Belytschko T. - Introduction (779635), страница 45

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The current angle of the corotational coordinate system is θ . We discuss how tocompute this angle subsequently.The motion can be expressed in terms of the triangular coordinates, as in Example 4.1. x   x1 =  y  y1x2y2ξ x3  1 ξ y3  2 ξ3 (E4.7.1)The displacement and velocity fields in the element are then given byξ uˆ x  ˆux1 uˆ x 2 uˆ x 3   1  =  ξ2 ˆuy  uˆ y1 uˆ y2 uˆ y 3   ξ3 (E4.7.2)ξ vˆ x  vˆ x1 ˆvx 2 vˆ x3  1  =ξ2 vˆ y  vˆ y1 vˆ y2 vˆ y3   ξ3 (E4.7.3)The derivatives of the shape functions with respect to the corotational coordinate system are givenby the counterpart of (E4.1.5):[yˆ 231ˆ∂N I ∂ˆx j ≡ ∂ξ I ∂xˆ j =y2A 31 yˆ 12] []xˆ 32 ˆx13  ≡ Bˆxˆ 21 4-57(E4.7.4)T.

Belytschko, Lagrangian Meshes, December 16, 1998The corotational components of the rate-of-deformation are given byˆ =D12(Bˆ I vTI + v IBˆ TI )(E4.7.5)The nodal internal forces are given by Eq. (4.6.10):∂ξ[ fIi ]int = ∫ Bˆ Ij σˆ jk RkiT dΩ= ∫ ∂xˆ I σˆ jk RkiT dΩΩΩ(E4.7.6)jWriting out the matrices using (E4.6.2) and (E4.7.4) gives f1x f2 x f3 xint ˆy23 xˆ 32 f1y 1 ˆ σˆ xˆf2 y  = ∫yx13   ˆ 31σ2AAˆy12 ˆx21  xyf3y σˆ xy  cosθ sinθ adAσˆ y − sinθ cos θ (E4.7.7)The rotation of the coordinate system can be obtained in several ways:1. by polar decomposition;2.

by rotating the corotational coordinate system with a material line in the element,e.g. a preferred direction in a composite;3. by rotating the corotational coordinate system with a side of the element (this is onlycorrect for small strain problems).To use polar decomposition, the same approach as described in Section 3, Example ?? is used.4.7. TOTAL LAGRANGIAN FORMULATION4.7.1. Governing Equations. The physical principles which govern the total Lagrangianformulation are the same as those for the updated Lagrangian formulation, which were given inSection 4.2. The form of the governing equations is different, but as has been seen in Chapter 3,they express the same physical principles and can be obtained by transforming the associatedconservation equations from Eulerian to Lagrangian form.Similarly, the finite element equations for the total Lagrangian formulation can be obtained bytransforming the equations for the updated Lagrangian formulation.

It is only necessary totransform the integrals to the reference (undeformed) domain and transform the stress and strainmeasures to the Lagrangian type. This approach is used in Section 4.7.2, and for most readersSection 4.7.2 and the following examples will suffice as an introduction to the total Lagrangianformulation. However, for readers who would like to see the entire structure of the totalLagrangian formulation or prefer to learn it first, Section 4.8 gives a development of the weak formin the total Lagrangian description, followed by the direct derivation of the finite element equationsfrom this weak form.The governing equations are given in both tensor form and indicial form in Box 4.5. We havechosen to use the nominal stress P in the momentum equation, because the resulting momentumequation and its weak form are simpler than for the PK2 stress.

However, the nominal stress isawkward in constitutive equations because of its lack of symmetry, so we have used the PK2stress for constitutive equations. Once the PK2 stress has been evaluated by the constitutive4-58T. Belytschko, Lagrangian Meshes, December 16, 1998equations, the nominal stress stress can then easily be obtained by a transformation given in Box3.2, Eq. (B4.5.5).

The constitutive equation can relate the Cauchy stress σ to the rate-ofdeformation D. The stress would then be converted to the nominal stress P prior to evaluation ofthe nodal forces. However, this entails additional transformations and hence additionalcomputational expense, so when the constitutive equations are expressed in terms of σ it isadvantageous to use the updated Lagrangian formulation.Box 4.5Governing Equations For Total Lagrangian Formulationconservation of massρJ = ρ 0 J0 = ρ 0(B4.5.1)conservation of linear momentum˙˙∇ X ⋅ P + ρ0b = ρ0u∂Pji+ ρ 0 bi = ρ 0u˙˙i∂X jor(B4.5.2)conservation of angular momentum:F ⋅P = PT ⋅ FTorFij Pjk = Fkj Pji(B4.5.3)conservation of energy˙ T :P −∇ ⋅ q + ρ s˙ int = Fρ0 wX0where q = JF −1q˙ int = F˙ ij Pji − ∂qi +ρ 0sρ0 w∂Xior(B4.5.4)constitutive equationS = S(E,..etc)P = S⋅F T(B4.5.5)measure of strainE=()1 TF ⋅F − I2orEij =()(B4.5.6)on Γt0i , ui = ui on Γu0i(B4.5.7)1F F − δ ij2 ki kjboundary conditionsn 0j Pji = ti0 or e i ⋅ n0 ⋅P = ei ⋅ t0Γt0i ∪Γu0i = Γ 0Γt0i ∩Γu0i = 0 for i = 1 to ns4-59(B4.5.8)T.

Belytschko, Lagrangian Meshes, December 16, 1998initial conditionsP(X,0) = P 0 (X) in(B4.5.9)u(X,0) = u0 (X)(B4.5.10)internal continuity conditions0n0j Pji = 0 on Γint(B4.5.11)The nominal stress is conjugate to the material time derivative of the deformation tensor, F˙ ,see Box 3.4. Thus in (B4.5.4) the internal work is expressed in terms of these two tensors. Notethat we have used the left divergence of P (see (B.5.1.2)), so n0 appears before P in the tractionexpression; if the order is reversed the resulting matrix corresponds to the transpose of P , which isthe PK1 stress; see Section 3.4.1.

The PK1 stress is also frequently used, so it is important tonote the distinction between these two stress tensors. The traction is obtained in terms of thenominal stress by putting the initial normal to the left, and the left divergence operator is used in themomentum equation. For the PK1 stress, the normal appears to the right and the right divergenceis used in the momentum equation.The deformation tensor F is not suitable as a measure of strain in constitutive equations sinceit does not vanish in rigid body rotation. Therefore constitutive equations in total Lagrangianformulations are usually formulated in terms of the Green strain tensor E, which can be obtainedfrom F.

In the continuum mechanics literature, one often sees constitutive equations expressed asP = P( F) , which gives the impression that the constitutive equation uses F as a measure of strain.In fact, when writing P(F), it is implicit that the constitutive stress depends on F T F (i.e., E + I ,where the unit matrix I makes no difference) or some other measure of deformation which isindependent of rigid body rotation. Similarly, the nominal stress P in constitutive equations isincorporated so it satisfies conservation of angular momentum, Eq.

(B4.5.3).As in any mechanical system, the same component of traction and displacement cannot beprescribed at any point of a boundary, but one of these must be prescribed; see Eqs. (B4.5.7B4.5.8). In the Lagrangian formulation, tractions are prescribed in units of force per undeformedarea.The total Lagrangian formulation can be obtained in two ways:1.

transforming the finite element equations for the updated Lagrangian fomulation to theinitial (reference) configuration and expressing it in terms of Lagrangian variables.2. by developing the weak form in terms of the initial configuration and Lagrangianvariables and then using this weak form to obtain discrete equations.We will begin with the first approach since it is quicker and more convenient. The secondapproach is only recommended for intensive courses or for those who prefer the total Lagrangianformulation.4.7.2. Total Lagrangian Finite Element Equations by Transformation. To obtain thediscrete finite element equations for total Lagrangian formulation, we will transform each of the4-60T.

Belytschko, Lagrangian Meshes, December 16, 1998nodal force expressions in the updated Lagrangian formulation, beginning with the internal nodalforces. The mass conservation equation (B4.5.1), ρJ = ρ0 , and the relationdΩ = JdΩ 0(4.7.1)will also be used. The internal nodal forces are given in the updated Lagrangian formulation byEq. (4.4.10)∫fiIint =Ω∂N Iσ dΩ∂x j ji(4.7.2)Using the transformation from Box 3.2, Jσ ji = Fjk Pki =∫fiIint =Ω∂N I ∂x jPki J −1dΩ∂x j ∂Xk∂x j∂XkPki , we convert (4.7.2) to:(4.7.3)Recognizing that the product of the first two terms is a chain rule expression of ∂NI ∂Xk andusing Eq.

(4.7.1), we get∂N I∫ ∂XfiIint =Ω0kPkidΩ 0 = ∫ B0 Ik Pki dΩ0(4.7.4)Ω0where∂NI∂XkB0kI =(4.7.5)In matrix form, the above can be written as(f intI ) = ∫ B0T PdΩ0T(4.7.6)Ω0The expression has been written in the above form to stress the analogy to the updated Lagrangianform: if B is replaced by B0 , Ω by Ω 0 , and σ by P, we obtain the updated Lagrangian formfrom the above.The external nodal forces are next obtained by transforming the updated Lagrangianexpression to the total Lagrangian form.

We start with Eq. (4.4.13)fiIext = ∫ N Iρbi dΩ+Ω∫ N t dΓ(4.7.7)I iΓtiSubstituting Eq. (3.6.1), ρbdΩ= ρ0 bdΩ 0 , and Eq. (3.4.4), t dΓ = t0 dΓ0 , into Eq. (4.7.7) gives4-61T. Belytschko, Lagrangian Meshes, December 16, 1998fiIext =∫ N ρ b dΩ + ∫ N tI0 i0I i0Ω0dΓ0(4.7.8)Γt0iwhich is the total Lagrangian form of the external nodal forces. The two integrals are over theintial (reference) domain and boundary; note that ρ0 b is the body force per unit of the referencevolume, see (3.6.1). This can be written in matrix form as:f Iext = ∫ NI ρ0 bdΩ 0 + ∫ NI e i ⋅t 0 dΓΩ0(4.7.9)Γ t0iThe inertial nodal forces and the mass matrix were expressed in terms of the initial configuration inthe development of the updated Lagrangian form, Eq. (4.4.50).

Thus, all of the nodal forces canbe expressed in terms of Lagrangian variables on the initial (reference) configuration by thetransformations. The equations of motion for the total Lagrangian discretization are identical to thatof the updated Lagrangian discretization, Eq. (4.4.48).4.8 TOTAL LAGRANGIAN WEAK FORMIn this Section, we develop the weak form from the strong form in a total Lagrangian format.Subsequently, we will show that the weak form implies the strong form. The strong form consistsof the momentum equation, Eq. (B4.5.2), the traction boundary condition, Eq. (B4.5.7), and theinterior continuity conditions, Eq.

(B4.5.11). We define the spaces for the test and trial functionsas in Section 4.3:δu(X ) ∈U0 ,u( X, t ) ∈U(4.8.1)where U is the space of kinematically admissible displacements and U0 is the same space with theadditional requirement that the displacements vanish on displacement boundaries.Strong Form to Weak Form. To develop the weak form, we multiply the momentumequation (B4.5.3) by the test function and integrate over the initial (reference) configuration: ∂Pjiδui + ρ0bi − ρ 0˙u˙ i  dΩ 0 = 0 ∂X jΩ0∫(4.8.2)In the above, the nominal stress is a function of the trial displacements via the consitutiveequation and the strain-displacement equation.

This weak form is not useful because it requires thetrial displacements to be C1 , since a derivative of the nominal stress appears in (4.8.2); seeSections 4.3.1-2.To eliminate the derivative of the nominal stress from Eq. (4.8.2), the derivative productformula is used:∂Pji∂∫ δui ∂X j dΩ 0 = ∫ ∂X j (δuiPji )dΩ 0 − ∫Ω0Ω0Ω0∂ ( δui )P dΩ∂X j ji 04-62(4.8.3)T. Belytschko, Lagrangian Meshes, December 16, 1998The first term of the RHS of the above can be expressed as a boundary integral by Gauss'stheorem (3.5.6):∫Ω0∂δui Pji dΩ 0 =− δui n0j Pji dΓ0 +∂X j()∫Γ0∫ δuin 0j Pji dΓ0(4.8.4)Γin0tFrom the strong form Eq.

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