FEM Isoparametric Concept

FEM Isoparametric Concept home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/cover_sheet.tex page of 25. p./25

Table of contents. Interpolation Functions for the Finite Elements 2. Finite Element Types 3. Geometry 4. Lagrange and Hermite Element Family 5. Interpolation Approach Function 6. Cartesian - Natural Coordinates 7. Isoparametric Concept 8. Example to the Isoparametric Concept home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/contents.tex page 2 of 25. p.2/25

Shape Functions for FEM For the finite elements method, the following is valid: The global function of a sought function consists of a sum of local functions: E e= V ω e i dv e Galerkin method: the interpolation function corresponds to weighted function Ritz method: the global variation principle is constructed from the sum of the local variation principles. home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/interpolation.tex page 3 of 25. p.3/25

Shape functions for FEM Critical step in the FEM: Selection of suitable interpolation functions,that are formed through the shape of the finite elements, the approximation order. Choice of Finite elements depends on the geometry of the global area, the desired exactness of the area, the simple integration through the area. home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/interpolation2.tex page 4 of 25. p.4/25

Finite Element Types In order to be able to formulate special physical problems, several elements are often necessary. They are distinguished by the geometry (-D, 2-D or 3-D), the selection of the interpolation function (polynomials; Lagrange or Hermite polynomials), the selection of the element coordinates (Cartesian or natural coordinates), the selection of the variables specified at the nodes (Lagrange group or Hermite group of variables). home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/finite element types.tex page 5 of 25. p.5/25

Geometry a) Square elements with straight sides b) Square elements with curved sidesx c) Cubic elements home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/geometry.tex page 6 of 25. p.6/25

Interpolation Approach Function Polynomials: One dimensional element approximation u = a 0 + a x + a 2 x 2 + a 3 x 3 +... or u = a i x i = i a i x i with i = linear variation i = 2 quadratic variation i = 3 cubic variation D-element with two nodes: 2 For two nodes we need a linear variability. home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/interpolation approach function.tex page 7 of 25. p.7/25

Lagrangian polynomials Lagrange interpolation functions save us from inverting the coefficient matrix as is necessary for a standard polynomial. They have the following form u(x) = L (x)u + L 2 (x) + + L n (x)u n with L N (x) chosen such that L N (x) takes the form L N (x m ) = δ NM. L N (x) = c N (x x )(x x 2 )... (x x N )(x x N+ )... (x x n ). home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/lagrangian_poly.tex page 8 of 25. p.8/25

Lagrangian polynomials c N can be found to be c N = (x N x )(x N x 2 )... (x N x N )(x N x N+ )... (x N x n ). The polynomials are therefore L N (x) = (x x )(x x 2 )... (x x N )(x x N+ )... (x x n ) (x N x )(x N x 2 )... (x N x N )(x N x N+ )... (x N x n ). For a quadratic approximation we yield L (x) = (x x 2)(x x 3 ) (x x 2 )(x x 3 ) L 3 (x) = (x x )(x x 2 ) (x 3 x )(x 3 x 2 ). L 2 (x) = (x x )(x x 3 ) (x 2 x )(x 2 x 3 ) home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/lagrangian_poly2.tex page 9 of 25. p.9/25

Lagrangian polynomials For the following element 0 we get L = x (x ), 2 L 2 = x 2, L L 3 L 2 L 3 = x (x + ). 2 0 home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/lagrangian_poly3.tex page 0 of 25. p.0/25

Hermite polynomials Hermite polynomials do not only assume the continuity of the variable at a common node, but also the derivative at that node. d element with two nodes. We need a cubic ansatz. ũ(ξ) = H 0 j (ξ)û j + H j (ξ) ( û ξ ũ(ξ) = N r w r r =, 2, 3, 4 ) j =, 2 N = H 0 = 3ξ 2 + 2ξ 3 N 2 = H2 0 = 3ξ2 2ξ 3 N 3 = H = ξ 2ξ 2 + ξ 3 N 4 = H2 = ξ 3 ξ 2 home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/hermite_poly.tex page of 25. p./25

Hermite polynomials The four functions have the following shape. 0.8 H 0 H 2 0 0.6 0.4 0.2 H 0 H 2-0.2 0 0.2 0.4 0.6 0.8 home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/hermite_poly2.tex page 2 of 25. p.2/25

Natural coordinates The more general approach is to use the natural coordinates, ξ, with the origin either at the left end of the domain (upper figure) or in the middle (lower figure). ξ=0 ξ= ξ=0 2 ξ= 2 ξ= home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/natural_coord.tex page 3 of 25. p.3/25

Natural coordinates The interpolation functions then read for the upper figure ϕ = ξ ϕ 2 = ξ, and for the lower figure ϕ = ( ξ) 2 aϕ 2 = ( + ξ). 2 home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/natural_coord22.tex page 4 of 25. p.4/25

Natural coordinates 2D Isoparametric elements Definition: The same parametric function which describes the geometry is used for the interpolation of the variables (shifting, water level, etc.) within an element. 2 s r 3 4 N = ( 2 + 2 r) ( = 4 N 2 = 4 N 3 = 4 N 4 = 4 2 + 2 s) ( + r)( + s) analogous: ( r)( + s) ( r)( s) ( + r)( s) home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/natural_coord3.tex page 5 of 25. p.5/25

Cartesian Natural Coordinates Further examples for interpolation functions:. Interpolation function for a 2-D element with a 4 to 9 variable node number y s s = + 2 5 Node 6 9 8 s = 0 r 3 r = - 7 r = 0 r = + 4 s = - x home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/cartesian_natural_coordinates2.tex page 6 of 25. p.6/25

Cartesian Natural Coordinates i = 5 i = 6 i = 7 i = 8 i = 9 h = ( + r)( + s) 4 h 2 5...... h 2 8 h 4 9 h 2 = ( r)( + s) 4 h 2 5 h 2 6 h 4 9 h 3 = ( r)( s) 4..... h 2 6 h 2 7 h 4 9 h 4 = ( + r)( s) 4.......... h 2 7 h 2 8 h 4 9 h 5 = ( 2 r2 )( + s).................... h 2 9 h 6 = ( 2 s2 )( r).................... h 2 9 h 7 = ( 2 r2 )( s).................... h 2 9 h 8 = ( 2 s2 )( + r).................... h 2 9 h 9 = ( r 2 )( s 2 ) home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/cartesian natural coordinates3.tex page 7 of 25. p.7/25

Isoparametric concept y 7 s G e 5 T ii + i R e 3 + r x iii iv 25 We can save a lot of effort, if we construct our shape functions on a reference element and then transfer these functions to the global elements. home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/isopara_concept.tex page 8 of 25. p.8/25

Isoparametric concept In technical applications (e.g. ground water flow W ĥ), terms have to be often differentiated or integrated according to the Cartesian coordinates. Since we formulate our shape functions in isoparametric coordinates, one looks for a transformation relation between the two coordinate systems. With chain rule: x = r y = r ( r ) x + s ( ) r y + s ( s ) x ( ) s y = r x r y s x s y r s home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/isopara_concept2.tex page 9 of 25. p.9/25

Isoparametric concept The calculation of r, is not easily possible. Therefore, the opposite transformation is calculated x first: r s = x r x s y r y s [ ] x y = J J ˆ= Jacobi Matrix, can be calculated more easily. x y home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/isopara_concept3.tex page 20 of 25. p.20/25

Isoparametric concept In order to calculate the Jacobian, we use the property of our shape functions. We can express the variable x with the shape functions and the nodal values. x = n e i= N i x i x = [N, N 2, N 3, N 4 ] (linear interpolation of the coordinates between the nodes, n e = number of nodes per element) x x 2 x 3 y = [N, N 2, N 3, N 4 ] y y 2 y 3 x 4 y 4 home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/isopara_concept4.tex page 2 of 25. p.2/25

Isoparametric concept Now we differentiate this expression. As the nodal values are no functions of x and y, we can take them out of the derivative and differentiate the local shape functions only. The product of these two matrices is the Jacobian matrix. r s = N r N s N 2 r N 2 s N 3 r N 3 s N 4 r N 4 s x y x 2 y 2 x 3 y 3 x 4 y 4 } {{ } J x y home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/isopara_concept5.tex page 22 of 25. p.22/25

Isoparametric concept Example We want to calculate the Jacobian of the element shown on the left. We can do this with the help of the previously introduced local shape functions. cm y cm P = (, 25) cm 2 x 0.75 cm P 2 = ( 0, 25) P 3 = ( 0, 75) P 4 = ( 0, 75) 3 2 cm 4 home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/isopara_example.tex page 23 of 25. p.23/25

Isoparametric concept Example The global coordinates can be transformed into local coordinates as follows. n x = = r n y = N i x i = N i y i = 4 {( + r)( + s)() + ( r)( + s)( ) + ( r)( s)( ) + ( + r)( s)()} { ( + r)( + s) 5 4 4 + ( r)( + s) 4 + ( r)( s) 3 } + ( + r)( s) 3 4 4 = 4 r + 3 4 s + 4 rs home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/isopara_example2.tex page 24 of 25. p.24/25

Isoparametric concept Example These terms have to be differentiated J = x r x s y r y s. This yields J = (r) r (r) s ( r + 3s + rs) r 4 4 4 ( r + 3s + rs) s 4 4 4 = [ ] ( + s) 4 4 0 ( 3 + r) 4 4. home/lehre/vl-mhs--e/folien/vorlesung/4_fem_isopara/isopara_example3.tex page 25 of 25. p.25/25