2D Elasticity Computational Issues

# 2D Elasticity Computational Issues - Computational Issues...

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Unformatted text preview: Computational Issues Related to Finite Element Analysis of Elasticity Problems Recall the numerical integration schemes used to calculate element matrices for two—dimensional rectangular elements. These methods incorporate transformations onto the master element in the natural coordinate system, require use of the J acobian matrix, and evaluate things at particular Gauss points within the element as shown in table below. These steps have inﬂuence on the accuracy of determining the displacements and stresses within an element. The following notes describe some of the issues related to this. Table 9.3.1 Selection of the integration order and location of the Gauss points for linear, quadratic, and cubic quadrilateral elements (nodes not shown) Maximum Order of Order of Location of Element polynomial integration the integration points* type degree (r X r) residual in master element ‘1 Constant (r = 1) 0 1 x l 0012) Linear (r = 2) 2 2 X 2 0014) Quadratic (r = 3) 4 (3 x 3) 00:6) Cubic (r = 4) 6 (4 x 4) 0(128 ) *See Table 6.1.2 for the integration points and weights for each coordinate direction. General Concepts on Reduced Integration It can be shown that the displacement formulation for linear elastic ﬁnite element analysis represents a lower bound on the strain energy of the system and thus this formulation results in overestimating the system stiffness. Thus it may turn out that by not evaluating the element stiffness matrices exactly in the numerical integration scheme, a more accurate result could be obtained. This would be the case if the error in the numerical routine compensates appropriately for the overestimation of the structural stiffness due to the ﬁnite element discretization. Therefore, a reduction in the usual numerical integration order used to calculate the element stiffness matrices can actually improve the ﬁnal results, and this concept is generally referred to as reduced integration. The method provides a softening effect because certain higher—order polynomials terms will vanish at Gauss points of a low—order rule, thus eliminating these terms from contribution to the system stiffness. With fewer sampling points, some of the more complicated displacement modes offer less resistance to deformation. This concept has lead to many studies aimed at determining the optimum integration order and scheme for various ﬁnite element types. In addition to simply using a reduced integration scheme, it has been found that optimum results for some problems can be obtained by using a selective integration method whereby different orders of integration are applied to different terms in the stiffness matrix. Although reduced integration can be useful it can also lead to ill-conditioned system matrices, and thus the choice of using such a scheme is not always straightforward. The following two criteria are commonly used to help ensure the proper selection of reduced integration methods: 1. the element should not contain any spurious zero energy modes; i.e. the rank of the element stiffness matrix should not be smaller than that evaluated exactly. 2. the element should contain the required constant strain states. Bathe (1982) provides additional general information on this concept, and some additional notes from: Concepts and Applications of Finite Element Analysis, by Cook, Malkus and Plesha, 1989 follows. 6.11 APPROPRIATE ORDER OF QUADRATURE For numerically integrated elements, we deﬁne “full integration” as a quadrature rule sufﬁcient to provide the exact integrals of all terms kg- in the element stiffness matrix if the element is undistorted (e. g., if a quadratic element has straight sides and midside nodes). The same “full integration” rule will not exactly integrate all kij if sides are curved or if side nodes are offset from the midpoints, for then J is not constant throughout the element. For example, in Eq. 6.2-6, [B J is linear in g and J is constant if node 3 is centered. Therefore, the integrand contains terms up to £2, which are integrated exactly by two Gauss points. Accordingly, for this element, even if node 3 is not centered, two-point Gauss quadrature is considered “full integration.” Use of full integration is the only sure way to avoid pitfalls such as mesh instabilities, which are discussed in Section 6.12. However, a lower—order quadrature rule, called “reduced integration,” may be desirable for two reasons. First, since the expense of generating a matrix [k] by numerical integration is proportional to the number of sampling points, using fewer sampling points means lower cost. Second, a low=order rule tends to soften an element, thus countering the overly stiff behavior associated with an assumed displacement ﬁeld. Softening comes about because certain higher-order poly-3 nomial terms happen to vanish at Gauss points of a low—order rule, so that these terms make no contribution to strain energy. In other words, with fewer sampling points, some of the more complicated displacement modes offer less resistance to deformation. In sum, our argument is that reduced integration may be able to simultaneously reduce cost, reduce accuracy in the evaluation of integral expres- sions, and increase the accuracy of a ﬁnite element analysis. Reduced integration should not be used if cost reduction is the only motivation. The number of Gauss points has a lower limit because in the limit of mesh reﬁnement, element volume must be integrated exactly. We argue as follows. As a mesh is indeﬁnitely reﬁned, a constant-strain condition is approached in each element, provided that the element is valid in the patch test sense. Thus strain energy density U0 becomes constant throughout each element. Strain energy in an element, for plane and solid problems, respectively, is Ue=fo0tJd§dn or U€=ffoOJd§dnd§ (6.11-1) If U0 is constant, then U6, will be correct if volume dV = t] a’f d7; (or (W = J dg dn d5”) is correctly integrated. In practice, we prefer to use exact volume integration for any shape and size of element. From Eq. 4.1—10 we see that integrals in Eq. 6.11-1 produce terms in the element stiffness matrix [k]. Accordingly, if [k] is produced by an integration rule adequate to compute element volume exactly, the element will be able to provide the correct strain energy in a constant-strain deformation mode. Thus, for an element of arbitrary geometry, the minimum quadrature require— ment is a rule that exactly integrates tJ (plane case) or J (solid case). In a plane bilinear element of constant thickness, U is linear in f and in n, so one Gauss point is required. In a plane quadratic element of constant thickness, tJ contains 53 and 773, so a 2 by 2 Gauss rule is required. The eightanode solid also requires an'order 2 Gauss rule (8 points). However, with rare practical exceptions, indeﬁnitely repeated subdivision of a mesh yields elements that become straight-sided parallelograms of constant thick- ness. Thus t and J cease to be functions of the coordinates and, in the limit, a single Gaus‘s point yields the correct element volume. For an isoparametric element based on an assumed displacement ﬁeld, the best quadrature rule is usually the lowest-order rule that computes volume correctly and does not produce instability. Numerical testing of any proposed rule is man— datory. Solution accuracy may be mesh-dependent and problem—dependent, but usually one quadrature rule will be clearly superior to others. For bilinear and eight-node plane elements, and for the eight-node linear solid element, an order 2 Gauss rule is favored (four and eight points for plane and solid elements, respectively). The quadratic serendipity solid, having eight corner nodes and twelve edge nodes, can be integrated with an order 3 rule (27 points), but a special 14-point rule may be preferred, especially if the element is made very thin in one direction [6.8—6. 10]. 6.12 ELEMENT AND MESH INSTABILITIES An instability may also be called a spurious singular mode. In structural me— chanics, an instability may be known as a mechanism, a kinematic made, an hourglass mode, or a zero-energy mode. The term “zero—energy mode” refers to a nodal displacement vector {D} that is not a rigid-body motion but nevertheless produces zero strain energy {D}T[K]{D}/2. lnstabilities arise because of shortcom— ings in the element formulation process, such as use of a low—order Gauss quad- rature rule. In the present context, an instability has nothing to do with buckling problems of structures. Y A structure that appears adequately constrained may yet have an instability that makes [K] singular. Or, unstable elements may combine to form a structure that is stable but unduly susceptible to certain load patterns, so that computed displacements are excessive. To explain the term “zero—energy mode” further and show how such a mode may arise, we substitute the relation {5} = [B]{d} into the expression for strain energy in an element, U... From Eq. 4.1—10 and the standard expression for [k], Eq. 4.1-5, we obtain Ue = %{d}T[k]{d} = %{d}T [V [BlTiEllBl dVld} = t [V {€}T[E]{é} dV (6-12-1) When [k] is formed by numerical integration, it contains only the information that can be sensed at the sampling points of the quadrature rule. If it happens that strains {6} = [B]{d} are zero at all sampling points for a certain mode {(1}, then Ue will vanish for that {d}, and, according to Eq. 6.12—1, [k] will be a zero-stiffness matrix in the sense that strain energy Ue = {d}T[k]{d}/2 is zero for this particular {(1}. We expect that U... = 0 if {d} is a rigid—body motion. If U6 = 0 when {d} is not a rigid—body motion, then an instability is present. An element that displays a mechanism is said to be rank deﬁcient. That is, the rank of [k] is less than the number of element d.o.f. minus the number of rigid- body modes. An instability in an existing [k] can be detected by means of an eigenvalue test (Section 18.8). In the present section we give examples of instabilities and brieﬂy discuss their prevention. Examples. Consider the four-node plane (bilinear) element, whose stiffness matrix is 8 by 8. Eight independent displacement modes {(1} can be identiﬁed (Fig. 6.12-1). The ﬁrst three are rigid-body modes, for which Ue = 0, as is correct, regardless of the quadrature rule used. The next three modes, numbers 4, 5, and 6, are constant-strain modes, for which U, > 0, regardless of the quadrature rule used. Modes 7 and 8 are bending modes. An order 1 rule, whose single Gauss point is at the element center, does not sense these modes, as ex = E), = 7,, = O at the center. Accordingly, Ue = O for modes 7 and 8, and the element displays two mechanisms. These two spurious modes disappear if the Gauss rule is order 2 or greater. The foregoing mechanisms can appear in a mesh of elements as well as in a single element (Fig. 6.12-2). In Fig. 6.12-2a’, modes 7 and 8 of Fig. 6.12—1 are combined with a rigid-body rotation of each element. The mechanisms of Fig. Figure 6.124. Independent displacement modes of a bilinear element. 6.12—2 are called hourglass modes because of their physical shape. Each of these distortions, as well as each straining mode in Fig. 6.12—1, would be considered the same mode if nodal d.0.f. were reversed—that is, if {d} of that mode were replaced by -{d}. Elements need not be rectangular in order to display a mechanism. Imagine, for example, that displacements are u = algn and u : azgn, where a 1 and a2 are constants. Then, at§ = n = 0, we have “,5 = u”, = v,§ = 0,1, = 0; hence, according to Eqs. 6.3—17 and 6.3—18, ex = e, = 31,, = 0 at the Gauss point of an order 1 rule, regardless of the shape of the element. Consider next the quadratic plane element, having either eight or nine nodes, and integrated with a 2 by 2 Gauss rule (Fig. 6.12-3). Displacements in the nine- node element of Fig. 6.12=3b are [6.11] i: f (3)52“ ‘ 52 ‘ 7'2 (6.122) At the Gauss points of a 2 by 2 rule—that is, where g and n are : 1/\/§——0ne ﬁnds Ll,§ = u”, = U“: = v”, = 0. Therefore, according to Eqs. 6.3-17 and 6.318, strains are zero at these points, for any geometric shape of the undeformed ele- .71,” (c) 7 " Figure 6.12-2. (a) Mesh of four bilinear elements, showing Gauss points of an order 1 rule in each element (squares). (b,c,d) Possible mechanisms (“hourglass” modes). Eight- or nine-node elements Nine-node element only Nine-node element only Eight- and nine-node elements (a) (b) (0) ((1) Figure 6.12-3. Possible mechanisms (“hourglass” modes) in quadratic elements integrated by an order 2 rule. Gauss points are shown by squares. ment. A similar 0 ﬁeld is possible (Fig. 6.12-3c). Thus we have identiﬁed two mechanisms. The foregoing two mechanisms are not possible in the eight—node element be= cause the £2712 term is not present (see Eqs. 6.6-4). However, yet another mech- anism is possible in both eight-node and nine-node elements (Fig. 6.12—3d). Its displacement ﬁeld is simple to state for a square element; it is u = 5(3772 — 1) and I) = n(1 ~— 38) (6.12—3) Again, strains are zero at the Gauss points of a 2 by 2 rule. This mechanism is usually not of great concern because two adjacent elements cannot both have such a mode, as may be seen by trying to connect two deformed elements. Thus an instability present in individual elements is not present in the mesh. Summing up, Fig. 6.12-3 identiﬁes three element instabilities in quadratic ele— ments arising from a 2 by 2 Gauss quadrature rule. The element stiffness matrix has rank 12 for both eight—node and nine—node elements (rank equals order less the number of rigid-body and instability modes). None of these instabilities exists if the Gauss rule is 3 by 3 or greater. A mesh that has no mechanisms may yet behave badly because restraints on the mechanisms are weak. Consider Fig. 6.12a4a. Elements may be the four-node elements of Fig. 6.12—2 or the nine—node elements of Fig. 6.123, respectively integrated by one—point and four-point rules. Load P is concentrated and applied centrally rather than being distributed across the right end. Mechanisms are not possible because all nodes at the left support are ﬁxed. However, this restraint ﬁc [<— —3% LT ——>=| (a) Figure 6.124. Problems that involve “near mechanisms,” if reduced integration is used. In (b), elements that were initially rectangular are here shown deformed. becomes weaker with increasing distance from the support. Nearthe load, dis— tortions of the type shown in Figs. 6.12—219 and 6.12—3b become pronounced. Indeed, for a 2 by 24 mesh, the computed displacement of load P may be over 500 times the displacement predicted by the elementary formula PL/AE [6.12]. A similar situation is depicted in Fig. 6.12-4b [4.6]. A 2 by 2 Gauss rule is used to integrate [k] of each element. The stiff element, shown shaded, is weakly restrained by soft elements connected to the ﬁxed boundary, allowing the mode of Fig. 6.12—3d (with signs of {d} reversed) to become pronounced, although not unbounded. Elements for solids, for plate bending, and for nonstructural problems can also suffer from instabilities. Methods for detecting and controlling these modes are similar to methods used for plane elements. A conservative analyst will avoid using any element that contains a possible instability because its dangers may not be foreseen. Control of Instabilities. Various control methods have been proposed. Their goal is to eliminate instability by providing restraint, but without simultaneously stiff~ ening the element’s response to “legitimate” modes that are already working well. In what follows we summarize an effective method, with particular reference to a rectangular bilinear element. The method adds “hourglass stiffness” to an ele- ment integrated by one—point quadrature. The resulting element is inexpensive to formulate and works very well. For simplicity, consider only the x—direction nodal displacements {d,‘.} of the eight modes shown in Fig. 6.12—1. For modes 1 and 8, {dx} = {0}. An arbitrary combination of modes 2 through 6 is 1 1 ~ 1 —1 1 1 1 —1 —1 {dx} = a2 1 + a3 _1 + a4 1 + a5 _1 + a6 1 (612-4) 1 -1 —1 1 1 where the a,- are constants. Mode 7 is {dx}7 = a7 ’1 1 " IJT To provide mode 7 with the stiffness it lacks under one-point quadrature, we form the “stabilization matrix” 7‘ = A similar matrix [k]8, containing a constant as, serves to restrain mode 8. To the stiffness matrix computed by one—point quadrature, we now add [k]7 and [k]g. It is possible to choose values of a7 and a8 such that a rectangular element displays the exact strain energy in states of pure bending. Note that mode 7 is orthogonal to all other modes—that is, {dx}7T{dx},- = {0} for i = 1, 2, 3, 4, 5, 6, 8 (6.12—7) Orthogonality prevents [k]7 from stiffening modes other than mode 7. That this is so may be seen by computing nodal forces {’1‘}, associated with matrix [k]7, = = = = for i = 1 through 6 and for i = 8. The foregoing control method can be generalized to elements having more than four nodes and to elements of arbitrary shape [6.11, 6.13, 13.49, 13.52—13.54]. 6.13 REMARKS ON STRESS COMPUTATION Element stresses follow from Eq. 4.7-1, with the substitution {6} = [B]{d}: {0'} = [El(lBl{d} — {60}) + {0‘0} (613-4) Here, in isoparametric elements, [B] is a function of the natural coordinates and {0'} contains stresses referred to the global coordinate system xyz. Where in the element should stresses be calculated? For isoparametric elements, it often hap— pens that stresses (especially shear stresses) are most accurate at Gauss points of a quadrature rule one order less than that required for full integration of the element stiffness matrix. Consider Fig. 6.13—1. Sides of a bilinear element remain straight during defor- mation. A typical element, deformed by bending moment but with rigid—body motion removed, is shown in Fig. 6.13-11). Displacements in the element are u = mag»? and v = 0, where a, is a positive constant. Thus shear strain yxy is pro= portional to 5. On the neutral surface of bending, 34,, displays the sawtooth pattern seen in Fig. 6.13-1c. Only at g = 0 in each element is 34., correctly computed (as zero) under pure bending deformation. In a general problem of plane stress anal- ysis, where bending can occur in both directions (modes 7 and 8 of Fig. 6.12-1 simultaneously), the best computation point for yxy in a'bilinear element is at E = n = O. This is the Gauss point location of an order .1 rule, which is one order less than the order 2 rule of full integration. A similar circumstance occurs with the eight-node and nine—node quadratic elements. In the beam of Fig. 6.13-2, the exact 31,, is constant along the x axis. In the quadratic element, 'yxy along the x axis displays the parabolic distributions shown. However, one ﬁnds that the quadratic element displays the correct yxy at the Gauss points of a 2 by 2 quadrature rule. In other problems of stress analysis, normal strains can also display parabolic variations, and again the most accurate strains are to be found at the Gauss points of an order 2 rule. 'YmJ Finite element (0) Figure 6.13-1. (a) Beam loaded in bending. (b) Bending distortion of a typical bilinear element. (0) Shear strain along the x axis. Remarks on Stress Computation 195 Finite element y 7W Exact , l l / . 1 l 1 iv 0 «l— L—x l “a | ‘ | 0 L rLlL» (a) (b) Figure 6.132. ((1) Beam loaded by transverse tip force V. (b) Shear strain along the x axis. In elements based on displacement ﬁelds, one expects stresses to be less ac“ curate than displacements, as explained in Section 3.5. However, in the foregoing examples, stresses are “superaccurate” or “superconvergent” at the Gauss points because there they have the same degree of accuracy as displacements. Indeed, in unusual situations it may happen that stresses are more accurate than displace- ments. For example, in Fig. 6.l2~4a, stresses may be substantially correct at Gauss points (of an order 1 or order 2 rule, for four— and nine—node elements, respectively), although displacements are grossly in error. This is possible because the modes that permit excessive displacements produce zero strain at the Gauss points. The theory of locating error—minimal points for stress computation is explained elsewhere [6.]4,6.15]. One discovers that these points are Gauss points: at g = 77 = O in bilinear (plane) and trilinear (solid) elements, and where E, 77, and g are : l/\/§ in eight— or nine—node quadratic (plane) and 20— or 21-node quadratic (solid) elements. These conclusions are rigorously true for rectangular elements. For distorted elements, Gauss points may not be optimal locations but they remain very good choices. Stresses at Gauss points can be interpolated or extrapolated to other points in the element. The result obtained is usually more accurate than the result of eval- uating Eq. 6.13—1 directly at the point of interest. The interpolation—extrapolation process is explained as follows. ’ Imagine that stresses have been computed at the four Gauss points of a plane element (points 1, 2, 3, and 4 in Fig. 6.13-3). We now Wish to interpolate or extrapolate these stress values to other points in the element. In Fig. 6.13—3, coordinate r is proportional to f and s is proportional to 7;. At (say) point 3, r = s = 1 and E = 7; = 1/\/§. Therefore, the factor of proportionality is W; that is, Figure 6.13-3. Natural coordinate systems used in extrapolation of stresses from Gauss points. 196 THE ISOPARAMETRIC FORMULATION r = §\/§ and s : n\/§ (6.13-2) Stresses at any point P in the element are found by the usual shape functions, 0"}? 2 ZNiCTf for = 1, 2, 3, 4 where 0" is 0x, cry, or 730,. The N,~ are the bilinear shape functions given by Eq. 6.3-2, but now written in terms of r and 5 rather than 5 and n; that is, N,- : 2411(1 : r)(l : s) (6.13—4) In Eq. 6.13-3, the N,- are evaluated at the r and s coordinates of point P. For example, let point P coincide with corner A. To calculate stress aim at corner A from 0:, values at the four Gauss points, we substitute r = s = —\/§ into the shape functions, and obtain 0,, = 1.86601.1 — 0.5000,.2 + 0.13401.3 —- 0.5000,.4 (6.13—5) For solids, an interpolation~extrapolation formula similar to Eq. 6.13—3 is based on stresses at eight Gauss points and the trilinear N, of Eq. 6.7—6. In Section 4.7 we advised that usually the temperature ﬁeld used for thermal stress analysis should have the same competence as the element strain ﬁeld. Accordingly, if element stresses are based on Gauss point values, thermal strains {50} in Eq. 6.13—1 should also be based on Gauss point values. 6.14 EXAMPLES. EFFECT OF ELEMENT GEOMETRY Simple test problems show how accuracy is affected by element distortion, changes in Gauss quadrature rule, and changes in element aspect ratio. Our examples are two—dimensional, but the trends displayed pertain to three—dimensional elements as well. Example Problems. Table 6.14—1 illustrates the behavior of the bilinear element when its [k] is formed by four—point Gauss quadrature. Results are expressed as TABLE 6.144. STRESSES AND DEFLBCTIONS IN CANTILEVER BEAMS OF CONSTANT THICKNESS UNDER TRANSVERSE TIP LOAD P. LENGTH = 10, DEPTH = 2, V : 0.25. VALUES BY BEAM THEORY = 1.000, OF WHICH 3% OF 0,. IS DUE To TRANSVERSE SHEAR DEFORMATION . 1W luv lyw %, 9 E. g :g «»~v.w.~+.e-—:+,~,vwxr;rwgl w» J a.»“we-«waves?Hm...”Why‘d/my Examples. Effect ofElement Geometry 197 the ratio of computed value to the value given by beam theory. We see that square elements are better than elongated elements, and that geometric distortion stiffens the element and makes answers less accurate. The principal failing of the bilinear element is that under pure bending loads, for which yxy should be zero, the element displays substantial values of yxy except at its center, as noted in connection with Fig. 6.13-1. This defect, known as parasitic shear, makes the element too stiff in bending. An improved form of the element discussed in Section 8.3. Table 614—2 illustrates the behavior of eight—node and nine—node versions of the quadratic element [6.12]. All nodes at the left end are ﬁxed. Load P on the right end is allotted to nodes in the proportion 1—8—1, which is consistent with a parabolic distribution of shear stress. Side nodes are midway along the sides. Point B is a Gauss point of a 2 by 2 quadrature rule. Results are expressed as the ratio of computed value to the value given by beam theory. When elements are rectangular, we see that eight-node and nine-anode elements have comparable accuracy. Both become stiffer if the quadrature rule used to generate [k] is changed from 2 by 2 to 3 by 3. Next in Table 614—2, elements are made trapezoidal by moving nodes C and D horizontally to positions U4 and 3L/4, where L is the length of the beam. The ﬁnal mesh in Table 614-2 introduces one curved interelement boundary by moving node E left of center an amount L/20. We see that 2 by 2 is the preferred integration rule, and that the eight-node element is much more sensitive to geometric dis»= tortion than the nine=node element. In one case, stress 0,3 in the eightnnode element is not even of the correct sign. The obvious lesson is that an ideal element is compact, straight—sided, and has equal corner angles. Of course, elements must be distorted to some extent in modeling an actual structure, but gratuitous distortion is to be avoided. In par— ticular, if an element side is curved to model the curved boundary of a structure, other element sides that form interelement boundaries should be straight. Quadratic triangles (Sections 5 .5 and 6.8) have approximately the same accuracy as the nine-node element. For example, in the second case in Table 614—2, let TABLE 6.142. STRESSES AND DEFLECTIONS IN TWO—ELEMENT CANTILEVER BEAMS OF CONSTANT THICKNESS UNDER TRANSVERSE TIP LOAD P. LENGTH = 100, DEPTH = 10, v = 0.30. VALUES BY BEAM THEORY = 1.000 (IN WHICH THE TRANSVERSE-SHEAR CONTRIBUTION To 0,, Is NEGLECTED). SKETCHES ARE NOT TO SCALE. lam) Element \ \ Type DA 0' xB DA 0 x3 DA 8 node 2 X 2 1.000 0.968 0.051 0.362 —0.048 0.430 8 node 3 X 3 1.129 0.930 0.048 0.161 0.050 0.221 9 node 2 X 2 1.000 1.006 1.125 1.109 0.958 0.955 9 node 3 X 3 1.141 0.954 0.687 0.791 0.705 0.737 ...
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## This note was uploaded on 10/03/2011 for the course MCE 561 taught by Professor Sadd during the Spring '11 term at Rhode Island.

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2D Elasticity Computational Issues - Computational Issues...

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