Triangular Plate Displacement Elements

Transcription

1 Triangular Plate Displacement Elements

2 Chapter : TRIANGULAR PLATE DISPLACEMENT ELEMENTS TABLE OF CONTENTS Page. Introduction Triangular Element Properties Triangle Geometry Triangular Coordinates Triangular CoordinatesProperties Linear Interpolation Coordinate Transformations Partial Derivatives Six Node Quadratic Interpolation Cubic Interpolation Cubic Interpolation Choice : 0 Nodes Cubic Interpolation Choice : 4 Nodes plus 6 Side Slopes Cubic Interpolation Choice : 4 Nodes plus 6 Cartesian Slopes Cubic Interpolation Choice 4: 4 Nodes plus 6 Cartesian Rotations 9.4 The BCIZ Plate Bending Element The Kinematic Constraint The Curvature Displacement Matrix The Element Stiffness Matrix Notes and Bibliography Exercises

3 . TRIANGULAR ELEMENT PROPERTIES.. Introduction This Chapter presents an overview of finite element models for thin plates using the Kirchhoff Plate Bending (KPB) model and the Total Potential Energy (TPE) vaariational principle.the derivation of shape functions for the triangle geometry is covered in the next Chapter. The purpose of this Chapter is to explain the construction of displacement-based triangular Kirchhoff plate bending elements through the development of the necessary interpolation formulas. The resulting element is complete but does not satisfy full normal-slope conformity... Triangular Element Properties We recall the following properties of a straight-sided triangular element, taken from Chapter of IFEM.... Triangle Geometry The geometry of the -node triangle shown in Figure.(a) is specified by the location of its three corner nodes on the {x, y} plane. The nodes are labeled,, while traversing the sides in counterclockwise fashion. The location of the corners is defined by their coordinates: x i, y i, i =,, (.) The area of the triangle is denoted by A and is given by [ A = det x x x = (x y x y ) + (x y x y ) + (x y x y ). (.) y y y It is important to realize that the area given by formula (.) is a signed quantity. It is positive if the corners are numbered in counterclockwise order as shown in Figure.(b). This convention is followed in the sequel. (a) (x,y ) (b) Area A > 0 y (x,y ) x z up, towards you (x,y ) Figure.. Geometry of stright-sided triangular element.

4 Chapter : TRIANGULAR PLATE DISPLACEMENT ELEMENTS... Triangular Coordinates Points of the triangle may also be located in terms of a parametric coordinate system: ζ,ζ,ζ. (.) In the literature these three parameters receive many names. In the sequel the name triangular coordinates will be used to stress its close association with this particular geometry.... Triangular CoordinatesProperties Equations ζ i = constant (.4) represent a set of straight lines parallel to the side opposite to the i th corner. See Figure.. The equation of sides, and are ζ = 0, ζ = 0 and ζ = 0, respectively. The three corners have coordinates (,0,0), (0,,0) and (0,0,). The three midpoints of the sides have coordinates (,, 0), (0,, ) and (, 0, ), the centroid (,, ), and so on. The coordinates are not independent because their sum is unity: ζ + ζ + ζ =. (.) /4 ζ = 0 /4 ζ = / / /4 / /4 /4 /4 ζ = ζ = 0 ζ = ζ = 0 Figure.. Triangular coordinates...4. Linear Interpolation Consider a function w(x, y) that varies linearly over the triangle domain. In terms of Cartesian coordinates it may be expressed as w(x, y) = a 0 + a x + a y, (.6) where a 0, a and a are coefficients to be determined from three conditions. In finite element work such conditions are often the nodal values taken by w at the corners: w,w,w (.7) 4

5 . TRIANGULAR ELEMENT PROPERTIES (a) (b) 6 4 Figure.. Nodal configurations for: (a) linear interpolation of w by three values w i, at corners i =,, ; (b) quadratic interpolation of w by six values w i at corners i =,, and midpoints i = 4,, 6. The expression in triangular coordinates makes direct use of these three values: w(ζ,ζ,ζ ) = w ζ + w ζ + w ζ = [ w w w [ ζ ζ ζ = [ ζ ζ ζ Expression (.8) is called a linear interpolant for w. See Figure.(a).... Coordinate Transformations [ w w w. (.8) Quantities which are closely linked with the element geometry are naturally expressed in triangular coordinates. On the other hand, quantities such as displacements, strains and stresses are often expressed in the Cartesian system x, y. We therefore need transformation equations through which we can pass from one coordinate system to the other. Cartesian and triangular coordinates are linked by the relation [ x y [ = x x x y y y [ ζ ζ. (.9) ζ The first equation says that the sum of the three coordinates is one. The second and third express x and y linearly as homogeneous forms in the triangular coordinates. These simply apply the linear interpolant formula (.8) to the Cartesian coordinates: x = x ζ + x ζ + x ζ and y = y ζ + y ζ + y ζ. Inversion of (.9) yields [ ζ ζ ζ [ [ = x y x y y y x x x y x y y y x x x = A x y x y y y x x y A [ [ A y x A y x x. A y x y (.0) Here x jk = x j x k, y jk = y j y k, A is the triangle area given by (.) and A jk denotes the area subtended by corners j, k and the origin of the x y system. If this origin is taken at the centroid of the triangle, A = A = A = A/.

6 Chapter : TRIANGULAR PLATE DISPLACEMENT ELEMENTS..6. Partial Derivatives From equations (.9) and (.0) we immediately obtain the following relations between partial derivatives: x y = x ζ i, = y i ζ i, (.) i A ζ i x = y jk, A ζ i y = x kj. (.) In (.) j and k denote the cyclic permutations of i. For example, if i =, then j = and k =. The derivatives of a function w(ζ,ζ,ζ ) with respect to x or y follow immediately from (.) and application of the chain rule: w x = ( w y + w y + w ) y A ζ ζ ζ w y = ( w x + w x + w ) (.) x A ζ ζ ζ which in matrix form is..7. Six Node Quadratic Interpolation w w ζ x w = [ y y y w A x x x ζ. (.4) y w ζ Consider next the six-node triangle shown in Figure.(b). This element still has straight sides, and nodes 4, and 6 are located at the midpoint of the sides. In Chapter 6 of the IFEM course it was shown a function w(x, y) that varies quadratically over the element and takes on the node values w i, i =,,, 4,, 6, can be interpolated in terms of the triangular coordinates by the formula N q N q N q w = [ w w w w 4 w w 6 N q. (.) 4 N q N q 6 where the N q i are the quadratic shape functions N q = ζ (ζ ) N q = ζ (ζ ) N q = ζ (ζ ) N q 4 = 4ζ ζ N q = 4ζ ζ N q 6 = 4ζ (.6) ζ The geometry is still defined by (.9) because we are not permitting nodes 4,,6 to be away from the midpoint positions. Similarly, the partial derivative expressions of..6 remain valid. 6

7 . CUBIC INTERPOLATION s (a) 9 8 s 7 (b) s s s s Figure.4. Nodal configurations for two forms of cubic interpolation of w(x, y) over triangle: (a) interpolation by ten values w i at corners i =,,, thirdpoints i = 4,, 6, 7, 8, 9 and centroid i = 0; (b) interpolation by four w i values at corners i =,, and centroid i = 0, plus six side slopes ( w/ s ), ( w/ s ), ( w/ s ), ( w/ s ), ( w/ s ) and ( w/ s )... Cubic Interpolation Cubic interpolants for w are fundamental in the construction of the simplest Kirchhoff plate bending elements using the TPE functional. The complete two-dimensional cubic polynomial has 0 terms. There are several choices for the selection of nodal values that determine that interpolation. Several important ones are examined next.... Cubic Interpolation Choice : 0 Nodes Next consider the ten-node triangle shown in Figure.4(a). This element still has straight sides. Nodes 4 through 9 are placed at the thirdpoints of the sides as indicated. The tenth node is placed at the centroid and is labeled 0. In a Homework of the IFEM course it was shown that a function w(x, y) that varies cubically over the element and takes on the node values w i, i =,,, 4,, 6, 7, 8, 9, 0, can be interpolated in terms of the triangular coordinates by the formula w = [ w w w w 4 w w 6 w 7 w 8 w 9 w 0 7 N c N c N c N c 4 N c N c 6. N0 c. (.7)

8 Chapter : TRIANGULAR PLATE DISPLACEMENT ELEMENTS where the N c i are the 0-node cubic shape functions N c = ζ (ζ )(ζ ) N c = ζ (ζ )(ζ ) N c = ζ (ζ )(ζ ) N4 c = 9 ζ ζ (ζ ) N c = 9 ζ ζ (ζ ) N6 c = 9 ζ ζ (ζ ) N7 c = 9 ζ ζ (ζ ) N8 c = 9 ζ ζ (ζ ) N9 c = 9 ζ ζ (ζ ) N0 c = 7ζ ζ ζ (.8) The geometry is still defined by (.9) because we are not permitting nodes 4 through 9 to be away from the thirdpoint positions as well as forcing 0 to be exactly at the centroid. Similarly, the partial derivative expressions of..6 remain valid.... Cubic Interpolation Choice : 4 Nodes plus 6 Side Slopes Consider now a variant of the cubic interpolation over the triangle as shown in Figure.4(b). This element still has straight sides. In addition to values at the corners,, and the centroid 0, we we specify the side-slope corner derivatives w jk = ( w/ s j ) k. There are six combinations: w = ( w/ s ), w = ( w/ s ), w = ( w/ s ), w = ( w/ s ), w = ( w/ s ) and w = ( w/ s ). The resulting interpolation is w = [ w w w w w w w w w w 0 N c N c N c N c 4 N c N c 6. N0 c. (.9) in which the N c i are the shape functions N c = ζ (ζ + ζ + ζ ) 7ζ ζ ζ N c = L (ζ ζ ζ ζ ζ ) N c = L (ζ ζ ζ ζ ζ ) N4 c = ζ (ζ + ζ + ζ ) 7ζ ζ ζ N c = L (ζ ζ ζ ζ ζ ) N6 c = L (ζ ζ ζ ζ ζ ) N c 7 = ζ (ζ + ζ + ζ ) 7ζ ζ ζ N c 8 = L (ζ ζ ζ ζ ζ ) N c 9 = L (ζ ζ ζ ζ ζ ) N c 0 = 7ζ ζ ζ (.0) and L jk denotes the length of the side joining corners j and k. The function of the corrective terms ζ ζ ζ is to make the first nine functions (.0) vanish at 0. 8

9 The c and c formulas are related by the transformation matrix. CUBIC INTERPOLATION w w w w 4 w w 6 w 7 w 8 w 9 w 0 = L 7 L L 0 4L L 7 L L 0 4L L L 0 0 4L L w w w w w w w w w w 0 (.)... Cubic Interpolation Choice : 4 Nodes plus 6 Cartesian Slopes This choice is similar to the previous one but the corner derivatives are taken with respect to the directions x, y of the Cartesian reference system. The notation used for the slopes is w x = ( w/ x), w y = ( w/ y), w x = ( w/ x), w y = ( w/ y), w x = ( w/ x) and w y = ( w/ y). The resulting interpolation is in which the N c i w = [ w w x w y w w x w y w w x w y w 0 are the shape functions N c N c N c N c 4 N c N c 6. N0 c. (.) N c = ζ (ζ + ζ + ζ ) 7ζ ζ ζ N c = ζ (x ζ x ζ ) + (x x )ζ ζ ζ N c = ζ (y ζ y ζ ) + (y y )ζ ζ ζ N c 4 = ζ (ζ + ζ + ζ ) 7ζ ζ ζ N c = ζ (x ζ x ζ ) + (x x )ζ ζ ζ N c 6 = ζ (y ζ y ζ ) + (y y )ζ ζ ζ N c 7 = ζ (ζ + ζ + ζ ) 7ζ ζ ζ N c 8 = ζ (x ζ x ζ ) + (x x )ζ ζ ζ N9 c = ζ (y ζ y ζ ) + (y y )ζ ζ ζ N0 c = 7ζ ζ ζ (.) in which x ij = x i x j and y ij = y i y j...4. Cubic Interpolation Choice 4: 4 Nodes plus 6 Cartesian Rotations This interpolation is the same as the previous one, with some rearrangements. The rotational degrees of freedom at the corners are introduced using the relations θ x = w/ y and θ y = w/ x.. The 9

10 Chapter : TRIANGULAR PLATE DISPLACEMENT ELEMENTS resulting interpolation is in which the i w = [ w θ x θ y w θ x θ y w θ x θ y w (.4) are the same shape functions of Choice with some sign and index changes: = N c, = N c, = N c,, 8, 4 = N c 4, = N c 6, 6 = N c 7 = N c 7, 8 = N c 9, 9 = N c 0 = N c The BCIZ Plate Bending Element (.) One of the simplest Kirchhoff plate bending elements was presented in 966 by Bazeley, Cheung, Irons and Zienkiewicz [66. This is called the BCIZ element after the authors initials. This element can be derived from the cubic interpolation choice 4 (.4). The technique appears a bit mysterious at first. Basically one can construct a 0 0 plate stiffness matrix. Freedom w 0 at the centroid can be statically condensed out and the resulting 9 9 stiffness used in finite element analysis. Unfortunately the static condensation destroys curvature completeness and the solutions will not generally converge, as emperically shown in Tocher s thesis [770. The idea behing the BCIZ element is that elimination of the centroidal DOF is done in such a way that completeness is maintained, using a kinemtic constraint..4.. The Kinematic Constraint As explained above, we seek a kinematic constraint to eliminate w 0 in terms of the nine connector DOFs: w 0 = [ a w a θ x a θy a w a θ x a θy a w a θ x a θy w θ x θ y w θ x θ y w θ x θ y (.6) Historically the first triangular plate bending element that satisfied completeness and invariance. This element is also important as the motivation for development of the patch test. 0

11 .4 THE BCIZ PLATE BENDING ELEMENT Substituting this into any of the 4 choices of the previous section, w 0 is eliminated. For choice 4: w = [ w θ x θ y w θ x θ y w θ x θ y + a wn 0 + a θ xn 0 + a θyn 0. N9 c4 + a θyn 0. (.7) in which N 0 = 7ζ ζ ζ is the same for all choices. To determine the a s in (.7) we impose the condition that all constant curvature states must be exactly represented. This is a completeness condition. Constant curvature states are associated with quadratic variations of w, which are defined by the 6-node interpolation (.6). The procedure is as follows: (i) Use (.6) to express w 0 in terms of w through w 6 by setting ζ = ζ = ζ = /. (ii) Take the cubic interpolation choice 4 (.4) and express the value of w 4, w and w 6 in terms of the nine connection DOF w through θ y by setting ζ, ζ and ζ to appropriate midpoint coordinates. (iii) Eliminate w 4, w and w 6 from (i) and (ii) to get (.6) and hence (.7). The resulting formula is called the BCIZ interpolation. The associated N s are the BCIZ shape functions. It can be shown that the constraint is q = [/,( x + x + x )/8,( y + y + y )/8, /,(x x + x )/8,(y y + y )/8, /,(x + x x )/8,(y + y y )/8 (.8) The derivation of the entries of (.8) is the subject of a homework Exercise.

12 Chapter : TRIANGULAR PLATE DISPLACEMENT ELEMENTS (a) (b) Figure.. The two -point Gauss integration rules for triangles, which are useful in the computation of the plate element stiffness matrix K. Numbers annotated near sample points are the weights..4.. The Curvature Displacement Matrix Once we have an interpolation formula for w, as in (.7), the curvatures over the plate element can be evaluated by double differentiation of the shape functions with respect to x and y. The resulting relation can be expressed in matrix form as κ xx w/ x B B B B 4 B B 6 B 7 B 8 B 9 κ yy = w/ y = B B B B 4 B B 6 B 7 B 8 B 9 κ xy w/ y y B B B B 4 B B 6 B 7 B 8 B 9 w θ x θ y w θ x θ y w θ x θ y (.9) or κ = Bu (.0) The entries B ij are linear in the triangular coordinates ζ, ζ and ζ. The derivation of the entries of the B matrix is the subject of a homework Exercise..4.. The Element Stiffness Matrix For this displacement element, the stiffness matrix is given by the usual formula K = B T DB d (.) (e) where (e) is the element area, D is the plate rigidity matrix that relates moment to curvatures (see 9..) and B is the curvature-displacement matrix defined in (.0). If D is constant over the element, as frequently assumed, the integrand of K is quadratic in the triangular coordinates. If so the integral can be done exactly using one of the two -point Gauss rule presented in Chapter of IFEM. The two rules are reproduced below for convenience: F(ζ,ζ,ζ ) d = A F(, 6, 6 ) + F( 6,, 6 ) + F( 6, 6, ). (.) (e)

13 .4 THE BCIZ PLATE BENDING ELEMENT F(ζ,ζ,ζ ) d = A F(,, 0) + F(0,, ) + F(, 0, ). (.) (e) The second one is called the midpoint rule because the three sample points are at the triangle midpoints. These rules are depicted in Figures.(a) and (b), respectively. Both are exact up to quadratic polynomials in the triangular coordinates, which is what is needed for the plate stiffness matrix.

14 Chapter : TRIANGULAR PLATE DISPLACEMENT ELEMENTS Notes and Bibliography The foregoing shape functions were first developed and presented in terms of triangular coordinates in the 966 thesis [. 4

15 Exercises Homework Exercises for Chapter Triangular Plate Displacement Elements EXERCISE. [A: Check that the shape functions given for cubic interpolation choice 4 satisfy the conditions of being for the associated DOF, and zero for all others. For example N, which is associated with θ x, must satisfy N (, 0, 0) = 0, N y (, 0, 0) = = θ x, N x (, 0, 0) = 0 = θ y, etc. (E.) Note: It is sufficient to verify N, N, and N because the next six shape functions are obtained by cyclic permutation of subscripts. N 0 is easy to check since its corner value and corner slopes are zero. EXERCISE. [A: Using the chain rule and the results of..6, show that the following formula can be used to compute the plate curvatures from an interpolation formula in w: w x w y w x y y x x y = y x x y y x x y 4A y y x x (x y + x y ) y y x x (x y + x y ) y y x x (x y + x y ) T w ζ w ζ w ζ w ζ ζ w ζ ζ w ζ ζ, (E.) EXERCISE. [A: Derive the expression for the a s in (.7) following the procedure outline in that section, and derive the modified shape functions appearing in (.8). EXERCISE.4 [A: Program the BCIZ element starting from the interpolation formula provided by the previous exercise, the expression (E.) to form the curvature-displacement matrix B and one of the -point Gauss rules given in. to evaluate K (e) = (e) B T DBd. Assume that D = (h /)E in which both the thickness h and the stress-strain matrix E are constant over the element, and supplied through arguments. EXERCISE. [A: Evaluate K (e) of the BCIZ element for a triangle with corners at (0, 0), (, ) and (, ), fabricated with isotropic material of E = 0 and ν = 0, and constant thickness h =. As a check, compute the 9 eigenvalues; three of them should be zero and six positive.