I Sulem J (2014b) Chemically induced compaction bandsTriggering conditions and band thickness Journal of Geophysical Re-600

search Solid Earth 119(2)880ndash899 DOI 1010022013JB010342 URLhttpdoiwileycom1010022013JB010342

Stefanou I Sulem J Rattez H (2017) Cosserat approach to localization in geo-materials In Handbook of Nonlocal Continuum Mechanics for Materials andStructures Springer605

Sulem J Famin V (2009) Thermal decomposition of carbonates in fault zones Slip-weakening and temperature-limiting effects Journal of Geophysical Research114(B3)1ndash14 DOI 1010292008JB006004

Sulem J Stefanou I Veveakis M (2011) Stability analysis of undrained adia-batic shearing of a rock layer with Cosserat microstructure Granular Matter610

13(3)261ndash268 DOI 101007s10035-010-0244-1Tung R Poulet T Alevizos S Veveakis M Regenauer-Lieb K (2017) Shear heating

in creeping faults changes the onset of convection Geophysical Journal Interna-tional 211(1)270ndash283

Unterreiner P (1994) Contribution a lrsquoetude et a la modelisation numerique des615

sols cloues application au calcul en deformation des ouvrages de soutenementPhD thesis Ecole Nationale des Ponts et Chaussees

Vardoulakis I (1985) Stability and bifurcation of undrained plane rectilinear de-formations on water-saturated granular soils International Journal for Numer-ical and Analytical Methods in Geomechanics 9(5)399ndash414 DOI 101002nag620

1610090502 URL httpdoiwileycom101002nag1610090502

Vardoulakis I (1986) Dynamic stability analysis of undrained simple shear onwater-saturated granular soils International Journal for Numerical and Ana-lytical Methods in Geomechanics 10(2)177ndash190 DOI 101002nag1610100206URL httpdoiwileycom101002nag1610100206625

Vardoulakis I Sulem J (1995) Bifurcation Analysis in Geomechanics Blackie Glas-cow

Vardoulakis I Unterreiner P (1995) Interfacial localisation in simple shear testson a granular medium modelled as a Cosserat continuum Studies in AppliedMechanics 42(C)487ndash512 DOI 101016S0922-5382(06)80023-1630

Vardoulakis I Shah KR Papanastasiou P (1992) Modelling of Tool-Rock ShearInterfaces Using Gradient-dependent Flow Theory of Plasticity InternationalJournal of Rock Mechanics and Mining Sciences amp Geomechanics Abstracts


29(6)573ndash582Wentworth CK (1922) A Scale of Grade and Class Terms for Clastic Sediments635

The Journal of Geology 30(5)377ndash392Wibberley CA Shimamoto T (2003) Internal structure and permeability of major

strike-slip fault zones the Median Tectonic Line in Mie Prefecture SouthwestJapan Journal of Structural Geology 25(1)59ndash78 DOI 101016S0191-8141(02)00014-7640

Wibberley CAJ Yielding G Toro GD Wibberley CAJ Yielding G Toro GDIEinstein A House NB Lane NB Pe L (2008) Recent advances in the under-standing of fault zone internal structure a review Geological Society LondonSpecial Publication pp 5ndash33 DOI 101144SP2992

Zervos A Papanastasiou P Vardoulakis I (2001) A finite element displacement645

formulation for gradient elastoplasticity International Journal for NumericalMethods in Engineering 50(6)1369ndash1388 DOI 1010021097-0207(20010228)506〈1369AID-NME72〉30CO2-K

x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21

Fig 1 Representation of stress and couple-stress components in a Cosserat continuum

x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h

Fig 2 Boundary conditions for the simple shear of a layer consisting of linear elastic Cosseratmaterial (Boundary layer effect)


(a) (b)

Fig 3 (a) Results of the FEM simulation for the elasticity test with 50 elements in the verticaldirection (Deformation Scale Factor 20) (b) Comparison of the profiles of the Cosserat rotationωc3 obtained from the analytical solution and the FEM simulation with 50 points in the x2-

direction

ωc3(x2=0)

5 100 1000

00

13

10

1

number of mesh elements in x2-direction

Relativeerror(

)

Fig 4 Relative error of the results obtained by FEM for the Cosserat rotation ωc3(x2 = 0 9)

as a function of the number of elements in the x2-axis direction

h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c

Fig 5 Pure shear of an infinite layer with Cosserat microstructure Notations and boundaryconditions for the plasticity tests


(a) (b)

Fig 6 Results of the FEM simulation for the example of the infinite layer under simpleshear considering an elasto-plastic Cosserat continuum and von Mises yield criterion (a)TheCosserat rotation is plotted on the deformed mesh with 80 elements in the vertical direction(τ12 = 26MPa Deformation Scale Factor 1) (b)Stress-strain response of the layer plotted fordifferent number of elements in the vertical direction

200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)

Fig 7 Shear strain profile γ12 from FEM simulations of an elasto-plastic infinite layer plottedfor different number of elements in the vertical direction (τ12 = 26MPa)


d hard amp soft MampV hard

MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)

YZ hard [ soft MampV hard

MampV soft

]^_` abce fhij klmnγp

op

qr

st

uv

wxy

z|

τ (~)

(b)

Fig 8 FEM results for the shearing of one element (a) Stress-strain response and (b) Shearstress invariant as a function of the plastic shear strain invariant with coefficients of Muhlhausand Vardoulakis (1987) (ldquoMampVrdquo) and de Borst (1991) (ldquodBrdquo)

Fig 9 Geometry and boundary conditions for the rod used in the first test for Thermo-Mechanical couplings

Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)

Fig 10 Results of the FEM simulations for the heated rod (cth = 1 mm2s) (a) Temperatureevolution in the middle of the rod with time compared to the analytical solution (b) Normaldisplacement at the end of the rod u3(x3 = L) due to thermal dilation


Fig 11 Geometry and boundary conditions for the second rod used in the tests for Thermo-Mechanical couplings

(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11

iexcl centpoundcurren yenbrvbarsect

-umlcopyordflaquonot

-shyregmacrdegplusmn

-sup2sup3acutemicropara

-middotcedilsup1ordmraquo

-frac14frac12frac34iquestAgrave

AacuteAcircAtildeAumlAring

time (s)

stress(AElig

CcedilEgrave )

Fig 12 Results for the heated elastic confined rod (a) stress τ11 as a function of time (b)displacement u3 at the end of the rod as a function of time

(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )

uIgrave (IacuteIcircIumlL)

ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)

Fig 13 Comparison of results for (a) the displacement u3(x3 = L) and (b) the stress τ11 forthe heated confined rod in elasticity and plasticity


Fig 14 Comparison of the temperature increment as a function of time for the shearing testwith the mechanical dissipation term considered in the energy balance equation

Fig 15 Geometry and boundary conditions used for Hydro-Mechanical oedometer tests


FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)

Fig 16 Comparison of numerical and analytical results for the pore pressure in a undrainedoedometric test (βlowast=2MPaminus1)

τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)

Fig 17 Evolution of the stresses with time in a undrained oedometric test (a)Comparison ofFEM and analytical results for an elastic material (b)Comparison of elastic and elasto-plasticmaterials

T

p

10 20 30 40time (s)

05

10

15

20

25

30

T(degC)Oacute Ocirc(OtildeOumltimes)

Fig 18 Evolution of the pore pressure and temperature for the sheared layer of 60mm witha mesh of only one element considering simplified THM couplings (no diffusion)


OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2

Fig 19 Geometry and boundary conditions for the biaxial test

ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)

Fig 20 Stress-strain responses of undrained biaxial tests on sandstones for different meshesidentified by the number of elements in the vertical direction ny (R=2mm K1 = 10)


K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)

Fig 21 Stress-strain responses of undrained biaxial tests on sandstones for different valuesof the parameter Hs in the parabolic hardening evolution (R=2mm ny = 30)

R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)

Fig 22 Stress-strain responses of undrained biaxial tests on sandstones for different valuesof the characteristic size of the microstructure R (K1 = 10 ny = 30)

UacuteUcircUumlYacutem THORNszligagraveaacuteacircatildem

Fig 23 Deformed mesh of biaxial tests on sandstones for different values of the characteristicsize of the microstructure R


M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)

Fig 24 Comparison of the stress-strain responses of biaxial tests for sandstones taking intoaccount THM couplings or only the mechanical behavior (M) (K1 = 10 R=2mm and ny = 30)

M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)

Fig 25 Comparison of the stress-strain responses of fault zones modeled as infinite shearedlayers taking into account THM couplings or only the mechanical behavior considering perfectplasticity and the parameters summarized in Table 2

χigrave10-iacuteicircmiuml

χethntildeograve-oacuteocircmotilde

χoumldivideoslash-ugraveuacutemucirc

uumlyacutethorn yuml0

u1

h0

2

313

40

5

60

7

τ (MPa)

Fig 26 Stress-strain response of a fault zone modeled as an infinite sheared layer taking intoaccount THM couplings for different values of the permeability χ


=10-17

m2

=10-19

m2

Fig 27 FEM results of the plastic strain profile plotted on the deformed mesh for two differentvalues of the permeability χ

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Introduction

Constitutive model formulation

Finite element implementation and validation tests

Examples of numerical simulations

Concluding Remarks

See discussions stats and author profiles for this publication at httpswwwresearchgatenetpublication325905084

Numerical analysis of strain localization in Rocks with Thermo-Hydro-

Mechanical couplings using Cosserat continuum

Article in Rock Mechanics and Rock Engineering middot June 2018

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continuum



Received 2018







List of Symbols







1 Introduction

















ij by


k (1)



εij =1

2(uij + uji) Ωij =

1

2(uij minus uji)


ij


ck (2)







τijj = 0 (3)






ekl and microij =Me

ijkl κekl (5)



tensor

Meijkl = (Lminus 2



Ceijkl = (K minus 2






MG








τ =

radic



γp =radic

g1 epij e

pij + g2 e

pij e

pji +R2(g3 k

pij k

pij + g4 k

pij k

pji) (9)







partp

partt= chy pii +

λlowast

βlowast

partT

parttminus 1

βlowast

partεvpartt

(10)



of the medium



ρC(partT


pij + microijκ

pij (11)

where cth = kT











minusint

Ω

τijψijdΩ +

int

partΩΣ


minusint

Ω

microijψijdΩ +

int

partΩΣ


Ω



forms

int

Ω

pψdΩ + chy(

int

Ω

piψidΩ minusint

partΩ


βlowast

int

Ω

T ψdΩ +1

βlowast

int

Ω

εvψdΩ = 0

(14)int

Ω

T ψdΩ + cth(

int

Ω

TiψidΩ minusint

partΩ

TiniψdS)minus1

ρC

int

Ω


pij)ψidΩ = 0

(15)






32 Mechanical tests











partτ12partx2


partmicro32



3


+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)


ωc3 = K1 e


2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M





2

partu1partx2

(x2 = 0) (22)











c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks




continuum



Received 2018







List of Symbols







1 Introduction

















ij by


k (1)



εij =1

2(uij + uji) Ωij =

1

2(uij minus uji)


ij


ck (2)







τijj = 0 (3)






ekl and microij =Me

ijkl κekl (5)



tensor

Meijkl = (Lminus 2



Ceijkl = (K minus 2






MG








τ =

radic



γp =radic

g1 epij e

pij + g2 e

pij e

pji +R2(g3 k

pij k

pij + g4 k

pij k

pji) (9)







partp

partt= chy pii +

λlowast

βlowast

partT

parttminus 1

βlowast

partεvpartt

(10)



of the medium



ρC(partT


pij + microijκ

pij (11)

where cth = kT











minusint

Ω

τijψijdΩ +

int

partΩΣ


minusint

Ω

microijψijdΩ +

int

partΩΣ


Ω



forms

int

Ω

pψdΩ + chy(

int

Ω

piψidΩ minusint

partΩ


βlowast

int

Ω

T ψdΩ +1

βlowast

int

Ω

εvψdΩ = 0

(14)int

Ω

T ψdΩ + cth(

int

Ω

TiψidΩ minusint

partΩ

TiniψdS)minus1

ρC

int

Ω


pij)ψidΩ = 0

(15)






32 Mechanical tests











partτ12partx2


partmicro32



3


+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)


ωc3 = K1 e


2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M





2

partu1partx2

(x2 = 0) (22)











c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks


List of Symbols







1 Introduction

















ij by


k (1)



εij =1

2(uij + uji) Ωij =

1

2(uij minus uji)


ij


ck (2)







τijj = 0 (3)






ekl and microij =Me

ijkl κekl (5)



tensor

Meijkl = (Lminus 2



Ceijkl = (K minus 2






MG








τ =

radic



γp =radic

g1 epij e

pij + g2 e

pij e

pji +R2(g3 k

pij k

pij + g4 k

pij k

pji) (9)







partp

partt= chy pii +

λlowast

βlowast

partT

parttminus 1

βlowast

partεvpartt

(10)



of the medium



ρC(partT


pij + microijκ

pij (11)

where cth = kT











minusint

Ω

τijψijdΩ +

int

partΩΣ


minusint

Ω

microijψijdΩ +

int

partΩΣ


Ω



forms

int

Ω

pψdΩ + chy(

int

Ω

piψidΩ minusint

partΩ


βlowast

int

Ω

T ψdΩ +1

βlowast

int

Ω

εvψdΩ = 0

(14)int

Ω

T ψdΩ + cth(

int

Ω

TiψidΩ minusint

partΩ

TiniψdS)minus1

ρC

int

Ω


pij)ψidΩ = 0

(15)






32 Mechanical tests











partτ12partx2


partmicro32



3


+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)


ωc3 = K1 e


2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M





2

partu1partx2

(x2 = 0) (22)











c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks



1 Introduction

















ij by


k (1)



εij =1

2(uij + uji) Ωij =

1

2(uij minus uji)


ij


ck (2)







τijj = 0 (3)






ekl and microij =Me

ijkl κekl (5)



tensor

Meijkl = (Lminus 2



Ceijkl = (K minus 2






MG








τ =

radic



γp =radic

g1 epij e

pij + g2 e

pij e

pji +R2(g3 k

pij k

pij + g4 k

pij k

pji) (9)







partp

partt= chy pii +

λlowast

βlowast

partT

parttminus 1

βlowast

partεvpartt

(10)



of the medium



ρC(partT


pij + microijκ

pij (11)

where cth = kT











minusint

Ω

τijψijdΩ +

int

partΩΣ


minusint

Ω

microijψijdΩ +

int

partΩΣ


Ω



forms

int

Ω

pψdΩ + chy(

int

Ω

piψidΩ minusint

partΩ


βlowast

int

Ω

T ψdΩ +1

βlowast

int

Ω

εvψdΩ = 0

(14)int

Ω

T ψdΩ + cth(

int

Ω

TiψidΩ minusint

partΩ

TiniψdS)minus1

ρC

int

Ω


pij)ψidΩ = 0

(15)






32 Mechanical tests











partτ12partx2


partmicro32



3


+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)


ωc3 = K1 e


2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M





2

partu1partx2

(x2 = 0) (22)











c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks









ij by


k (1)



εij =1

2(uij + uji) Ωij =

1

2(uij minus uji)


ij


ck (2)







τijj = 0 (3)






ekl and microij =Me

ijkl κekl (5)



tensor

Meijkl = (Lminus 2



Ceijkl = (K minus 2






MG








τ =

radic



γp =radic

g1 epij e

pij + g2 e

pij e

pji +R2(g3 k

pij k

pij + g4 k

pij k

pji) (9)







partp

partt= chy pii +

λlowast

βlowast

partT

parttminus 1

βlowast

partεvpartt

(10)



of the medium



ρC(partT


pij + microijκ

pij (11)

where cth = kT











minusint

Ω

τijψijdΩ +

int

partΩΣ


minusint

Ω

microijψijdΩ +

int

partΩΣ


Ω



forms

int

Ω

pψdΩ + chy(

int

Ω

piψidΩ minusint

partΩ


βlowast

int

Ω

T ψdΩ +1

βlowast

int

Ω

εvψdΩ = 0

(14)int

Ω

T ψdΩ + cth(

int

Ω

TiψidΩ minusint

partΩ

TiniψdS)minus1

ρC

int

Ω


pij)ψidΩ = 0

(15)






32 Mechanical tests











partτ12partx2


partmicro32



3


+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)


ωc3 = K1 e


2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M





2

partu1partx2

(x2 = 0) (22)











c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks





ekl and microij =Me

ijkl κekl (5)



tensor

Meijkl = (Lminus 2



Ceijkl = (K minus 2






MG








τ =

radic



γp =radic

g1 epij e

pij + g2 e

pij e

pji +R2(g3 k

pij k

pij + g4 k

pij k

pji) (9)







partp

partt= chy pii +

λlowast

βlowast

partT

parttminus 1

βlowast

partεvpartt

(10)



of the medium



ρC(partT


pij + microijκ

pij (11)

where cth = kT











minusint

Ω

τijψijdΩ +

int

partΩΣ


minusint

Ω

microijψijdΩ +

int

partΩΣ


Ω



forms

int

Ω

pψdΩ + chy(

int

Ω

piψidΩ minusint

partΩ


βlowast

int

Ω

T ψdΩ +1

βlowast

int

Ω

εvψdΩ = 0

(14)int

Ω

T ψdΩ + cth(

int

Ω

TiψidΩ minusint

partΩ

TiniψdS)minus1

ρC

int

Ω


pij)ψidΩ = 0

(15)






32 Mechanical tests











partτ12partx2


partmicro32



3


+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)


ωc3 = K1 e


2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M





2

partu1partx2

(x2 = 0) (22)











c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks






partp

partt= chy pii +

λlowast

βlowast

partT

parttminus 1

βlowast

partεvpartt

(10)



of the medium



ρC(partT


pij + microijκ

pij (11)

where cth = kT











minusint

Ω

τijψijdΩ +

int

partΩΣ


minusint

Ω

microijψijdΩ +

int

partΩΣ


Ω



forms

int

Ω

pψdΩ + chy(

int

Ω

piψidΩ minusint

partΩ


βlowast

int

Ω

T ψdΩ +1

βlowast

int

Ω

εvψdΩ = 0

(14)int

Ω

T ψdΩ + cth(

int

Ω

TiψidΩ minusint

partΩ

TiniψdS)minus1

ρC

int

Ω


pij)ψidΩ = 0

(15)






32 Mechanical tests











partτ12partx2


partmicro32



3


+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)


ωc3 = K1 e


2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M





2

partu1partx2

(x2 = 0) (22)











c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks






minusint

Ω

τijψijdΩ +

int

partΩΣ


minusint

Ω

microijψijdΩ +

int

partΩΣ


Ω



forms

int

Ω

pψdΩ + chy(

int

Ω

piψidΩ minusint

partΩ


βlowast

int

Ω

T ψdΩ +1

βlowast

int

Ω

εvψdΩ = 0

(14)int

Ω

T ψdΩ + cth(

int

Ω

TiψidΩ minusint

partΩ

TiniψdS)minus1

ρC

int

Ω


pij)ψidΩ = 0

(15)






32 Mechanical tests











partτ12partx2


partmicro32



3


+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)


ωc3 = K1 e


2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M





2

partu1partx2

(x2 = 0) (22)











c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks







partτ12partx2


partmicro32



3


+ 2Gcpartωc

3

partx2= 0 (18)

Mpart2ωc

3

partx22minus 4Gcω

c3 = 2Gc

partu1partx2

(19)


ωc3 = K1 e


2G(20)

u1 = minus 2Gc

G+Gc(K1

δeδx2 minus K2

δeminusδx2) +

τcGx2 +K3 (21)

where δ = 2radic

G Gc

(G+Gc)M





2

partu1partx2

(x2 = 0) (22)











c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks





c1 ω

c2 ω



u1 = 0 u2 = 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = Ω21 (24)

Top edge (x2=h)

u1h

= 001partu2partx2

= 0 u3 = 0 ωc1 = 0 ωc

2 = 0 ωc3 = minus01 (25)








where σ = τkk









4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks






4 h3 = 3


3185


2 andg1 + g2 = 2





















(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks





(Fig 9)230




γ33 =3K

K + 43G

α∆T (29)


τ11 = τ22 = (K minus 2









ρCpartT

partt= σij ε

pij + τ[ij]γ

p




p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks



p

12

∆tat each265



p

12∆t

ρC (with





partp

partt= minus 1

βlowast

partεvpartt

(32)




γ33 = minusv3tL

(33)



3)γ33 (34)

τ prime33 = (K +4G

3)γ33 (35)



(36)







τ =

radic

2


2 (37)



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks



t =

radic3cL

2Gv3radic

2(h1 + h2)(38)





partp

partt=λlowast

βlowast

partT

partt(39)





pressurization)




















g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks













g1 = g3 = 85 and g2 = g4 = 2



c = c0 +(1minusK1γ

p)γp

K2 +K3γp(41)













































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks







































References























































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks











































































x1

x2

x3

σ33

σ23

σ13

σ32

σ22

σ12 σ21

σ31

σ11

μ22

μ32μ12

μ23

μ33μ13

μ31μ11

μ21


x2

x1

h

x2=0

21

12

m32w3

u1

c

u1=0 w3

c=W21

u1=001mmw3

c

=-01x2=h



(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks


(a) (b)


direction

ωc3(x2=0)

5 100 1000

00

13

10

1


Relativeerror(

)



h21

12

m32

u1

τ22=0

τ

τ

u2=0

u2

=0w3

c

u1=Vt

u1=0

=0w3

c

w3

c



(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks


(a) (b)


200

8

32

5 1 15 2γ12

1

2

3

4

5

6$

height (mm)




MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks



MampV soft

()+ -3 4678 9lt =gt ABCD

uEh

FG

HI

JK

LM

NOP

QRS

τTU (VWX)

(a)


MampV soft


op

qr

st

uv

wxy

z|

τ (~)

(b)



Analytic

100 200 300 400 500time(s)

02

04

06

08

10

T (degC)

(a)

u3(x3=L)

0 100 200 300 400 500

0000

0005

0010

0015

0020

0025

time (s)

(

mm)

(b)




(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks



(a) (b)

u3 3=L

0 100 200 300 400 500000

001

002

003

time (s)

cement(mm)

τ11








time (s)

stress(AElig

CcedilEgrave )


(a) (b)

τ11 elastic

elasto-plastic

0 100 200 300 400 500

-0010

-0008

-0006

-0004

-0002

0000

time (s)

stre

ss(Eacute

EcircEuml )


ETHNtildeOgravestic

elasto-plastic

0 100 200 300 400 500000

001

002

003

004

time (s)

displacement

(mm

)






FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks





FEM

Analytical

1 2 3 4 5time (s)

0005

0010

0015

0020

0025

p (MPa)


τ11 num

τ11 analytical

τ33 num

τ33 analytical

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(a)

τ11 elast

τ11 plast

τ33 elast

τ33 plast

1 2 3 4 5time (s)

-020

-015

-010

-005

τ (MPa)

(b)


T

p

10 20 30 40time (s)

05

10

15

20

25

30




OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks


OslashUgrave

h

m32

w3

u1

c

u

h2

x2

u2


ny=10

ny=20

ny=30

ny=40

02 04 06 08 10 12 14u2(mm)

10

20

30

40

50

F

S(MPa)



K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks


K1=5

K1=10

K1=20

05 10 15 20u2(mm)

10

20

30

40

50

F

S(MPa)


R= 2 mm

R=1 mm

R=02 mm

02 04 06 08 10 12u2

10

20

30

40

50

F(MPa)

S

(mm)





M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks


M

aumlaringaelig

02 04 06 08 10 12u2 (mm)

10

20

30

40

50

F

S(MPa)


M

ccedilegraveeacute

01 02 03 04 05

u1

h

10

20

30

40

50

60

ecirceuml

τ12 (MPa)






u1

h0

2

313

40

5

60

7

τ (MPa)



=10-17

m2

=10-19

m2



Introduction




Concluding Remarks


=10-17

m2

=10-19

m2



Introduction




Concluding Remarks

Numerical analysis of strain localization in Rocks with ... · The full stress-strain response of...

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Transcript of Numerical analysis of strain localization in Rocks with ... · The full stress-strain response of...