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Families of Shape Functions, Numerical Integration, Physical and Master Element, Concept and Mapping, Element Stiffness and Load Vector Calculations PM Mohite Department of Aerospace Engineering Indian Institute of Technology Kanpur TEQIP School on Computational Methods in Engineering Application Knowledge Incubation for TEQIP Two Dimensional Problems Two Dimensional Problems: Single variable or multivariable problems The phenomenon is represented through partial differential equations Finite element formulation involves same steps as one dimensional case Boundary of a two dimensional domain Ω, in general, is a curve Two dimensional shapes like triangle and rectangle/quadrilaterals are used to approximate the geometry. Sample Boundary Value Problems: • Poisson equation ku f gradient operator in ˆ ˆ i j x y These equation represent • Heat conduction • Electrostatics • Stream function • Magnetic statics • Torsion of non-circular sections, Transverse deflection of elastic membranes Sample Boundary Value Problems: Heat Conduction u u u u k11 k12 k22 f x, y k21 x y y x x x in with k12 k21 0and k11 k22 k it becomes Poisson’s equation. Finite element discretization: • The number, shape and type (linear, quadratic,……….) of elements should be such that the geometry of the domain is accurately represented. • Density of elements should be such that the regions of large gradients of the solution are adequately modeled. Heat Conduction Equation: Weak Formulation Weighted residual form: u u u u x k11 x k12 y y k22 x k22 y w dA w dA We know that, q x w q x w w q x x x x Therefore, Similarly, w qx w qx w qx x x x w q y w q y w q y y y y Heat Conduction Equation: Weak Formulation Now using the Divergence Theorem, x qx wdA qx w nx ds and y q y wdA q y w ny ds where nx and ny are the direction cosines or the components of the unit normal vector n n nx iˆ ny ˆj cos iˆ sin ˆj Heat Conduction Equation: Weak Formulation writing the weak form over an element w u u w u u k 21 wf dA 0 k11 k12 k 22 x x y y x y e u u u u n y k 21 ds e wnx k11 k 22 k 22 x y x y Looking at boundary term: • wu Primary variable – essential BC and • coefficient of weight function in boundary term form the natural BC. Heat Conduction Equation: Weak Formulation u u u u n y k21 qn nx k11 k12 k22 y x y x on e qn - is the secondary variable. qn - projection of the vector k u along the unit normal n - positive outward from the surface as the as we more counter-clock wise along the boundary Thus, w u u w u u 0 k11 k12 k21 k22 wf dA wqn ds x x y y x y e e Heat Conduction Equation: Weak Formulation Or Bw, u F w where, w u u w u u k 21 k 22 dA Bw, u k11 k12 x x y y x y e F w e w f dA e wq n ds Bw, u - bilinear from, is not symmetric. It is symmetric only when k12 k21 F w - linear from Quadratic Potential: I u B u, u F u Heat Conduction Equation: Finite Element Formulation Approximation of u over e as u x, y uh x, y NODE j 1 u ej x, y e j u ej - are the nodal values of at node. e j - are the Lagrange interpolation functions with the property ie x j , y j ij Substitute uh x, y replace w 0 by ie to NODE j 1 i e x f e i NODE u ej ej x, y j 1 in the weak form and to get equation give j i j k12 k11 x y y dA i qn ds e j j k22 k21 dA x y i, j 1, 2, , NODE Heat Conduction Equation: Finite Element Formulation This leads to: NODE kije u ej fie Qie j 1 where, i j j i i j k11 k 21 dA k k12 k 22 x x y y x y e e ij fi e and f ie dA, e In matrix notation, k e Qie qn ie ds e k u f Q e e e is symmetric only when k12 k 21 e Interpolation Functions in Two Dimensions Interpolation Functions: Requirements The approximate solution u h x, y over an element, for the assurance of convergence to the actual solution as the number of elements is increased and their size is decreased, must satisfy: 1) It must be continuous over the element and differentiable as required by the weak form. - It ensures a nonzero coefficient matrix 2) It must not allow a strain to appear if the nodal displacements are compatible with a rigid-body displacements. Interpolation Functions: Requirements 3) If the nodal displacements are compatible with a uniform strain in the elements, then the interpolation functions must yield this strain for the nodal displacements. In other fields: If the nodal variables are compatible with uniform states of the variable and any of its derivatives up to the highest in the appropriate quadratic functional I , then these uniform states must be preserved in the element as it shrinks to zero size. This condition is called completeness. - This requirement is necessary in order to capture all possible states of the actual solution Interpolation Functions: Requirements Example: If a linear polynomial without the constant term is used to represent the temperature distribution in a one-dimensional system, the approximate solution can never be able to represent a uniform state of temperature in the element. Interpolation Functions: Requirements 4) The interpolation functions should be chosen so that the strain remains finite at the boundary. Note, however, that the strain can be indeterminate at a boundary. For example, if the strains involve only the first derivatives, then the displacement field must be continuous across a boundary. (strain energy must be finite) In other fields: The dependent variable and all its derivatives up to a one less than the highest order derivative in the appropriate quadratic functional should be continuous at the element interfaces. (The quadratic potential must be finite) This requirement is called compatibility condition. Elements satisfying these requirements are called conforming elements. Interpolation Functions: Requirements Degree of Compatibility achieved by interpolation functions at the element: • C 0 continuity is achieved when the field variable only and none of its derivatives maintain continuity at an interelement interface. • C1 continuity is achieved when the field variable and its first derivatives maintain continuity at an interelement interface. • C 2 continuity is achieved when the field variable, its first derivatives and second derivatives all maintain continuity at an interelement interface. Interpolation Functions: Requirements Continuity Requirement for completeness and compatibility: For assurance of finite element convergence for a functional having p as the highest order of a derivative – • C p continuity – requirement for completeness • C p 1 continuity – requirement for compatibility Interpolation Functions: Types of Families Lagrange Family: • Interpolation functions derived using the dependent unknown only and not its derivatives, at nodes. • C 0 - continuity Hermite Family: • Interpolation functions derived using the dependent unknown and its derivatives as well, at nodes. • C1 continuity – continuity of the first derivative of the dependent unknown. Cubic interpolation functions • C 2 continuity – continuity of the first and second derivatives of the dependent unknown. Quintic interpolation functions Linear Interpolation Functions: ie must be at least linear in x and y. Let uh x, y c1 c2 x c3 y Three linearly independent terms, linear in both x and y. ci to be expressed terms of nodal values of u h . Hence, we must have three nodes. This is possible for a triangle with its vertices as nodes. Linear Interpolation Functions: • Consider a polynomial uh x, y c1 c2 x c3 y c4 xy • Four linear independent terms, linear in x, y with a bilinear term in x and y. • This requires an element with four nodes. • Two possible geometries: - A triangle with the three nodes at its vertices and the fourth node at its centre or centroid. - A rectangle with the nodes at the vertices Linear Interpolation Functions: • A triangle with the fourth node at its centre does not provide a single valued variation of primary variable at interelement boundaries. • This results in incompatible variations of primary variable there. Therefore, this is not admissible. • Thus, the only possible element with assumed polynomial approximation is rectangle with the nodes at the four vertices Quadratic Interpolation Functions: • Consider a polynomial with five constants uh x, y c1 c2 x c3 y c4 xy c5 x 2 y 2 • This is an incomplete quadratic polynomial • Note that the terms x2 and y2 can not be varied independently. • The polynomial requires an element with five nodes. • The possible geometry is a rectangle with a node at each vertex and the fifth node at its centre. • This again does not give single valued variation of primary variable. Quadratic Interpolation Functions: • Consider a polynomial with six constants uh x, y c1 c2 x c3 y c4 xy c5 x 2 c6 y 2 • This is a complete quadratic polynomial • The polynomial requires an element with six nodes. • The possible geometry is a triangle with a node at each vertex and a node at the midpoint of each side. Lagrange Interpolation Functions in Two Dimensions Linear Interpolation Functions in Two Dimensions Interpolation Functions: Linear Triangular Element A triangular element with three nodes at vertices 3 1 2 Approximate solution over an element is of the form uh x, y c1 c2 x c3 y We need three conditions to find the unknowns Ci ‘s. Using nodal values of solution we get three conditions Interpolation Functions: Linear Triangular Element Thus, we have Or in a matrix form u1 uh x1 , y1 c1 c2 x1 c3 y1 u2 uh x2 , y2 c1 c2 x2 c3 y2 u3 uh x3 , y3 c1 c2 x3 c3 y3 u1 1 x1 u2 1 x2 u 1 x 3 3 y1 c1 c1 y2 c2 Ac2 c y3 c 3 3 unknowns Ci ‘s are found as u1 c1 A1 u2 c2 u c 3 3 where, 1 2 3 1 1 A 1 2 3 2 Ae 1 2 3 2 Ae 1 2 3 Interpolation Functions: Linear Triangular Element Solving for Ci ‘s: 1 1u1 2u2 3u3 c1 2 Ae 1 1u1 2u2 3u3 c2 2 Ae 1 1u1 2u2 3u3 c3 2 Ae - determinant of A Ae - area of the triangle 2 Ae where, i x j y k xk y j i y j yk i j k and i, j , k permute in a natural order i x j xk Interpolation Functions: Linear Rectangular Element Substituting values of Ci ‘s: u h x, y 1 1u1 2u2 3u3 1u1 2u2 3u3 x 1u1 2u2 3u3 y 2 Ae 3 uie ie x, y i 1 where, 1 x, y ie ie ie i 1,2,3 2 Ae e i With the properties: e i x e j ,y e j ie 0, x i 1 3 3 ij , e i 1, i 1 ie 0 y i 0 3 Lagrange Interpolation Functions: Linear Triangular Element Linear interpolation functions for the three-noded triangular element 3 3 1 1 2 1 2 1 ψ2 ψ1 1 3 2 1 ψ3 Lagrange Interpolation Functions: Triangular Geometries to Avoid • The nodes are almost in-line. • The resulting coefficient matrix is nearly singular or non-invertible. a L>>a L L>>h h L Lagrange Interpolation Functions: Linear Triangular Element Shapes that should be avoided in the meshing. Interpolation Functions: Linear Rectangular Element A rectangular element with four nodes at the corner and of sides a and b. y y b 4 3 1 2 a x x Approximate solution over an element is of the form uh x , y c1 c2 x c3 y c4 x y We need four conditions to find the unknowns Ci ‘s. Interpolation Functions: Linear Rectangular Element Using the nodal solution values, we get four conditions: uh 0,0 u1 c1 uh a,0 u2 c1 a c2 uh a, b u3 c1 c2 a c3 b c4 ab uh 0, b u4 c1 c3 b Gives the values of constants as u3 u4 u1 u2 u2 u1 u4 u1 c1 u1 , c2 , c3 , c4 a b ab Putting these back and re-arranging, x y xy x x y x y y x y uh x , y u1 1 u2 u3 u4 a b ab a a b a b b a b Interpolation Functions: Linear Rectangular Element uh x , y u1 2 u2 2 u3 3 u4 4 where, x y e x x 1 1 , 2 1 a b a b e 1 3e x y x y , 4e 1 a b a b And can be given as x , y 1 e i With the properties: i 1 x xi y yi 1 1 a b ie x j , y j ij , 4 e i 1 i 1 Lagrange Interpolation Functions: Linear Rectangular Element Linear interpolation functions for the four-noded rectangular element 4 4 1 1 1 3 3 ψ1 2 2 1 ψ2 4 1 4 1 3 2 ψ3 1 1 3 2 ψ4 Interpolation Functions: Linear Rectangular Element • Can also be obtained from tensor product of 1D linear interpolation functions associated with sides 1-2 and 2-3 x y e x x e x y x e 1 1 , 2 1 , 3 , 4 1 a b ab a b a e 1 1 x a y b 1 x a x b a x 1 a y x 1 b a y 1 b 2 4 3 y b y y b Interpolation Functions: Linear Rectangular Element Another approach: Derivation of 1e x , y We know that 1e x1 , y1 1 and 1e xi , yi 0 for i 2,3,4 And zero on the lines x a and y b Therefore, 1e is of the form: Putting first condition that Hence, 1e is 1e x , y c1 a x b y e 1 x , y 1 at x 0, y 0 1 x y a x b y 1 1 x , y ab b a e 1 In a similar way, the other interpolation functions can be derived. 1 c1 ab Higher Order Interpolation Functions For Triangular Element Interpolation Functions: Higher Order Element • The interpolation functions of higher degree can be systematically developed with the help of Pascal’s Triangle. • It contain the terms of polynomials of various degrees in the two coordinates x and y. • x and y denote the local coordinate (and not the global coordinates of the problem). • Shape of the triangle need not be equilateral triangle. • p th order triangular element has n nodes given by the relation 1 n p 1 p 2 2 th p • A complete polynomial of degree is given by n n u ( x, y ) ai x r y s u j j i 1 j 1 with r s p Interpolation Functions: Pascal’s Triangle 1 y x x2 x x4 x5 x6 x7 x6 y x5 y 2 3 x y x3 y x4 y x4 y2 y3 xy 2 x y 3 xy 5 x2 y4 3 x y y5 xy 4 x2 y3 x3 y3 4 y4 xy 3 x2 y2 x3 y 2 x5 y 2 y2 xy 4 2 x y 5 y6 xy 6 y7 Interpolation Functions: Pascal’s Triangle Polynomial Degree 1 We choose the terms y in the polynomial x x2 x x4 x5 x6 x7 x6 y x5 y x y x3 y x4 y x4 y2 x5 y 2 y2 x y 3 y4 xy 3 xy 5 x2 y4 3 x y y5 xy 4 x2 y3 x3 y3 4 u h x, y c1 y3 xy 2 x2 y2 x3 y 2 1 0 xy 2 3 Element 4 2 x y 5 y6 xy 6 y7 Interpolation Functions: Pascal’s Triangle Polynomial Degree Element 1 y x x2 x x4 x5 x6 x7 x6 y x5 y x y x3 y x4 y x4 y2 x y 3 y4 xy 3 xy 5 x2 y4 3 x y y5 xy 4 x2 y3 x3 y3 4 uh x, y c1 c2 x c3 y y3 xy 2 x2 y2 x3 y 2 x5 y 2 y2 xy 2 3 1 4 2 x y 5 y6 xy 6 y7 3 Interpolation Functions: Pascal’s Triangle Polynomial Degree Element 1 y x x2 x x x5 x6 x7 x6 y x5 y 4 2 x y x4 y x y 2 x3 y 2 x4 y2 x5 y 2 y3 xy 2 x y 3 xy x y 3 3 y x y c4 xy c5 x 2 c6 y 2 y5 xy 5 x2 y4 3 uh x, y c1 c2 x c3 y 4 xy 4 x2 y3 x3 y3 4 6 y2 xy 2 3 2 4 2 x y 5 y6 xy 6 y7 Interpolation Functions: Pascal’s Triangle Polynomial Degree 1 We choose the terms y in the polynomial x x2 x x4 x5 x6 x7 x6 y x5 y 2 3 x y x3 y x4 y x4 y2 x5 y 2 y2 y3 xy 2 x y 3 xy 5 x2 y4 3 x y y5 xy 4 x2 y3 x3 y3 4 y4 xy 3 x2 y2 x3 y 2 1 1 3 2 6 3 10 4 15 5 21 28 0 xy 4 2 x y 5 y6 xy 6 6 y7 Element 7 1 n p 1 p 2 2 Lagrange Interpolation Functions: Quadratic Triangular Element Geometric variation of Lagrangian interpolation functions Interpolation Functions For Triangular Element using Area Coordinates Interpolation Functions: Using Area Co-ordinates P is an arbitrary point inside a linear triangular element k i (j) P (i) (k) j Let Ai, Aj and Ak be the area of triangles formed by point P with an edge of the triangle. For example, Ak is the area formed by edge i-j and point P. Define: Natural /triangular/ barycentric coordinate Aj Ai A Li , Lj , Lk k A A A and Li A Ai Aj Ak Interpolation Functions: Using Area Co-ordinates • For P at node i: Li 1, L j 0, Lk 0 • For P at node j: Li 0, L j 1, Lk 0 • For P at node k: Li 0, L j 0, Lk 1 Connection with triangular element: L3 1, L1 0, L2 0 L3 L1 1, L2 0, L3 0 1 (2) P (1) (3) L1 2 3 L2 L2 1, L1 0, L3 0 Interpolation Functions: Using Area Co-ordinates • We needed to invert coefficient matrix when we derived interpolation functions. This can be avoided in this approach. 6 3 1 Consider a quadratic triangular element (2) P(1) (3) 5 We express natural co-ordinate Li at any point 4 by using superscript to identify the nodes. 2 Example: Point P moves to node 6. 1 L16 A1 1 , A 2 L26 A2 0 0, A A L36 A3 1 A 2 4 6 3 P (3) (1) 5 2 Interpolation Functions: Using Area Co-ordinates • Point P moves to node at 4. A 1 L1 1 , A 2 4 A 1 L2 2 , A 2 4 6 A 0 L3 3 0 A A 4 1 (2) 4 P • Point P moves to node at 5. A1 0 L1 0, A A 5 L25 A2 1 , A 2 3 (1) 5 2 L35 A3 1 A 2 6 (2) 1 4 3 P (3) 2 5 Interpolation Functions: Using Area Co-ordinates • For a cubic triangular element 1 4 Example: Point P moves to node 10. 10 L1 A1 1 , A 3 10 L2 A2 1 , A 3 L310 A3 1 A 3 1 8 9 (2) 4 P 5 10 7 (1) 6 2 4 3 L1 4 (2) P 10 7 (3) (1) 6 5 2 9 1 Example: Point P moves to node 4. A1 2 , A 3 L24 A2 1 , A 3 A3 0 L3 0 A A 4 8 9 3 8 (2) P 10 7 (3) (1) 6 5 2 3 Interpolation Functions: Using Area Co-ordinates To get the interpolation functions, we first transform the Lagrangian interpolation formula from Cartesian to natural coordinates. L Lj r r nL j i 1 L Lj 1 r where nL j i 1 i r for nL j 1 r for nL j 0 j = one of the three natural coordinates n (=p) = order of interpolation polynomial r = free index from 1 up to the total number of nodes Thus, for each node: N r L L1 L L2 L L3 r r r Interpolation Functions: Using Area Co-ordinates Example: For quadratic triangular element (n=p=2) For N1, put r=1 and considering j=1 first r 1 nL j 2 L1 2 1 2 Thus, L L1 1 1 2 L1 i 1 2 L1 i 1 i 2 i 1 2 L1 i 1 i 2 L1 1 1 2 L1 2 1 L1 2 L1 1 1 2 and L L2 1 0, L L3 1 0 . Therefore, N1 L1 2L1 1 Interpolation Functions: Using Area Co-ordinates Example: For quadratic triangular element (n=p=2) For N4, put r=4 and considering j=1 first r 4 nL j 2 L1 Thus, L L1 4 4 2 L1 i 1 Now, considering j=2 2 L1 i 1 i r 1 i 1 4 nL j 2 L2 1 2 1 2 2 L1 i 1 2 L1 1 1 2 L1 i 1 1 2 1 2 L L2 2 L2 4 Interpolation Functions: Using Area Co-ordinates Example: For quadratic triangular element (n=p=2) For N4, considering j=3 r 4 nL j 2 L3 2 0 0 L L3 1 4 Thus, N 4 4 L1 L2 For remaining nodes, N 2 L2 2 L2 1 , N 3 L3 2 L3 1 , N 5 4 L2 L3 , N 6 4 L1 L3 Higher Order Interpolation Functions For Rectangular Element Interpolation Functions: Pascal’s Triangle Polynomial Degree = 1 1 y x x2 x x x x7 x y x3 y3 x4 y3 xy y4 3 x2 y3 x3 y 2 2 y3 2 x2 y2 x4 y2 5 xy x y x4 y x5 y x6 y 2 x3 y 4 5 x6 3 uh x, y c1 c2 x c3 y c4 xy y2 xy x2 y4 x3 y 4 y5 xy 4 y6 xy 5 2 x y 5 xy 6 y7 Interpolation Functions: Pascal’s Triangle Polynomial Degree = 2 1 x2 x x x x7 x6 y 5 x y x3 y3 x4 y3 2 y4 3 x2 y3 x3 y 2 2 xy y5 xy 4 x2 y4 x3 y 4 2 c7 x y c8 xy c9 x y y3 2 x2 y2 x4 y2 x5 y xy x y x4 y c4 xy c5 x c6 y 2 y2 xy 2 x3 y 4 5 x6 3 uh x, y c1 c2 x c3 y y x y6 xy 5 2 x y 5 xy 6 y7 2 2 2 Interpolation Functions: Pascal’s Triangle Polynomial terms can be chosen Polynomial Degree = 3 1 x x2 x x6 x7 5 x y 2 x4 y3 c10 x 3 c11 x 3 y c12 x 3 y 2 y xy 3 x2 y3 x3 y3 c7 x 2 y c8 xy 2 c9 x 2 y 2 y3 2 2 x3 y 2 x4 y2 x5 y x6 y x y x y x4 y 5 xy x y 2 c4 xy c5 x 2 c6 y 2 y2 xy 2 3 x4 x 3 uh x, y c1 c2 x c3 y y x3 y 4 y5 xy 4 x2 y4 c13 x 3 y 3 c14 y 3 c15 xy3 c16 x 2 y 3 4 y6 xy 5 2 x y 5 xy 6 y7 Interpolation Functions: Pascal’s Triangle Polynomial terms can be chosen Polynomial Degree Element 1 1 y x x2 x x x x7 x4 y 2 x y x y 5 x y 2 2 y3 xy 3 3 2 x y x3 y 4 y 3 4 3 4 y5 xy 4 x2 y3 x y x4 y3 2 2 2 x3 y 2 4 x y xy x y x y 5 x6 y 2 3 4 5 x6 3 y2 xy xy 2 x y 5 4 5 y xy 6 No. of Nodes n p 1 6 y7 2 Interpolation Functions: Pascal’s Triangle in Rectangular Array Form Polynomial Degree Element Rectangular array 0 1 x x2 x3 y xy x 2 y x 3 y 2 2 2 2 3 2 y xy x y x y y 3 xy3 x 2 y 3 x 3 y 3 1 2 3 No. of Nodes n p 1 2 Interpolation Functions: Higher Order Rectangular Elements The Lagrange interpolation functions associated with rectangular elements can be obtained from corresponding one-dimensional Lagrange interpolation functions by taking tensor product. 1 2 3 4 5 6 a x ( x a ) 2 T b a ( a ) y 2 ( y a ) b2 2 2 7 x ( x a ) y ( y b ) 8 1 b2 1 a a) a ( 4 9 2 x( x2 a ) b y( y 2) 2 b2 a( 12 a) 2 Interpolation Functions: Master Rectangular Elements Tensor product of 1D functions: 1,1 4 1,1 1,1 3 1 1,1 1 1 1 1 1 , 2 1 1 , 4 4 1 1 1 1 1 , 4 1 1 4 4 2 1,1 7 8 4 5 3 1,1 1 2 4 1 2 6 2 1 2 9 4 , 1 , 1,1 1 2 1 2 , 2 1 2 4 2 1 2 1 4 1 2 , 5 1 2 2 4 1 2 1 7 2 , 8 1 2 4 2 1 , 1 , , 2 2 2 6 2 1 1,1 9 3 2 2 2 Lagrange Interpolation Functions: Quadratic Rectangular Element Geometric variation of Lagrangian interpolation functions Lagrange Interpolation Functions: Quadratic Rectangular Element Geometric variation of Lagrange interpolation functions: Hermite Interpolation Functions: Master Rectangular Element • The shape function presented earlier are for the interpolation of primary variables. • The Hermite cubic interpolation functions based on the interpolation of can be generated. u u 2u u, Interpolation for Variable u derivative u x , y , xy 1 2 2 i i 2 i i 2 16 1 2 2 i i i 1 i i 2 16 Hermite Interpolation Functions: Master Rectangular Element Derivative u 2u Derivative 1 2 2 i i 2 i i i 1 16 1 2 2 i i 1 1i i i 1 16 where, i=1,….4 are the nodes of the element. Serendipity Family of Interpolation Functions For Rectangular Element Interpolation Functions: Pascal’s Triangle Polynomial Degree Element 1 y x x2 x x x x7 x4 y 2 x y x y 2 y3 xy y 3 x2 y3 x3 y3 x4 y3 2 2 2 x3 y 2 x4 y2 5 xy x y x y x5 y x6 y 2 3 4 5 x6 3 y2 xy x3 y 4 y5 xy 4 x2 y4 3 4 4 y6 xy 5 2 x y 5 xy 6 y7 Terms inside this cone are not included Interpolation Functions: Pascal’s Triangle Polynomial Degree = 2 1 x2 x x x x7 x6 y 5 x y 2 x3 y3 x4 y3 xy 2 y4 3 x2 y3 x3 y 2 c7 x y c8 xy y3 2 x2 y2 x4 y2 x5 y xy x y x4 y c4 xy c5 x c6 y 2 y2 xy 2 x3 y 4 5 x6 3 uh x, y c1 c2 x c3 y y x x2 y4 x3 y 4 y5 xy 4 y6 xy 5 2 x y 5 xy 6 y7 2 2 Interpolation Functions: Pascal’s Triangle Polynomial Degree = 3 1 x x x x x6 x7 x6 y x y xy x3 y3 x4 y3 y xy c7 x y c8 xy c9 x 2 3 x2 y4 x3 y 4 y5 xy 4 y6 xy 5 2 x y 5 2 3 c10 x 3 y c11 y 3 c12 xy3 y4 3 x2 y3 x3 y 2 2 c4 xy c5 x 2 c6 y 2 2 2 x2 y2 x4 y2 5 y x y x4 y x5 y xy 2 x3 y 4 5 2 3 uh x, y c1 c2 x c3 y y x xy 6 y7 Interpolation Functions: Pascal’s Triangle Polynomial terms can be chosen Polynomial Degree Element 1 y x x2 x x x x7 x4 y 2 x y x y 2 y3 xy y 3 x2 y3 x3 y3 x4 y3 2 2 2 x3 y 2 x4 y2 5 xy x y x y x5 y x6 y 2 3 4 5 x6 3 y2 xy x3 y 4 y5 xy 4 x2 y4 3 4 4 y6 xy 5 2 x y 5 xy 6 y7 Interpolation Functions: Derivation of Serendipity Functions Derivation of 1 : 1 vanishes on the lines 1,1 6 1 0 1,1 7 8 1 0 1 0 1 0 1 0, 1 0, 1 0 1 0 4 2 1 1,1 1 0 5 1 0 3 1 0 1,1 Interpolation Functions: Derivation of Serendipity Functions Therefore, 1 is of the from 1 , C 1 1 1 It has a value of 1 at 1, 1 c 1 4 1 1 1 1 1 4 2 vanishes along the lines 1 0,1 0,1 0 Let 2 c 1 1 1 2 1 at 0,-1 c 1 2 2 1 1 2 2 1 Interpolation Functions: Derivation of Serendipity Functions Similarly, 3 14 1 1 1 4 1 4 5 1 2 1 1 1 1 2 2 1 1 1 7 12 1 2 1 8 14 1 1 1 6 1 4 Note: The serendipity interpolation functions are different than Lagrange interpolation functions as internal nodes are not present. Hierarchic Interpolation Functions For Rectangular Element Interpolation Functions: Hierarchic Functions • The set of interpolation functions of polynomial degree p should be in the set of interpolation functions of polynomial degree (p+1). • The number of interpolation functions which do not vanish at vertices and sides should be smallest possible. • Example: 1D Hierarchic interpolation functions Hierarchic Shape Functions: Rectangular Elements • Nodal Interpolation functions: - Four nodal interpolation functions - Exactly same as the functions for four noded quadrilateral element 1 1 1 , 2 1 1 1 , 4 4 1 1 1 1 1 , 4 1 1 4 4 1 Hierarchic Shape Functions: Rectangular Elements • Side Nodes Interpolation functions: - There are 4 p 1 interpolation functions associated with the sides of finite elements with the order of approximation p 2 - For Side 1 - For Side 2 - For Side 3 - For Side 4 1 1 i 2 i 2, , p 2 1 1 i 2 i 2, , p 3 1 1 i 2 i 2, , p 1 1 i 2 i 2, , p i1 i i i4 Hierarchic Shape Functions: Rectangular Elements • Internal Nodes Interpolation functions: - There are p 2 p 3 2 interpolation functions associated with the internal nodes of finite elements with the order of approximation p 4 0 1 0 2 2 , 2 3 2 , 0 0 0 0 3 2 3 , 4 4 2 , 5 3 3 , 6 2 4 , and so on. Hierarchic Shape Functions: Rectangular Elements where, polynomial of degree j is: j 2 j 1 2 Pj 1 t dt , j 2,3, 1 Pj - is Legendre polynomial. Using the properties of Legendre polynomials j 1 2 2 j 1 Pj Pj 2 Hierarchic Shape Functions: Rectangular Elements Legendre polynomial is given by Bonnet Recursion formula: n 1 Pn1 2n 1 Pn nPn1 , with P0 1, P1 and 2 Pi Pj d 2i 1 1 0 1 for i j for i j n 1, 2, Hierarchic Shape Functions: Rectangular Elements Hierarchic Interpolation Functions For Triangular Element Hierarchic Shape Functions: Rectangular Elements • Nodal Interpolation functions: - Three nodal interpolation functions - L1 , L2 , L3 - Exactly same as the functions for three noded triangular element • Side Modes: - There are 3 p 1 side modes and vanish on the other two sides 1 Sides 1: i L2 L1i L2 L1 , where, 1 j 1 2 j , 4 i 2, j 2, ,p ,p Hierarchic Shape Functions: Rectangular Elements For example: • Internal Modes: - There are p 1 p 2 2 internal modes Mode 1: 0 1 L1L2 L3 Other modes: 0 0 2 L1L2 L3 P1 L2 L1 , 3 L1L2 L3 P1 2L3 1 and so on. 7 2 2 6,3 10 ,4 5 1 8 Physical to Master Element Mapping Mapping: • Non rectangular domains cannot be represented using rectangular elements. However, it can be represented more accurately by quadrilateral elements. • Interpolation functions are easily available for a rectangular element • And it is easy to evaluate integral over rectangular geometric • Therefore, we transform the finite element integral statement over quadrilaterals to a rectangular. • Similarly, the finite element integral statement over a triangular element is transformed to an isosceles triangle. • Caution: Transformation is for numerical integration purpose only ! The actual domain is not mapped to another domain. Mapping: • Then numerical integration schemes like Gauss-Legendre are used to evaluate the integrals. • These schemes require that the integral be evaluated on a specific domain or with respect to a specific coordinate system. • In general it is a coordinate system , such that 1 , 1 • Transformation between physical element e and master element ̂ m x j 1 xejˆ ej , , y m j 1 y ejˆ ej , e ˆ where, j interpolation functions over master element ̂ Mapping: In general, the dependent variables(s) are approximated by expressions of the form u x, y n u ej ie x, y j 1 The degree of approximations used for the geometry and the dependent variables(s) are different. Depending upon it, the finite element formulations are classified as: - Sub-parametric - Iso-parametric - Super parametric Mapping: a) super- parametric m n : The approximation used for the geometry is of higher order than used for the dependent variable (s). b) iso-parametric m n dependent variable(s). : Equal degree of approximation for geometry and - This is extensively used in h- version of finite element programs. Mapping: c) Sub-parametric m n : Approximation order used for geometry is lower than that used for primary variable (s). • Coordinate transformations is only for the purpose of numerical integration. • When an element is transformed to its master element for the purpose of numerical integration, the integrand must also be expressed in terms of the coordinates , of the master element. Mapping : For example, consider the stiffness coefficient e e e e j j e e e i i kij a x, y b ( x, y ) c ( x, y ) i j dxdy x x y y e Here, we need to relate ie ie , x y with ie and ie using the transformation. • ie ( x, y) can be expressed in terms of local coordinate and . Hence, the chain rule for partial differentiation gives ie ie x ie y x y In matrix form, ie e i x x y ie x y ie Jacobian Matrix ie ie x ie y x y Mapping : In matrix form, ie e i x x y ie x y ie where, the Jacobian matrix is given as x J x y y e Mapping : e e We need to relate i , i with ˆ ie x ˆ ie and y Hence, we need the inverse of the Jacobian matrix. J must be non singular. ie x e i y ˆ ie 1 J e ˆ i Now, using the transformation, we can write x x ˆ ej m x j j 1 ˆ ej m x j 1 j , , y y m j 1 m j 1 ˆ ie yj , ˆ ie yj Mapping : m xj j 1 J m xj j 1 ˆ ie ˆ1 yj j 1 m ˆ1 ˆ ie yj j 1 ˆ ie ˆ ie m ˆ 2 ˆ 2 ˆ m ˆ m x1 x 2 xm y1 y2 ym A necessary and sufficient condition for J 1to exist is that the determinant J, called the Jacobian, be non negative at every point , in ̂ . x y x y J det J 0 Thus, the function ( x, y ), ( x, y ) must be continuous, differentiable and invertible. Mapping : • In case of higher order triangular and rectangular elements the placement of edge and interior nodes is restricted. • It can be shown that the side nodes should be placed at a distance greater than a quarter of the length of the side form either corner node. x 4(a 1 2) y 4(b 1 2) 0,1 0 6 5 1 4 0,0 0 2 1,0 0.25 3 1 3 y 6 5 1 4 J 1 2 2b 1 2 2a 1 0 a, b 0.25 2 x Mapping : Mapping the elemental area dA from x,y to d d ˆ dA dx dy dx dy k x y x y ˆ ˆ ˆ d i d j d i d We have ˆj kˆ x y y y d d J d d Now, the derivatives of the shape functions with respect to x,y can be given as ie x e i y ˆ ie 1 J e ˆ i ˆ ie * J e ˆ i Mapping : Now, using this we can write the element calculations over master element as kije i j i j a b c i j dxdy x x y y e ˆ ˆ ˆ i ˆ i j j aˆ J11 J12 J12 J11 ˆ i i i i ˆ b J 21 J 22 J 22 J 21 cˆ i j J d d or kije F , d d ˆ Mapping of Physical Elements to Master Elements Affine Mapping Physical to Master Element Mapping: Linear Triangular Element 3 3 j 1 j 1 x x jˆ ej , y y jˆ ej 0,1 3 0 1 y 3 1 2 1 0,0 2 0 1,0 Linear Master Element x Linear Physical Element • The affine mapping essentially stretches, translates, and rotates the triangle. • Straight or planar faces of the reference cell are therefore mapped onto straight or planar faces in the physical coordinate system. • Preferred when all the edges of an element are straight. Numerical Integration: Triangular Element For three noded triangular element the transformation of co-ordinates is given by 3 x x ˆ , , i i 1 i 3 y y ˆ , i i i 1 • where, ˆ i are the interpolation functions over the master three noded triangular element. ˆ1 1 ˆ 2 ˆ 3 • Master triangle is an isosceles triangle with unit length and right angle Mapping: Triangular Element The Jacobian matrix, [J] for the linear triangular element is x2 x1 J x3 x1 y2 y1 y3 y1 Inverse transformation from the element e to ̂ is: 1 x x1 y3 y1 y y1 x3 x1 2A 1 x x1 y1 y2 y y1 x2 x1 2A e where, A the area of the element Physical to Master Element Mapping: Linear Rectangular Element y 1,1 4 1,1 3 3 2 4 1 1 1,1 2 x 1,1 Linear Master Element Linear Physical Element 4 4 j 1 j 1 x x jˆ ej , y y jˆ ej Mapping of Physical Elements to Master Elements Iso-parametric Mapping Physical to Master Element Mapping: Quadratic Triangular Element 6 6 x x jˆ ej , y y jˆ ej j 1 0,1 y j 1 3 0 6 1 0,0 6 1 3 5 4 0 5 1 4 2 1,0 Quadratic Master Element 2 x Quadratic Physical Element • The straight faces of the reference triangle are mapped onto curved faces of parabolic shape in the physical coordinate system. Physical to Master Element Mapping: Quadratic Rectangular Element 1,1 4 y 1,1 8 8 3 7 3 2 4 6 1 1,1 7 5 5 6 1 2 x 1,1 Quadratic Master Element Quadratic Physical Element 8 8 j 1 j 1 x x jˆ ej , y y jˆ ej Mapping of Physical Elements to Master Elements Blending Function Mapping Physical to Master Element Mapping: Quadratic Rectangular Element Curved Quadratic Physical Element with Side 2 is the only curved side. x x2 y y y2 3 Curve x x2 , y y2 given in parametric form so that x2 1 X 2 , y2 1 Y2 , x2 1 X 3 , y2 1 Y3 X i ,Yi are the coordinates of node i 2 4 1 x Physical to Master Element Mapping: Quadratic Rectangular Element We have 1 1 1 1 1 1 X 1 1 X 1 1 X 1 2 3 1 1 X 4 4 4 4 4 1 1 1 x2 X2 X3 2 2 2 x The first four terms are the linear mapping terms. The fifth term is the product of two functions: The bracketed term represents the difference between x2 and the x-coordinates of the chord that connects X 2 , Y2 and X 3 , Y3 The other is the linear blending function 1 2 , which is unity along side 2 and zero along side 4. Physical to Master Element Mapping: Quadratic Rectangular Element Therefore, we simplify as 1 1 1 x 1 1 X 1 1 1 X 4 x2 4 4 2 Similarly, 1 1 1 y 1 1 Y1 1 1 Y4 y2 4 4 2 In the general case all sides may be curved. We write the curved sides in the parametric form: x xi , y yi , 1 1 i 1,2,3,4 Mapping functions are: 1 1 1 1 x 1 1 X1 1 1 X 2 1 1 X 3 1 1 X 4 4 4 4 4 1 1 1 1 x1 x2 x3 x4 2 2 2 2 Physical to Master Element Mapping: Quadratic Rectangular Element Mapping functions for y coordinate are: 1 1 1 1 y 1 1 Y1 1 1 Y2 1 1 Y3 1 1 Y4 4 4 4 4 1 1 1 1 y1 y2 y3 y4 2 2 2 2 Inverse mapping cannot be given explicitly. Newton-Raphson method or similar procedure is used. One can see the quadratic iso-parametric mapping as a special application of the blending function method. Numerical Integration over a Master Rectangular Element Numerical Integration: kije F , d d ˆ 1 1 F , d d F , d d ˆ 1 1 1 1 M , j j 1 w j d F i , j wi w j M,N - Number of quaddrature points in and i ,i - Gauss points, wi , w j - corresponding weights. M N i 1 j 1 directions. The M N Gauss point locations are given by the tensor product of one dimensional Gauss points. Numerical Integration: Rectangular Element Selection of integration order and location of the Gauss points: Numerical Integration over a Master Triangular Element Numerical Integration: kije F , d d ˆ NINT F , dd j , j w j j 1 0 1 1 F , d d ˆ 0 NINT – Number of quadrature points i ,i - Gauss points, wi - corresponding weight. Numerical Integration: Triangular Element Numerical Integration: Using Area Co-ordinates Line integration: b L m 1 L2 k a m !k ! ds b a m k 1! Area integration: A L1 m L2 L2 k l m ! k !l ! dA 2A m k l 2 ! Transformation of the co-ordinates to natural co-ordinates: n x x L , i i i 1 where, n y y L i i i 1 xi , yi are the global coordinates of ith node. Thank you. Physical to Master Element Mapping: Quadratic Rectangular Element 1,1 7 y 1,1 8 8 3 9 2 4 4 1 1,1 5 6 2 7 5 6 1 3 1,1 x Quadratic Master Element Quadratic Physical Element 8 8 j 1 j 1 x x jˆ ej , y y jˆ ej • This is also an Iso-parametric mapping of quadratic element. Physical to Master Element Mapping: Quadratic Triangular Element y 0,1 3 6 1 3 0 6 5 1 4 0,0 0 1 4 5 2 2 1,0 x Quadratic Master Element Quadratic Physical Element • This is also an Iso-parametric mapping of quadratic element. 6 6 j 1 j 1 x x jˆ ej , y y jˆ ej