The proper generalized decomposition for advanced numerical simulations ch16

The proper generalized decomposition for advanced numerical simulations ch16 Many problems in scientific computing are intractable with classical numerical techniques. These fail, for example, in the solution of high-dimensional models due to the exponential increase of the number of degrees of freedom. Recently, the authors of this book and their collaborators have developed a novel technique, called Proper Generalized Decomposition (PGD) that has proven to be a significant step forward. The PGD builds by means of a successive enrichment strategy a numerical approximation of the unknown fields in a separated form. Although first introduced and successfully demonstrated in the context of high-dimensional problems, the PGD allows for a completely new approach for addressing more standard problems in science and engineering. Indeed, many challenging problems can be efficiently cast into a multi-dimensional framework, thus opening entirely new solution strategies in the PGD framework. For instance, the material parameters and boundary conditions appearing in a particular mathematical model can be regarded as extra-coordinates of the problem in addition to the usual coordinates such as space and time. In the PGD framework, this enriched model is solved only once to yield a parametric solution that includes all particular solutions for specific values of the parameters. The PGD has now attracted the attention of a large number of research groups worldwide. The present text is the first available book describing the PGD. It provides a very readable and practical introduction that allows the reader to quickly grasp the main features of the method. Throughout the book, the PGD is applied to problems of increasing complexity, and the methodology is illustrated by means of carefully selected numerical examples. Moreover, the reader has free access to the Matlab© software used to generate these examples.

16 The Ten Node Tetrahedron 16–1 16–2 Chapter 16: THE TEN NODE TETRAHEDRON TABLE OF CONTENTS Page §16.1 INTRODUCTION 16–3 §16.2 THE QUADRATIC TETRAHEDRON §16.3 PARTIAL DERIVATIVE CALCULATIONS §16.3.1 Implementation Considerations §16.4 GAUSS RULES OVER TETRAHEDRA 16–3 16–4 16–6 16–8 §16.5 THE ELEMENT STIFFNESS MATRIX 16–8 §16.6 THE CONSISTENT NODE FORCE VECTOR 16–9 EXERCISES 16–2 16–10 16–3 §16.2 THE QUADRATIC TETRAHEDRON §16.1 INTRODUCTION The technique used in Chapter 15 for the derivation of the linear triangle is now extended to the 10-node tetrahedron, also called the quadratic tetrahedron This element can have curved faces and edges The extension is based on the isoparametric technique and the partial-derivative construction procedure followed the IFEM course for the 6-node triangle §16.2 THE QUADRATIC TETRAHEDRON The ten node tetrahedron shown in Figure 16.1 is the next complete-polynomial member of the isoparametric tetrahedron family The element has four corners with local numbers through 4, which must be traversed following the same convention as the four node tetrahedron It has six side nodes, with local numbers through 10; nodes 5,6,7 are located on sides 1-2, 2-3 and 3-1, whereas nodes 8,9,10 are located on sides 1-4, 2-4, and 3-4 The side nodes may be located arbitrarily subjected to positive-Jacobian-determinant constraints Each element face is defined by six nodes, which are not necessarily on a plane The isoparametric element definition is  1  x1   x   y1     y  =  z1    ux  u x1  uy u y1 u z1   x2 y2 z2 u x2 u y2 u z2 x3 y3 z3 u x3 u y3 u z3 x4 y4 z4 u x4 u y4 u z4   (e)  N1 x10   N2(e)    y10   N (e)    z 10   N (e)    u x10       u y10 (e) N10 u z10 (16.1) The conventional (non-hierarchical) shape functions are N1(e) = ζ1 (2ζ1 − 1), N2(e) = ζ2 (2ζ2 − 1) N3(e) = ζ3 (2ζ3 − 1), N4(e) = ζ4 (2ζ4 − 1) N5(e) = 4ζ1 ζ2 , N6(e) = 4ζ2 ζ3 N7(e) = 4ζ3 ζ1 , N8(e) = 4ζ1 ζ4 N9(e) = 4ζ2 ζ4 , (e) N10 = 4ζ3 ζ4 (16.2) These shape functions are similar in form to those of the six node quadratic triangle discussed in IFEM If the element is curved (that is, the six nodes that define each face are not on a plane), the tetrahedron coordinates no longer fall on planes, but form a curvilinear system 16–3 16–4 Chapter 16: THE TEN NODE TETRAHEDRON z 10 x y Figure 16.1 The ten-node (quadratic) tetrahedron §16.3 PARTIAL DERIVATIVE CALCULATIONS The main task involved in writing the shape function subroutine for the 10-node tetrahedron is the computation of the partial derivatives of the functions in (16.1) with respect to x, y and z at any point in the element For this purpose consider a generic scalar function, w(ζ1 , ζ2 , ζ3 , ζ4 ), that is quadratically interpolated over the ten-node tetrahedron with the shape functions (16.2):  w = [ w1 w2 w3 w4 w5 w6  ζ1 (2ζ1 − 1)  ζ2 (2ζ2 − 1)     ζ3 (2ζ3 − 1)     ζ4 (2ζ4 − 1)   w10 ]  4ζ1 ζ2      4ζ2 ζ3      (16.3) 4ζ3 ζ4 Symbol w may represent x, y, z, u x , u y or u z in the isoparametric representation (7.1), or other element-varying quantities such as body forces or temperatures Taking partials with respect to x, y and z, and applying the chain rule twice we get ∂w = ∂x ∂w = ∂y ∂w = ∂z ∂ Ni = ∂x ∂ Ni wi = ∂y ∂ Ni wi = ∂z wi wi wi wi ∂ Ni ∂ζ1 ∂ Ni ∂ζ1 ∂ Ni ∂ζ1 ∂ζ1 ∂ Ni ∂ζ2 ∂ Ni ∂ζ3 ∂ Ni + + + ∂x ∂ζ2 ∂ x ∂ζ3 ∂ x ∂ζ4 ∂ζ1 ∂ Ni ∂ζ2 ∂ Ni ∂ζ3 ∂ Ni + + + ∂y ∂ζ2 ∂ y ∂ζ3 ∂ y ∂ζ4 ∂ζ1 ∂ Ni ∂ζ2 ∂ Ni ∂ζ3 ∂ Ni + + + ∂z ∂ζ2 ∂z ∂ζ3 ∂z ∂ζ4 16–4 ∂ζ4 ∂x ∂ζ4 ∂x ∂ζ4 ∂x , , , (16.4) 16–5 §16.3 PARTIAL DERIVATIVE CALCULATIONS where all sums are understood to run from i = through 10, and element superscripts on the shape functions have been suppressed for clarity In matrix form:  ∂w ∂x  ∂w   ∂y ∂w ∂z   ∂ζ1 ∂x     =  ∂ζ1   ∂y ∂ζ1 ∂z ∂ζ2 ∂x ∂ζ2 ∂y ∂ζ2 ∂z ∂ζ3 ∂x ∂ζ3 ∂y ∂ζ3 ∂z   ∂ζ4  ∂x    ∂ζ4    ∂y     ∂ζ4  ∂z wi ∂∂ζNi wi ∂∂ζNi ∂ N wi ∂ζ i wi ∂∂ζNi          (16.5) Transposing both sides of (16.5) while switching the left and right hand sides, yields a form exploited below:  ∂ζ1 ∂ζ1 ∂ζ1  ∂x ∂y ∂z    ∂ζ2 ∂ζ2 ∂ζ2   ∂y ∂z   ∂ Ni  ∂ x ∂w ∂w wi ∂∂ζNi wi ∂∂ζNi wi ∂∂ζNi = ∂w  ∂x ∂y ∂z ∂ζ4  ∂ζ3 ∂ζ3 ∂ζ3     ∂x ∂y ∂z   ∂ζ4 ∂ζ4 ∂ζ4 ∂x ∂y ∂z (16.6) Now make w ≡ x, y, z and stack the results row-wise:      xi ∂∂ζNi yi ∂∂ζNi z i ∂∂ζNi xi ∂∂ζNi yi ∂∂ζNi z i ∂∂ζNi xi ∂∂ζNi yi ∂∂ζNi z i ∂∂ζNi  ∂ζ1 ∂x  xi ∂∂ζNi     ∂ζ   ∂x ∂ N yi ∂ζ i     ∂ζ3  ∂x ∂ N z i ∂ζ i  ∂ζ4 ∂x ∂ζ1 ∂y ∂ζ2 ∂y ∂ζ3 ∂y ∂ζ4 ∂y ∂ζ1 ∂z ∂ζ2 ∂z ∂ζ3 ∂z ∂ζ4 ∂z    ∂ζ1 ∂x  ∂ N i   xi ∂ζ   ∂ζ2   ∂x  ∂ N i yi ∂ζ   ∂ζ3   ∂x z i ∂∂ζNi ∂ζ4 ∂x ∂ζ1 ∂y ∂ζ2 ∂y ∂ζ3 ∂y ∂ζ4 ∂y ∂ζ1 ∂z ∂ζ2 ∂z ∂ζ3 ∂z ∂ζ4 ∂z    ∂x    ∂x   ∂y  =  ∂x   ∂z  ∂x ∂x ∂y ∂y ∂y ∂z ∂y ∂x ∂z ∂y ∂z ∂z ∂z     (16.7) This is a linear system with the required unknowns in the second matrix, but its coefficient matrix is not square To achieve that, differentiate both sides of the identity ζ1 + ζ2 + ζ3 + ζ4 = with respect to x, y and z, and insert as first row:  ∂1 ∂x    ∂x     ∂x  =  ∂y     ∂x  ∂z ∂x ∂1  ∂z  ∂ x   ∂z    ∂N ∂y  yi ∂∂ζNi yi ∂∂ζNi   yi ∂ζ i  ∂z  ∂ N ∂ N ∂ N ∂z i i i z i ∂ζ z i ∂ζ z i ∂ζ ∂z (16.8) But ∂ x/∂ x = ∂ y/∂ y = ∂z/∂z = and ∂1/∂ x = ∂1/∂ y = ∂1∂z = ∂ x/∂ y = ∂ x/∂z = ∂ y/∂ x = ∂ y/∂z = ∂z/∂ x = ∂z/∂ y = because x, y and z are independent coordinates Consequently we  xi ∂∂ζNi xi ∂∂ζNi xi ∂∂ζNi 16–5 ∂1 ∂y ∂x ∂y ∂y ∂y ∂z ∂y 16–6 Chapter 16: THE TEN NODE TETRAHEDRON arrive at a system of linear equations of order with three right-hand sides convention to get rid of sum symbols the system is  ∂ζ1 ∂ζ1 ∂ζ1   1 1  ∂x ∂y ∂z    ∂ N ∂ N ∂ N ∂ N i i i i   ∂ζ2 ∂ζ2 ∂ζ2   xi x x x i i i  ∂ζ1 ∂ζ2 ∂ζ3 ∂ζ4   ∂ x ∂y ∂z   1   =   ∂ Ni yi ∂∂ζNi yi ∂∂ζNi yi ∂∂ζNi   ∂ζ3 ∂ζ3 ∂ζ3   yi ∂ζ    ∂ x ∂ y ∂z   ∂ N ∂ N ∂ N ∂ N i i i i z i ∂ζ z i ∂ζ z i ∂ζ z i ∂ζ ∂ζ4 ∂ζ4 ∂ζ4 ∂x ∂y ∂z Using the summation 0  0  (16.9) In compact matrix notation JP = Iaug where (16.10)  1 1  1 1  x i ∂ Ni x i ∂ Ni x i ∂ Ni x i ∂ Ni  ∂ζ1 ∂ζ2 ∂ζ3 ∂ζ4  Jx2 Jx3 Jx4   J J =  x1 (16.11) = ∂ N ∂ N ∂ N ∂ Ni  i i i   Jy1 Jy2 Jy3 Jy4  yi ∂ζ1 yi ∂ζ2 yi ∂ζ3 yi ∂ζ4  Jz1 Jz2 Jz3 Jz4 z i ∂∂ζNi z i ∂∂ζNi z i ∂∂ζNi z i ∂∂ζNi and Iaug is the × identity matrix augmented with a zero first row Taking the partials of the shape functions (16.2) with respect to the tetrahedron coordinates and substituting into the above yields   Jx1 = x1 (4ζ1 − 1) + 4x5 ζ2 + 4x7 ζ3 + 4x8 ζ4 , Jx2 = x2 (4ζ2 − 1) + 4x6 ζ3 + 4x5 ζ1 + 4x9 ζ4 , Jx3 = x3 (4ζ3 − 1) + 4x7 ζ1 + 4x6 ζ2 + 4x10 ζ4 , Jx4 = x4 (4ζ4 − 1) + 4x8 ζ1 + 4x9 ζ2 + 4x10 ζ3 (16.12) For Jyi and Jzi replace xi by yi and z i , respectively, into the above formulas Summarizing, to compute the x-y-z partials for a function w interpolated as per (16.3) the recipe is: form the linear system (16.10) from the geometric data, solve for the 12 tetrahedron coordinates partials, and substitute these into (16.5) REMARK 16.1 By analogy with the isoparametric brick elements, matrix J of (16.10) may be called a Jacobian matrix However, the factor J that appears in the element-of-volume transformation d V (e) = J dζ1 dζ2 dζ3 ζ4 , (16.13) is not det J = |J|, but J = 16 |J If the element has straight sides with side nodes at the midpoints, J is constant and equal to the volume V of the tetrahedron, which is given by the usual formula J =V= x1 y1 z1 x2 y2 z2 as in Chapter In this case J is constant over the element 16–6 x3 y3 z3 x4 , y4 z4 (16.14) 16–7 §16.3 PARTIAL DERIVATIVE CALCULATIONS §16.3.1 Implementation Considerations To speed up computations in the shape function subroutine it is important to observe that the Jacobian matrix has the special structure (16.11) We can take advantage of this by subtracting the first column of J from the last three columns This manipulation reduces (16.10) to the solution of a × linear system: Jx2 − Jx1 Jy2 − Jy1 Jz2 − Jz1 Jx3 − Jx1 Jy3 − Jy1 Jz3 − Jz1 Jx4 − Jx1 Jy4 − Jy1 Jz4 − Jz1  ∂(ζ − ζ ) ∂ x   ∂(ζ3 − ζ1 )  ∂x  ∂(ζ4 − ζ1 ) ∂x or ∂(ζ2 − ζ1 ) ∂y ∂(ζ3 − ζ1 ) ∂y ∂(ζ4 − ζ1 ) ∂y ∂(ζ2 − ζ1 )  ∂z  ∂(ζ3 − ζ1 )  = ∂z  ∂(ζ4 − ζ1 ) ∂z J¯ P¯ = I 0 0 (16.15) (16.16) where I is the × identity matrix, and the modified Jacobian matrix is shown on the left hand ¯ which can be readily calculated by Cramer’s rule Finally, side Therefore P¯ is just the inverse of J, the first row of P is recovered from the constraints ∂ζ2 ∂ζ3 ∂ζ4 ∂ζ1 + + + = 0, ∂x ∂x ∂x ∂x (16.17) and similarly for the y and z partials This is done as follows: take the column sum sx of the ¯ then computed first column of P; ∂ζ1 = − 14 sx , (16.18) ∂x from which the other x can be recovered This operation is repeated over the other columns To tie up with the notation used in Chapter for the four-node tetrahedron, we may denote the partials as  ∂ζ1 ∂x   ∂ζ2  ∂x    ∂ζ3  ∂x  ∂ζ4 ∂x ∂ζ1 ∂y ∂ζ2 ∂y ∂ζ3 ∂y ∂ζ4 ∂y ∂ζ1 ∂z ∂ζ2 ∂z ∂ζ3 ∂z ∂ζ4 ∂z    a1   a  =   J a3  a4  b1 b2 b3 b4  c1 c2   c3 c4 (16.19) For the 4-node tetrahedron, J = 16 V was constant over the element, but now it generally will vary with position unless the tetrahedron has planar faces 16–7 16–8 Chapter 16: THE TEN NODE TETRAHEDRON §16.4 GAUSS RULES OVER TETRAHEDRA We mention the first two numerical integration rules, which find applications in the evaluation of element stiffness matrix and consistent force vector One point rule (exact for constant and linear polynomials over plane-face tetrahedra): V V (e) F(ζ1 , ζ2 , ζ3 , ζ4 ) d V (e) ≈ F( 14 , 14 , 14 , 14 ) (16.20) Four-point rule (exact for constant through quadratic polynomials over plane-face tetrahedra): V V (e) F(ζ1 , ζ2 , ζ3 , ζ4 ) d V (e) ≈ 14 F(α, β, β, β)+ 14 F(β, α, β, β)+ 14 F(β, β, α, β)+ 14 F(β, β, β, α), (16.21) where α = 0.58541020, β = 0.13819660 More details and exact values for the integration rules are posted in Chapter 17: A Compendium of Gauss Integration Rules for FEM §16.5 THE ELEMENT STIFFNESS MATRIX In three dimensional elasticity this element has 10 × = 30 degrees of freedom For the following derivations we assume that they are arranged as u(e) = [ u x1 u y1 u z1 u x2 u y2 u z2 u x10 u y10 The × 30 strain-displacement matrix for the 10-node tetrahedron is  qx1 qx2 qx1 0 0 q q 0 0 q  y1 y2 y10  0 qz1 qz2  B=  q y1 q y2 q y10 qx1 qx2 qx10  0 qz1 qz2 q y10 q y1 q y1 qx1 qx2 qx10 0 qz1 qz1 u z10 ]T (16.22)    qz10     q y10 qz10 (16.23) where (summation convention implied on j = 1, 2, 3, 4): ∂ Ni ∂ζ j ∂ Ni q yi = ∂ζ j ∂ Ni qzi = ∂ζ j qxi = ∂ζ j ∂ Ni = J −1 aj, ∂x ∂ζ j ∂ζ j ∂ Ni bj, = J −1 ∂y ∂ζ j ∂ζ j ∂ Ni cj, = J −1 ∂z ∂ζ j (16.24) The stiffness matrix K(e) is evaluated by numerical integration K(e) = p wk BT (ζik )EB(ζik ) J (ζik ) k=1 16–8 (16.25) 16–9 §16.6 THE CONSISTENT NODE FORCE VECTOR where p is the number of Gauss points, (ζik ) denotes the coordinate quartet (ζ1 ,ζ2 ,ζ3 , ζ4 ) at the k th integration point, and wk are the corresponding weights The stress-strain matrix E is the same as in Chapter The four-point rule integrates this element with the correct rank §16.6 THE CONSISTENT NODE FORCE VECTOR Consider a body force field over the element defined by its components b= bx by bz (16.26) NT b d V (e) , (16.27) The consistent node force vector is given by f(e) = V (e) where N is the × 30 matrix of shape functions that relates element displacements to node displacements: ux (16.28) u y = Nu uz For this element and the node displacement ordering (16.22) N= N1 0 N10 0 N1 where the shape functions are given by (16.2) 16–9 N10 0 N1 0 N10 , (16.29) Chapter 16: THE TEN NODE TETRAHEDRON 16–10 Homework Exercises for Chapter 16 EXERCISE 16.1 [A:15] The 10-node tetrahedron element is converted into an 11-point tetrahedron by adding node point 11 (e) ? (You not need to write located at the centroid ζ1 = ζ2 = ζ3 = ζ4 = 1/4 What is the shape function N11 the full element definition) EXERCISE 16.2 [A:15] The next full-polynomial, isoparametric member of the tetrahedron family is the cubic tetrahedron, which has 21 node points Where you think the nodes are located? EXERCISE 16.3 [A/C:25] Derive the shape functions for the 21-node tetrahedron EXERCISE 16.4 [A:15] Justify the rule (16.18) EXERCISE 16.5 [A/C:20] Compute f(e) for a straight-face 10-node tetrahedron if the body forces bx = b y = and bz is constant, using the 4-point rule (16.21) to evaluate the integral in (16.27) (You need to give only the z force components) 16–10 ... (16.12) For Jyi and Jzi replace xi by yi and z i , respectively, into the above formulas Summarizing, to compute the x-y-z partials for a function w interpolated as per (16.3) the recipe is: form the. .. from which the other x can be recovered This operation is repeated over the other columns To tie up with the notation used in Chapter for the four-node tetrahedron, we may denote the partials... extension is based on the isoparametric technique and the partial-derivative construction procedure followed the IFEM course for the 6-node triangle §16.2 THE QUADRATIC TETRAHEDRON The ten node tetrahedron