Belytschko T. - Introduction (779635), страница 91
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The displacement field is written in terms of a referencecoordinate system (ξ,η). Within the reference system, the element domain is a bi-unitsquare as shown in Fig. 8.6.1.η23(-1, +1)1(+1, +1)4312y4xξ(-1, -1)(+1, -1)Figure 8.6.1. Element domain in the physical and reference coordinate systems18The transformation (or mapping) between the physical domain and element or parentdomain is given by4x ξ,η = ∑ NI ξ,η xI(1.2.3a)I=14y ξ,η = ∑ NI ξ,η yI(1.2.3b)I=1where xI and yI are the nodal coordinates. Equations (3a) and (3b) can also be written inthe form4xi = ∑ NIxiI = Nx i(1.2.3c)I=1where N is a row matrix consisting of the 4 shape functionsandN = (N 1 , N 2 , N 3 , N 4 )x t1 = x t = (x1 , x 2 , x 3 , x 4 )x t2 = y t = (y1 , y 2 , y 3 , y 4 )Because the same shape functions are used for both the mapping and the displacementinterpolation, this element is called an isoparametric element.The interpolants and mapping, Eq.
(2), are bilinear in ξ,η , that is, they contain thefollowing monomials: 1, ξ, η, ξη ; the last term is called the bilinear term. Thus ux canbe written asux ξ,η = α 0x + α 1x ξ + α 2x η + α 3x ξη(1.2.4)where α ix are constants. It can easily be verified that the interpolants are linear along eachof the edges of the element as follows. Along any of the edges, either ξ or η is constant,so the monomial ξη is linear along the edges.
Therefore, while the bilinear term isnonlinear within the element, it is linear on the edges. Therefore compatibility, orcontinuity, of the displacement is assured when elements share two nodes along any edge.QUAD4 can be mixed with linear displacement triangles without any discontinuities.1.2.1 Strain Field. The strain field is obtained by using Eq. (1). Implicit differentiation isused to evaluate the derivatives because the shape functions are functions of ξ and η andthe relation (3) can not be inverted explicitly to obtain ξ and η in terms of x and y. Writingthe chain-rule for a shape function in matrix form gives:19∂NI∂NI∂xJ=∂NI∂ξ(1.2.5a)∂NI∂y∂ηwhere J is the Jacobian matrix given byJ=∂x∂y∂ξ∂ξ∂x∂y∂η∂η(1.2.5b)Its determinant is denoted by J, i.e.J = det(J)If we invert (5b), and multiply both sides of (5a) by the inverse, we obtain∂NI∂x∂NI=1J∂y-∂y∂NI∂η∂ξ∂ξ-∂x∂x∂NI(1.2.6)(1.2.7a)∂y∂η ∂ξ∂ηfrom which we see by the chain rule that∂ξ∂η∂x∂x∂η∂η∂x∂y=1J∂y-∂y∂η∂ξ-∂x∂x∂η∂ξ(1.2.7b)The derivatives of the spatial coordinates with respect to ξ and η can be obtainedfrom (3) and (2).
First∂x=1∂ξ 4∂x=1∂η 44∑ xIξI 1 + ηIη(1.2.8a)I=14∑ xIηI 1 + ξIξI=1(1.2.8b)20∂y=1∂ξ 4∂y=1∂η 44∑ yIξI (1 + ηIη(1.2.8c)I=14∑ yIηI 1 + ξIξ(1.2.8d)I=1Using the definitions of J and J given in Eqs. (5) and (6), respectively, then gives1J = 8[x24 y31 +x31 y42 + (x21 y34 +x34 y12 )ξ + (x14 y32 +x32 y41 )η]xIJ ≡ xI - xJyIJ ≡ yI - yJNote that the bilinear term is absent in J.Using the definition of the linear strain gives the following∂NIεxεy2εxy∂x4=∑0I=1∂N0∂NI∂y∂xuxI=uyI0(1.2.9b)(1.2.9c)0∂N∂y∂NI ∂NI∂N∂N∂y∂y∂x∂x(1.2.9a)ux≡ Bduy(1.2.10)utx = ux1 , ux2 , ux3 , ux4uty = uy1 , uy2 , uy3 , uy41.2.2 Linear Reproducing Conditions of Isoparametric Elements.
It will now be shownthat isoparametric elements of any order reproduce the complete linear velocity(displacement) field. This property is called linear completeness. It guarantees that theelement will pass the linear patch test and is essential for the element to be convergent.A general isoparametric element with neN nodes is considered because it is easy todemonstrate this property for any isoparametric element. The number of spatial dimensionsdenoted by neD. The isoparametric transformation isneNxi =∑ NI x xiI(1.2.11)I=1where i = 1 to neD.
The dependent variable is denoted by u. In the case of two or threedimensional solids, u may refer to any displacement component. For an isoparametric21element, the displacement field is interpolated by the same shape functions used in themapping (12), soneNu=∑ NI x uI(1.2.12)I=1Consider the situation where the displacement field is linearneDu = α o + ∑ α i xi(1.2.13)i=1so the nodal displacements are given byneDuI = α o + ∑ α i xiI(1.2.14)i=1where α o and α i are constants. This can also be written asneDuI = α o s I + ∑ α i xiI(1.2.15a)i=1orneDu = αos + ∑ αix i(1.2.15b)i=1where u and x i are column vectors of the nodal unknowns and coordinates; s is a columnvector of the same dimension consisting of all 1's. Substituting (14) into (12) yieldsneNneDu = ∑ α o s I + ∑ α i xiI NI xI=1and rearranging the terms(1.2.16)i=1neNneDneNu = α o ∑ s INI x + ∑ α i ∑ xiINI xI=1i=1neNneDI=1i=1(1.2.17)I=1It is recognized from (11) that the coefficients of α i on the right hand side of Eq (17)correspond to xi sou = α o ∑ s INI x + ∑ α i xi(1.2.18)We now make use of the fact thatneNneNI=1I=1∑ s INI = ∑ NI x=1(1.2.19)22The first equality is obvious since sI=1.
To obtain the second equality consider an elementwhose nodes are coincident: xi I = xi o for I = 1 to neN . The mapping (11) must then yieldneNxio = ∑ NI x xio(1.2.20a)I=1neN= xio∑ NI x(1.2.20b)I=1Since the above must hold for arbitrary xi o, the second equality in (19) follows.Making use of (19) then reduces (18) toneDu = α o + ∑ α i xi(1.2.21)i=1which is precisely the linear field (13) from which the nodal values uI were defined in Eq.(15).
Thus any isoparametric element contains the linear field and will exhibit constantstrain fields when the nodal displacements emanate from a linear field. As a consequence,it satisfies the linear patch test exactly.Although this attribute of isoparametric elements appears at first somewhat trivial, itssubtlety can be appreciated by noting that the bilinear terms xy will not be representedexactly in a 4-node isoparametric element. Consider for example the case when the nodaldisplacements are obtained from the bilinear field u(x,y)=xy:u(x,y) =44I=1I=1∑ uINI ξ,η = ∑ xIyINI ξ,η(1.2.22)It is impossible to extricate xy from the right hand side of Eq.
(22) unless x t = a(-1,+1, +1, -1), y t = b(-1, -1, +1, +1) where a and b are constants, i.e. when the element isrectangular. Therefore, for an arbitrary quadrilateral, the displacement field is not bilinearwhen the nodal values are determined from a bilinear field, i.e., when uI = xIyI,u(x,y) ≠ xy.Similarly, for higher order isoparametrics, such as the 9-node Lagrange element, thedistribution within the element is not quadratic when the nodal values of u are obtainedfrom a quadratic field unless the element is rectangular with equispaced nodes. For curvededges, the deviation of the field from quadratic is substantial, and the accuracy diminishes.The convergence proofs of Ciarlet and Raviart (1972) show that the order of convergencefor the 9-node element is better than the 4-node quadrilateral only when the elementmidpoint nodes are displaced from the midpoint of the side by a small amount.The linear completeness of subparametric elements can be shown analogously.
In asubparametric element, the mapping is of lower order than the interpolation of thedependent variable. For example, consider the element in Fig. 2 that has a 4-node bilinearmapping with a 9-node interpolation for u(x,y). This is written9u x,y = ∑ uINI ξ,ηI=1(1.2.23)234xIx =NI ξ,η∑yIyI=1(1.2.24)The 9-node Lagrange interpolant for the dependent variable u is distinguished from the 4node interpolant used for the element mapping by a superposed bar.
We now define a setof 9 nodes xI, yI , I = 1 to 9, which are obtained by evaluating (x,y) at the 9-nodes usedfor interpolating u(x,y) by Eq. (24). Then the mapping can be expressed as9x =∑ xyII NI ξ,ηyI=1(1.2.25)Using (23) and (25), the arguments invoked in going from Eqs. (13) to (21) can berepeated to establish the linear completeness of the subparametric element.Superparametric elements, in which the mapping is of higher order than theinterpolation of the dependent variable, are not complete. This can by shown byconsidering the element in Fig.
2 with 9-node mapping and 4-node bilinear interpolation foru(x,y). In order to use the previous argument, we would have to use the 4 nodes used forinterpolation to do a bilinear mapping, but such a mapping would be unable to reproducethe domain of the element unless it has straight edges with the nodes at the midpoints of thenodes.Nodes used for mappingNodes used forinterpolation of u(x,y)SubparametricSuperparametricFigure 2.
Examples of subparametric and superparametric elementsIn summary, it has been shown that isoparametric and subparametric elements arelinearly complete and consistent in that they represent linear fields exactly. This impliesthat when the nodal values are prescribed by a linear field, the interpolant is an identicallinear field and the derivative of the interpolant has the correct constant value throughout theelement. Therefore, for these elements, the correct constant strain state is obtained for alinear displacement field, and the patch test will be satisfied.
The element will alsorepresent rigid body translation and rotation exactly. The 4-node quadrilateral consideredhere is isoparametric, so it possesses these necessary features. A superparametric elementdoes not have linear completeness, and will therefore fail the patch test.1.2.3 Element Rank and Rank Deficiency. In order to perform reliably, an element musthave the proper rank. If its rank is too small, the global stiffness may be singular or nearsingular; in the latter case, it will exhibit spurious singular modes. If the rank of an elementis too large, it will strain in rigid body motion and either fail to converge or converge veryslowly.24The proper rank of an element stiffness is given byproper rank Ke = dim Ke - nRBrank deficiency(K e) = proper rank(K e) - rank(K e)(1.2.26a)(1.2.26b)where nRB is the number of rigid body modes.
Another way of expressing this is that ifthe element is of proper rank, thendim ker Ke = nRBwhere the kernel of K e is defined byx ∈ ker Ke if Kex = 0(1.2.27)To determine the rank of an element stiffness which is evaluated by numericalquadrature, consider the quadrature formula+1+1tKe =B t CB JdξdηB CB dΩ =Ωe-1-1nQ=∑w α J ξα B t ξα C ξα B ξα(1.2.28)α=1where C is a constitutive matrix, Ω e is the element domain, wα are the quadrature weights,and ξα are the nQ quadrature points.
In Gauss quadrature, wα correspond to the productsof the one-dimensional weight factors and ξα are the quadrature points in the referencecoordinates. The element domain in (28) and throughout this discussion is assumed tohave unit thickness. The above form can be written aso o owhereKe = B t CB(1.2.29a)B x1oB=B x2B xnQ(1.2.29b)25 w J(x 0)C(x ) w J(x 0)C(x )C= . .. 001112o22wn J(x n )C(x n ) 00QQ(1.2.29c)QoA special form of the product-rank theorem is now used, which states that when C ispositive definiteorank Ke = rank B(1.2.30)oNote that C is positive definite if and only if J and C are positive definite at all quadraturepoints.
If a material loses ellipticity, as for example in strain softening or non-associativeplastic materials, Eq. (30) no longer holds. Similarly, if the element is so distorted that J <0, the above may not hold.Assuming an element domain of unit thickness, the nodal forces are obtained directlyfrom stress field byfintxfinte =finty+1+1tB t s J dξdηB s dΩ ==Ωe-1(1.2.31a)-1where the stress is written asσxσyσxys=andtinttintfint= fx1 , fx2 , fx2 , fx4xfint= fy1 , fy2 , fy2 , fy4yApplying numerical quadrature, this becomesnQfinte =α=1o o= B tswhere∑wα J xα B t xα s xα(1.2.31b)(1.2.31c)26os t = w1 J x1 s x1 , w2 J x2 s x2 ,ownQJ xnQ s xnQ(1.2.31d)The rank of B can be estimated by the followingoorank B ≤ min(rows in B , dimDu)where D is the symmetric gradient operator given by∂∂xD=0(1.2.32)0∂∂y∂∂∂y∂xand the dimension of Du (or dimDu) is equal to the number of independent functions inDu.
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