Higher Order FEM

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Text of Higher Order FEM

  • MANE 4240 & CIVL 4240Introduction to Finite ElementsHigher order elementsProf. Suvranu De

  • Reading assignment:

    Lecture notesSummary:

    Properties of shape functions Higher order elements in 1D Higher order triangular elements (using area coordinates) Higher order rectangular elements Lagrange familySerendipity family

  • Recall that the finite element shape functions need to satisfy the following properties

    1. Kronecker delta property

    Inside an elementAt node 1, N1=1, N2=N3==0, henceFacilitates the imposition of boundary conditions

  • 2. Polynomial completeness

    Then

  • Higher order elements in 1D2-noded (linear) element:12x2x1In local coordinate system (shifted to center of element)12xaax

  • 3-noded (quadratic) element:12x2x1In local coordinate system (shifted to center of element)12xaa3x3x3

  • 4-noded (cubic) element:12x2x1In local coordinate system (shifted to center of element)122a/3xaa3x3xx44342a/32a/3

  • Polynomial completeness2 node; k=1; p=23 node; k=2; p=34 node; k=3; p=4Convergence rate (displacement)Recall that the convergence in displacementsk=order of complete polynomial

  • Triangular elementsArea coordinates (L1, L2, L3)1A=A1+A2+A3Total area of the triangleAt any point P(x,y) inside the triangle, we defineNote: Only 2 of the three area coordinates are independent, sinceL1+L2+L3=1

  • Check that

  • xy23PA11L1= constantPLines parallel to the base of the triangle are lines of constant L

  • We will develop the shape functions of triangular elements in terms of the area coordinates

  • For a 3-noded triangle

  • xy231L1= 0L1= 1L2= 0L2= 1L3= 0L3= 1For a 6-noded triangleL1= 1/2L2= 1/2L3= 1/2456

  • How to write down the expression for N1?Realize the N1 must be zero along edge 2-3 (i.e., L1=0) and at nodes 4&6 (which lie on L1=1/2) Determine the constant c from the condition that N1=1 at node 1 (i.e., L1=1)

  • xy231L1= 0L1= 1L2= 0L2= 1L3= 0L3= 1For a 10-noded triangleL1= 2/3L2= 2/3L3= 1/3L3= 2/3L2= 1/3L1= 1/354678910

  • NOTES:1. Polynomial completeness3 node; k=1; p=26 node; k=2; p=310 node; k=3; p=4Convergence rate (displacement)

  • 2. Integration on triangular domain1xy23l1-2

  • 3. Computation of derivatives of shape functions: use chain rulee.g.,Bute.g., for the 6-noded triangle

  • Rectangular elementsLagrange familySerendipity familyLagrange family4-noded rectanglexyaa1234bbIn local coordinate system

  • 9-noded quadraticxyaa1234bbCorner nodes56789Midside nodesCenter node

  • NOTES:1. Polynomial completeness4 node; p=29 node; p=3Convergence rate (displacement)Lagrange shape functions contain higher order terms but miss out lower order terms

  • Serendipity familyThen go to the corner nodes. At each corner node, first assume a bilinear shape function as in a 4-noded element and then modify: xyaa1234bb5678First generate the shape functions of the midside nodes as appropriate products of 1D shape functions, e.g.,4-noded same as Lagrange8-noded rectangle: how to generate the shape functions?bilinear shape fn at node 1:actual shape fn at node 1:

  • Corner nodesMidside nodes8-noded rectangle

  • NOTES:1. Polynomial completeness4 node; p=28 node; p=3Convergence rate (displacement)12 node; p=416 node; p=4More even distribution of polynomial terms than Lagrange shape functions but p cannot exceed 4!