an orthotropic continuum model for the analysis of masonry structures

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an orthotropic continuum model for the analysis of masonry structures

Delft University of TechnologyFaculty of Civil EngineeringAN ORTHOTROPIC CONTINUUM MODEL FORTHE ANALYSIS OF MASONRY STRUCTURESAuthor : P. B. LOURENÇODate : June 1995TU-DELFT report no. 03-21-1-31-27TNO-BOUW report no. 95-NM-R0712TNO Building and Construction ResearchComputational Mechanics


SummaryA continuum model for the analysis of masonry structures subjected to in-plane loading is proposed.The model combines anisotropic elastic behaviour with anisotropic plastic behaviour.The proposed composite yield surface includes a Hill type failure criterion and a Rankine typefailure criterion. Different uniaxial strengths and post-peak behaviour are predicted by the modelalong the material axes both in tension and compression. The formulation of the elastoplasticalgorithm is made in modern plasticity concepts, including implicit Euler backward return mappingschemes and consistent tangent operators for all regimes of the model. The problem of localizationis tackled, in an engineering way, by the introduction of an equivalent length h relatedto the element size.The performance of the implementation and behaviour of the model is assessed by means of singleelement tests. A comparison between numerical results and experimental results available inthe literature shows good agreement both for ductile and brittle failure modes, providing that thesize of the structure is large enough to permit a macro-modelling strategy.AcknowledgementsThe author wishes to express his gratitude to Dr. ir. J.G. Rots and Dr. ir. P. Feenstra for their supportduring the course of this work.The financial support by the Netherlands Technology Foundation (STW) under grantDCT-33.3052 is gratefully acknowledged.The calculations have been carried out with the Finite Element Package DIANA of TNO Buildingand Construction Research on a Silicon Graphics Indigo R4000 workstation of the DelftUniversity of Technology.


TNO-95-NM-R0712 1995 11. INTRODUCTIONAn accurate analysis of masonry structures in a macro-modelling (or composite) perspectiverequires a material description for all stress states. The difÞculties are, however, quite strong. Thisis due, not only, to the fact that almost no comprehensive experimental results (including pre- andpost-peak behaviour) are available, but also to intrinsic difÞculties in the formulation of orthotropicinelastic behaviour. It is noted that a representation of an orthotropic yield surface in terms of principalstresses or stress invariants is not possible. For plane stress situations, which is the case of thepresent report, a graphical representation in terms of the full stress vector (σ x , σ y and τ xy ) is necessary.The material axes are assumed to be deÞned by the bed joints direction (x direction) and thehead joints direction (y direction). In some of the pictures shown below, the yield surface and theexperimental results are plotted in terms of principal stresses and a angle θ . The angle θ measuresthe rotation between the principal stress axes and the material axes. Clearly, different principalstress diagrams are found according to different values of θ .Tw o different strategies for the macro-modelling of masonry can be used, namely:¥ Extend the conventional formulation of isotropic quasi brittle materials in order to describeorthotropic behaviour. Current approaches consider different inelastic criteria for compressionand tension. A possible extension of conventional models is to use a Hill type yield criterionfor compression and a Rankine type à yield criterion for tension (see Fig. 1). Thisapproach will be discussed in this report;¥ Describe the material behaviour with a single yield criterion. The Hoffman yield criterion isquite ßexible and attractive to use, see Schellekens and de Borst (1990) and Scarpas andBlaauwendraad (1993), but yields a non-acceptable Þt of the masonry experimental values(see Fig. 2). A least squares Þt of the experimental results from Page (1981) with a Hoffmantype yield criterion turns out to show no tensile strength in uniaxial behaviour and a manualÞt through the different uniaxial strengths plus the compressive failure obtained upon loadingwith σ 1 = σ 2 and θ = 0 giv es a very poor representation of the diagrams for the other θ values.In fact a single surface Þt of the experimental values would lead to an extremely complexyield surface with a mixed hardening/softening rule in order to describe properly theinelastic behaviour. The author believes that this strategy is practically non-feasible.( ) The word type is used here because the original authors, see Hill (1948) and Hoffman (1967), assumed a3-dimensional formulation. The inßuence of the out-of-plane direction is generally unknown and will be notconsidered in the present report. The proposed yield surface for compression should in fact be considered as aparticular case of the complete quadratic formulation from Tsai and Wu (1971).(à)The word type is used here because the Rankine yield criterion represents the material strength along the maximumprincipal stress. For an anisotropic material such deÞnition is clearly not possible. The proposed yield surface fortension represents only a Þt of the experimental results.


2 1995 TNO-95-NM-R0712σ2σ2σ2σ 1σ 1σ 1σσ 2σ22θθσ1σ1σ 1θ =0° b) θ = 22.5° c) θ = 45.0°Fig. 1 - Comparison between a Hill type + Rankine type composite yield surfaceand experimental results from Page (1981)Through axes + biaxial compression fitBest fit with least squaresσ2σ 1σ 2σ1Through axes + biaxial compression fitBest fit with least squaresσ2σ 1θσ2σ1Through axes + biaxial compression fitBest fit with least squaresσ2σ 1θσ 2σ 1θ =0° b) θ = 22.5° c) θ = 45.0°Fig. 2 - Comparison between a Hoffman type single yield surfaceand experimental results from Page (1981)The composite yield surface to be presented features anisotropic behaviour in tension and compressionas well as non-isotropic softening. The formulation of the model is given in modern plasticityconcepts including fully implicit Euler backward return mapping, a local Newton-Raphson methodto solve the return mapping, proper handling of the corners and tangent operators consistent withthe integration of the update equations for all modes of the model, including the apex and the cornerregimes.The application of the model is limited, at present, to a plane stress conÞguration. The model isimplemented in the DIANA Þnite element package. For the purpose of compatibility with the currentcode and for simpler future extension of the model to a 3-dimensional stress-state, the formulationwill be given in 4 stress components (i.e. plane strain). The expansion/compression mechanismfrom 3 stress components to 4 stress components is described in de Borst (1991) and will not bereviewed here.Only a few authors tried to develop speciÞc macro-models for the analysis of masonry structures, inwhich anisotropic elasticity is combined with anisotropic inelastic behaviour. To the knowledge ofthe author of this report only Dhanasaker et al. (1985,1986) and Seim (1994) dealt with the implementationof a speciÞc numerical model for masonry. Both of these authors fail to include rationallysoftening in the model: brittle softening was included for tension, which leads to mesh sensitiveresults and numerical instabilities, and compressive softening was either absent or brittle.Moreover, the numerical analyses of masonry walls carried out by these authors included interfaceelements in the boundaries. This yields a weak assessment of the material macro-model because theinterface elements were responsible for most of the inelastic behaviour. Finally, the complex yield


TNO-95-NM-R0712 1995 3surfaces suggested by the above authors almost preclude the use of modern plasticity concepts. Themodel proposed by Dhanasaker et al. (1985,1986) for solid units masonry consists of three ellipticalcones, see Fig. 3. The model is based on the experimental Þndings of Page (1981) and Þts thedata extremely well, see Fig. 4. However, it is difÞcult to handle this model in modern plasticityconcepts. Not only the composite yield surface contains several corners and apexes but also theinclusion of ÒrealisticÓ inelastic behaviour is practically impossible - how to control the expansion/shrinkageof the yield surface?σyσxσy0.10.20.3σ xyσxFig. 3 - Yield surface for solid units masonry proposed by Dhanasekar (1986) with iso-shear stress linesContour spacing: 0.1 f mxσ2σ 1σ 2σ1σ2σ 1θσ2σ1σ2σ 1θσ 2σ 1θ =0° b) θ = 22.5° c) θ = 45.0°Fig. 4 - Comparison between DhanasekarÕs (1986) masonry yield surfaceand experimental results from Page (1981)The model proposed by Seim (1994) is based on the failure surface proposed by Ganz (1989) forhollow masonry units, see Fig. 5. The failure surface was derived from the theorems of Limit Analysis,assuming rigid-perfectly plastic behaviour for the masonry components. The assumptions forthe material behaviour of the components include a Mohr-Coulomb yield surface for the units and aCoulomb friction law for the joints. The head joints do not feature any strength. This yield surfaceis even more difÞcult to handle numerically in a consistent way, specially when softening isincluded in the model with different softening parameters for all regions. Seim (1994) assumedideally plastic behaviour in compression and brittle tension failure.


4 1995 TNO-95-NM-R07120.50.10.20.30.4σyσxσyσ xyσxFig. 5 - Yield surface for hollow units masonry proposed by Ganz (1989) with iso-shear stress linesContour spacing: 0.1 f mxFinally, it should be realised that a masonry macro-model always includes some degree of approximation.The basic features of a two-material composite cannot be reproduced but only smeared outin the continuum. The Þeld of applications of these models are indeed large structures where thestate of stress and strain across a macro-length can be assumed uniform. It is noted that, due to thedifÞculties of carrying out experiments in large structures, the examples used in the present reportfor the assessment of the model performance (extracted from available literature) are, in general,ÒsmallÓ for a macro-modelling strategy.The model proposed in the present report seems however capable of reproducing the globalbehaviour of the analysed structures. Satisfactory predictions of collapse loads are also found providedthat the structures do not show a highly localized failure mode. In such cases the interactionbetween units and mortar can be of capital importance and a micro-model, in which both masonrycomponents are modelled separately, should be used instead.


TNO-95-NM-R0712 1995 52. TENSION - A RANKINE TYPE ANISOTROPIC YIELD SURFACEIn this section a possible extension of the standard Rankine yield criterion to an orthotropic formulationis given. It is clear that the yield surface obtained cannot be derived from the materialstrength in the maximum principal stress direction. The proposed yield surface has to be regardedas pure curve Þtting from existing experimental results. The yield surface is coined as Rankine typeas the derivation is based on the original Rankine yield surface.The difÞculties of formulating the Rankine yield criterion in the principal stress state are addressed,for example, in Feenstra (1993). Consider a plane-stress situation in which the major principalstress σ 1 and the minor principal stress σ 2 are deÞned by means of a MohrÕs circle asσ 1, 2 = σ x + σ y22⎛σ ±√⎺⎺⎺⎺⎺x − σ y ⎞⎝ 2 ⎠+ τ 2 xy . (1)The hardening behaviour is assumed to be described by two internal variables κ 1 and κ 2 which governthe corresponding principal stresses. The yield functions are then given by the principal stressσ j and an equivalent stress σ j as a function of an internal variable κ j according to⎧⎪⎪⎪⎨⎪⎪⎪⎩f I = σ x + σ y2f II = σ x + σ y22⎛σ +√⎺⎺⎺⎺⎺x − σ y ⎞⎝ 2 ⎠2⎛σ +√⎺⎺⎺⎺⎺x − σ y ⎞⎝ 2 ⎠+ τ 2 xy − σ I (κ I )+ τ 2 xy − σ II (κ II )The yield function is depicted in Fig. 6 in the principal stress space. The problem which occurswith the two yield functions is that the transformation between the stress space and the principalstress space has to be deÞned uniquely for two yield functions with different hardening models..(2)σ 2σ 1Fig. 6 - Rankine yield surface in the principal stress spaceIn Feenstra (1993), the formulation is given by a single function which is governed by the Þrst principalstress and one equivalent stress which describes the softening behaviour of the material. Theassumption of isotropic softening is not completely valid for a material such as concrete or masonrywhich can be loaded to the tensile strength even if in the perpendicular direction the strength has


6 1995 TNO-95-NM-R0712been reduced due to softening of the material. This problem is partially solved in Feenstra (1993)by using kinematic softening such that the yield surface is shifted in the direction of the Þrst principalstress. It is noted that the above formulation of kinematic softening is also not quite realistic: letus assume that the material is loaded initially along a certain direction until softening is completed.If now the material is loaded in a direction orthogonal to the crack previously open, ideal plasticbehaviour is found. This is due to the fact that all the fracture energy has been consumed during theopening of the Þrst crack. An elegant solution is found if two independent softening parameterscontrol the shifting of the yield surface. Such a formulation for the Rankine yield surface is givenin Louren•o et al. (1995) and reproduces exactly the material feature in tension just described. It isshown by Louren•o et al. (1995) that the response of the model seems to lie between the Þxed androtating crack models and, therefore, comprises the beneÞt of a model with memory and a ßexibleshear response. Unfortunately, for certain values of the trial stress the return mapping becomes illposedand an almost singular Jacobian is found close to the solution. This precludes the use of sucha yield surface in large scale computations due to the lack of robustness and need to considerextremely small steps.Here a different approach is used aiming at an orthotropic Rankine type yield surface controlled byonly one scalar that measures the amount of softening simultaneously in the two material axes but,still, two corresponding different fracture energies are considered. This approach is less attractivefrom a physical point of view but leads to a more robust algorithm and should be preferred in practice.The Rankine yield surface reads, cf. eqs (1,2),f 1 = σ x + σ y22⎛σ +√⎺⎺⎺⎺⎺x − σ y ⎞⎝ 2 ⎠This expression can however be rewritten asf 1 = (σ x − σ t (κ t )) + (σ y − σ t (κ t ))2+ τ 2 xy − σ t (κ t ) . (3)2⎛ (σ +√⎺⎺⎺⎺⎺⎺⎺⎺⎺⎺x − σ t (κ t )) − (σ y − σ t (κ t )) ⎞⎝2⎠+ τ 2 xy , (4)where coupling exists between the stress components and the equivalent stress. Setting forth aRankine type yield surface for an orthotropic material, with different tensile strengths along the x, ydirections, see Þg. 7, is now straightforward if eq. (4) is modiÞed tof 1 = (σ x − σ t1 (κ t )) + (σ y − σ t2 (κ t ))22⎛ (σ +√⎺⎺⎺⎺⎺⎺⎺⎺⎺⎺⎺x − σ t1 (κ t )) − (σ y − σ t2 (κ t )) ⎞⎝2⎠+ ατ 2 xy . (5)Note that a parameter α is introduced to calibrate the shear strength. The parameter α readsα = f tx f tyτ 2 u, (6)where f tx , f ty and τ u are, respectively, the uniaxial tensile strengths in the x and y directions and thepure shear strength. Note that the material axes are now Þxed and it shall be assumed that allstresses and strains for the elastoplastic algorithm are given in the material reference axes.


TNO-95-NM-R0712 1995 7τxyτ uσxftxftyσyFig. 7 - Orthotropic Rankine type yield surface (plotted for τ xy ≥ 0)Eq. (5) can be recast in a matrix form asf 1 = ( 1 / 2 ξ T P t ξ ) 1 / 2+ 1 / 2 π T ξ , (7)where the projection matrix P t reads⎡1/ 2 − 1 / 2 0 0 ⎤⎢ −P t =1 / 21/ 2 0 0 ⎥⎢⎥ , (8)⎢0 0 0 0⎥⎣ 0 0 0 2α ⎦the projection vector π readsπ = { 1, 1, 0, 0 } T , (9)the reduced stress vector ξ readsξ = σ − η (10)and the back stress vector η readsη = { σ t1 (κ t ), σ t2 (κ t ), 0, 0 } T . (11)Exponential tensile softening is considered for both equivalent stress-equivalent strain diagrams,with different fracture energies (G fx and G fy ) for the yield values, and readsσ t1 = f tx exp ⎛ ⎝ − h f tx ⎞κG tfx ⎠andσ t2 = f ty exp ⎛ ⎝ − h f tyG fyκ t⎞⎠ . (12)Here, the scalar κ t controls the amount of softening and the equivalent length h, see Baºant and Oh(1983), is assumed to be related to the area of an element A e by, see Feenstra (1993),⎛h = α h √⎺⎺A e = α h ⎜⎝n ξΣ n ηξ = 1 η = 11 /2⎞Σ det(J) w ξ w η ⎟ , (13)⎠in which w ξ and w η are the weight factors of the Gaussian integration rule as it is tacitly assumed


8 1995 TNO-95-NM-R0712that the elements are always integrated numerically. The local, isoparametric coordinates of theintegration points are given by ξ and η. The factor α h is a modiÞcation factor which is equal to onefor quadratic elements and equal to √⎺2 for linear elements, see Rots(1988). With this approach theresults which are obtained in the analyses are reasonably objective with regard to mesh reÞnement.It is however possible that the equivalent length of an element results in a snap-back at the constitutivemodel if the element size is large. Then, the concept of fracture energy which has beenassumed is no longer satisÞed. In such a case, the strength limit has to be reduced in order toobtain an objective fracture energy by a sudden stress drop, resulting at a certain stage in brittlefailure, see Rots (1988). The condition of maximum equivalent length is given byh ≤ G f Eft2 , (14)where E is the YoungÕs modulus in the respective material axis. If the condition is violated, for anyof the material axes, the tensile strength in the respective axis is reduced according tof t = ⎛ 1G f E/2⎞. (15)⎝ h ⎠It is noted that eq. (15) yields a reduction on the material strength without any physical ground. Theidea is solely to obtain an energy release independent of the mesh size but this objective should beaccomplished by means of a mesh reÞnement and not with a strength reduction.Finally, for an orthotropic material with different yield values along the material axes, it wouldseem only natural to assume two different equivalent lengths along the material axes (this is ofcourse irrelevant in the special case of a mesh with square elements). However, the equivalentlength depends on so many factors that more complex assumptions are disregarded.The ßow rule is written in a standard fashion (non-associated softening) asúεúε p = ú∂g 1 λ t∂σ , (16)where the plastic potential g readsg 1 = ( 1 / 2 ξ T P g ξ ) 1 / 2+ 1 / 2 π T ξ (17)and the projection matrix P g is given by⎡1/ 2 − 1 / 2 0 0 ⎤⎢ −P g =1 / 21/ 2 0 0 ⎥⎥⎥⎦⎢. (18)⎢0 0 0 0⎣ 0 0 0 2α gThe parameter α g is taken equal to the unit value (Rankine plastic ßow), unless otherwise stated.The inelastic behaviour is described by a modiÞed strain softening hypothesis given byúκ t = úε p α = úε p x + úε p y2+ 1 / 2 (úε x√⎺⎺⎺⎺⎺⎺p − úε y) p 2 + 1 (úγ xy) p 2 . (19)α gIt is noted that the above expression represents a ÒmodiÞedÓ maximum principal plastic strain, inwhich the shear component has been averaged by the inverse of α g . The expression for the softeningscalar rate úκ t can be recast in a matrix form and reads


TNO-95-NM-R0712 1995 9whereúκ t = úε p α = ⎛ 1 /2⎝ 1 / 2 (úεúε p ) T Q úεúε p ⎞+⎠1 / 2 π T úεúε p , (20)⎡ 1/⎢ 2⎢ − 1 / 2Q = ⎢⎢ 0⎢ 0⎣− 1 / 21/ 20000000⎤⎥⎥⎥⎥⎥⎦0. (21)012α g2.1 Return mapping algorithm - Tension regimeThe integration of the constitutive equations given above is a problem of evolution that can beregarded as follows. At a stage n the total strain and plastic strain Þelds as well as the hardeningparameter (or equivalent plastic strain) are known:{ε n , ε p n, κ t, n } given data . (22)Note that the elastic strain and stress Þelds are regarded as dependent variables which can bealways be obtained from the basic variables through the relationsε e n = ε n − ε p n and σ n = D ε e n . (23)Therefore, the stress Þeld at a stage n+1is computed once the strain Þeld is known. The problemis strain driven in the sense that the total strain ε n+1 is trivially updated according to the exact formulaε n+1 = ε n +∆ε n+1 . (24)It remains to update the plastic strains and the hardening parameter. These quantities are determinedby integration of the ßow rule and hardening law over the step n → n+1. In the frame of afully implicit Euler backward integration algorithm this problem is transformed into a constrainedoptimization problem governed by discrete Kuhn-Tucker conditions as shown by Simo et al.(1988). It has been shown in different studies, e.g. Ortiz and Popov (1985) and Simo and Taylor(1986), that the implicit Euler backward algorithm is unconditionally stable and accurate. Thisalgorithm results in the following discrete set of equations:ε n+1 = ε n +∆ε n+1σ n+1 = σ trial −∆λ t, n+1 D ∂g ⎪1⎪∂σ⎪ n+1, (25)ε p n+1 = ε n p ∂g 1⎪+∆λ t, n+1 ⎪∂σ⎪ n+1κ t, n+1 = κ t, n +∆κ t, n+1in which ∆κ t, n+1 results from the integration of the rate equation, eq. (19), and the elastic predictorstep returns the value of the elastic trial stress σ trial asσ trial = σ n + D ∆ε n+1 . (26)The above equations must be satisÞed and simultaneously the yield criterion must be fulÞlled


10 1995 TNO-95-NM-R0712f 1 = ( 1 / 2 ξ T n+1P t ξ n+1 ) 1 / 2+ 1 / 2 π T ξ n+1 = 0 . (27)It is noted that the update of the softening scalar ∆κ t, n+1 reduces to the particularly simple expression∆κ t, n+1 =∆λ t, n+1 . (28)The above equations can be reduced to the following set of Þve equations containing 5 unknowns(σ n+1 and ∆κ t, n+1 =∆λ t, n+1 )⎧⎪D −1 (σ n+1 − σ trial ∂g) 1⎪+∆λ t, n+1 ⎪ = 0⎪∂σ⎪⎨n+1⎪f⎪ 1 = ( 1 / 2 ξ n+1P T t ξ n+1 ) 1 / 2+ 1 / 2 π T ξ n+1 = 0⎩Due to the coupling of the σ n+1 and κ t, n+1 values it is not possible to obtain an explicit one variablenon-linear equation. The system of non-linear equations is therefore solved with a regular Newton-Raphson method. The Jacobian necessary for this procedure reads (note that the subscript n+1 isdropped in the derivatives and matrices for convenience).(29)where⎡D⎢−1 ∂ 2 g 1+∆λ t, n+1∂σ 2J = ⎢ −−−−−−−−−−−−⎢T⎢ ⎛ ∂ f 1 ⎞⎣ ⎝ ∂σ ⎠|+|∂g 1∂σ +∆λ ∂ 2 g 1 ⎤t, n+1∂σ ∂κ t ⎥⎥⎥⎥⎦−−−−−−−−−− , (30)∂ f 1∂κ t∂ f 1∂σ = P t ξ n+12( 1 / 2 ξ T n+1P t ξ n+1 ) 1 / 2+ 1 / 2 π ;∂g 1∂σ = P g ξ n+12( 1 / 2 ξ T n+1P g ξ n+1 ) 1 / 2+ 1 / 2 π∂ 2 g 1∂σ 2 = P gP g ξ n+1 ξ n+1P T g2( 1 / 2 ξ n+1P T −g ξ n+1 ) 1 / 24( 1 / 2 ξ n+1P T .g ξ n+1 ) 3 / 2∂g 1∂κ t=− ⎛ ∂g 1 ⎞T ∂η;⎝ ∂σ ⎠ ∂κ t∂ 2 g 1=− ∂2 g 1∂σ ∂κ t ∂σ 2∂η∂κ t;∂η= ⎧ ∂σ t1⎨ , ∂σ t2,0,0 ⎫ T⎬∂κ t ⎩∂κ t ∂κ t ⎭(31)The numerical algorithm explained above is howev er not stable through all the stress domain. In theapex of the yield surface the gradient of the plastic potential, cf. eq. (31.1), is not deÞned. It is furthernoted that the proposed plastic potential can be written in a quadratic form asg 1 = 1 / 2 ξ T P g ξ + 1 / 4 ξ T π |ξ T π | . (32)However, this formulation does not overcome the problem of a non-deÞned gradient in the apex.The new expression for the gradient reads∂g 1∂σ = P gξ + 1 / 2 π |ξ T π | , (33)which degenerates to a point in the apex ( ∂g 1= 0) due to the singularity of yield surface.∂σFor the apex regime, the stress update, cf. eq. (25.2), in case of plane stress conÞguration, is independentof the trial stress. It is simply a return mapping to the apex and reads


TNO-95-NM-R0712 1995 11σ n+1 = η n+1 . (34)However, the present yield surface is implemented in the expansion/compression concept for planestrain (four stress components), see de Borst (1991). This concept yields several advantages formulti-purpose Þnite elements packages but for the present yield surface no advantage is found. Onthe contrary, eq. (34) cannot be used because the third normal stress component is not zero duringthe global/local iteration procedure. In this case (σ n+1 ) z must be calculated so that (∆ε p n+1 ) z = 0.This can be easily done fromand results in(∆ε p n+1 ) z = ⎛ ⎝ D−1 (σ trial − σ n+1 ) ⎞ ⎠ z= 0 (35)(σ n+1 ) z = (σ trial ) z + d−1 31 ⎛d −133⎝ σ trial ⎞− σ n+1+ d−1 32⎠ xd −133⎛⎝ σ trial ⎞− σ n+1 . (36)⎠ yHere the values d −1ij are terms from the compliance matrix D −1 . The others stress components aregiven from eq. (34) and read(σ n+1 ) x = σ t1 (κ t ) , (σ n+1 ) y = σ t2 (κ t ) and (τ n+1 ) xy = 0 . (37)The above expression can be advantageously recast in a matrix format and, after some manipulations,the stress update readsσ n+1 = A 1 η n+1 + A 2 σ trial , (38)where the auxiliary matrices A 1 and A 2 are given by⎡⎢⎢A 1 = ⎢⎢⎢⎣10− d−1 31d −133001− d−1 32d −133000100 ⎤⎥0 ⎥⎥0 ⎥⎥1 ⎦and A 2 =⎡⎢⎢⎢⎢⎢⎣00d −131d −133000d −132d −133000100 ⎤⎥0 ⎥⎥0 ⎥. (39)⎥0 ⎦The stress update given is sufÞcient to fulÞll f n+1 = 0. It remains to update the softening scalaraccording to eq. (20). For this purpose a non-linear equation in one variable can be written asF = F(∆κ t, n+1 ) =∆κ t, n+1 −∆ε p α ,n+1 = ⎛ 1 /2⎝ 1 / 2 (∆ε p n+1 )T Q∆ε p ⎞n+1+⎠1 / 2 π T ∆ε p n+1 = 0 , (40)where the increment of the plastic strain vector ∆ε p n+1can be calculated from∆ε p n+1 = D−1 (σ trial − σ n+1 ) (41)and the update of the stress vector σ n+1 is given by eq. (38). The secant method is used to solve thisnon-linear equation instead of the regular Newton method. This has proven robust and fast, seeLouren•o (1994) and the Appendix A.2.2 Consistent tangent operator - Tension regimeIn order to obtain quadratic convergence when making use of a Newton-Raphson iterative solvingprocedure at the structural level, a tangent operator consistent with the integration algorithm mustbe used, see Simo and Taylor (1985). For the standard part of the yield criterion, differentiation of


12 1995 TNO-95-NM-R0712the update equations and the consistency condition (d f 1, n+1 = 0) results inJ ⎧ dσ n+1⎫⎨ ⎬⎭ = ⎧ dε n+1⎫⎨ ⎬ . (42)⎩dλ t, n+1 ⎩0⎭Then, the consistent tangent operator is given byD ep = ∂σ ⎪⎪ = J −14×4 (43)∂ε⎪ n+1in which J −14x4 is the top-left 4×4 submatrix of the inverse of J. The consistent tangent operator canalso be written in other fashion by means of a condensation of the matrix J. Let us deÞne the modi-Þed compliance matrix H t and the modiÞed ßow direction vector γ t asH t = D −1 ∂ 2 g 1+∆λ t, n+1∂σ 2 and γ t = ∂g 1∂σ +∆λ ∂ 2 g 1t, n+1 . (44)∂σ ∂κ tCondensation of the Jacobian J and the Sherman-Morrison formula yield, after algebraic manipulation,D ep = ∂σH⎪−1 ⎛ ∂ f 1 ⎞⎪ = H −1t γ t⎝ ∂σ ⎠t −∂εT⎪ n+1 ⎛ ∂ f 1 ⎞⎝ ∂σ ⎠TH −1tH −1t γ t − ∂ f 1∂κ t. (45)The consistent tangent for the apex regime is obtained from differentiation of eqs. (38,40-41). Thisresults in the following system⎡I|⎢⎢ −−−−−−−−−−− +⎢T⎢ ⎛ ∂∆κ t ⎞⎣ ⎝ ∂ε p D −1 |⎠∂η−A⎤1∂κ t⎥−−−−−⎥⎥1 ⎥⎦⎧ dσ n+1⎫⎨ ⎬⎭ =dκ⎩ t, n+1⎧⎫⎪ A 2 D dε n+1 ⎪⎬⎪⎭⎨⎛∂∆κ t ⎞, (46)⎪Tdε⎩⎝ ∂ε p ⎠ n+1with∂∆κ t∂ε p = Q∆ε p n+12( 1 / 2 (∆ε p n+1 )T Q∆ε p n+1 )1 / 2+ 1 / 2 π . (47)Let us call the matrix above A. Then, the consistent tangent operator is given byD ep = A −14×4A 2 D + A −14×1⎛⎝T∂∆κ t ⎞∂ε p ⎠, (48)where the A −14×4 and A −14×1 are submatrices of the inverse of A. The consistent tangent operator canalso be written in other fashion by means of a condensation of the matrix A and readsD ep = ∂σ ⎪⎪ = ⎡ −1∂η ⎛ ∂∆κ⎢I + A t ⎞1∂ε⎪ n+1 ⎣∂κ t ⎝ ∂ε p D −1⎤ ⎡⎢⎣ ∂η ⎛ ∂∆κ t ⎞⎠⎥ A 1⎦∂κ t ⎝ ∂ε p ⎠TT+ A 2 D ⎤ ⎥⎦. (49)An investigation on the performance of the numerical implementation is given in Appendix A.


TNO-95-NM-R0712 1995 132.3 Features of the model - Tension regimeOne plane stress element with unit dimensions is loaded under different conditions in order to discussthe behaviour of the model. Before considering orthotropic material behaviour, a Þrst examplewith isotropic material behaviour is presented. This is extremely relevant for a comparison with differentmodels and an assessment of the assumptions made in the previous section.2.3.1 Isotropic material behaviour. Tension-shear model problemThe elementary problem proposed by Willam et al. (1987) introduces biaxial tension and shearloading in the element. This causes a continuous rotation of the principal strain axes after cracking,as it is typical of crack propagation in smeared Þnite element analysis. The element is subjected totensile straining in the x-direction accompanied by lateral Poisson contraction in the y-direction tosimulate uniaxial loading. Immediately after the tensile strength has been reached, the element isloaded in combined biaxial tension and shear strain, see Fig. 8. The ratio between the differentstrain components is given by ∆ε x : ∆ε y : ∆γ xy = 0. 5 : 0. 75 : 1. The material properties are givenin Table 1.Table 1 - Material properties (isotropic - α = 1.0)Material propertiesE 10000 N/mm 2ν 0.2f t 1.0 N/mm 2G f 0.00015 N.mm/mm 2The behaviour of different crack models for this problem can be found in Rots (1988). A comparisonwith different smeared ÒcrackingÓ formulations (total, tangential and Rankine plasticity) can befound in Feenstra (1993). The analyses from Rots (1988) of this problem with the multi-directionalcrack model showed that the shear response becomes softer with decreasing threshold angle, resultingin the limiting case of the rotating crack model with zero threshold angle as the most ßexibleresponse. The analyses from Feenstra (1993) showed that the Rankine model with kinematic hardeningis in very good agreement with the rotating crack model. A comparison between the proposedmodel and the commonly used smeared crack models is relevant to assess the adequacy ofthe model to describe cracking behaviour. The results for the different stress and strain componentsare depicted in Fig. 9-Fig. 11.ε yε y =-νε xε xγ xyyxγ xyε xa) Tension up to cracking b) Biaxial tension with shear after crackingFig. 8 - Tension-shear model problem


14 1995 TNO-95-NM-R0712Initial shear modulusProposed modelxy[MPa]τ0.1Fixed crack model( = 0.05) β0.0Rotating crack model-0.10.0 1.0 2.0-4γxy[10 ]Fig. 9 - Comparison of ÒcrackingÓ models.τ xy - γ xy response1.0σ x [MPa]0.80.60.4Fixed crack modelRotating crack modelProposed model0.20.00.0 1.0 2.0 3.0-4εx[10 ]Fig. 10 - Comparison of ÒcrackingÓ models.σ x - ε x response1.0Fixed crack modely [MPa]σ0.80.60.40.2Rotating crack modelProposed model0.0-1.0 0.0 1.0 2.0 3.0-4εy[10 ]Fig. 11 - Comparison of ÒcrackingÓ models.σ y - ε y responseThe main conclusions from the above results are:Ñ The shear stress-strain behaviour gives a good impression about the model because a ßexibleresponse is obtained. This is clearly in opposition with the Þxed crack model with constantshear retention factor but is close to the rotating crack model;


TNO-95-NM-R0712 1995 15Ñ The normal stress-strain response in the x-direction shows an implicit coupling between normalstress and shear stress. This is also in opposition with the Þxed crack model, where nocoupling is found, but also characterizes the rotating crack model, though to a less extent;Ñ The normal stress-strain response in the y-direction shows the implicit coupling between normalstresses. This is also in opposition with the Þxed crack model, where no coupling isfound, but also characterizes the rotating crack model, though to a less extent. The largeramount of coupling found in the proposed model is due to the isotropic softening.2.3.2 Orthotropic material behaviourThe orthotropic behaviour of the model is now discussed in a single element test under pure uniaxialtension. The material properties given in Table 2 are assumed, in which the y-direction is penalizedby a factor 2. Two different fracture energies are considered for the y-direction: G fx /2 and500 × G fx (almost ideally plastic behaviour).Table 2 - Material properties (orthotropic - α = 1.0)Material propertiesE x 10000 N/mm 2 E y 5000 N/mm 2ν xy 0.2 G xy 3000 N/mm 2f tx 1.0 N/mm 2 f ty 0.5 N/mm 2G fx 0.0002 N.mm/mm 2 G fyCase 1 Case 20.0001 N.mm/mm 2 0.1 N.mm/mm 2The values chosen for the material properties conÞrm the fact that completely different behaviouralong the two material axes can be reproduced, see Fig. 12. In the Þrst example isotropic softeningis considered. This means that the ratio of the material strength along the material axes is constant,see Fig. 13a, during any load history. It is important that this deÞnition is not confounded with thedeÞnition of isotropy used in damaged models. Isotropic softening is related to the current yieldstrength values and, not necessarily, to all the components the current stress vector. When all thefracture energy is exhausted a no-tension material is recovered, see Fig. 14a. In the second exampleideally plastic behaviour in the y-direction is considered. This means that the ratio of the materialstrength along the material axes ( f tx / f ty) tends to zero, see Fig. 13b. The yield surface is onlyallowed to shrink along the x-axis, see Fig. 14b.


16 1995 TNO-95-NM-R0712[MPa]1.00.80.6x-directiony-direction[MPa]1.00.80.6x-directiony-directionσσ0.40.40.20.20.00.00.0 1.0 2.0 3.0 4.00.0 1.0 2.0 3.0 4.0-4-4ε [10 ]ε [10 ]a) Case 1 (isotropic softening) b) Case 2 (ideally plastic behaviour in y-direction)Fig. 12 - Stress-strain response in uniaxial tension along the two material axes/ f tyf tx2.52.01.51.00.5/ f tyf tx2.52.01.51.00.50.00.00.0 2.0 4.0 6.0 8.00.0 2.0 4.0 6.0 8.0-4-4κ [10 ]κ [10 ]a) Case 1 (isotropic softening) b) Case 2 (ideally plastic behaviour in y-direction)Fig. 13 - Equivalent plastic strain vs. strength ratio along the material directions(tension regime)σ yInitial yieldsurfaceσ xσ yInitial yieldsurfaceσ xResidual yieldsurfaceResidual yieldsurfacea) Case 1 (isotropic softening) b) Case 2 (ideally plastic behaviour in y-direction)Fig. 14 - Trace of the yield surface in the plane τ xy = 0 (tension regime)


TNO-95-NM-R0712 1995 173. COMPRESSION - A HILL TYPE ANISOTROPIC YIELD SURFACEIn this section, a possible Þt of the experimental results in the compression regime is given, see alsoFig. 1. The Hill type yield criterion here introduced is capable of reproducing different behaviouralong two orthogonal material axes. The yield surface is coined Hill type because the formulation islimited to plane stress material properties. The properties in the out-of-plane direction are usuallyunknown and are not included in the model, in opposition to the original formulation from Hill(1948). The proposed yield surface should, in fact, be considered a particular form of the completequadratic formulation from Tsai and Wu (1971).τxyτ uσxfmxfmyσyFig. 15 - The Hill type yield surface (plotted for τ xy ≥ 0)The simplest yield surface that features different compressive strength along the material axes is arotated centered ellipsoid in the full plane stress vector (σ x , σ y and τ xy ), see Fig. 15. The expressionfor such a quadric isf 2 = Aσ 2 x + Bσ x σ y + Cσ 2 y + Dτ 2 xy − 1 = 0 , (50)where A, B, C and D are four material parameters such that B 2 − 4AC < 0, in order to ensure convexity.For the numerical implementation the yield surface will be recast in a square root matrixform and the variables will be rewritten in a more amenable way. Thus, the proposed yield surfaceis given byf 2 = ( 1 / 2 σ T P c σ ) 1 / 2− σ c (κ c ) , (51)where the projection matrix P c readsP c =⎡ 2 σ c2(κ c )⎢ σ c1 (κ c )⎢⎢β⎢⎢⎢⎢ 0⎢ 0⎣β2 σ c1(κ c )σ c2 (κ c )00the yield value σ c is given by00000 ⎤⎥⎥⎥0⎥⎥⎥0 ⎥2γ ⎥⎦, (52)


18 1995 TNO-95-NM-R0712σ c = √⎺⎺⎺⎺⎺ σ c1 σ c2 (53)and the single scalar κ c controls the amount of hardening and softening. The current yield stressvalues along the materials axes (σ c1 (κ c )and σ c2 (κ c )) follow the inelastic law giv en below as a functionof the material strength along the material axes (respectively f mx and f my as shown in Fig. 15).It is noted that the β and γ values introduced in eq. (52) are additional material parameters thatdetermine the shape of the yield surface. The parameter β controls the coupling between the normalstress values and must be obtained from one additional experimental test, e.g. biaxial compressionwith a unit ratio between principal stresses. If this test is used to obtain the parameter β , thecollapse load under biaxial compression (σ x = σ y =− f 45° and τ xy = 0) leads toβ = ⎡ ⎢⎣1f 2 45°− 1f 2 mx− 1f 2 my⎤⎥ f mx f my . (54)⎦The parameter γ , which controls the coupling between the normal stress values and the shearstrength, can be obtained fromγ = f mx f myτ 2 u, (55)where τ u is the material pure shear strength.Parabolic hardening followed by parabolic/exponential softening is considered for both equivalentstress-equivalent strain diagrams, with different compressive fracture energy (G fcx and G fcy ) alongthe material axes. The problem of mesh objectivity of the analyses of strain softening materials is awell debated issue, at least for tensile behaviour. Due to localization of deformation in a single elementor row of elements the stress-strain diagram must be adjusted according to a characteristiclength h to provide an objective energy dissipation. Here, the same expression for h is used as forthe tension regime even if it is recognized that tensile fracture is a surface driven process and compressivefailure is a volume driven process. The inelastic law shown in Fig. 16 features hardening,softening and a residual plateau of ideally plastic behaviour. The compressive fracture energy G fc(shaded area in Fig. 16) is deÞned only as the non-local contribution of the stress-strain diagram.The basis for the present deÞnition is only numerical, so that objective analyses with regard tomesh reÞnement are obtained, see also Section 3.4. Howev er some experimental evidence exists ona local and non-local component for the total compressive fracture energy, see Vonk (1992). Thepeak strength value is assumed to be reached simultaneously on both materials axes, i.e. isotropichardening, followed by anisotropic softening as determined by the different fracture energies. Aresidual strength value is considered to avoid a cumbersome code (precluding the case when thecompressive mode falls completely inside the tension mode) and a more robust code (precludingdegeneration of the yield surface to a point).


TNO-95-NM-R0712 1995 19σ cσ pσ mσ iσ I (κ c )σ II (κ c )G fcσ III (κ c )σ I (κ c ) = σ i + (σ p − σ i)√⎺⎺⎺2κ cκ p− κ 2 cκ 2 pσ II (κ c ) = σ p + (σ m − σ p ) ⎛ 2κ c − κ p ⎞⎝κ m − κ p ⎠σ III (κ c ) = σ r + (σ m − σ r )exp ⎛ ⎝ m κ c − κ m ⎞σ m − σ r ⎠σ rκ p κ m κ cwith m = 2 σ m − σ pκ m − κ pFig. 16 - Hardening/softening law for compressionFor practical reasons, it is assumed that all the stress values for the inelastic law are determinedfrom the peak value σ p = f m as following: σ i = 1 / 3 f m , σ m = 1 / 2 f m and σ r = 1 / 10 f m . The equivalentplastic strain corresponding to the peak compressive strength, κ p , is assumed to be an additionalmaterial parameter. In the case that no experimental results are available, this material parametercan be calculated assuming a total peak stress equal to 2 × 10 −3 . Then, in order to obtain a meshindependent energy dissipation the parameter κ m is given byκ m = 7567G fch f m+ κ p . (56)To avoid a possible snap-back at constitutive lev el, the conditionκ m ≥ f mE + κ p (57)must be fulÞlled. Otherwise, the strength limit, in order to obtain an objective fracture energy, isreduced tof m = ⎛ 67⎝ 751G fc E/2⎞. (58)h ⎠The ßow rule is written in a standard fashion (associated softening) asúεúε p = ú∂ f 2 λ c∂σ . (59)The inelastic behaviour is described by a work hardening hypothesis given byúκ c = 1 σ T úεúε p = ú λ c . (60)σ c3.1 Return mapping algorithm - Compression regimeThe return mapping algorithm in the frame of a implicit Euler backward integration scheme isgiven in Section 2.1 and, for the present yield surface, results in the following set of Þve equationscontaining 5 unknowns (σ n+1 and ∆κ c, n+1 =∆λ c, n+1 ), cf. eqs. (29),


20 1995 TNO-95-NM-R0712⎧⎪D −1 (σ n+1 − σ trial ∂ f) 2⎪+∆λ c, n+1 ⎪ = 0⎪∂σ⎪⎨n+1⎪f⎪ 2 = ( 1 / 2 σ n+1P T c, n+1 σ n+1 ) 1 / 2− σ c, n+1 = 0⎩This set of equations can be reduced to one non-linear equation, namely f 2 (∆λ c, n+1 ) = 0, if thestress update is manipulated to obtain.σ n+1 = ⎡ ⎢⎣I + ∆λ c, n+12σ c, n+1DP c, n+1⎤⎥⎦−1σ trialn+1 . (62)This approach will, however, not be followed here so that a constant framework is obtained for theseveral modes of the composite yield surface. Alternatively, the system of 5 non-linear equations issolved with a regular Newton-Raphson method. The Jacobian necessary for this procedure reads(note that the subscript n+1 is dropped in the derivatives and matrices for convenience)(61)where⎡D⎢−1 ∂ 2 f 2+∆λ c, n+1∂σ 2J = ⎢ −−−−−−−−−−−−⎢T⎢ ⎛ ∂ f 2 ⎞⎣ ⎝ ∂σ ⎠∂ f 2∂σ = P c σ n+12( 1 / 2 σ T n+1 P cσ n+1 ) 1 / 2;|+|∂ f 2∂σ +∆λ ∂ 2 f 2 ⎤c, n+1∂σ ∂κ c ⎥⎥⎥⎥⎦−−−−−−−−−− , (63)∂ f 2∂κ c∂ 2 f 2∂σ 2 = P c2( 1 / 2 σ T n+1 P cσ n+1 ) 1 / 2−P c σ n+1 σ T n+1P c4( 1 / 2 σ T n+1 P cσ n+1 ) 3 / 2∂ f 2∂κ c=σ n+1T ∂P cσ n+1∂κ c4( 1 / 2 σn+1 T P −cσ n+1 ) 1 / 2∂σ c1∂κ cσ c2, n+1 + ∂σ c2∂κ cσ c1, n+12σ c, n+1.∂ 2 f 2∂σ ∂κ c=∂P cσ n+1 (σ T ∂P cn+1 σ n+1 )P c σ n+1∂κ c ∂κ2( 1 / 2 σn+1 T P −ccσ n+1 ) 1 / 28( 1 / 2 σn+1 T P cσ n+1 ) 1 / 2(64)⎧ ⎡ ∂σ c2σ∂P c⎪ ⎢ c1, n+1 − ∂σ c1∂κ= diag⎨2c⎢∂κ c ⎪ ⎢σc1, 2 n+1⎩ ⎣⎤ ⎡σ c2, n+1∂κ c⎥ ⎢⎥,2⎢⎥ ⎢⎦ ⎣3.2 Consistent tangent operator - Compression regime∂σ c1∂κ cσ c2, n+1 − ∂σ c2σ 2 c2, n+1⎤ ⎫σ c1, n+1∂κ c⎥ ⎪⎥,0,0⎬⎥ ⎪⎦ ⎭Differentiation of the update equations and the consistency condition (d f 2, n+1 = 0) results, afteralgebraic manipulation, in


TNO-95-NM-R0712 1995 21[MPa]-10.0-8.0-6.0D ep = dσH⎪−1 ⎛ ∂ f 2 ⎞⎪ = H −1c γ c⎝ ∂σ ⎠c −dεT⎪ n+1 ⎛ ∂ f 2 ⎞⎝ ∂σ ⎠TH −1cH −1c γ c − ∂ f 2∂κ c, (65)where the modiÞed compliance matrix H c and the modiÞed ßow direction vector γ c readH c = D −1 ∂ 2 f 2+∆λ c, n+1∂σ 2 and γ c = ∂ f 2∂σ +∆λ ∂ 2 f 2c, n+1 . (66)∂σ ∂κ cAn investigation on the performance of the numerical implementation is given in Appendix A.3.3 Features of the model - Compression regimeOne plane stress element with unit dimensions is loaded under different conditions in order to discussthe behaviour of the model.The material properties given in Table 3 are assumed, in which the material strength and YoungÕsmodulus in the y-direction are penalized by a factor 2. Three different fracture energies are consideredfor the y-direction: 0. 3G fcx ,G fcx / 2 (isotropic softening) and 500 × G fcx (almost ideally plasticbehaviour).Table 3 - Material properties (β = -1.0, γ = 3.0 and κ p = 0.0005)Material propertiesE x 10000 N/mm 2 E y 5000 N/mm 2ν xy 0.2 G xy 3000 N/mm 2f mx 10.0 N/mm 2 f ty 5.0 N/mm 2G fcx 0.05 N.mm/mm 2 G fcyCase 1 Case 2 Case 30.015 N.mm/mm 2 0.025 N.mm/mm 2 7.5 N.mm/mm 2The orthotropic behaviour of the model is now discussed in a single element test under pure uniaxialcompression. The uniaxial stress-strain responses for the different cases considered are illustratedin Fig. 17. As shown in this picture, the model is capable of reproducing different behaviouralong the two material axes.x-directiony-direction[MPa]-10.0-8.0-6.0x-directiony-directionσ-4.0-2.0σ-4.0-2.00.00.00.0 -2.5 -5.0 -7.5 -10.0 0.0 -2.5 -5.0 -7.5 -10.0-3-3ε [10 ]ε [10 ]a) Case 1 b) Case 2 (isotropic softening)Fig. 17 - Stress-strain response in uniaxial compression along the two material axes (cont.)


22 1995 TNO-95-NM-R0712-10.0[MPa]-8.0-6.0x-directiony-directionσ-4.0-2.00.00.0 -2.5 -5.0 -7.5 -10.0-3ε [10 ]c) Case 3 (ideally plastic behaviour in y-direction)Fig. 17 - Stress-strain response in uniaxial compression along the two material axes (contd.)Further insight on the behaviour of the model can be obtained from Fig. 18 and Fig. 19. Fig. 18gives the ratio of the material strength along the material axes and Fig. 19 shows the trace of theyield surface in the plane τ xy = 0. In the Þrst case, the material strength in y-direction degradesfaster than the material strength in x-direction. In the second case, degradation of the materialstrength in both directions occurs with the same rate and isotropic softening is obtained. Finally, inthe third case, degradation only occurs in the material strength in x-direction.Note that, in the case of isotropic softening, the post-peak stress-strain diagrams under uniaxialloading conditions along the two material axes, see Fig. 17b, is not scaled by a factor 2. This issolely due to the deÞnition of the softening scalar and the fact that the yield value σ c is not equal tothe uniaxial strength along each material axis. This also means that the deÞnition of the Òcompressivefracture energyÓ can be argued because a perfect equivalence to the stress-strain diagram is notobtained. This limitation of the model can be solved by e.g. if a unit norm is used for the plasticßow vector and a strain softening hypothesis is used for the softening scalar. The additional difÞcultyintroduced in the formulation ( úκ c ≠ ú λ c ) is not particularly difÞcult to solve but, with the notoriouslack of experimental results on the material behaviour, the initial assumptions are kept inorder to simplify the implementation.8.08.0/ f myf mx6.04.0/ f myf mx6.04.02.02.00.00.00.0 5.0 10.0 15.0 20.0 0.0 5.0 10.0 15.0 20.0-3-3κ c [10 ]κ c [10 ]a) Case 1 b) Case 2 (isotropic softening)Fig. 18 - Equivalent plastic strain vs. strength ratio along the material directions(compression regime) (cont.)


TNO-95-NM-R0712 1995 238.0/ f myf mx6.04.02.00.00.0 5.0 10.0 15.0 20.0-3κ c [10 ]c) Case 3 (ideally plastic behaviour in y-direction)Fig. 18 - Equivalent plastic strain vs. strength ratio along the material directions(compression regime) (contd.)Residual yield surfaceσyResidual yield surfaceσyPeak yield surfacePeak yield surfaceσxσxa) Case 1 b) Case 2 (isotropic softening)Residual yield surfaceσyPeak yield surfaceσxc) Case 3 (ideally plastic behaviour in y-direction)Fig. 19 - Trace of the yield surface in the plane τ xy = 0 (compression regime)3.4 About the definition of a mesh independent energy releaseFor strain-softening materials, the need to introduce an equivalent length h in the stress-strain diagramto obtain analyses which are objective with respect to mesh reÞnement is a well debated issue


24 1995 TNO-95-NM-R0712since the original work of Baºant and Oh (1983). As stated in Section 2., in the present work h isassumed to be related to the area of an element, cf. eq. (13). However, this approach is generallyused in engineering practice only for modelling tensile behaviour with linear elastic pre-peakbehaviour followed by inelastic softening until total degradation of strength.The constitutive relation shown in Fig. 16 features pre-peak hardening and a residual plateau.Clearly, the hardening branch of the constitutive relation is stable and should not be adjusted as afunction h but also the residual plateau is constant and independent of the h value. To demonstratethe veracity of the deÞnition of a mesh independent release of energy upon mesh reÞnement anexample of a simple bar loaded in uniaxial is given. The problem is similar to the well-known problemof a simple bar loaded in tension proposed by CrisÞeld (1982).Consider the bar shown in Fig. 20 which is divided in n elements with n = 10, 20 and 40 elements.The length of the bar is 50 mm and the tranversal section of the bar has unit dimensions(1. 0 mm 2 × 1. 0 mm 2 ). The compressive fracture energies are assumed to equal G fcx = 10. 0 N/mmand G fcy = 5. 0 N/mm. For the rest of the material properties the values used in the previous sectionare assumed. One element is slightly imperfect (10%) to trigger the localization: f cx = 9. 0 N/mm 2 ,f cy = 4. 5 N/mm 2 ,G fcx = 9. 0 N/mm and G fcy = 4. 5 N/mm. The other material parameters remainthe same.Imperfect elementFig. 20 - Simple bar with imperfect element loaded in compressionThe load-displacement response of the bar is depicted in Fig. 21a for the energy-based regularizationmethod (note that ÒdisplacementÓ is understood as the relative displacement between the endsof the bar). It can be observed that the response is totally independent from the number of elements.The response of the bar with a constitutive model which has not been modiÞed by the size of theÞnite element mesh, see Fig. 21b, shows a dramatic mesh-dependent behaviour in the post-peakresponse. The brittleness of the response increases with an increasing number of elements.10.010.0Load [MN]8.06.04.0n = 10, 20 and 40Load [MN]8.06.04.0n = 20n = 102.02.0n = 400.00.0 10.0 20.0 30.0 40.0 50.0-3Displacement [10 mm]0.00.0 10.0 20.0 30.0 40.0 50.0-3Displacement [10 mm]a) Energy-based regularization b) No regularizationFig. 21 - Load-displacement diagram for simple bar with imperfect element


TNO-95-NM-R0712 1995 254. A COMPOSITE YIELD SURFACE FOR MASONRYThe two yield surfaces detailed in the previous section are now combined in a composite yield surfaceas illustrated in Fig. 22.σyσxσy0.1 0.2 0.3σ xyσxFig. 22 - Proposed composite yield surface with iso-shear stress linesf tx = 1. 0, f ty = 0. 5, f mx = 10. 0, f my = 5. 0 [N/mm 2 ] − α = 1. 0, β =−1. 0, γ = 3. 0Contour spacing: 0.1 f mxA full description of the compressive and tensile parts of the composite yield surface are given inthe previous sections and will not be repeated here. Only the aspects relative to the corner regimewill be addressed in the present section.One of the most important issues of multi-surface plasticity is to deÞne the number of active yieldsurfaces. Simo et al. (1988) have proposed an algorithm in which the assumption is made that thenumber of active yield surfaces in the Þnal stress state is less than or equal to the number of activeyield surfaces in the trial stress state. This implies that it is not possible that a yield surface, whichis inactive in the trial state, becomes active during the return mapping. This is not valid for the proposedyield surface. Due to the small number of yield surfaces (two), a trial and error iterative procedureis used, in which the return mapping process is restarted if a non-admissible solution isfound, see Louren•o (1994) for a complete description. From the experience of the author, thisleads to a robust and always convergent algorithm. The disadvantage is that the return mappingalgorithm might have to be restarted before the correct solution is obtained.4.1 Return mapping algorithm - Corner regimeThe return mapping algorithm in the frame of a implicit Euler backward integration scheme isgiven in Section 2. for single surface plasticity. For multisurface plasticity, the most importantassumption is KoiterÕs (1953) generalisation of the plastic strain rate asúεúε p = ú∂g 1 λ t∂σ+ ú∂ f 2 λ c∂σ . (67)Note that no coupling is assumed between the compressive and tensile regimes.Upon algebraic manipulation, the return mapping algorithm for the corner regime results in the followingset of six equations containing 6 unknowns (σ n+1 , ∆κ t, n+1 =∆λ t, n+1 and ∆κ c, n+1 =∆λ c, n+1 ),cf. eqs. (29,61),


26 1995 TNO-95-NM-R0712⎧⎪D −1 (σ⎪ n+1 − σ trial ∂g) 1⎪∂ f 2⎪+∆λ t, n+1 ⎪ +∆λ c, n+1 ⎪ = 0∂σ∂σ⎪⎪ n+1 ⎪ n+1⎨ f 1 = ( 1 / 2 ξ n+1P T t ξ n+1 ) 1 / 2+ 1 / 2 π T ξ n+1 = 0 .⎪⎪ f⎪ 2 = ( 1 / 2 σ n+1P T c, n+1 σ n+1 ) 1 / 2− σ c, n+1 = 0⎩This system of non-linear equations is solved with a regular Newton-Raphson method and the Jacobiannecessary for this procedure reads (note that the subscript n+1 is dropped in the derivativesand matrices for convenience)(68)⎡⎢D −1 ∂ 2 g 1+∆λ t, n+1∂σ⎢2 +∆λ ∂ 2 f 2c, n+1∂σ 2−−−−−−−−−−−−−−−−⎢TJ = ⎢⎛ ∂ f 1 ⎞⎢⎝ ∂σ ⎠⎢ −−−−−−−−−−−−−−−−⎢T⎢⎛ ∂ f 2 ⎞⎣⎝ ∂σ ⎠|+|+|∂g 1∂σ +∆λ ∂ 2 g 1t, n+1∂σ ∂κ t−−−−−−−−−−∂ f 1∂κ t−−−−−−−−−−0|+|+|∂ f 2∂σ +∆λ ∂ 2 f 2c, n+1∂σ ∂κ c−−−−−−−−−−⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦0 . (69)−−−−−−−−−−∂ f 2∂κ c4.2 Consistent tangent operator - Corner regimeDifferentiation of the update equations and the consistency conditions (d f 1, n+1 = 0 and(d f 2, n+1 = 0) results inJ⎧dσ⎫ ⎧⎪ n+1 dε⎫⎪⎬⎪⎭ ⎪ n+1⎪⎨dλ t, n+1 = ⎨ 0 ⎬⎪dλ c, n+1⎪ 0 ⎪⎩⎩ ⎭. (70)Then, the consistent tangent operator is given byD ep = ∂σ ⎪⎪ = (J −1 ) 4×4∂ε⎪ n+1, (71)in which (J −1 ) 4x4 is the top-left 4×4 submatrix of the inverse of J. The consistent tangent operatorcan also be written in other fashion by means of a condensation of the matrix J. The derivation ofan expression equivalent to eqs. (45,65) is given in Louren•o (1994).An investigation on the performance of the numerical implementation is given in Appendix A.


TNO-95-NM-R0712 1995 275. EXAMPLESIn the present section the masonry model proposed in this report is used to analyse different structuresfound in the literature.5.1 TU Eindhoven tests on shear wallsSeveral tests on masonry shear walls were carried out within the scope of the STRUCTURALMASONRY program from the Netherlands, see Raijmakers and Vermeltfoort (1992) and Vermeltfoortand Raijmakers (1993). In Louren•o (1994) some of the specimens are analysed with a micromodeland very good agreement is found with the experimental results. The parameters necessaryto characterize the micro-material model are available from micro-experiments but almost noexperimental data are available on the behaviour of the composite.The purpose of this report is not a sharp reproduction of the experimental results and the main concernof the author is to reproduce the behaviour observed in the experiments. For this purpose theexact knowledge of the macro-material parameters is not capital (and, in any case, it is not knownfor this particular structure).The Þrst shear wall analysed by Louren•o (1994) is used for comparison between the micro- andmacro-model. The specimen consists of a pier with a width/height ratio of one (990 × 1000 mm 2 ),built up with 18 layers (16 layers unrestrained) of Joosten solid clay bricks (dimensions 204 x 98 x50 mm 3 ) and 10 mm thick mortar (1:2:9, cement:lime:sand by volume). The pier was subjected toa vertical precompression force P = 30 kN before a horizontal load was monotonically increasedunder top displacement control d until failure (see Fig. 23). The cracking patterns are depicted inFig. 24 for the two tests carried out.Pda) Phase 1 - Vertical loading b) Phase 2 - Horizontal loadingFig. 23 - Loads for Vermeltfoort shear wall


28 1995 TNO-95-NM-R0712Fig. 24 - Experimental failure patterns for Vermeltfoort shear wallsFor the numerical analysis linear plane stress continuum elements (4-noded) with full Gauss integrationare utilised. A regular mesh of 15 × 15 elastoplastic elements is used. It is noted that the topand bottom layers of bricks are almost completely restrained and, therefore, two additional rows oflinear elastic elements are also included in the model. The same assumption is used for micromodellingand yield good results, see Louren•o (1994). The elastic properties of the composite areobtained from homogenisation, see Louren•o (1995), of the elastic properties of brick(E b = 16700 N/mm 2 and ν b = 0. 15) and mortar (E m = 780 N/mm 2 and ν m = 0. 125) and read:E x = 7520 N/mm 2 , E y = 3960 N/mm 2 , ν xy = 0. 090 and G xy = 1460 N/mm 2 . The only inelasticparameters available from experimental results are the vertical tensile strength f ty , the vertical stackbond compressive strength f my and the vertical fracture energy G fy . The rest of the material parametersare estimated and the full parameters list reads: f tx = 0. 35 N/mm 2 , f ty = 0. 25 N/mm 2 ,α = α g = 1. 0, G fx = 0. 05 N/mm, G fy = 0. 018 N/mm, f mx = 10. 0 N/mm 2 , f my = 8. 8 N/mm 2 ,β =−1. 0, γ = 2. 5, G fcx = 20. 0 N/mm, G fcy = 15. 0 N/mm and κ p = 0. 0012.The results of the numerical analysis are given in Fig. 25 to Fig. 29. The comparison betweenexperimental and numerical results, in terms of load-displacement diagrams, is given in Fig. 25.Fig. 25a shows the comparison between macro-, micro- and experimental results and Fig. 25bshows the agreement between the collapse load values. Clearly, the macro-model yields an enormousover-prediction of strength. The difference found cannot be explained solely by a possiblywrong estimation of the material parameters but must be due to some deÞciency of the model ormodelling strategy. A physical reason for this phenomena does exist and is given further below inthe text.The behaviour of the wall is depicted in Fig. 26 to 29 in terms of total deformed meshes, incrementaldeformed meshes, cracked and crushed Gauss points and principal stresses. Note that the cracksare plotted normal to the tensile principal plastic strain directions with a thickness proportional tothe tensile equivalent plastic strain (the lowest 5% values are discarded in order to obtain legiblepictures). Masonry crushing is represented by dotted triangles with a size proportional to the compressiveequivalent plastic strain (only the points in the post-peak regime are plotted, i.e. criticalpoints in the softening regime). Finally, note that the principal stresses in the elastic bricks are notshown. In the elastoplastic elements, the negative principal values are represented by a dotted lineand the positive principal values by a solid line.These pictures demonstrate that the behaviour of the structure is quite well captured by the model.Initially, two horizontal cracks develop at the top and bottom of the wall, see Fig. 26. Upon


TNO-95-NM-R0712 1995 29Horizontal force F [kN]100.075.050.025.0ExperimentalMicro-model, Lourenco (1994)Macro-model0.00.0 1.0 2.0 3.0 4.0 5.0Horizontal displacement d [mm]Horizontal force F [kN]200.0150.0100.050.0Experimental collapse loadMacro-model0.00.0 10.0 20.0 30.0Horizontal displacement d [mm]a) Vs. experimental result and micro-model prediction b) Full responseFig. 25 - Load-displacement diagram for Vermeltfoort shear wallincreasing deformation, a diagonal crack arises, see Fig. 27. The diagonal crack progresses in thedirection of the compressed corners of the wall, accompanied by crushing of the toes, see Fig. 28,until total degradation of strength in the compressive strut, see Fig. 29. This behaviour agrees wellwith Fig. 24 and the micro-analysis of Louren•o (1994).a) Total deformed mesh b) Incremental deformed meshc) Cracks and crushing d) Principal stresses σ min = -1.85 N/mm 2Fig. 26 - Numerical results for a horizontal displacement d = 0.5 mm. 15 × 15 elements


30 1995 TNO-95-NM-R0712a) Total deformed mesh b) Incremental deformed meshc) Cracks and crushing d) Principal stresses σ min = -3.52 N/mm 2Fig. 27 - Numerical results for a horizontal displacement d = 1.0 mm. 15 × 15 elementsa) Total deformed mesh b) Incremental deformed meshc) Cracks and crushing d) Principal stresses σ min = -10.1 N/mm 2Fig. 28 - Numerical results for a horizontal displacement d = 6.9 mm (peak). 15 × 15 elements


TNO-95-NM-R0712 1995 31a) Total deformed mesh b) Incremental deformed meshc) Cracks and crushing d) Principal stresses σ min = -9.65 N/mm 2Fig. 29 - Numerical results for a horizontal displacement d = 30.0 mm (ultimate). 15 × 15 elementsTw o other points of the performance of the model deserve special attention. A very good impressionabout the robustness of the model is obtained from the load-displacement diagram because it ispossible to follow the complete load path until total degradation of strength. Another importantissue is the dependency of the model upon mesh reÞnement. Fig. 30 shows the comparison betweenthe initial mesh and a 2 × Þner mesh. The results are, practically, mesh insensitive and thebehaviour encountered for the Þner mesh, see Fig. 31 and Fig. 32, shows no difference from thebehaviour of the coarser mesh.200.0Horizontal force F [kN]150.0100.050.030 x 30 elements15 x 15 elements0.00.0 10.0 20.0 30.0Horizontal displacement d [mm]Fig. 30 - Mesh dependency of the solution


32 1995 TNO-95-NM-R0712a) Total deformed mesh b) Incremental deformed meshc) Cracks and crushing d) Principal stresses σ min = -10.2 N/mm 2Fig. 31 - Numerical results for a horizontal displacement d = 6.9 mm (peak). 30 × 30 elementsa) Total deformed mesh b) Incremental deformed meshc) Cracks and crushing d) Principal stresses σ min = -10.5 N/mm 2Fig. 32 - Numerical results for a horizontal displacement d = 30.0 mm (ultimate). 30 × 30 elements


TNO-95-NM-R0712 1995 33The explanation for the over prediction of the failure load will now be detailed. Fig. 25a shows thatthe experimental load-displacement diagram is well reproduced until a horizontal displacement dequal to 1.0 mm, which coincides with the opening of the diagonal shear crack. After this point, itseems that the behaviour of the structure is not captured by the model. The crucial question iswhether the masonry macro-model here proposed is not acceptable or the modelling strategy isinadequate. The author believes the latter to be true.Fig. 24 clearly shows an extremely localized diagonal failure type in which all inelastic phenomenaconcentrate, basically, in a single stepped crack. A complete understanding of the behaviour of thestructure is obtained by the distribution of internal forces calculated by the micro-model, see Fig.33. Fig. 33a shows that, initially, the direction of the compressive strut is determined by the geometryof the bricks and deviates from the strut direction found in the macro-analysis. The discrepancybetween the micro- and macro-model gets only larger under increasing deformation. Fig. 33bshows that in reality two independent struts occur once the diagonal crack is fully open and a welldeÞned band completely unloads, with approximately the width of a brick at the center of the specimen.This phenomenon cannot be captured by a (smeared) macro-model unless some internallength scale, that reßects the masonry micro-structure, is incorporated in the model. This problemis outside the scope of the present report and, presently, no simple solution is envisaged. It is notedthat the task is quite complex because it is not clear yet if the above length scale is a geometricalproperty dependent on the unit size or a structural property that depends on the micro-structure,geometry of the structure and loading conditions.a) d = 1.0 mm d) d = 4.0 mmFig. 33 - Minimum principal stresses for a horizontal displacement d obtained by a micro-modelLouren•o (1994)For the sake of completeness a new analysis is carried out with the macro-model formulated insuch a way that degradation of tensile strength induces degradation of compressive strength, i.e. acoupled model. Coupling is introduced via an equation of the following type:


34 1995 TNO-95-NM-R0712(σ peak ) c = f mf tσ t , (72)Horizontal force F [kN]where the compressive yield value is made dependent of the current tensile yield value. This formulationleads to additional terms in the derivations of the numerical algorithm given in the previoussections but details about the coupled formulation will not be given here. By making use of Sections2. to 4., it is relatively straightforward to obtain the new algorithm. The results obtained withthe coupled model are given in Fig. 34 and Fig. 35. A much better agreement is found between thenew results and the experimental values. Fig. 34a shows that the collapse load is well predicted bythe model (note that the Þrst peak in the load-displacement diagram is obtained when the top andbottom horizontal bending cracks occur) and Fig. 35a shows that the expected two independentstruts can be reproduced. Therefore, for this particular structure, with a good estimate of the widthof the inactive shear band, it seems possible to reproduce the experimental results, at least to someextent. In fact, the coupled model introduces an internal length that is related to the element sizebecause the release of compressive fracture energy is coupled with tensile softening. This leads to aresponse totally inobjective with regard to the mesh size, see Fig. 34b and Fig. 35b. Upon meshreÞnement the solution in terms of load-displacement diagram converges to the uncoupled modeland, simultaneously, the width of the inactive shear band tends to zero.75.050.025.0ExperimentalIsotropic model - 8x8 elements0.00.0 1.0 2.0 3.0 4.0 5.0Horizontal displacement d [mm]Horizontal force F [kN]75.050.025.030 x 30 elements15 x 15 elements8 x 8 elementsExperimentalCoupled modelUncoupled model0.00.0 1.0 2.0 3.0 4.0 5.0Horizontal displacement d [mm]a) Vs. experimental results b) Mesh dependency of the solutionFig. 34 - Load-displacement diagram for Vermeltfoort shear wall (isotropic model)a) 8 × 8 elements b) 30 × 30 elementsFig. 35 - Principal stresses at peak for isotropic model


TNO-95-NM-R0712 1995 355.2 ETH Zurich tests on shear wallsLarge shear wall tests made with hollow clay bricks were carried out at ETH Zurich, Switzerlandby Ganz and ThŸrlimann (1984). These experiments are very well suited for validation of themodel presented here not only because they are large and feature well distributed cracking patternsbut also because experimental tests on the strength of the composite material are available, seeGanz and ThŸrlimann (1982). Fig. 36 shows the geometry of the specimens, which consist of amasonry panel of 3600 × 2000 × 150 mm 3 and two ßanges of 150 × 2000 × 600 mm 3 . Additionalboundary conditions are given by two concrete slabs placed in the top and bottom of the specimen.Initially, the wall is subjected to a vertical load p uniformly distributed over the length of the wall.This is followed by the application of a force F on the top slab along a horizontal displacement d.pd1501602000200 3600 2001806009001400Fig. 36 - Geometry for Ganz specimensFor the Þrst specimen analysed here (Wall W1) the resultant of the conÞning pressure p equals415 kN and for the second specimen (Wall W2) the resultant of the conÞning pressure p equals1287 kN. The Þrst specimen shows a very ductile response with tensile and shear failure along thediagonal stepped cracks, see Fig. 37a,b, and the second specimen shows a brittle failure with explosivepost-peak behaviour due to crushing of the compressed toes, see Fig. 37c,d. Note that thesepictures present a back view of the structure, i.e. the horizontal load shown is applied from left toright.


36 1995 TNO-95-NM-R0712a) Wall W1 - Peak b) Wall W1 - Ultimatec) Wall W2 - Peak d) Wall W2 - UltimateFig. 37 - Experimental failure patterns for Ganz walls(pictures rotated 90°)


TNO-95-NM-R0712 1995 37For the numerical analysis linear plane stress continuum elements (4-noded) and constant strain trianglesin a cross diagonal patch with full Gauss integration are utilised. A regular mesh of 24 × 154-noded elements is used for the panel and 2 × 15 cross diagonal patches of 3-noded triangles areused for each ßange. The properties of the composite material are obtained from Ganz andThŸrlimann (1982). The elastic properties read E x = 2460 N/mm 2 ,E y = 5460 N/mm 2 , ν xy = 0. 18and G xy = 1130 N/mm 2 . The inelastic parameters for the panel read f tx = 0. 05 N/mm 2 ,f ty = 0. 25 N/mm 2 , α = 1. 66, α g = 1. 0, G fx = 0. 02 N/mm, G fy = 0. 02 N/mm, f mx = 1. 87 N/mm 2 ,f my = 7. 61 N/mm 2 , β =−1. 05, γ = 1. 2, G fcx = 5. 0 N/mm, G fcy = 10. 0 N/mm and κ p = 0. 0008.The inelastic parameters that control the shape of the yield surface are obtained from a least squaresÞt from the experimental results but no data are available for the post-peak range. It is noted that, inthe ßanges, a stack bond masonry is used and different material properties must be considered. Theinelastic parameters for the ßanges read f tx = 0. 68 N/mm 2 , f ty = 0. 25 N/mm 2 , α = α g = 1. 0,G fx = 0. 05 N/mm, G fy = 0. 02 N/mm, f mx = 9. 5 N/mm 2 , f my = 7. 61 N/mm 2 , β =−1. 05,γ = 3. 0, G fcx = 10. 0 N/mm, G fcy = 10. 0 N/mm and κ p = 0. 0008. Finally, note that the selfweightof wall and top slab is also considered in the analysis.5.2.1 Wall W1The results of the numerical analysis are given in Fig. 38 to Fig. 44. The comparison betweenexperimental and numerical results, in terms of load-displacement diagrams, is given in Fig. 38. Inopposition to the Þrst example shown in the present report, good agreement is found between bothresults. This is due to the distributed nature of the process prior to collapse. The behaviour of thewall is depicted in Fig. 39 to 44 in terms of total deformed mesh, incremental deformed mesh,cracked Gauss points and minimum principal stresses contour. Note that the center node of thecrossed diagonal patch is not shown in the meshes to obtain a more legible picture. For the samereason the contour of minimal principal stresses is shown instead of the representation at eachGauss point. The comparison between experimental and numerical behaviour results is more difÞcultbut reasonable agreement seems to be found. Immediately after starting loading the structure,extensive diagonal cracking of the panel is found, see Fig. 39. Upon increasing deformation crackingtends to concentrate in a large shear band going from one corner of the specimen to the other,see Fig. 40 and Fig. 41. This is accompanied by ßexural cracking of the right ßange and, at a laterstage, also the left ßange, see Fig. 42 and Fig. 43. At ultimate stage, see Fig. 44, a well deÞned failuremechanism is formed with a Þnal shear band going from one corner of the specimen to theother and intersecting the ßanges. This means that cracks rotate signiÞcantly since initiation governedby MohrÕs circle to failure in a sort of shear band, which agrees extremely well with theexperiments (see Fig. 37a,b).300.0Horizontal force F [kN]200.0100.0ExperimentalNumerical0.00.0 4.0 8.0 12.0Horizontal displacement d [mm]Fig. 38 - Load-displacement diagram for wall W1 (low conÞning pressure)


38 1995 TNO-95-NM-R0712a) Total deformed mesh b) Incremental deformed mesh-2.40-2.05-1.70-1.36-1.01-0.67-0.320.03[MPa]c) Cracks d) Minimum principal stressesFig. 39 - Numerical results for a horizontal displacement d = 1.0 mm. Wall W1a) Total deformed mesh b) Incremental deformed mesh-3.94-3.37-2.79-2.22-1.64-1.06-0.490.09[MPa]c) Cracks d) Minimum principal stressesFig. 40 - Numerical results for a horizontal displacement d = 2.0 mm. Wall W1


TNO-95-NM-R0712 1995 39a) Total deformed mesh b) Incremental deformed mesh-5.43-4.64-3.85-3.06-2.27-1.49-0.700.09[MPa]c) Cracks d) Minimum principal stressesFig. 41 - Numerical results for a horizontal displacement d = 4.0 mm. Wall W1a) Total deformed mesh b) Incremental deformed mesh-6.48-5.53-4.58-3.64-2.69-1.75-0.800.14[MPa]c) Cracks d) Minimum principal stressesFig. 42 - Numerical results for a horizontal displacement d = 6.0 mm. Wall W1


40 1995 TNO-95-NM-R0712a) Total deformed mesh b) Incremental deformed mesh-6.86-5.86-4.86-3.87-2.87-1.87-0.870.12[MPa]c) Cracks d) Minimum principal stressesFig. 43 - Numerical results for a horizontal displacement d = 8.0 mm. Wall W1a) Total deformed mesh b) Incremental deformed mesh-7.00-5.98-4.95-3.93-2.90-1.87-0.850.18[MPa]c) Cracks d) Minimum principal stressesFig. 44 - Numerical results for a horizontal displacement d = 11.5 mm (ultimate). Wall W1


TNO-95-NM-R0712 1995 415.2.2 Wall W2The results of the numerical analysis are given in Fig. 45 to Fig. 51. Fig. 45 shows the comparisonbetween experimental and numerical results, in terms of load-displacement diagrams. Again, goodagreement is found between both results. The collapse load value found in the numerical analysis is20% higher than the experimental value but apart from this difference the same tendency in foundin both diagrams (note that the sharp reproduction of the collapse load value is not the main issuehere). As in the experiment, a brittle collapse is obtained shortly after the peak load due to compressivefailure of the specimen.The behaviour of the wall is depicted in Fig. 46 to 51 in terms of total deformed mesh, incrementaldeformed mesh, cracked Gauss points and minimum principal stresses contour. Again, reasonableagreement seems to be found between the experimental and numerical behaviour. Initially, extensivecracking leads to some separation between the left ßange and the panel while the rest of thestructure remains elastic, see Fig. 46. Shortly afterwards, three diagonal cracks occur, see Fig. 47.Upon mode I cracking of the right ßange, the previous diagonal cracks in the panel concentrate intwo well deÞned diagonal crack bands starting from each corner of the panel and progressing to theopposite side of the panel, see Fig. 48 and Fig. 49. This agrees well with the experimental crackingpatterns of Fig. 37c. Finally, crushing of the compressed bottom toe leads to the explosive type offailure shown in Fig. 50 and Fig. 51.600.0Horizontal force F [kN]400.0200.0ExperimentalNumerical0.00.0 2.0 4.0 6.0 8.0 10.0Horizontal displacement d [mm]Fig. 45 - Load-displacement diagram for wall W1 (high conÞning pressure)


42 1995 TNO-95-NM-R0712c) Cracks d) Minimum principal stressesFig. 46 - Numerical results for a horizontal displacement d = 1.0 mm. Wall W2a) Total deformed mesh b) Incremental deformed mesha) Total deformed mesh b) Incremental deformed meshc) Cracks d) Minimum principal stressesFig. 47 - Numerical results for a horizontal displacement d = 2.0 mm. Wall W2


TNO-95-NM-R0712 1995 43c) Cracks d) Minimum principal stressesFig. 48 - Numerical results for a horizontal displacement d = 4.0 mm. Wall W2a) Total deformed mesh b) Incremental deformed mesha) Total deformed mesh b) Incremental deformed meshc) Cracks d) Minimum principal stressesFig. 49 - Numerical results for a horizontal displacement d = 6.0 mm (peak). Wall W2


44 1995 TNO-95-NM-R0712c) Cracks d) Minimum principal stressesFig. 50 - Numerical results for a horizontal displacement d = 8.0 mm (before unloading). Wall W2a) Total deformed mesh b) Incremental deformed mesha) Total deformed mesh b) Incremental deformed mesh-10.7-9.19-7.64-6.09-4.55-3.00-1.450.10[MPa]c) Cracks d) Minimum principal stressesFig. 51 - Numerical results for a horizontal displacement d = 8.0 mm (ultimate). Wall W2


TNO-95-NM-R0712 1995 455.2.3 Comparison of the responses from specimens W1 and W2A comparison between the behaviour of both walls in terms of load-displacement diagrams isshown in Fig. 52. Wall W1, with a low conÞning pressure, exhibits a extremly ductile behaviourwhereas Wall W2, with a high conÞning pressure, exhibits a relatively small plateau followed bybrittle failure, which agree well with the experiments. This gives a good impression of the modelbecause both types of failure can be captured.600.0Wall W2Horizontal force F [kN]400.0200.0ExperimentalNumericalWall W10.00.0 5.0 10.0 15.0Horizontal displacement d [mm]Fig. 52 - Comparison of load-displacement diagrams for walls W1 and W2


46 1995 TNO-95-NM-R07126. CONCLUSIONSAn anisotropic plane stress continuum model for the analysis of masonry structures has been presented.The model combines orthotropic elasticity with orthotropic plasticity. A composite yieldsurface that includes a Hill type yield criterion for compression and a Rankine type yield criterionfor tension has been developed. The model is novel in the sense that a relatively simple compositeyield surface is proposed and hardening and softening are included in a rational way. Modern plasticityconcepts as unconditionally stable implicit Euler backward return mappings, local Newton-Raphson iterative procedures and consistent tangent operators are used for all the modes of themodel. The performance of the implementation is assessed by means of single element tests. It isshown that the numerical algorithm is robust and numerically efÞcient.The model predicts different yield strengths and fracture energies along the material axes, both intension and compression. These features deÞne an extremely ßexible model capable of accommodatingthe behaviour obtained from experiments. The behaviour of the model is detailed by meansof single element tests, in which the response is evaluated upon different choices of the materialparameters. Unfortunately the number of test results of the composite material is relatively scarceand almost non-existent in the post-peak regime. The setup of experimental programs coordinatedby experimentalists and analysts seem therefore crucial.A comparison between numerical results and experimental results available is also included. Previousattempts to use macro-models, speciÞcally developed for masonry structures, included interfaceelements at the boundaries. The interface elements were responsible for most of the non-linear phenomenaobserved and, thus, only a poor validation of previously proposed macro-models exists. Itis shown that the model proposed in the present report is able to predict well the behaviour ofmasonry structures, with both ductile and brittle failure modes, as well as sufÞciently accurate collapseload values. However, this statement is only true if the structure is sufÞciently large that amacro-modelling strategy can be applied.


TNO-95-NM-R0712 1995 477. REFERENCESde Borst, R., The zero-normal stress condition in plane-stress and shell elasto-plasticity, Comm. in.Appl. Numer. Methods, Vol. 7, pp. 29-33 (1991)de Borst, R. and Feenstra, P.H., Studies in anisotropic plasticity with reference to the Hill criterion,Int. J. Numer. Meth. Engrg., Vol.29, pp. 315-336 (1990)Baºant, Z.P. and Oh, B.H., Crack band theory for fracture of concrete, Materials and Structures,RILEM, Vol. 93(16), pp. 155-177 (1983)CrisÞeld, M.A., Accelerated solution techniques and concrete cracking, Comp. Meth. Appl. Mech.Engrg., Vol. 33, pp. 585-607 (1982)Dhanasekar, M., Kleeman, P.W. and Page, A.W., Biaxial stress-strain relations for brick masonry, J.Struc. Engrg., Vol. 111(5), pp. 1085-1100 (1985)Dhanasekar, M. and Page, A.W., The inßuence of brick masonry inÞll properties on the behaviourof inÞlled frames, Proc. Intsn. Civ. Engrs., Part 2, Vol.81, pp. 593-605 (1986)Feenstra, P.H., Computational aspects of biaxial stress in plain and reinforced concrete, Dissertation,Delft University of Technology, Delft, The Netherlands (1993)Ganz, H.-R., Failure criteria for masonry, 5 th Canadian Masonry Symposium, Vancouver, Canada,pp. 65-77 (1989)Ganz, H.-R. and ThŸrlimann, B., Experiments about the strength of biaxially loaded masonry panels,Report 7502-3, Institute of Structural Engineering, ETH Zurich, Switzerland (in German)(1982)Ganz, H.-R. and ThŸrlimann, B., Experiments of masonry walls under normal and shear loading,Report 7502-4, Institute of Structural Engineering, ETH Zurich, Switzerland (in German) (1984)Hill, R., A theory of the yielding and plastic ßow of anisotropic metals, Proc. Roy. Soc. (London)A, Vol. 193, pp. 281-288 (1948)Hoffman, O., The brittle strength of orthotropic materials, J. Comp. Mat., Vol. 1, pp. 200-206(1967)Koiter, W.T., Stress-strain relations, uniqueness and variational problems for elastic-plastic materialswith a singular yield surface, Q. Appl. Math., Vol. 11, pp. 350-354 (1953)Louren•o, P.B., Analysis of masonry structures with interface elements: Theory and applications,Delft University of Technology, Report 03-21-22-0-01, Delft, The Netherlands (1994)Louren•o, P.B., The elastoplastic homogenisation of masonry structures: With an extension tomasonry structures, Delft University of Technology, Report 03-21-1-31-02, Delft, The Netherlands(1995)Louren•o, P.B., Rots, J.G. and Feenstra, P.H., A ÕtensileÕ Rankine type orthotropic model formasonry, Proc. 3 rd Int. Symp. on Computer Methods in Structural Masonry, Lisbon, Portugal(1995)Ortiz, M. and Popov, E.P., Accuracy and stability of integration algorithms for elastoplastic constitutiverelations, Int. J. Numer. Methods Engrg., Vol. 21, pp. 1561-1576 (1985)Page, A.W., The biaxial compressive strength of brick masonry, Proc. Intsn. Civ. Engrs., Part 2,Vol. 71, pp. 893-906 (1981)


48 1995 TNO-95-NM-R0712Raijmakers, T.M.J. and Vermeltfoort, A.Th., Deformation controlled meso shear tests on masonrypiers, Report B-92-1156, TNO-BOUW/TU Eindhoven, Building and Construction Research, TheNetherlands (in Dutch) (1992)Rots, J.G., Computational modelling of concrete fracture, Dissertation, Delft University of Technology,Delft, The Netherlands (1988)Scarpas, A. and Blaauwendraad, J., Non-local plasticity softening model for brittle materials, in:Fr acture and Damage of Concrete and Rock - FDCR-2, Eds. H.P. Rossmanith, E & FN. Spon.,London, U.K., pp. 44-53 (1993)Schellekens, J.C.J. and de Borst, R., The use of the Hoffman yield criterion in Þnite element analysisof anisotropic composites, Comp. Struct., Vol. 37(6), pp. 1087-1096 (1990)Seim, W., Isotropic or anisotropic? Simulation of in-plane loaded masonry structures close to reality,10 th Int. Brick/Block Masonry Conf., Calgary, Canada, pp. 77-86 (1994)Simo, J.C. and Taylor, R.L., Consistent tangent operators for rate-independent elastoplasticity,Comp. Meth. Appl. Mech. Engrg., Vol. 48, pp. 101-118 (1985)Simo, J.C. and Taylor, R.L., A return mapping for plane stress elastoplasticity, Int. J. Numer. MethodsEngrg., Vol. 22, pp. 649-670 (1986)Simo, J.C., Kennedy, J.G. and Govindjee, S., Non-smooth multisurface plasticity and viscoplasticity.Loading/unloading conditions and numerical algorithms, Int. J. Numer. Methods Engrg., Vol.26, pp. 2161-2185 (1988)Tsai, S.W. and Wu, E.M., A general theory of strength of anisotropic materials, Comp. Materials,Vol. 5, pp. 58-80 (1971)Vermeltfoort, A.Th. and Raijmakers, T.M.J., Deformation controlled meso shear tests on masonrypiers, Part 2, Draft report, TU Eindhoven, The Netherlands (in Dutch) (1993)Vonk, R.A., Softening of concrete loaded in compression, Dissertation, Eindhoven University ofTechnology, Eindhoven, The Netherlands (1992)Willam, K.J., Pramono, E. and Sture, S., Fundamental issues of smeared cracked models, Proc.SEM/RILEM Int. Conf. Fracture of Concrete and Rock, eds. S.P.Shah and S.E. Swartz,Springler-Verlag, New-York, U.S.A., pp. 142-157 (1987)


TNO-95-NM-R0712 1995 49APPENDIX APERFORMANCE OF THE NUMERICAL IMPLEMENTATIONIn this appendix the performance of the numerical implementation is investigated.One plane stress element with unit dimensions is loaded under different conditions. The load isintroduced in the element under direct displacement control and a number of equally spaced plasticincrements of the displacement follow one elastic increment. If the direction of the plastic ßow isconstant, no dependence is found on the step size, i.e, the integration is exact because the algorithmis strain driven. Otherwise an extremely small (< 1%) dependence on the mesh size was found butno discussion will be given here about the accuracy of the return mapping algorithm. For this purposesee e.g. Ortiz and Popov (1985), de Borst and Feenstra (1990) and Schellekens and de Borst(1990). The tolerance for the local return mapping algorithm equals an absolute value of 1 × 10 −9for all residuals and the tolerance for the force norm of the structural response equals 1 × 10 −8 .Aregular Newton-Raphson is used for the global iterative procedure.A.1 TENSION REGIMEThe material properties given in Table A.1 are assumed.A.1.1 Uniaxial tensionTable A.1 - Material properties (isotropic - α = 1.0)Material propertiesE 10000 N/mm 2ν 0.2f t 1.0 N/mm 2G f 0.00015 N.mm/mm 2The element is subjected to uniaxial tension along the x axis as shown in Fig. A.1. The results areillustrated in Fig. A.2.1.00.8yxdx [MPa]σ0.60.40.2Fig. A.1 - Uniaxial tension test0.00.0 1.0 2.0 3.0 4.0-4εx[10 ]Fig. A.2 - Stress-strain response in uniaxial tension


50 1995 TNO-95-NM-R0712The performance of the model is given in Table A.2 to Table A.4. Quadratic convergence is indeedfound at local and global level.Table A.2 - Number of local iterations per global iteration for uniaxial tensionPlastic steps Av erage Maximum100 3.24 410 4.24 5A.1.2 General pathTable A.3 - Number of global iterations for uniaxial tensionPlastic steps Av erage Maximum100 2.08 310 2.40 3Table A.4 - Convergence for selected steps (uniaxial tension)100 plastic steps10 plastic stepsPlastic step1Force norm0. 453 × 10 00. 370 × 10 −30. 270 × 10 −8Plastic step1Force norm0. 371 × 10 00. 310 × 10 −30. 876 × 10 −8501000. 930 × 10 −40. 564 × 10 −110. 219 × 10 −40. 537 × 10 −135100. 406 × 10 −30. 703 × 10 −80. 853 × 10 −40. 886 × 10 −10The general path shown in Fig. A.3 is now considered. The response in terms of a forcedisplacementdiagram is given in Fig. A.4.0.51F, d1.5yxσ y = -1.0MPaF [MN]1.00.5Fig. A.3 - General test0.00.0 2.0 4.0 6.0-4d [10 mm]Fig. A.4 - Force-displacement response for general testThe performance of the model until peak (thicker line in Fig. A.4) is given in Table A.5 to TableA.7. After this point two integration points reach the apex and the apex algorithm is used. This caseis analysed independently in the next section. Note that quadratic convergence is indeed found.


TNO-95-NM-R0712 1995 51Plastic step150100A.1.3 ApexTable A.5 - Number of local iterations per global iteration for general path100 plastic stepsPlastic steps Av erage Maximum100 3.15 410 4.08 5Table A.6 - Number of global iterations for general pathPlastic steps Av erage Maximum100 3.01 410 3.80 5Table A.7 - Convergence for selected steps (general path)Force norm0. 212 × 10 20. 712 × 10 −10. 190 × 10 −40. 344 × 10 −120. 113 × 10 −20. 956 × 10 −70. 000 × 10 −∞0. 233 × 10 −30. 172 × 10 −60. 126 × 10 −12Plastic step151010 plastic stepsForce norm0. 816 × 10 1 0. 419 × 10 −50. 776 × 10 0 0. 407 × 10 −120. 134 × 10 −10. 103 × 10 10. 113 × 10 −30. 952 × 10 −80. 498 × 10 −20. 213 × 10 −30. 284 × 10 −60. 512 × 10 −12The apex algorithm performance is now evaluated. The test illustrated in Fig. A.5 is considered.The obtained force-displacement diagram is given in Fig. A.6.2.0F, d11F [MN]1.51.0yx0.5Fig. A.5 - Apex test0.00.0 2.0 4.0 6.0-4d [10 mm]Fig. A.6 - Force-displacement response for apex testThe performance of the model is given in Table A.8 to Table A.10. Quadratic convergence at globallevel is indeed found. At local level quadratic convergence is not found but supra-linear convergenceis found. As the apex is a well-deÞned area of the yield surface the number of local iterations


52 1995 TNO-95-NM-R0712is approximately the same as in the former cases. Note that the inconvenience of using the expansion/compressionmechanism, de Borst(1991), for a basically plane stress yield criterion is againevident. In a plane stress algorithm global convergence would be found in one step because all thedisplacements are prescribed whereas, with the expansion/compression mechanism, a non-zero outof-planenormal stress component must be reduced to zero.Table A.8 - Number of local iterations per global iteration for the apex algorithmPlastic steps Av erage Maximum100 3.40 410 4.32 5Table A.9 - Number of global iterations for the apex algorithmPlastic steps Av erage Maximum100 2.01 310 2.10 3Table A.10 - Convergence for selected steps (apex algorithm)100 plastic steps10 plastic stepsPlastic step1Force norm0. 515 × 10 00. 480 × 10 −50. 210 × 10 −11Plastic step1Force norm0. 516 × 10 00. 491 × 10 −40. 557 × 10 −1250100A.2 COMPRESSION REGIME0. 529 × 10 −50. 168 × 10 −120. 101 × 10 −50. 139 × 10 −135100. 693 × 10 −30. 500 × 10 −110. 120 × 10 −30. 112 × 10 −13The material properties given in Table A.11 are assumed, in which the material strength andYoungÕs modulus in they-direction are penalized by a factor 2. Three different fracture energiesare considered for the y-direction: 0. 3G fcx ,G fcx / 2 (isotropic softening) and 500 × G fcx (almostideally plastic behaviour).Table A.11 - Material properties (β = -1.0, γ = 3.0 and κ p = 0.0005)Material propertiesE x 10000 N/mm 2 E y 5000 N/mm 2ν xy 0.2 G xy 3000 N/mm 2f mx 10.0 N/mm 2 f ty 5.0 N/mm 2G fcx 0.05 N.mm/mm 2 G fcy 7.5 N.mm/mm 2A.2.1 Uniaxial compressionThe element is subjected to uniaxial compression along the x axis as shown in Fig. A.7. The resultsare illustrated in Fig. A.8.


TNO-95-NM-R0712 1995 53-10.0-8.0yxdx [MPa]σ-6.0-4.0-2.0Fig. A.7 - Uniaxial compression test0.00.0 -2.0 -4.0 -6.0 -8.0 -10.0-3εx[10 ]Fig. A.8 - Stress-strain response in uniaxial compressionThe performance of the model is given in Table A.12 to Table A.14. Note that, in the calculation ofthe maximum value of local iterations per global iteration, the Þrst plastic step is ignored (valuegiven inside brackets). The author believes that this value is misleading because the problem is illdeÞnedfor the Þrst plastic step, i.e. the derivative of the equivalent stress-equivalent strain diagramequals +∞ for κ c = 0. Note also that quadratic convergence is indeed found at local and globallevel.Table A.12 - Number of local iterations per global iteration for uniaxial compressionPlastic steps Av erage Maximum200 4.82 6 (10 )20 7.68 13First stepTable 13 - Number of global iterations for uniaxial compressionPlastic stepsAv e

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