Belytschko T. - Introduction (779635), страница 16
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v( X ,t) = N( X (ξ ) )ve (t ) = N I (X (ξ )) veI (t )(B2.3.2)note N,x = N,ξ x, −1ξin each element∂N I evI = BveI =1 ∂xmDx = ∑(B2.3.3)evaluate s by constitutive equationfei nt = ∫Ω∂NσdΩ ore ∂xfein t = ∫ΩB TσdΩ(B2.3.4)efeext = ∫Ω ρN T bdΩ+ (N T At x ) Γ et(B2.3.5)M e = ∫Ω ρ0N T NdΩ(B2.3.6)esame as total LagrangianeM ˙u˙ + f int = f ext(B2.3.7)Example 2.8.1.Updated Lagrangian Form of Two-Node LinearDisplacement Element. This element is the same as in Example 2.5.1, Fig. 3, exceptthe updated Lagrangian treatment is now used. Recall that A 0 and ρ 0 are assumed to beconstant in each element.
The velocity field is the same as for the updated Lagrangianelement, (2.5.19): v1( t) v( X, t) = l1 [ X 2 – X, X – X1] 104 442 4443 v2 (t )N( X )(2.8.31)In terms of element coordinates, the velocity field is2-47T. Belytschko, Chapter 2, December 16, 1998 v1 (t )v (ξ, t ) = [1– ξ , ξ ] 14243 v2 (t )N (ξ )ξ=X– X1l0(2.8.32)The displacement is the time integrals of the velocity, and sinceξ is independent of timeu(ξ , t ) = N(ξ )ue (t )(2.8.33)Therefore, since x = X + ux (ξ,t ) = N(ξ ) xe( t) = [1– ξ x1( t )ξ] x2( t) x, ξ = x2 − x1 = l(2.8.34)where l is the current length of the element. For this element, we can express ξ in termsof the Eulerian coordinates byx– xx– x1ξ = x – x1 = l 1 , l = x2 – x1 , ξ, x =21l(2.8.35)So ξ, x can be obtained directly, instead of through the inverse of x, ξ .
This is not the casefor higher order elements.The B matrix is obtained by the chain ruleB = N, x = N, ξ ξ, x = 1l [ –1 ,+1 ](2.8.36)so the rate-of-deformation is given byDx = B v e = 1l (v2 – v1)(2.8.37)Using (2.8.18) then givesx2feintx2 –1 = ∫ B σ Adx = ∫ 1l σ Adx+1xxT1(2.8.38)1If the integrand in (2.8.38) is constant, as if often is, then (2.8.38) yields–1f inte = Aσ +1(2.8.39)Thus the internal nodal forces for the element correspond to the forces resulting from thestress σ. Note that the internal nodal forces are in equilibrium.The external nodal forces are evaluated using (2.8.21)2-48T.
Belytschko, Chapter 2, December 16, 1998 1– ξ 1– ξ= ∫ ρbAdx + At x ξ ξ Γex1x2feext(2.8.40)twhere the last term makes a contribution only if a node of the element is on the tractionboundary. Since x is a linear function of ξ and t, Eq. (2.8.16), b(x,t) can always beexpressed as a function of ξ and t. It is conventional to fit the data for b(x,t) by linearinterpolants for linear displacement elements (the information in higher order interpolationswill be beyond the resolution of the mesh). So we letb(ξ,t ) = b1( 1−ξ ) + b2ξ(2.8.41)Substituting into (2.8.31) and integrating givesρAl 2b1 + b2 f ext=e6 b1 + 2b2 (2.8.42)Comparison to Total Lagrangian. We will now compare the nodal forces to thoseobtained by the total Lagrangian formulation.
Replacing σ in (2.8.39) by the nominalstress using Eq. (2.1.3a), we see that (2.8.39) and (2.5.27) are equivalent. It can easilybe shown that (2.8.29) and (2.8.21) lead to the same expression as (2.8.31).To compare the external nodal forces, we note that by the conservation of matter, ρAl= ρ 0 A0 l 0 . Using this in Eq. (2.8.42) gives (2.5.26), the total Lagrangian form of the nodalexternal forces. In the updated Lagrangian formulation, the mass from the total Lagrangianformulation is used, see Eq.
(2.8.25), so the equivalence need not be considered.Example 2.8.2.Updated Lagrangian of Three Node Element, QuadraticDisplacement Element The 3-node element is shown in Fig. 2.7. Node 2 can beplaced anywhere between the end-nodes, but we shall see there are restrictions on theplacement of this node if the one-to-one condition is to be met. We will also examine theeffects of mesh distortion.tx1x1(t)x2x3x2(t)x3(t)X, xX1X2X3Fig.
2.7. Three node, quadratic displacement element in original and current configurations.The displacement and velocity fields will be written in terms of the element coordinates2-49T. Belytschko, Chapter 2, December 16, 1998u(ξ, t ) = N(ξ ) ue (t ) , v (ξ, t ) = N(ξ ) ve (t ) ,x (ξ,t ) = N(ξ ) xe( t)(2.8.43)whereN(ξ) =[ (ξ –ξ )1221– ξ 212(ξ 2 +ξ)](2.8.44)anduTe = [u1, u2 ,u3 ]vTe = [v1 , v2 ,v3]x Te = [ x1 , x 2 , x3 ](2.8.45)The B matrix is given byB = N, x = x,−1ξ N, ξ1 [ 2ξ –1= 2x,ξ(2.8.46)– 4ξ2ξ + 1](2.8.47)wherex, ξ = N, ξ x e = (ξ– 12 ) x1 – 2ξ x 2 + (ξ + 12 ) x 3(2.8.48)The rate of deformation is given by1 [2ξ–1 – 4ξDx = N, x v e = B v e = 2x,ξ2ξ + 1] ve(2.8.49)This rate-of-deformation varies linearly in the element if x ,ξ is constant, which is the casewhen node 2 is midway between the other two nodes.
However, when node 2 movesaway from the midpoint due to element distortion, x ,ξ becomes linear and the rate-ofdeformation is a rational function. Furthermore, as node 2 moves from the center, itbecomes possible for x ,ξ to become negative or vanish. In that case, the mapping betweenthe current spatial coordinates and the element coordinates is no longer one-to-one.The internal forces are given by (2.8.18):ξ – 12 ξ– 12 +1= ∫ BTσ Adx = ∫ x1,ξ –2ξ σ Ax ,ξ dξ = ∫ σ A –2ξ dξξ + 1 ξ + 1 x1–1–122x3fei nt+1(2.8.50)where we have used dx = x , ξ dξ . Using (2.1.3), we can see that this expression isidentical to the internal force expression for the total Lagrangian formulation.Mesh Distortion.
We will now examine the effects of mesh distortion on this element.When x2 = 41 (x 3 + 3x1 ) , i.e. when node 2 of the element is one quarter of the elementlength from node 1, then x, ξ = 21 ( x3 – x1 ) (ξ +1) , so x ,ξ = 0 at ξ =–1. Examining theJacobian given by Eq. (2.2.3)2-50T. Belytschko, Chapter 2, December 16, 1998J=AAx, X =x X −1A0A0 , ξ , ξ(2.8.51)we see that it will also vanish. By E. (2.2.4a) this implies that the current density becomesinfinite at that point. As node 2 moves closer to node 1, the Jacobian becomes negative inpart of the element, which implies a negative density and a violation of the one-to-onecondition.
This corresponds to a violation of mass conservation and continuity of thedisplacement field. These situations are often masked by numerical quadrature, because thecondition must be more severe to appear at Gauss quadrature points.The failure to meet the one-to-one condition can also affect the rate-of-deformation,which is given by Dx = Bve . From (2.8.37) we can see the potential for difficulties whenthe denominator x ,ξ vanishes or becomes negative. When x2 = 41 ( x 3 + x1 ) , and x ,ξ =0 at ξ=–1, then the rate-of-deformation becomes infinite at node 1.
This property of quadraticdisplacement elements has been exploited in fracture mechanics to develop elements withsingular cracktip stresses called quarter-point elements, but in large displacement analysisthis phenomenon can be troublesome.In one-dimensional elements the effects of mesh distortion are not as severe as inmulti-dimensional problems. In fact, the effects of mesh distortion can be alleviatedsomewhat in this element by using F as a measure of deformation, see Eq. (2.5.40).
Thedeformation gradient F never becomes singular in the 3-node element if the initial positionof X 2 is at the midpoint. However, any constitutive equation expressed in terms of F willdiffer markedly from one expressed in terms of the rate-of-deformation Dx when thestrains are large.Example 2.8.3.
Axisymmetric 2-Node Element. As an example where theconcept of the principle of virtual power or work becomes quite useful, we consider theanalysis of an axisymmetric two dimensional disc of constant thickness, a, which is thincompared to its dimensions so σ z = 0 . The only nonzero velocity is vr (r) , which asshown, is only a function of the radial coordinate in an axisymmetric problem. Thenonzero Cauchy stresses and rate-of-deformations are written in cylindrical coordinatesusing Voigt notationσ σ θ D Dθ {D} = r {σ} = r (2.8.52)The rate-of-deformations are given byDr = vr ,rDθ =vrr(2.8.53)and the momentum equation is∂σ r σ r −σ θ++ ρbr = ρv˙ r∂rr(2.8.54)The boundary conditions are2-51T. Belytschko, Chapter 2, December 16, 1998σ r (b ) =σ bσ r (a) = σ aeθ(2.8.55)erarbzrar1 radiansegmentelementFig.
Schematic of axisymmetric discthe shaded area is considered in work termsIt is not necessary to integrate the momentum equation to obtain its weak form. By theprinciple of virtual power the weak form isδP =0∀δvr ∈U0(2.8.56)The internal virtual power is obtained from the rate-of-deformation and stressr2eδPeint=T∫ (δDr σ r + δDθσ θ )ardr = ∫Ω {δD} {σ}dΩ(2.8.57)er1ewhere dΩ = ardr because a segment of one radian in the circumferential direction has beenchosen to avoid the factor 2π in all terms. The external virtual power is given byδPeext = ∫Ω δvr ρbr dΩ + (art r ) Γ(2.8.58)tewhere ar in the last term is the area of a one radian segment. The virtual inertial power isgiven byδPeinert = ∫Ω δvr ρv˙ r dΩ(2.8.59)eConsider a two-node finite element with a linear velocity field written in terms of elementcoordinates v1( t ) v (ξ,t ) = [1− ξ ξ]v2 (t )(2.8.60)2-52T.
Belytschko, Chapter 2, December 16, 1998The rate-of-deformation is evaluated by Eq.() using the above velocity field andimmediately put into matrix form− 1 Dr rD = = 21 Dθ 1− ξ r1r21 v1( t ) = Bveξ v 2(t )r (2.8.61)The internal nodal forces are given by an expression identical to () except that the stress isreplaced by the column matrix− 1 r= ∫ BT {σ}dΩ = 21Ωe1−ξr1 rr2feint∫1 r21 σ r ardrξ σ θ r (2.8.62)The external nodal forces are given byr2feext1−ξ = ∫ρb ardr + ( art r ) Γtξ rr1 (2.8.63)The element mass matrix is given byr21−ξ Me = ∫ [1− ξ ξ ]ρardrξr1ρar21 3r1 + r2=12 r1 + r2(2.8.64)r1 + r2 r1 +3r2 The lumped mass matrix can be computed by the row sum technique or by lumping half themass at each node, which gives, respectivelyMe =ρar21 2r1 + r26 00 r1 + 2r2 row-sumMe =ρar21 (r1 + r2 ) 14001(2.8.65)lumpAs can be seen the two lumping procedures give slightly different results.2.9.
Governing Equations for Eulerian FormulationIn an Eulerian formulation, the nodes are fixed in space and the independent variablesare functions of the Eulerian spatial coordinate x and the time t. The stress measure is theCauchy (physical) stress σ (x, t) , the measure of deformation is the rate-of-deformationDx ( x,t ) . The motion will be described by the velocity v ( x, t ) . In Eulerian formulations,the deformation is not expressed as a function of the reference coordinates since an2-53T. Belytschko, Chapter 2, December 16, 1998undeformed, initial configuration cannot be established, and no counterpart of (2.2.1) isavailable.Box. 2.4.
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