Isoparametric Elements

Structural MechanicsDisplacement-based Formulations

Fundamental Dilemma• A primary reason engineers go to FEA is complex geometry

• But elements give the most accurate results when they have regular shapes (isosceles triangles, squares)

• You should always minimize element distortion when you create a mesh (more on this later …)

• It is also important to understand how element shape is managed …interpolation (shape) functions

Reduced Accuracy

• These elements work, but not well …

Isoparametric Elements• There are two roles of interpolation in FEA:

– Defining the location of interior points within an element in terms of nodal values (geometry interpolation)

– Defining the displacement of interior points within an element in terms of nodal values (result interpolation)

• There is no fundamental reason why both types of interpolation must be conducted in the same way

• But a common class of highly versatile elements does just that – Iso = same; the same basis for geometry and result interpolation

Bilinear Quadrilateral (Q4)• Interpolation involves the summation of nodal values

multiplied by corresponding shapes functions

€

xy ⎧ ⎨ ⎩

⎫ ⎬ ⎭=

N ix i∑N iy i∑

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= N[ ] c{ }

€

uv ⎧ ⎨ ⎩

⎫ ⎬ ⎭=

N iui∑N iv i∑

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= N[ ] d{ }

€

c{ } = x1 y1 x2 y2 x3 y3 x4 y4[ ]T

d{ } = u1 v1 u2 v2 u3 v3 u4 v4[ ]T

N[ ] =N1 0 N2 0 N3 0 N4 00 N1 0 N2 0 N3 0 N4

⎡ ⎣ ⎢

⎤ ⎦ ⎥

- where -geometry interpolation field variable interpolation

nodal coordinates

nodal displacements

shape functions

Shape Functions• Shape functions have a value of 1.0 at the “corresponding” node and a

value of 0.0 at all others (the function “belongs” to a node)• They span a normalized domain, typically [-1,1] over each spatial

dimension

€

N1 = 14

1− a( ) 1−b( )

€

N2 = 14

1+ a( ) 1−b( )€

N3 = 14

1+ a( ) 1+ b( )

€

N4 = 14

1− a( ) 1+ b( )

Element Geometry Interpolation• Edges of adjacent elements match (no overlaps, gaps) as long as common

nodes are shared• There are consistent interior point locations defined by the interpolation

functions (e.g. you can define the “center” of an element)

Example 2D Element

N1 = (3,2)N2 = (11,3)N3 = (10,10)N4 = (4,9)

€

xmid = 3( )14

1− 0( ) 1− 0( ) + 11( )14

1+ 0( ) 1− 0( ) + 10( )14

1+ 0( ) 1+ 0( ) + 4( )14

1− 0( ) 1+ 0( ) = 7

€

ymid = 2( )14

1− 0( ) 1− 0( ) + 3( )14

1+ 0( ) 1− 0( ) + 10( )14

1+ 0( ) 1+ 0( ) + 9( )14

1− 0( ) 1+ 0( ) = 6

N1N2

N3N4

X

Y

The shape functions establish a geometric equivalence between elements with different node locations…

XY

Field Quantity Interpolation• We assume the field quantity (e.g. x-component of displacement) varies within the element

as a sum of node-weighted shape functions (visualized here as height above the plane)• If the field quantity actually does vary in this way (or close to it) then the element choice

(size, order) is justified

u1 = 2u2 = 3u3 = 4u4 = 5

€

umid = 2( )14

1− 0( ) 1− 0( ) + 3( )14

1+ 0( ) 1− 0( ) + 4( )14

1+ 0( ) 1+ 0( ) + 5( )14

1− 0( ) 1+ 0( ) = 3.5

nodal values

At the element “center” …

XY

Multiple Elements

N1 = (3,2)N2 = (11,3)N3 = (10,10)N4 = (4,9)

N5 = (0,-5)N6 = (9,-3)N3 = (11,3)N4 = (3,2)

element 1

element 2

……

u1 = 2u2 = 3u3 = 4u4 = 5u5 = 1u6 = 2.5

C0 Continuous• A key feature of these elements is their continuity• The value (C0) of an interpolated quantity is continuous across

element boundaries– No geometric “gaps” or “overlaps” along the edges of shared elements– Field quantities transition without any “steps” in value

• But the slope (C1) (derivative) of a field quantity is not continuous along element edges– As element density increases the transitions become less abrupt– It is possible to construct C1 shape functions, but they require “slope

nodes” and are not as computationally efficient as mesh refinement

Higher-Order Elements• The overall scheme stays exactly the same• But there are more nodes and different shape function definitions

€

N1 = 14

1− a( ) 1−b( ) − 12N8 + N5( )

€

N2 = 14

1+ a( ) 1−b( ) − 12N5 + N6( )

€

N3 = 14

1+ a( ) 1+ b( ) − 12N6 + N7( )

€

N4 = 14

1− a( ) 1+ b( ) − 12N7 + N8( )

€

N5 = 12

1− a2( ) 1−b( )

€

N6 = 12

1+ a( ) 1−b2( )

€

N7 = 12

1− a2( ) 1+ b( )

€

N8 = 12

1− a( ) 1−b2( )

Multiple Elements