Hutton - Fundamentals of Finite Element Analysis (523155), страница 35
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Therefore, r and s are such that the values range from−1 to +1 , and the nodal coordinates are as in Figure 6.14b.Applying the conditions that must be satisfied by each interpolation functionat each node, we obtain (essentially by inspection)N 1 (r, s) =1(1 − r )(1 − s)4N 2 (r, s) =1(1 + r )(1 − s)4(6.56a)1N 3 (r, s) = (1 + r )(1 + s)4N 4 (r, s) =1(1 − r )(1 + s)4hence(x , y) = (r, s) = N 1 (r, s)1 + N 2 (r, s)2 + N 3 (r, s)3 + N 4 (r, s)4(6.56b)434 (⫺1, 1)rry3 (1, 1)ss–y121 (⫺1, ⫺1)2 (1, ⫺1)x–x(a)(b)Figure 6.14 A four-node rectangular element showing(a) the translation to natural coordinates, (b) the naturalcoordinates of each node.185Hutton: Fundamentals ofFinite Element Analysis1866.
Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element FormulationAs in the case of triangular elements using area coordinates, the interpolationfunctions are much simpler algebraically when expressed in terms of the naturalcoordinates. Nevertheless, all required conditions are satisfied and the functionalform is identical to that used to express the field variable in Equation 6.50.Also as with area coordinates, integrations involving the interpolation functionsexpressed in the natural coordinates are simplified, since the integrands are relatively simple polynomials (for rectangular elements) and the integration limits(when integrating over the area of the element) are –1 and +1 .
Further discussionof such integration requirements, particularly numerical integration techniques,is postponed until later in this chapter.To develop a higher-order rectangular element, the logical progression is toplace an additional node at the midpoint of each side of the element, as in Figure 6.15. This poses an immediate problem, however. Inspection of the Pascaltriangle shows that we cannot construct a complete polynomial having eightterms, but we have a choice of two incomplete, symmetric cubic polynomials:(x , y) = a0 + a1 x + a2 y + a3 x 2 + a4 x y + a5 y 2 + a6 x 3 + a7 y 3(x , y) = a0 + a1 x + a2 y + a3 x + a4 x y + a5 y + a6 x y + a7 x y222(6.57a)2(6.57b)Rather than grapple with choosing one or the other, we use the natural coordinates and the nodal conditions that must be satisfied by each interpolation function to obtain the functions serendipitously.
For example, interpolation functionN 1 must evaluate to zero at all nodes except node 1, where its value must beunity. At nodes 2, 3, and 6, r = 1 , so including the term r − 1 satisfies the zerocondition at those nodes. Similarly, at nodes 4 and 7, s = 1 so the term s − 1 ensures the zero condition at those two nodes. Finally, at node 5, (r, s) = (0, −1) ,and at node 8, (r, s) = (−1, 0) . Hence, at nodes 5 and 8, the term r + s + 1 isidentically zero. Using this reasoning, the interpolation function associated withnode 1 is to be of the formN 1 (r, s) = (1 − r )(1 − s)(r + s + 1)(6.58)Evaluating at node 1 where (r, s) = (−1, −1) , we obtain N 1 = −4 , so acorrection is required to obtain the unity value. The final form is then1N 1 (r, s) = (r − 1)(1 − s)(r + s + 1)(6.59a)4473sy8x6r152Figure 6.15 Eight-noderectangular element showingboth global and naturalcoordinate axes.Hutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.7 Three-Dimensional ElementsA parallel procedure for the interpolation functions associated with the otherthree corner nodes leads toN 2 (r, s) =1(r + 1)(1 − s)(s − r + 1)4(6.59b)N 3 (r, s) =1(1 + r )(1 + s)(r + s − 1)4(6.59c)N 4 (r, s) =1(r − 1)(1 + s)(r − s + 1)4(6.59d)The form of the interpolation functions associated with the midside nodes issimpler to obtain than those for the corner nodes.
For example, N 5 has a value ofzero at nodes 2, 3, and 6 if it contains the term r − 1 and is also zero at nodes 1,4, and 8 if the term 1 + r is included. Finally, if a zero value is at node 7,(r, s) = (0, 1) is obtained by inclusion of s − 1 . The form for N 5 isN5 =11(1 − r )(1 + r )(1 − s) = (1 − r 2 )(1 − s)22(6.59e)where the leading coefficient ensures a unity value at node 5. For the other midside nodes,N6 =1(1 + r )(1 − s 2 )2(6.59f)N7 =1(1 − r 2 )(1 + s)2(6.59g)N8 =1(1 − r )(1 − s 2 )2(6.59h)are determined in the same manner.Many other, successively higher-order, rectangular elements have been developed [1].
In general, these higher-order elements include internal nodes that,in modeling, are troublesome, as they cannot be connected to nodes of otherelements. The internal nodes are eliminated mathematically. The eliminationprocess is such that the mechanical effects of the internal nodes are assignedappropriately to the external nodes.6.7 THREE-DIMENSIONAL ELEMENTSAs in the two-dimensional case, there are two main families of three-dimensionalelements. One is based on extension of triangular elements to tetrahedrons andthe other on extension of rectangular elements to rectangular parallelopipeds187Hutton: Fundamentals ofFinite Element Analysis1886.
Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulation(often simply called brick elements). The algebraically cumbersome techniquesfor deriving interpolation functions in global Cartesian coordinates has beenillustrated for two-dimensional elements. Those developments are not repeatedhere for three-dimensional elements; the procedures are algebraically identicalbut even more complex.
Instead, we utilize only the more amenable approach ofusing natural coordinates to develop the interpolation functions for the two basicelements of the tetrahedral and brick families.6.7.1 Four-Node Tetrahedral ElementA four-node tetrahedral element is depicted in Figure 6.16 in relation to aglobal Cartesian coordinate system. The nodes are numbered 1–4 per the convention that node 1 can be selected arbitrarily and nodes 2–4 are then specifiedin a counterclockwise direction from node 1. (This convention is the same asused by most commercial finite element analysis software and is very important in assuring geometrically correct tetrahedrons.
On the other hand, tetrahedral element definition for finite element models is so complex that it isalmost always accomplished by automeshing capabilities of specific softwarepackages.)In a manner analogous to use of area coordinates, we now introduce the concept of volume coordinates using Figure 6.17. Point P (x , y, z) is an arbitrarypoint in the tetrahedron defined by the four nodes. As indicated by the dashedlines, point P and the four nodes define four other tetrahedra having volumesV1 = vol(P234)V2 = vol(P134)V3 = vol(P124)V4 = vol(P123)(6.60)33P442211Figure 6.16 A four-nodetetrahedral element.Figure 6.17 A four-nodetetrahedral element, showingan arbitrary interest pointdefining four volumes.Hutton: Fundamentals ofFinite Element Analysis6.
Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.7 Three-Dimensional ElementsThe volume coordinates are defined asV1VV2L2 =VV3L3 =VV4L4 =VL1 =where V is the total volume of the element given by 1 x1 y1 z 1 1 1 x2 y2 z 2 V = 6 1 x3 y3 z 3 1 x4 y4 z 4(6.61)(6.62)As with area coordinates, the volume coordinates are not independent, sinceV1 + V2 + V3 + V4 = V(6.63)Now let us examine the variation of the volume coordinates through theelement. If, for example, point P corresponds to node 1, we find V1 = V ,V2 = V3 = V4 = 0 . Consequently L 1 = 1, L 2 = L 3 = L 4 = 0 at node 1. As Pmoves away from node 1, V1 decreases linearly, since the volume of a tetrahedron is directly proportional to its height (the perpendicular distance from P tothe plane defined by nodes 2, 3, and 4) and the area of its base (the triangleformed by nodes 2, 3, and 4).
On any plane parallel to the base triangle of nodes2, 3, 4, the value of L 1 is constant. Of particular importance is that, if P lies in theplane of nodes 2, 3, 4, the value of L 1 is zero. Identical observations apply tovolume coordinates L 2 , L 3 , and L 4 . So the volume coordinates satisfy all required nodal conditions for interpolation functions, and we can express the fieldvariable as(x , y, z) = L 1 1 + L 2 2 + L 3 3 + L 4 4(6.64)Explicit representation of the interpolation functions (i.e., the volume coordinates) in terms of global coordinates is, as stated, algebraically complexbut straightforward. Fortunately, such explicit representation is not generallyrequired, as element formulation can be accomplished using volume coordinatesonly.
As with area coordinates, integration of functions of volume coordinates(required in developing element characteristic matrices and load vectors) is relatively simple. Integrals of the formL a1 L b2 L c3 L d4 dV(6.65)V189Hutton: Fundamentals ofFinite Element Analysis1906. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulationwhere a, b, c, d are positive integers and V is total element volume, appear inelement formulation for various physical problems. As with area coordinates,integration in volume coordinates is straightforward [7], and we have the integration formulaL a1 L b2 L c3 L d4 dV =Va!b!c!d !(6V )(a + b + c + d + 3)!(6.66)which is the three-dimensional analogy to Equation 6.49.As another analogy with the two-dimensional triangular elements, the tetrahedral element is most useful in modeling irregular geometries.
However, thetetrahedral element is not particularly amenable to use in conjunction with otherelement types, strictly as a result of the nodal configurations. This incompatibility is discussed in the following sections. As a final comment on the fournode tetrahedral element, we note that the field variable representation, as givenby Equation 6.64, is a linear function of the Cartesian coordinates.
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