Implicit stress integration procedure for small and large strains of the Gurson material model

INTERNATIONAL JOURNAL FOR NUMERICAL METHODS IN ENGINEERINGInt. J. Numer. Meth. Engng 2002; 53:2701–2720 (DOI: 10.1002/nme.410)

Implicit stress integration procedure for small andlarge strains of the Gurson material model

Milos Kojic∗;†, Ivo Vlastelica and Miroslav Zivkovic

Faculty of Mechanical Engineering; University of Kragujevac; 34000 Kragujevac; Serbia

SUMMARY

The Gurson material model has broad applications in fracture mechanics, large strain deformationsand failure of metals. Void growth and void nucleation are included in the model considered in thispaper. An implicit stress integration procedure with calculation of the consistent tangent moduli isdeveloped for the Gurson model. The general 3D deformations and the plane stress conditions areconsidered. The procedure is robust, simple and computationally e�cient, suitable for use within the�nite element method (FEM). It represents an application of the governing parameter method (GPM)for stress integration in case of inelastic material deformation. A large strain formulation, based on themultiplicative decomposition of the deformation gradient for material with plastic change of volumeand logarithmic strains, is used in the paper. The developed numerical procedure for stress integrationis applicable to small and large strains conditions. Solved examples illustrate the main features of thedeveloped numerical algorithm. Copyright ? 2002 John Wiley & Sons, Ltd.

KEY WORDS: stress calculation; Gurson model; large strains; �nite element method

1. INTRODUCTION

A commonly used assumption in metal plasticity is that plastic deformation is volume pre-serving. Material models representing this behaviour are of the von Misses and Hill’s typefor isotropic and orthotropic material, described for example, in Mendelson [1] and Hill [2].However, in case of large local plastic �ow occurring in, for instance, processes of ductilefracture, void nucleation and void growth is observed. Then, permanent volumetric plasticstrain develops and the hydrostatic stress independent plasticity models are not adequate todescribe behaviour of the porous metals.

1.1. Model formulation

The basic plasticity rate-independent model for porous metal was formulated by Gurson [3].The yield criterion is developed by considering some simpli�ed physical models of aggregates

∗Correspondence to: M. Kojic, Faculty of Mechanical Engineering, University of Kragujevac, 34000 Kragujevac,Serbia

†E-mail: [email protected]

Received 12 January 2000Copyright ? 2002 John Wiley & Sons, Ltd. Revised 4 June 2001

2702 M. KOJIC, I. VLASTELICA AND M. ZIVKOVIC

of voids and the ductile matrix. It was shown that normality rule holds for the formulatedyield locus. Tvergaard [4; 5] modi�ed the original Gurson yield condition and proposed theform, which will be used in this paper:

fy = 1=2S · S+ 13

[2f∗q1 cosh

(3q2�m2�y

)− 1− q23f∗2

]�2y = 0 (1)

where �y is the yield stress of the material, �m is the mean stress, f∗ is a function of thevoid volume fraction (porosity) f, q1, q2 and q3 are material constants, and

S · S= SijSij (2)

Here Sij are components of the stress deviator, and summation on the repeated indices isassumed (i; j=1; 2; 3). The function f∗ of the void volume fraction is given by

f∗=

∣∣∣∣∣f for f6fc

fc + Kf (f − fc) for f¿fc(3)

where fc is a critical void volume fraction when the onset of rapid volume coalescence begins,and

Kf =1=q1 − fcff − fc (4)

Here ff is value of f at material failure, since then f∗=1=q1, and from (1) it follows(for q1= q3) that the e�ective stress ��=(3=2 Sij Sij)1=2 is equal to zero ( ��=0) for �m =0.Note that in case of f∗=0 the yield condition (1) reduces to the von Mises yield conditionwith isotropic hardening behaviour. An extension of the yield condition (1) to a materialwith the kinematic or mixed hardening response was proposed by Lee and Zhang [6], andto an orthotropic-type model by Doege et al. [7], and Brunet and Sabourin [8]. The yieldcondition (1) was employed by Jha and Narasimhan [9] as the �ow potential in a rate-sensitiveviscoplastic model for dynamic fracture in steel.In general, rate of change of porosity can be written as [8]

f= fG + fN + fC (5)

where fG, fN, and fC correspond to void growth, nucleation and coalescence, respectively.Assuming that the material matrix is plastically incompressible and neglecting elastic part ofthe void volume, the rate of void volume growth fG can be expressed as

fG = (1− f)ePV (6)

where ePV is plastic volumetric strain rate. If the nucleation of voids is dominated by themaximum normal stress, the rate of void nucleation fN can be written in the form [6; 10]

fN =K�y(�y + �m) (7)

where K is a dimensionless material parameter. This parameter can also be expressed in ananalytical form that involves the yield stress �y and the mean stress �m, obtained through the

Copyright ? 2002 John Wiley & Sons, Ltd. Int. J. Numer. Meth. Engng 2002; 53:2701–2720

IMPLICIT STRESS INTEGRATION PROCEDURE 2703

statistical distribution of �y and �m. In case of nucleation dominated by the plastic strain, fNcan be written as [10]

fN =A �eP (8)

where A can be either a material constant, or it can be chosen such that the void nucleationfollows a normal distribution. In the latter case the coe�cient A is [10]

A=aN

sN√2�exp

[−12

(�eP − �eNsN

)2](9)

where aN is the amplitude, sN is the standard deviation, �eP is the accumulated e�ective plasticstrain, and �eN is the mean e�ective plastic strain of the normal distribution of the voidnucleation. Finally, the rate of void coalescence fC is proportional to the e�ective plasticstrain rate, as it is fN in Equation (8), i.e.

fC =B �eP (10)

where B is a material parameter.Besides the above e�ects on change of porosity, Ramaswamy and Aravas [11] introduced

a gradient-type term into the evolution equation for void change, to take into account �uxof voids. Also,Tvergaard and Needleman [12] considered the non-local e�ects on rate ofporosity f.

1.2. Stress integration

We further brie�y present stress calculation, usually called the stress integration, as a veryimportant part of the overall �nite element analysis of inelastic material deformation. Namely,the calculated material response is strongly dependent on the computational procedure forstress calculation. Implicit schemes have been generally favoured in the literature, becausethey provide better accuracy with respect to the explicit algorithms (usually simpler), andprovide high convergence rate within structural equilibrium iterations, cf. Bathe [13], Simoand Hughes [14]. The general strategy within an implicit stress integration procedure for therate-independent plasticity models is the so-called return mapping [14]. It consists of twosteps: (a) determination of the trial elastic state (elastic predictor), and (b) stress correctiondue to plastic �ow (plastic corrector). The trial elastic state is calculated under the assumptionthat there is no plastic �ow in the time step. In case of small strains, the trial elastic stresst+�t�E∗ (written in the matrix form) is

t+�t�E∗ =CE t+�teE∗ (11)

t+�teE∗ =t+�te − teP (12)

where CE is the elastic constitutive matrix (we assume here that CE is stress and temperatureindependent); t+�te and teP are the total strain at the end of time step, and the plastic strainat start of the time step �t, respectively. In the case of large strain deformation, the trial



elastic strains correspond to the local stress-free con�guration V�, with the left Cauchy–Greendeformation tensor t+�t�BE∗ determined from the relation [14]

t+�t�BE∗=

t+�ttF

t�BE t+�t

tFT (13)

where t+�ttF is the relative deformation gradient. In the case of logarithmic strains, we de-

compose the hyperelastic free energy into the deviatoric and volumetric parts [14–18], andinstead of t+�ttF, we use in (13) the modi�ed deformation gradient t+�tt �F:

t+�tt�F=(det t+�ttF)

−1=3 t+�ttF (14)

The trial elastic deviatoric strain t+�t◦e′E∗ is

t+�◦e

′E∗ =

3∑k=1ln(t+�t◦�

E∗k)

t+�t�pEkt+�t�pEk (15)

where t+�t◦�E∗k are the principal stretches, and

t+�t�pEk are the principal directions obtained byusing the tensor t+�t�BE∗. The trial elastic mean strain t+�t

◦eE∗m is

t+�t◦eE∗m=

13 ln[det(

t+�t◦F)]− t

◦ePm (16)

In the above expressions the lower left index denotes the con�guration to which we refer inde�ning the tensor. The trial elastic deviator and the mean stress, t+�tSE and t+�t�Em, followfrom (11) as

t+�tSE = 2Gt+�t◦e

′E∗ (17)

t+�t�Em = cmt+�t

◦eE∗m (no sum on m) (18)

where cm =E=(1+2�), with E and � being Young’s modules and Poisson’s ratio, respectively.Simo and Hughes [19; 14] proposed a general form of the return mapping, consisting of

the successive linearizations. A simpler procedure of the same type presents the cutting planealgorithm, introduced by Simo and Ortiz [20]. Another approach in the return mapping is thegoverning parameter method (GPM) formulated by Kojic [21]. In this procedure the stressintegration reduces to solving one non-linear equation with respect to a suitable selectedgoverning parameter. It represents a generalization of the ‘e�ective-stress-function’ algorithmintroduced for thermoplasticity and creep of metal by Kojic and Bathe [22; 23]. The GPM wasapplied to the orthotropic plasticity and to geologic materials, Kojic et al. [24–26; 33; 43; 44].A detailed description of the GPM, with applications to the broad �eld of inelastic materialdeformation, is presented in Kojic and Bathe [27].

1.3. The governing parameter method

We here brie�y summarize the main steps of the GPM [21], because we use the GPM toformulate the stress integration algorithm for the Gurson model. We suppose that the knownquantities are

t+�te; t�; teIN; t� (19)



while the unknowns are

t+�t�; t+�teIN; t+�t� (20)

where �, eIN, e and � are stresses, inelastic strains, strains and internal variables, with theleft upper indices t and t+�t indicating the start and end of the time step, respectively. Thebasic steps of the stress integration consist of the following.

1. Express all unknowns in terms of one, governing parameter, p, i.e.

t+�t�= t+�t�(t+�te; t�; teIN; t�; p)t+�teIN = t+�teIN(t+�te; t�; teIN; t�; p)t+�t�= t+�t�(t+�te; t�; teIN; t�; p)

(21)

2. Solve the governing equation

f(p)=0 (22)

3. Determine unknowns (21) by substituting t+�tp.

Determination of the consistent tangent constitutive matrix is straightforward within theGPM. Namely, by de�nition of the constitutive matrix we have

t+�tC=@t+�t�@ t+�te

∣∣∣∣p=const

+@t+�t�@ t+�tp

@t+�tp@t+�te

(23)

Derivatives @t+�tp=@ t+�te can be calculated by di�erentiation of the governing equationf(t+�tp)=0 with respect to strains t+�te, i.e.

@t+�tp@t+�te

=−a−1 @f@t+�te

(24)

where

a=@fT

@t+�t�@ t+�t�@ t+�tp

+@fT

@t+�teIN@t+�teIN

@t+�tp+

@fT

@t+�t�@ t+�t�@ t+�tp

+@f@t+�tp

(25)

We use here the matrix notation, where @fT=@� means transponse.Stress integration for the Gurson model have been performed in the past within applica-

tion of FEM. Aravas [28] proposed an implicit stress integration scheme for the general 3Ddeformations by forming a system of two simultaneous equations with the volumetric plas-tic strain and the equivalent plastic strain increments as the unknowns. In case of the planestress conditions the system of equations consists of three equations, with the increment ofthe through- the-thickness strain as the third unknown. The system was solved by using theNewton iteration method. The same approach was followed by other authors, e.g. Lee andZhang [29], Jha and Narasimhan [9], Doege et al. [7], Zhang [30; 31], Lee and Zhang [6]and Brunet and Sabourin [8].The paper is organized as follows. In the next section we give detailed derivation of the

relations for stress integration for Gurson model according to the GPM, for the general 3Ddeformations and for the plane stress conditions, and then in Section 3 we derive expressions



for the consistent tangent elastic–plastic matrix. Derivations are presented for small strains,with necessary additions=corrections for large strain problems. Application of the proposedalgorithm is illustrated through numerical examples in Section 4. Finally, in Section 5 someconcluding remarks are given.

2. STRESS INTEGRATION PROCEDURE

Before developing the stress integration procedure, we add two fundamental assumptions to therelations of the previous section. First, it is the �ow rule, as the generally accepted plasticityconstitutive law (derivation of this law based on the micromechanical considerations for theGurson model is given [3]),

eP= �@fy@�

(26)

where � is a positive scalar. The second assumption is the equivalence of plastic work in caseof general loading and uniaxial loading conditions, expressed in the form [32]

� · eP= (1− f)�y �eP (27)

2.1. General 3D deformations

We start from the constitutive relations in the form (18) and (17), and write

t+�t�m = t+�t�Em − cm�ePm (no sum on m) (28)

t+�tS= t+�tSE − 2G�e′P (29)

where �ePm and �e′P are increments of the mean and deviatoric plastic strains, respectively.For these increments we employ the Euler backward scheme, and from (26) and (1) obtain

�ePm =��3

t+�tf′y (30)

�e′Pij =��t+�tSij (31)

where

t+�tf′y =

@t+�tfy@�m

= q1q2 t+�t�y t+�tf∗ sinh(3q2 t+�t�m2 t+�t�y

)(32)

Substituting (31) into (29) we obtain

t+�tS=t+�tSE

1 + 2G��(33)



Further, we employ the evolution equations for porosity. We write (5) in the incrementalform and use (6), (8) and (10):

�f=3(1− t+�tf)�ePm + (A+ B)� �eP (34)

If the evolution equation (7) is used instead of (8), we have

�f=3(1− t+�tf)�ePm +K

t+�t�y(��y + ��m) + B��eP (35)

In further derivations we will employ (34), with use of A instead A+B. Analogous derivationswould follow if the relation (35) is taken instead of (34). From (34) we obtain

�f=[3(1− tf)�ePm + A��eP]=(1 + 3�ePm) (36)

Equation (27), corresponding to the end of time step, can be written in the form

t+�tfe =�� t+�tS · t+�tS+ 3 t+�t�m�ePm − (1− t+�tf)t+�t�y� �eP =0 (37)

where we have used (31).Finally, the yield condition (1) at the end of time step must be satis�ed, i.e.

t+�tfy(t+�tS; t+�t�m; t+�t�y; t+�tf)=0 (38)

where the yield stress t+�t�y is related to t+�t �eP by the yield curve of the material,

t+�t�y = t+�t�y(t �eP + ��eP) (39)

Now we can establish an implicit stress integration procedure according to the GPM.Namely, we can select �ePm as the governing parameter and �nd that all unknowns canbe calculated according to the steps 1–3 of Section 1. The computational path is as follows.Select �ePm and calculate

t+�t�m from (28); then, solve (37) with respect to � �eP: suppose� �eP, determine t+�t�y from (39), �f from (36), t+�tf∗ from (3), t+�tf′

y from (32), �� from(30), t+�tS from (33); check if |t+�tfy|6�—if not, suppose new �ePm and repeat calculations.Table I summarizes these computational steps. After the solution for the unknowns is reached,the increments �e′Pij can be calculated from (31). Trial values for �ePm and ��e

P are obtainedby use of the robust bisection method with an acceleration scheme (use of secants). In casewhen the porosity due to void volume growth is negative, it is considered that the materialbehaves according to the von Mises model [8].By inspecting the described algorithm we can reach the following conclusions. The al-

gorithm is computationally e�cient because the stress calculation reduces to solving onenon-linear equation with respect to �ePm, which includes numerical calculation of � �e

P. Thealgorithm is robust since it is applicable to large strain increments in one load step (of orderof several per cent), see Example 2. We have found that the governing function fy(�ePm) ismonotonic and that the simple and robust bisection procedure, with an acceleration scheme, isvery e�cient. Normally, we have obtained the solutions within order of 5–10 trials, even forlarge strain increments. It is also accurate (as an implicit procedure) and it provides satisfaction



Table I. Computational steps for stress integration.

1. Known valuesteP; t�y ; t �eP; tf; t+�te

2. Calculate trial elastic solution and check for yieldingt+�tSE; t+�t�Em from (11); or from (17) and (18).If t+�tfy(t+�t�E; tf∗; t�y)60 solution is elastic; end of calculations.Elastic–plastic solution∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

→ 3: Iteration for the governing parameter �ePmt+�t�m from (28)∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

→ 4: Iterations for � �eP

Calculate:t+�t�y from (39)�f from (36)t+�tf∗ from (3)t+�tf′

y from (32)�� from (30)t+�tS from (33)Check for |t+�tfe|¡�Check for |t+�tfy|¡�

5. Final calculations�e′P from (31)t+�t�= t+�tS+ t+�t�mt+�teP = teP + �ePt+�t �eP = t �eP + � �eP

of the governing relations—the yield condition and the equivalence of plastic work—withina selected tolerance. Our experience shows that these governing relations cannot be satis-�ed within tide tolerances (e.g. below 10−8) if the standard iterative scheme proposed byAravas [28] is employed. This is particularly true for large strain increments, preferable ingeneral FEM applications.

2.2. Plane stress conditions

In case of deformation under the plane stress conditions (plane x; y) we have to satisfy therelation

�zz=0 (40)

Then the normal components of the stress deviator can be expressed as (Kojic and Bathe [23])

Sxx=p1e′′xx + p2e

′′yy

p21 − p22; Syy=

p2e′′xx + p1e′′yy

p21 − p22(41)

and also

t+�tSzz=−t+�t�m =−t+�tSxx − t+�tSyy (42)



Here we have

p1 = c1��+ aE; p2 = c��

e′′xx = c1(t+�texx − tePxx)− c�(t+�teyy − tePyy) (43)

e′′yy =−c�(t+�texx − tePxx) + c1(t+�teyy − tePyy)

with

c�=1− �3(1− �) ; c1 = 1− c�; aE =

12G

=1+ �E

(44)

The shear stress component t+�tSxy is given by (33).The computational algorithm is in principle the same as given in Table I, but with few

corrections. Namely, we have the external iteration loop on �ePm (step 3), with calculation of�f according to (36); and the internal iteration loop on ��eP (step 4), with calculation ofthe deviatoric stress components and the mean stress using the expressions (41) and (42).In Example 4 we show applicability and accuracy of the proposed algorithm for the plane

stress problems.

2.3. Extension to large strains

The above procedure is directly applicable to large strain problems. Update of the left Cauchy–Green tensor t+�t�BE is performed by calculation of t+�t�BE as

t+�t�B

E=3∑k=1

exp(2t+�t �e′Ek )t+�t �pEk

t+�t �pEk (45)

where t+�t �e′Ek are elastic deviatoric strains in the principal directions calculated as

t+�t �e′Ek =12G

t+�t �Sk (46)

Here �Sk are deviatoric stresses in the principal directions t+�t �pk .The proposed algorithm for the large strain conditions displays the characteristics of the

algorithm for small strains: e�ciency robustness and accuracy. In comparison with a Newton-type procedure involving a system of equations with several unknowns and possible divergencein the solutions process (e.g. Lee and Zhang [29] formed the system of three equations), theproposed algorithm has the obvious advantages.The above-derived expressions within the stress integration algorithm correspond to general

3D deformation, and to 2D-plane strain and axisymmetric conditions. Plane stress, shell orbeam conditions are not considered here. They require a speci�c approach, analogous to thatgiven in Kojic et al. [33], and will be the subject of another study.

3. ELASTIC–PLASTIC MATRIX

Here we derive the consistent tangent elastic–plastic matrix according to Equation (23). Thematrix is consistent with the stress integration procedure [14] and it is tangent because it



corresponds to the stresses and strains at the end of time step. For simpler writing we usehere the one-index notation. The stress and strain matrix columns are de�ned as

�T = {�11; �22; �33; �12; �23; �31}T

eT = {e11; e22; e33; �12; �23; �31}T (47)

where �ij =2eij are engineering shear strains. Therefore, the terms t+�tCEPij are

t+�tCEPij =t+�t�m; j + t+�tSi; j ; i=1; 2; 3; j=1; 2; : : : ; 6

t+�tCEPij =t+�tSi; j ; i=4; 5; 6; j=1; 2; : : : ; 6

(48)

where we have used t+�t�m; j= @t+�t�m=@ t+�tej and t+�tSi; j= @t+�tSi=@ t+�tej to simplify thenotation. In the subsequent derivations we use notation ;m ≡ @=@(�ePm) and the conventionthat there is no summation on the repeated index m.Starting with (28) and (29) we have

t+�t�m; j= cm(13 −�ePm; j

); j=1; 2; 3

t+�t�m; j=−cm�ePm; j ; j=4; 5; 6(no sum on m) (49)

and

t+�tSi; j=2G[�Ai; j −

t+�tSi1 + 2G��

(��; kAkj +��;m�ePm; j)]/

1 + 2G�� (no sum on m)

(50)

where we have used (17) and (18). The matrix �A is

�A=

[A 0

0 12I3

](51)

with

A=13

2 −1 −1−1 2 −1−1 −1 2

(52)

and I3 is 3× 3 identity matrix. From (30) we obtain

��;m = (3−��t+�tf′y;m)=

t+�tf′y (53)

Further, from (32) we write t+�tf′y;m in the form

t+�tf′y;m= a1 + a2� �e

P;m (54)

where coe�cients a1 and a2 follow directly from di�erentiation; they are given in Appendix A.



From (3) it follows that

t+�tf∗;m =

t+�tf;m; f6fct+�tf∗

;m =Kft+�tf;m; f¿fc

(55)

With the use of (36) we obtain

t+�tf∗;m = [3�f + (A

′ + A)�eP;m]=(1 + 3�eP;m) (56)

where A′=dA=d(�ePm) follows from (9). Hence, in order to calculate the derivative ��;mwe need the derivative � �eP;m ≡ @(� �eP)=@(�ePm). The last derivative can be determined bydi�erentiation with respect to �ePm of either Equation (37) or (38). If we employ (38), thenwe �nd

��eP;m =−km (57)

where the expressions for km is given in Appendix A.Finally, we determine the derivatives �ePm; j. If we di�erentiate (37) with respect to

t+�tej,we obtain

b1�ePm; j + 2��t+�tSit+�t Si; j + 3�ePm

t+�t�m; j=0 (58)

where b1 is the term following from the di�erentiation; and t+�t Si; j is given as

t+�t Si; j= t+�tSi; j ; i=1; 2; 3t+�t Si; j=2t+�tSi; j ; i=4; 5; 6

(59)

In writing the expression (53) we have taken into account that t+�t�y; j= t+�tEP� �eP; j wheret+�tEP = @t+�t�y\@t+�t �eP is the plastic modulus at the end of time step. Substituting (49) and(50) into (58) we obtain the equation which can be solved for �ePm; j in the form

�ePm; j=−wj=w0 (60)

with w0 and wj as the coe�cients depending on stresses, porosity and yield stress at the endof time step. With �ePm; j determined, from (49) and (50) then follow derivatives t+�t�m; j andt+�tSi; j for t+�tCEPij in (48).The above expressions are also applicable to large strains. Then we can write (e.g. [34; 35])

t+�tCEP = t+�tCEPmat +t+�tCkin (61)

where the matrix t+�tCEPmat depends on the material model and is given above for the smallstrain assumptions, while t+�tCkin takes into account the kinematics of large strains and isindependent of the material behaviour. Detailed derivation of t+�tCkin is given in References[34–38]. Addition of t+�tCkin to t+�tCEPmat increases the convergence rate, but has no e�ecton the solution (see e.g. Bathe [13]). We have included the geometrically non-linear matrixKNL according to the updated Lagrangian formulation [13], and not t+�tCkin in our FE codePAK [39]. The number of iterations for the examples in Section 4 was below 10, and it willbe reduced with the use of t+�tCkin.



Figure 1. One quarter of the axisymmetric cross-section of smooth bar; dimensions,co-ordinate system and boundary conditions.

4. EXAMPLES

4.1. Example 1. Tension of smooth bar

A standard smooth tensile specimen used to characterize the material behaviour and to identifythe critical damage parameters for the ductile tearing at room temperature, is analysed as the�rst example. Geometry, loading and boundary conditions are shown in Figure 1. One quarterof the bar is modelled due to symmetry. Two-dimensional 8-node axisymmetric elements (168elements) are used and the loading is applied through the prescribed displacements at the topof the mesh [39].Values for Young’s modulus and Poisson’s ratio are 250 GPa and 0.3, respectively. The

uniaxial stress–plastic strain dependence is described by the Ramberg–Osgood formula:

�y = 468:5 + 445( �eP)0:361

Material constants of the Gurson model are q1 = 1:5; q2 = 1; q3 = 1:5; the initial void volumefraction is f0 = 0:002, the failure void fraction is ff = 0:315, and the critical void volumefraction is fc = 0:05.The �nal end displacement of 3:625 mm is reached in 29 equal load steps. The deformed

mesh with the e�ective stress and e�ective plastic strain �elds is shown in Figure 2(a) and (b).Force-elongation and force reduction of diameter curves are shown in Figure 3. As can be seenfrom the �gure, good agreement of FE solution with experimental results [40] is obtained.

4.2. Example 2. Extension of a notched bar

This specimen is used for determination of the critical damage parameters for the ductiletearing at room temperature. Due to symmetry only one-quarter of the bar is modelled (seeFigure 4), two-dimensional 8-node axisymmetric elements (416 elements) are used and theloading is applied through the prescribed displacements at the top of the mesh.



Figure 2. Extension of smooth bar: (a) e�ective stress �eld; and (b) e�ective plastic strain �eld.

Figure 3. Extension of smooth bar: (a) load vs elongation; and (b) load vs reduction of diameter.

Values for Young’s modulus and Poisson’s ratio are 97GPa and 0.3, respectively. The yieldcurve given by

�y = 94:4− 1594( �eP)2 + 645( �eP)3

Material constants are q1=1:25; q2=0:95; q3 = 1:25, the initial void volume fractionf0 = 0:025, the failure void volume fraction is ff = 0:315, and critical void volume fractionis fc = 0:05.The �nal end displacement of 1:075mm is reached in 215 equal load steps. The deformed

mesh with the e�ective plastic strain �eld is shown in Figure 5. The axial stress in terms ofthe axial strain for the von Mises and for Gurson material models is shown in Figure 6. Radialdistribution of the void volume fraction across the middle cross-section, and axial distributionof void volume fraction along the axis, for strain at the specimen end, ey(log) = 0:3, are shownin Figure 7(a) and (b), respectively. Good agreement between the results of this paper andthe results of Worswick and Pick [41] is obtained.



Figure 4. Axisymmetric section of specimen,co-ordinate system and boundary conditions.

Figure 5. E�ective plastic strain �eld.

Figure 6. Axial stress �y vs logarithmic strain elog.

To illustrate the accuracy and robustness of the algorithm we have solved this exampleusing 30 steps and found that the maximum e�ective strain di�ers from the basic solution(215 steps) by 3.14 per cent.Table II shows unbalanced energy and unbalanced force during equilibrium iterations. High

convergence rate (for a typical step) demonstrates the tangent character of the elastic–plasticmatrix derived in Section 3.

4.3. Example 3. Compression of a cylinder

Geometry, loading and boundary conditions are shown in Figure 8. One quarter of the cylinderis modelled due to symmetry. Two-dimensional 8-node axisymmetric elements (127 elements)are used and the loading is applied through the prescribed displacements at the top of themesh [39]. A special contact elements of program PAK [42] are employed for modelling ofcontact between the plate and the material.



Figure 7. Void volume fraction distribution for ey(log) = 0:3: (a) radial distribution ofvoid volume fraction f across minimum section; and (b) axial distribution of void

fraction f along axis of the notch specimen.

Table II. Unbalanced energy during equilibrium iterations(step 49).

Iteration Unbalanced energy Unbalanced force

1 1:03× 101 3.0932 1:66× 10−5 6:77× 10−53 1:10× 10−10 1:137× 10−6

Values for Young’s modulus and Poisson’s ratio are 207GPa and 0.3, respectively. Theuniaxial stress–plastic strain dependence is described by the formula [6]

�y302:

=(�y302:

+3G302:

�eP)0:1

assuming isotropic hardening. Material constants of the Gurson model are q1=1:5, q2 = 1,q3=1:5; the initial void volume fraction is f0 = 0:, the failure void fraction is ff = 0:279,the critical void volume fraction is fc = 0:15; the standard dimensionless deviation of �y issN =0:1, with the amplitude aN =0:04 and the mean equivalent plastic strain eN =0:3, as givenin Reference [6].The �nal end displacement of 6mm is reached in 120 equal load steps. The undeformed

and deformed meshes are shown in Figure 9(a) and (b). Figure 10 shows dependence of thevoid fraction on the reduction parameter ln(H=H0), where H is the current cylinder heightand the initial height H0 = 20mm, for four Gauss integration points of the edge elements inthe middle of the cylinder. The results obtained here agree well with those reported in Leeand Zhang [6].



Figure 8. Geometry, loading and boundary conditions for porous cylinder.

Figure 9. Undeformed mesh (a) and deformed mesh (b) of one-quarter of the cylinder.

4.4. Example 4. Uniaxial tension of a plate

A plate with dimensions, boundary conditions and loading is shown in Figure 11. The plateis modelled by one plane stress element [39], and it is extended uniaxially by imposing thedisplacement increments �u=0:025l0.The Young’s modulus and Poison’s ratio are 210GPa and 0.3, respectively. The yield curve

is given by the Ramberg–Osgood formula,

�y700:

=(�y700:

+3G700:

�eP)0:1

The material constants of the Gurson model are q1 = 1:5, q2 = 1, q3 = q21; the initial voidvolume fraction is f0 = 0:; the standard dimensionless deviation of �y is sN =0:1, with theamplitude aN =0:04 and the mean equivalent plastic strain eN =0:3, as given in Reference [28].



Figure 10. Void fraction vs reduction parameterln(H0=H) for four integration points.

Figure 11. Uniaxial tension plane element.

Figure 12. Uniaxial stress–strain curve. Figure 13. Porosity as a function ofuniaxial strain.

The numerical solutions for the axial stress � and the void volume fraction f, in termsof the logarithmic strain elog, is shown in Figures 12 and 13, respectively. Note that theappropriate correction of the plate thickness during the deformation is included. The resultsobtained here agree with those reported in Aravas [28].

5. CONCLUDING REMARKS

We have presented an e�cient stress integration scheme for the Gurson material, and alsoevaluation of the tangent constitutive matrix consistent with the stress calculation procedure.The usual e�ects relevant for the model description, such as the void volume fraction growth,nucleation and coalescence, are taken into account in the numerical algorithm. The procedureis extended to the large strain problems based on the multiplicative decomposition of thedeformation gradient and on the logarithmic strains.



The developed numerical procedure is applicable to general problems where the Gursonmodel is adequate for �nding the material response, including general 3D deformations (andplane strain and axisymmetric deformations as special cases) and the plane stress conditions.

APPENDIX A

A.1. Coe�cients a1 and a2 in Equation (49)

By di�erentiating t+�tf′y given by (32) with respect to �e

Pm, we obtain

@t+�tf′y

@(�ePm)=�yq1q2 sinh

(3q2�m2�y

)f∗;m − 1:5�yq1f∗Cm

�ycosh

(3q2�m2�y

)

+[q1q2f∗ @�y

@( �eP)sinh

(3q2�m2�y

)− 1:5�yq1f∗ @�y

@( �eP)�m�y

]��eP;m (A1)

from which the expressions for a1 and a2 are obvious.

A.2. Coe�cient km in Equation (52)

We di�erentiate the expression (1) for the yield condition t+�tfy = 0, and obtain

@t+�tfy@(�ePm)

=d1 + d2� �eP;m =0 (A2)

where

d1=2GS · S3(1−��a1=3)f′y;m

+ c1(1 + 2G��)

d2=−2GS · S��b2f′y;m

+ c2(1 + 2G��)

with

c1=13�2y

[2q1f∗

;m cosh(3q2�m�y

)− 3q1q2f∗ sinh

(3q2�m�y

)Cm�y

− 2q23f∗f∗;m

]

c2=−�2yq1q2f∗ sinh(3q2�m�y

)�m�2y

@�y@( �eP)

+@�y@( �eP)

2�y

[2q1f∗ cosh

(3q2�m�y

)− 1− (q3f∗)2

]3

In this di�erentiation we have used Equation (30). From (A2)

��eP;m =−km =−d1d2

(A3)

follows.



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Implicit stress integration procedure for small and large strains of the Gurson material model

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