Belytschko T. - Introduction (779635), страница 63

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for hyperelasticmaterials and for associated rate-independent plasticity based on the Kirchhoffstress; see Chapter 5. This tangent modulus tensor for non-associated rateindependent plasticity is not symmetric. We will also show in the following thatthe tangent modulus for associated plasticity based on the Jaumann rate of theCauchy stress also does not have major symmetry.6-30T.

Belytschko & B. Moran, Solution Methods, December 16, 1998An expression for the material tangent stiffness matrix is now derivedusing the general form (6.4.38) of the constitutive relation for a rate-independentmaterial. Specific examples of this relation and the associated tangent moduli aregiven at the end of this subsection. In this derivation, instead of immediatelychanging to Voigt notation, we will continue with indicial notation to the finalexpression and then translate that to Voigt notation.Substituting (6.4.38) into (6.4.32) givesmat ˙KIJijd Jj = ∫Ωο∂NI τC D dΩ∂xk kijl jl ο(6.4.39)Recall from (4.4.7b), that the rate of deformation tensor is the symmetric part ofthe spatial velocity gradient, Dkl = v( k, l) = sym (vkI NI , l ) .

Substituting this and therate form of the constitutive equation (6.4.38) into (6.4.39) we obtain˙KtanIJij d Jj ==τ∫Ωο N I ,k Ckijl v j, l dΩο=∫ΩοτN I , kCkijlNJ , lu˙ jJ dΩ οτ= ∫Ω NI , k CkijlN J , l dΩοu˙ jJ(6.4.41)οτwhere in the second of the above we have used the result Cτkijl Djl = Ckijlvj ,l whichτfollows the minor symmetry of the tangent modulus matrix, Cτkijl = Ckijl.(6.4.41), the material tangent stiffness matrix is defined bymatKIJij= ∫Ω NI , k CτkijlN J , ldΩοFrom(6.4.42)οThis expression can also be written as an integral over the current domain, i.e.,matKIJij= ∫Ω NI , kο1 τσTCkijl N J , ldΩ = ∫Ω NI , k CkijlN J ,ldΩJ(6.4.43)where we have used (????) to write the second expression.We now convert the above to Voigt notation.

Equation (6.4.43) is nowwritten asmatKIJrs= ∫Ω NA, kδ riCσkijlT N B ,lδ sj dΩ(6.4.46)Noting that Cτkijl = Cτikjl = Cτkilj , (6.4.46) and using (4.5.19b), the above can bewritten as6-31T. Belytschko & B. Moran, Solution Methods, December 16, 1998matKIJrs= ∫Ω BikAr1 τC B dΩJ kijl jlBs(6.4.47)which is given in matrix form asT σTK matBJ dΩ= ∫Ω J −1BTI Cτ BJ dΩIJ = ∫Ω B I C(6.4.49)\This form is identical to the linear stiffness matrix except that the materialresponse matrix J −1Cτ relates the convected rate of Kirchhoff stress to the rateof-deformation (or alternatively the response matrix Cσ T relates the Truesdellrate of Cauchy stress to the rate-of-deformation).Some examples of tangent moduli for different materials are given in thefollowing.

Detailed derivations of the tangent material stiffness matrices forspecific finite elements are given in Section 6.4.2 below.Tangent Modulus for Hyperelastic Material The rate form of theconstitutive relation for a hyperelastic material is given by (5.x), i.e,τ ∇c = CijklDklorσ ∇T = Cijkl Dkl(6.4.50)Thus from (6.4.50), the tangent modulus for a hyperelastic material is given byσTSECτijkl = JCijkl= Fim Fjn FkpFlqCmnpq(6.4.51)where from (5.y) Cmnpq is derived from the hyperelastic potential, i.e.,SECmnpq=∂Smn∂S∂ 2 W (C )= 2 mn = 2∂Epq∂Cpq∂Cmn∂Cpq(6.4.52)An interesting feature of the expression (6.4.51) is that for a hyperelasticmaterial the rate form of the material response is expressed naturally in terms ofthe convected rate of Kirchhoff stress.

The expression (6.4.51) contains nogeometric terms consisting of the product of the current stress the spatial velocitygradient (or its symmetric or antisymmetric parts)??????MORAN???. In manymaterials, (including the elastic-plastic material considered in the following) thetangent moduli are functions of the initial stresses.Tangent Modulus for Hypoelastic-Plastic Material We will nowdevelop the tangent modulus for hypoelastic-plastic materials based on i) theKirchhoff stress and ii) the Cauchy stress.

The elastic response is assumed to begiven by relating the Jaumann rate of stress to the elastic part of the rate ofdeformation tensor.i) Kirchhoff Stress Formulation Recalling the relation ( ) which relatesthe Jaumann rate of Kirchhoff stress to the rate of deformation tensor, we haveττ ij∇c = CijklDkl(6.4.53)6-32T. Belytschko & B.

Moran, Solution Methods, December 16, 1998where Cτijkl is the elastic-plastic tangent modulus given in (5.??) Using therelationship (5.???) between the convected rate and the Jaumann rate givesτ ij∇c = τij∇ J − Dik τ kj −τ ik Dkjep= CijklDkl − Dik τkj −τ ikDkj()ep= Cijkl− δilτ kj − τik δ jl Dkl(6.4.54)The terms involving the stress tensor are a result of the use of the Jaumann rate inthe hypoelastic relation and the difference between the Jaumann rate and theconvected rate which appears in the expression for the tangent stiffness matrix.These are traditionally interpreted as part of the material tangent stiffness matrixalthough they can also be regarded as geometric terms due to the convection ofthe stress.Because of the symmetry of Dkl , the last expression can be written as()1 ττ ij∇c =  Cijkl− δ ilτ kj + τik δ jl +δ ikτ lj +τ ilδ jk  Dkl2tan= CijklDkl(6.4.43)whereepCtanijkl = Cijkl −(1δ τ + τ δ +δ ikτ lj +τ ilδ jk2 il kj ik jl)(6.4.55)is the tangent modulus.

Note that this tangent modulus has major and minorsymmetries for an associated law, so the tangent stiffness is symmetric.ii) Cauchy Stress Formulation We will now develop the tangent stiffnessfor a hypoelastic-plastic material based on the Cauchy stress and we will showthat the resulting stiffness is not symmetric. The constitutive relation in elastoplasticity relates the Jaumann rate of the Cauchy stress to the rate-of-deformation:σJσ ij∇J = CijklDkl(6.4.56)where CσijklJ is the elastic-plastic tangent modulus.

The Jaumann rate is usedbecause the invariants of the Cauchy stress tensor remain constant when theJaumann rate vanishes, see section 5.??. Using the relationship (5.??) betweenthe Jaumann rates of Kirchhoff and Cauchy stresses, the convected rate is writtenasτ ij∇c = τij∇ J − Dikτ kj − τik Dkj6-33T. Belytschko & B. Moran, Solution Methods, December 16, 1998= Jσ ij∇ + Dkkτ ij − Dikτ kj − τ ik Dkj(6.4.57)Linearization with Directional DerivativesThree difficulties arise in applying the traditional Newton-Raphson method tosolid mechanics problems:1. the nodal forces are not continuously differentiable functions of thenodal displacements for materials such as elastic-plastic materials;2.

for path-dependent materials, the classical Newton method pollutes theconstitutive models since the intermediate solutions to the linearproblem in the iterative procedure, dν , are not part of the actual loadpath;3. for large incremental rotations and deformation, the linearizedincrements introduce a substantial errorIn order to overcome these difficulties, the Newton-Raphson method isoften modified as follows:1.

directional derivatives, also called Frechet derivatives, are used todevelop the tangent stiffness;2. a secant method is used instead of a tangent method and the lastconverged solution is used as the iteration point.3. formulas depending on increment size are used to relate the incrementsforces and displacements.It should be pointed out that the third difficulty only arises when the secantmethod is used to circumvent the second difficulty. If a tangent Newton methodis used for a smooth material, there is no advantage to carrying higher order termsin the geometric terms.To illustrate the need for directional derivatives with elastic-plasticmaterials, consider the following example.

The two-bar truss shown in Fig. 6.???has been loaded so that the stresses in both bars are compressive and equal, andboth bars are at the compressive yield stress. For simplicity, we consider onlymaterial nonlinearities and neglect geometric nonlinearities. If an arbitrary loadincrement ∆f1ext is now applied to node 1, the tangent stiffness matrix will dependon the incremental displacement ∆u1 because the derivatives of the internal nodaldisplacements depend on the direction of the displacement increment.

Theresidual is not a continuoiusly differentiable function of the incremental nodaldisplacments at this point, because the change in nodal internal forces depends onwhether the response of the either rod is elastic or plastic. In this case, there arefour lines of discontinuity for the derivatives, as shown in Figure ???. Theseresult from the fact that if the displacment increment results in a tensile strainincrement, then the bar unloads elastically, so the tangent modulus changes fromthe elastic modulus E to the plastic modulus Hp .The internal nodal forces in the current configuration are6-34T. Belytschko & B.

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