Belytschko T. - Introduction (779635), страница 13

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(20):fei nt=∫ 2ξ– 11  2ξ −11= ∫ 2X ,  –4ξ  PA0 X, ξ dξ = ∫ 12  −4ξ  PA0dξξ2ξ +1−1−1 2ξ +11BT0 PdΩ 0Ω e0(2.5.40)The above integral is generally evaluated by numerical integration. For the purpose ofexamining this element further, let P(ξ) be linear in ξ : 1+ ξ 1– ξ P(ξ ) = P1 2 + P3 2 (2.5.41)where P 1 and P 3 are the values of P at nodes 1 and 3, respectively. If X, ξ is constant, thisis an exact representation for the stress field in a material which is governed by a linearstress-strain relation in these measures, Eq. (2.2.14), since F is linear in ξ by (2.5.40).The internal forces are then given by2-32T.

Belytschko, Chapter 2, December 16, 1998intfeint−5P1 − P2  f1 A0 =  f2  =4P − 4P2 6  1f  3 e P1 + 5P2 (2.5.42)When P is constant, the nodal force at the center node vanishes and the nodal forces at theend nodes are equal and opposite with magnitude A 0 P, as in the two node element. Inaddition, for any values of P1 and P2 , the sum of the nodal forces vanishes, which can beseen by adding all the nodal forces.

Thus this element is also in equilibrium.The external nodal forces arefeext+ 1 1ξ (ξ − 1) 12ξ(ξ − 1) =  1−ξ 2 ρ0 bA0 X,ξ dξ +  1−ξ2 A0 t x0 Γ e11tξ ξ + 1) 2 ξ (ξ + 1) −1  2 (∫2(2.5.43)where the shape functions in the last term are either one or zero at a traction boundary.1Using X, ξ =ξ ( X1 + X3 − 2X2 ) + 2 ( X3 − X1) , thenfeext  1ξ ξ − 1 ρ 0bA0  L 0 −2 ( X1 + X3 − 2 X2 ) 2 ( 2 ) =4 L0 +  1− ξ  A0 tx0 eΓt6 L +2 ( X + X − 2 X )  1ξ (ξ + 1)01322(2.5.44)Element Mass Matrix. The element mass matrix isMe =+1  1 ξ ξ −1 ( )2 2  1−ξ  1ξ (ξ +1)−1 2∫[ ξ(ξ −1) 1− ξ122ρ0 A0  4 L0 −6 (X1 + X3 − 2 X2 )=30 sym1ξ2(ξ +1)]ρ0 A0 X, ξ dξ2 L0 − 4( X1 + X3 − 2 X2 )− L016L02 L0 + 4( X1 + X3 − 2 X2 )4 L0 − 6( X1 + X3 − 2 X2 )(2.5.45)If the node 2 is at the midpoint of the element, i.e., X1 + X3 = 2 X2 , we have 4 2 −1ρ0 A0 L0 Me = 2 16 2 30  −1 2 4 (2.5.46)If the mass matrix is diagonalized by the row-sum technique, we obtain 1 0 0ρ0 A0 L0 Me =0 4 0 6  0 0 1(2.5.47)This results displays one of the shortcomings of diagonal masses for higher order elements:most of the mass is lumped in the center node.

This results in rather strange behavior when2-33T. Belytschko, Chapter 2, December 16, 1998high order modes are excited. Therefore, high order elements are usually avoided when alumped mass matrix is necessary for efficiency.2 . 6 Governing Equations for Updated Lagrangian FormulationIn the updated Lagrangian formulation, the discrete equations are formulated in thecurrent configuration. The stress is measured by the Cauchy (physical) stress σ given byEq. (2.1.1). In the updated Lagrangian formulation, variables need to be expressed interms of the spatial coordinates x and the material coordinates X in different equations.

Thedependent variables are chosen to be the stress σ(X,t) and the velocity v(X,t). This choicediffers from the total Lagrangian formulation, where we have used the displacement u( X , t)as the independent variable; this is only a formal difference since the displacement andvelocities are both computed in a numerical implementation.In developing the updated Lagrangian formulation, we will need the dependentvariables to be expressed in terms of the Eulerian coordinates. Conceptually this is asimple matter, for we can invert (2.2.1) to obtainX = φ −1 ( x,t ) ≡ X ( x,t )(2.6.1)Any variable can then be expressed in terms of the Eulerian coordinates; for exampleσ ( X,t) can be expressed as σ ( X( x, t), t) .

While the inverse of a function can easily bewritten in symbolic form, in practice the construction of an inverse function in closed formis difficult, if not impossible. Therefore the standard technique in finite elements is toexpress variables in terms of element coordinates, which are sometimes called parentcoordinates or natural coordinates. By using element coordinates, we can always express afunction, at least implicitly, in terms of either the Eulerian and Lagrangian coordinates.In updated Lagrangian formulations, the strain measure is the rate-of-deformationgiven byDx =∂v∂x(2.6.2a)This is also called the velocity-strain or stretching. It is a rate measure of strain, asindicated by two of the names.

It is shown in Chapter 3 that∫0 Dx ( X,t ) dt =ln F( X, t)t(2.6.2b)in one dimension, so the time integral of the rate-of-deformation corresponds to the"natural" or "logarithmic" strain in one dimension; as discussed in Chapter 3, this does nothold for multi-dimensional states of strain.The governing equations for the nonlinear dimensional continuum are:1.

conservation of mass (continuity equation)ρJ = ρ 0 or ρFA = ρ0 A02. conservation of momentum∂ ( Aσ ) + ρAb = ρAv˙ or ( Aσ ) + ρAb = ρA ˙v,x∂x2-34(2.6.3)(2.6.4)T. Belytschko, Chapter 2, December 16, 19983. measure of deformation∂vDx =or Dx = v, x∂x(2.6.5)4. constitutive equationin total formtσ( X, t) = Sσ D (Dx ( X,t ) Dx ( X, t),... ), ∫ Dx ( X, t )dt ,σ ( X,t ), t ≤t, etc.)(2.6.6a)0in rate formσ ,t ( X, t) = StσD (Dx ( X,t ),σ ( X, t ), etc., t ≤ t )energy conservation˙ in t =σDx −q x , x + ρs, qx = heatflux, s = heatsourceρw(2.6.6b)(2.6.7)The mass conservation equation in the updated Lagrangian formulation is the same as in thetotal Lagrangian formulation. The momentum equation in the updated formulation involvesderivatives with respect to the Eulerian coordinates, whereas in the total Lagrangianformulation, derivatives were with respect to Lagrangian coordinates; in addition, thenominal stress is replaced by the Cauchy stress, and that the current values of the crosssectional area A and density ρ are used.

The constitutive equation as written here relatesthe rate-of-deformation Dx ( X ,t ) or its integral, the logarithmic strain, to the Cauchy stressor its rate. Note that the constitutive equation is written in terms of material coordinates.The subscript "t" on (2.6.6b) indicates that the constitutive equation is a rate equation. Wecan also use a constitutive equation expressed in terms of the nominal stress and the stretchε . It would then be necessary to transform the stress to the Cauchy stress before using themomentum equation and use a different measure of strain.

Thus in the updated Lagrangianformulation, some of the system equations are in terms of Eulerian coordinates, whileothers (mass conservation and constitutive equations) are in terms of Lagrangiancoordinates.The subscripts have been appended to the constitutive function to indicate which stressand strain measures are related by the constitutive equation. The constitutive equationdepends on the stress and strain measures which are involved. For example, theconstitutive equation for a hypoelastic material in terms of the Cauchy stress and rate-ofdeformation isσ ,t ( X, t) = E σDDx ( X, t)(2.6.8)where E PF ≠ E σD .

To see the relationship between the two moduli, we use the relationDx =∂v ∂v ∂X ∂v −1 ˙ −1==F = FF∂x ∂X ∂x ∂X(2.6.9)where the first equality is the definition (2.6.5), the second stems from the chain rule, andthe third from the definition of F, Eq. (2.2.4). Then inserting Eqs. (2.2.9) and (2.6.9) in(2.6.8) gives2-35T. Belytschko, Chapter 2, December 16, 1998A0d  P˙ F−1  = Eσ D Fdt A(2.6.10)which after some manipulation yields˙ = A E σDF˙ + σ A˙PA0 FA0(2.6.11)In general, constitutive equations are not easily converted from one stress-strain pair toanother. For the above, the cross-sectional area must be known as a function of theelongation to make the conversion.The boundary conditions arev( X, t) = v ( t) onΓv(2.6.12)nσ ( X, t) = t x ( t) on Γ t(2.6.13)where v ( t) and t x ( t) are the prescribed velocity and traction, respectively, and n is thenormal to the domain. While the boundary condition is specified as applying to thevelocity, any velocity boundary condition is also a displacement boundary condition.

Notethat the traction always carries a subscript to distinguish it from the time t . The relationbetween the traction and velocity boundaries is the same as in (2.2.30):Γ v ∪ Γt = ΓΓv ∩ Γt = 0(2.6.14)The boundary over which the velocity is prescribed is denoted by Γ v ; it is an essentialboundary condition and it plays the same role as Γ u in the total Lagrangian formulation.The tractions in (2.6.13) are physical tractions, force per current area.

They are related tothe tractions on the undeformed area byt x A = t x0 A0(2.6.15)In addition we have the stress jump conditions, the counterpart of (2.2.33):σA = 0(2.6.16)The initial conditions areσ ( X,0 ) = σ 0 ( X )(2.6.17)v ( X,0 ) = v 0 ( X )(2.6.18)Since we have chosen the velocity and the stresses as the dependent variables, the initialconditions are imposed on these variables. In most practical problems, this choice of initialconditions is more practical than conditions on velocities and displacements, as indicated inChapter 4.2-36T. Belytschko, Chapter 2, December 16, 19982 . 7 Weak Form for Updated Lagrangian FormulationIn this Section, the weak form or variational form for the momentum equation isdeveloped.

Recall that the dependent variables are the velocity v( X, t) and the stressσ( X, t) .The conditions on the trial functions v(X,t) and the test functions δv(X) are:{}v( X, t) ∈UU = v( X,t ) v ∈ C 0( X), v = v on Γvδv( X) ∈U0U 0 = δv( X ) δv ∈ C 0 ( X ), δv = 0 on Γv{(2.7.1)}(2.7.2)These admissibility conditions are identical to those for the test and trial displacements inthe total Lagrangian formulation. As in the total Lagrangian formulation, the stress σ(X,t)is assumed to be a C–1 function in space. The current domain is [ xa ( t), xb ( t)] , wherexa = φ ( X a , t ) , xb = φ ( Xb , t ) .The strong form consists of the momentum equation (2.6.4), the traction boundaryconditions and the jump conditions.

The weak form is developed by multiplying themomentum equation (2.6.12) by the test function δv(X) and integrating over the currentdomain. The current domain of the body is appropriate, since the momentum equationinvolves derivatives with respect to the spatial (Eulerian) coordinates. This givesxbDv ∫ δv( Aσ ), x + ρAb − ρA Dt  dx = 0(2.7.3)xaIntegration by parts is performed as in Section 2.3 (see Eqs. (2.3.2) to (2.3.4)), whichgivesxbxbxaxa∫ δv( Aσ ), x dx = ∫ [(δvAσ ), x – δv, x Aσ ]dx= (δvAnσ ) Γ – ∑ δv AσtixbΓi –∫ δv, x Aσ dx(2.7.4)xawhere Γ i are the points of discontinuity of Aσ; see Eq. (2.6.16).

We have used thefundamental theorem of calculus to convert a line (domain) integral to a sum of point(boundary and jump) values, with Γ changed to Γ t because δv(X) = 0 on Γ v ; see Eq.(2.7.2). Since the strong form holds, the traction boundary condition (2.6.13) givesnσ = t x and the jump condition nσ = 0 , which when substituted into the above givexb∫xaDv  δv, x Aσ – δv ρAb– ρA Dt   dx– (δvAt x ) Γt = 02-37(2.7.5)T.

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