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konacni elementi viseg reda

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!