Belytschko T. - Introduction (779635), страница 64
Текст из файла (страница 64)
Moran, Solution Methods, December 16, 1998int f x1 f y1 σ 0 cosθ −σ 0 cosθ = Aσ 0 sinθ + σ 0 sin θ (6.4.58)where σ 0 is the current yield stress; the above is obtained by assembling theinternal nodal forces for rod elements as given by Eq. (E.4.6.7). For each rodelement, there are two possibilities depending on the direction of the force: eitherthe rod continues to load witha plastic modulus, or it unloads with an elasticmodulus. As a result, the tangent stiffness in this configuration can take on fourdifferent values.1σ = −σ Yσ = −σ y122θθ3∆u x 2E1 = E2 = Eθ θE1 = H pE2 = EE1 = EE2 = H pθ θ∆u x1line of discontinuityfor directional derivativeE1 = E2 = H pFigure ??.
A two-bar truss in a state with both bars in compressive yield and the four quadrants ofdirectional derivatives.The nodal force f1x is shown as a function of the two components of thenodal displacement increment in Figure ???, where the discontinuity in thederivatives is clearly apparent. Obviously, a standard derivative cannot beevaluated since it has four different values.These regimes of the four different responses are illustrated in Fig.
6.??,which shows the four quadrants in the space of the components of the nodal6-35T. Belytschko & B. Moran, Solution Methods, December 16, 1998displacement. The tangent stiffness for dkisplacement increments in the fourquadrants is given by the following:in quadrant 1:K int =AE 2cos 2 θl 00 2sin 2 θ (6.4.59a)in quadrant 2:Kint()2A E + H p cos θ=l E − Hp sin θcos θ()( E − Hp ) sinθ cosθ ( E + Hp ) sin 2 θ (6.5.59b)in quadrant 3:KintAH p 2cos 2 θ0 =l 02sin 2 θ (6.5.59c)in quadrant 4:Kint()2A E + H p cos θ= l H p − E sin θcos θ()( H p − E) sinθ cosθ ( E + Hp ) sin 2 θ 6-36(6.5.59d)T.
Belytschko & B. Moran, Solution Methods, December 16, 1998Wequilibriumsolutiond2d1d2d1Figure 6.4. Schematic of potential energy, a stable equilibrium solution, and the contours for thepotential with their gradient -r.6-37T. Belytschko & B. Moran, Solution Methods, December 16, 1998To deal with this type of behavior in a methodical manner, a Frechetderivative, often called a directional derivative, must be used. The Frechetderivative is defined bydf (d)d= limf ( d+ ε∆d)dd[∆ d] ε → 0 dεε =0(6.4.60)The subcript on the lower term gives the direction in which the derivative is taken.The notation Df (d) ⋅[ ∆d] is often used for the directional derivative in the finiteelement literature..The value of directional derivative depends on the direction of theincrement of the independent variable.
The tangent stiffnesses in (??) are basedon directional derivatives for the nodal forces have been given in (???).In the computation of the tangent stiffness matrix, and in particular thematerial tangent stiffness, directional derivatives are used for elastic-plasticmaterials. The direction is based on the last displacement increment in theiterative procedure.
If the load increment suddenly reverses, the last increment isnot in the direction of the next solution increment, and the directional derivativemay be quite erroneous,. However, after several iterations, the correct direction isdetermined for the displacement increment and the directional derivative gives thecorrect rate of change of the nodal forces.Since the directional derivative to a specific value of the nodaldisplacements, this approach cannot be used with the standard tangent Newtondescribed in Box ???. Instead, a secant Newton method must be used. The secantNewton method is given in Box ???.External Load Stiffness.
An important class of loads are follower loads,which change with the configuration of the body. Examples of follower forces areshown in Figure ??. Pressure loading is a common example of a follower load.Since a pressure loading is always normal to the surface, as the surface moves, thenodal external forces change even if the pressure is constant. These effects areaccounted for in the Jacobian matrix K ext , which is also called the load stiffness.The load stiffness K ext is obtained by relatinng the time derivative (orincrement) of the external nodal forces to the time derivative (or increment) ofnodal displacements.
We first consider loading by pressure, p( x, t ) . Theexternal nodal forces on a surface of element e are given by letting t =− pn in Eq.(4.9.13):f Iext =− ∫ N I pndΓ(6.4.61)ΓLet the surface Γ be described in terms of two variables ξ and η . For aquadrilateral surface element, these independent variables are the parent elementcoordinates on the biunit square. As in Eq. (E4.3.1b), since ndΓ = x,ξ ×x,η dξdηbecomes6-38T. Belytschko & B. Moran, Solution Methods, December 16, 19981 1f Iext =∫ ∫ pNI x,ξ ×x,η dξdη(6.4.62)−1−1Taking the time derivative of the above gives1 1˙f ext =I∫ ∫ NI (p˙ x,ξ ×x,η +pv,ξ ×x,η +px,ξ ×v,η )dξdη(6.4.63)− 1−1The first term is the rate of change of the external forces due to the rate of changeof the pressure.
In many problems the rate of change of pressure is prescribed aspart of the problem. In some problems, such as in fluid-structure interactionproblems, the pressure may arise from changes of the geometry; these effectsmust then be linearized and added to the load stiffness. The second two termsrepresent the changes in the external nodal forces due to the change in thedirection of the surface and the area of the surface. These are the terms which arereflected in the external load stiffness, soK extIK v K =∫ ∫ pNI ( v,ξ ×x,η +x,ξ ×v,η )dξdη11(6.4.64)−1−1At this point, it is convenient to switch partially to indicial notation.Taking the dot product of the above with the unit vector e i givesexte i ⋅K extIK v K ≡ KijIK v jK≡11∫ ∫ pNI−1−1[NJ , ξei ⋅( ek × x,η ) + N J , η ei ⋅( x,ξ ×ek )] dξdη(6.4.65)where we have expanded the velocity field in terms of the shape functionsbyv,ξ = v K N K, ξ .
We now defineHikη ≡ eikl x l, ηHikξ = eikl xl, ξ(6.4.66)Using these definitions and Eq. (6.4.65), we obtainextKijIJ=∫ ∫ pN I ( N J, ξ Hij − N J , ηHij ) dξdη11ηξ−1−1orK extIJ =∫ ∫ pNI ( NJ , ξH11−1−1η)− NJ , ηHξ dξdηIf we write out the matrices H ξ and H η we have6-39(6.4.67)T. Belytschko & B. Moran, Solution Methods, December 16, 1998 0K extIJ = ∫ ∫ pN I N J, ξ −z,η−1−1 y,ηz,η0−x,η−y,η x,η 0 0−N J, η −z,ξ y, ξz,ξ0−x,ξ− y,ξ x,ξ dξdη0 1 1(6.4.68)which is the load stiffness of any surface which is generated from a biunit squarein the parent element loaded by a pressure p.
The load surface for a surfaceoriginating can be similarly expressed in terms of the area coordinates, althoughthe limits of integration need to be changed. This load stiffness reflects the effectof the change in geometry on the nodal forces: both alterations in the direction ofthe loaded surfaces and size of the surface will changes in the nodal forces. It isimmediately apparent from (6.4.68) that the submatrices of the load stiffnessmatrix are not symmetric, so the complete matrix is not symmetric. However, itcan be shown that for a closed structure in a constant pressure field, the assembledexternal load stiffness is symmetric.Example 6.1.Three-Node Triangle Element.We first consider the three-node triangle element in two dimensions as inExample 4.1.
The tangent stiffness matrix is derived and explicit forms forhyperelastic and rate-independent hypoelastic-plastic materials are given. Thegeometric tangent stiffness matrix, which is independent of material response, isthen derived. Finally, the external load matrices are derived for a pressure loadalong any edge.Material Tangent Stiffness Matrix.
We consider the case of plane straindeformation (using the x-y plane). The only velocity components are v x and v yand derivatives with respect to z vanish. The tangent stiffness matrix for a rateindependent material given by Eq. (6.3.36):Ktan = ∫ BT [C]BdA(E6.1.1)where A is the current area of the element and we have assumed a unit thickness(see Eq. (4.??)).σTσTσT C1111C1122C1112σTσT C2212[ Cabtan ] = Cσ2211T C2222σTσTσT C1211C1222 C1212 (E6.1.2)The B matrix is given by Eq. (E4.1.45):6-40T. Belytschko & B. Moran, Solution Methods, December 16, 1998 y231 B=02A x320x 32y23y310x130x13y310x21 y12 y120x21(E6.1.3)The material tangent stiffness matrix, Eq.
(6.4.81), is rewritten, using Eqs.(6.4.82 - 6.4.84) as y 23 02 1 y31tanK =∫0A0 2 A y12 0 y230 x320x320x130x21x32 y23 σ TσT CC1122x13 1111σTTC2211Cσ2222y31σT Cσ T C1222x 21 1211y12 0x32y23y310x130x13y31y120x21σTC1112σT C2212 σTC1212(E6.1.4)0x21 dA0y12 Assuming the integrand to be constant, we obtain, by multiplying theintegrand by the element area A0 (note that, for plane strain, a unit thickness isassumed and the element volume is given by V0 = A0 (1) = A0 ).KtanAB y230A y31= 02 4A 0 y12 00x320x130x21 y230 x32x32 y23 σ TσT C1111 C1122x13 σ TσTCC2222y31 2211σTσT C1211C1222x 21y12 0x32y23y310x130 y12x13 0y31 x 21σTC1112σTC2212σTC1212 (E6.1.5)0x21 y12 Neo-Hookean Material. For a Neo-Hookean material (see Section 5.??),CσijklT = λδ ijδ kl + µ ( J )(δ ikδ jl +δ ilδ jk )(E6.1.6)µ( J ) = µ0 −λ log J ,(E6.1.7)whereJ =det F ,and Eq.
(6.4.88) is written in Voigt notation as6-41T. Belytschko & B. Moran, Solution Methods, December 16, 1998λ + 2µλ0[ C ] = λ λ + 2µ 0 00µ tanab(E6.1.8)Thus, for a Neo-Hookean material, the material tangent stiffness matrixhas the same form as the stiffness matrix for small strain linear elasticity exceptfor the dependence of the moduli on the deformation (through Eq.
(6.4.90)) andthe geometry factor A0∆A .Rate-Independent Elastoplasticity. For a rate-independent elastic-plastic model interms of the Kirchoff stress, with associated plastic flow and a von-Mises yieldcondition, the tangent modulus is given by Eq. (5.??)epCtanijkl = Cijkl −1(δ τ +τ δ + δ ikτ jl + τ ilδ jk )2 il jk ik jl(E6.1.9)The elastoplastic tangent modulus is given byepCijkl= Cijkl −Cijmn pmnCklrs prsh + pmnCmnrs prs(E6.1.10)3τij′is the plastic flow direction, τ ij′ is the2σdeviatoric part of the Kirchoff stress and σ is the effective stress defined by (??).Assuming constant isotropic elastic moduli, Eq.