View
222
Download
0
Category
Preview:
Citation preview
8/9/2019 FEM Zabaras 2DFiniteElements
1/15
CC OO RR NN EE LL LL U N I V E R S I T Y 1
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
MAE4700/5700Finite Element Analysis for
Mechanical and Aerospace DesignCornell University, Fall 2009
Nicholas ZabarasMaterials Process Design and Control Laboratory
Sibley School of Mechanical and Aerospace Engineering101 Rhodes Hall
Cornell UniversityIthaca, NY 14853-3801
http://mpdc.mae.cornell.edu/Courses/MAE4700/MAE4700.htmlhttp://mpdc.mae.cornell.edu/http://mpdc.mae.cornell.edu/http://mpdc.mae.cornell.edu/Courses/MAE4700/MAE4700.html8/9/2019 FEM Zabaras 2DFiniteElements
2/15
CC OO RR NN EE LL LL U N I V E R S I T Y 2
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Isoparametric finite elements
The basis functions used in the definition of themapping T e , do not have to be the same as thoseused for the approximation of functions.
Let M be the number of basis functions used to defineT e and let N e be the number of basis functions
(nodes) used in the approximation of functions.
Polynomials used to define geometry can be of higherorder ( M>N e ), equal ( M=N e ) or lower ( M
8/9/2019 FEM Zabaras 2DFiniteElements
3/15
CC OO RR NN EE LL LL U N I V E R S I T Y 3
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Sub-, iso- and super-parametric finite elements
8/9/2019 FEM Zabaras 2DFiniteElements
4/15
8/9/2019 FEM Zabaras 2DFiniteElements
5/15
CC OO RR NN EE LL LL U N I V E R S I T Y 5
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Quadrilateral elements: Bi-quadratic
We use tensor product ofpolynomials as discussed in1D (Lagrange family)
Note that here we have oneinternal node (9).
2 2 2 21 5
2 2 2 22 6
2 2 2 2
3 7
2 2 2 24 8
2 29
1 1( , ) ( )( ), ( , ) (1 )( )4 21 1
( , ) ( )( ), ( , ) ( )(1 )4 21 1
( , ) ( )( ), ( , ) (1 )( )4 21 1
( , ) ( )( ), ( , ) ( )(1 )4 2
( , ) (1 )(1 )
N N
N N
N N
N N
N
= =
= + = +
= + + = +
= + =
=
8/9/2019 FEM Zabaras 2DFiniteElements
6/15
CC OO RR NN EE LL LL U N I V E R S I T Y 6
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Quadratic eight node element
These type of elementsare not derived fromtensor product of 1Dpolynomials. They arecalled serendipityelements.
To derive the shapefunction for node 1, weneed a polynomial that
vanishes on thefollowing lines:1 ,1 ,1 + +
21 5
22 6
23 7
2
4 8
1 1( , ) (1 )(1 )( 1 ), ( , ) (1 )(1 )
4 2
1 1( , ) (1 )(1 )( 1 ), ( , ) (1 )(1 )4 21 1
( , ) (1 )(1 )( 1 ), ( , ) (1 )(1 )4 21 1
( , ) (1 )(1 )( 1 ), ( , ) (1 )(1 )4 2
N N
N N
N N
N N
= =
= + + = +
= + + + + = +
= + + =
8/9/2019 FEM Zabaras 2DFiniteElements
7/15
CC OO RR NN EE LL LL U N I V E R S I T Y 7
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Quadrature rules
Quadrature rules are defined from the 1D Gauss rulespresented earlier as follows:
Here we re-labeled(m,n) with a singleindex
( )'
1 11 1
1 1 1
( , ) ( , )
( , ) ( , )i i l N N N
n m n m l l lm n l
G d d G d d
G w w G w
= = =
=
= =
1l=
8/9/2019 FEM Zabaras 2DFiniteElements
8/15
CC OO RR NN EE LL LL U N I V E R S I T Y 8
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Triangular elements
We first consider triangular with straight sides . Weconsider the mapping from a right-isosceles mastertriangle. By inspection, we can write the basisfunctions as:
The coordinate mapping is then defined from:
1
2
3
( , ) 1
( , )
( , )
N
N
N
=
=
=
3
1
3
1
( , )
( , ) ,
j j j
j j
j
x x N
y y N
=
=
=
=
8/9/2019 FEM Zabaras 2DFiniteElements
9/15
CC OO RR NN EE LL LL U N I V E R S I T Y 9
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Triangular elements
Inverting this mapping gives:
Can you recognize these as the linear shape functions ofthe 4 node quadrilateral element? (take nodes )
1
2
3
( , ) 1( , )
( , )
N N
N
= =
=
3
1
3
1
( , )
( , ) ,
j j j
j j j
x x N
y y N
=
=
=
=
{ }
{ }
3 1 1 3 1 1
2 1 1 2 1 1
1( )( ) ( )( )
2
1 ( )( ) ( )( )2
e
e
e
areaof
y y x x x x y y A
y y x x x x y y A
=
= +
2
3
1
( , )
( , )
1 ( , )
e
e
e
N x y
N x y
N x y
=
=
=
Using these, one can now easily computeand thus the element stiffness and load
, , , ,| | J x y x y
3 4
8/9/2019 FEM Zabaras 2DFiniteElements
10/15
CC OO RR NN EE LL LL U N I V E R S I T Y 10
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Area coordinates
The expressions for caneasily be interpreted as ratios ofareas . We will see this interpretationto be useful in deriving higher ordertriangular elements.
Let us join the points and (x,y)
with the vertices of the trianglesand respectively. We denoteas the areas of the subtriangles
opposite node in and We define the area coordinates onas: where is the
area of the master element.
, ,1
( , )
,e,i ia a
,,e
, 1,2,3,iia i= = 1/ 2=
respectively.1
2
3
1
= ==
8/9/2019 FEM Zabaras 2DFiniteElements
11/15
CC OO RR NN EE LL LL U N I V E R S I T Y 11
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Area coordinates
Since |J| is constant (the ratio of theareas of and ), the map T e transforms areas uniformly, thus:
This is only true for triangles with
straight sides.
1
2
3
1
= ==
,e ,
, 1,2,3,i iie
aa i A A
= = =
8/9/2019 FEM Zabaras 2DFiniteElements
12/15
CC OO RR NN EE LL LL U N I V E R S I T Y 12
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Area coordinates
tani cons t =
0, 1,2,3.i i= =
Several interesting propertiesof area coordinates areshown in the figures.
At a given point, the lineis parallel to theside of the element
opposite node i. The boundary segments ofthe element are defined by
The vertices of the triangleare (1,0,0), (0,1,0) and
(0,0,1).
8/9/2019 FEM Zabaras 2DFiniteElements
13/15
CC OO RR NN EE LL LL U N I V E R S I T Y 13
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Higher-degree shape functions
The area coordinates on can be used todetermine higher degree shape functionsi
,( ).i
Quadratic shapefunctions
Cubic shapefunctions
1 41 1 1 2
2 52 2 2 3
3 63 3 3 1
12 ( ), 42
12 ( ), 4
21
2 ( ), 4
2
N N
N N
N N
= =
= =
= =
1 1 1 1
4 1 2 1
3 1 2 2
101 2 3
9 2 1( )( )
2 3 3
27 1( )2 3
27 1( )
2 3
27
N
N
N
N
=
=
=
=
8/9/2019 FEM Zabaras 2DFiniteElements
14/15
CC OO RR NN EE LL LL U N I V E R S I T Y 14
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Shape functions on triangles using area coordinates
The coordinate transformation is now having the form:
The calculations here defer from those used inquadrilaterals because of the redundant areacoordinate
Calculation of derivatives proceeds as:
Alternatively, one can use andproceed exactly as was done before for quadrilaterals.
1 2 3 1 2 31
1 2 3 1 2 31
( , ) ( , , ) ( , , )
( , ) ( , , ) ( , , ),
e
e
N j j
j
N j j
j
x x x N
y y y N
=
=
=
=
1 2 31 . =
1 1 1 1 31 2
1 2 3
N N N N x x x x
= + +
1 2 3 2 31 , , , =
8/9/2019 FEM Zabaras 2DFiniteElements
15/15
CC OO RR NN EE LL LL U N I V E R S I T Y 15
MAE 4700 FE Analysis for Mechanical & Aerospace Design N. Zabaras (10/06/2009)
Quadrature integration formulas for triangles
We use particular quadrature rules appropriate for thearea coordinates introduced earlier.
int
1 2 3 2 3 1 2 3 1 2 31
int
int s in
( , , ) ( 1 ) ( , , ) N
l l l ll
egrationquadratureweights po
G d d note G w =
= =
Linear Quadratic Cubic Polynomialsup to thisdegree are
integratedexactly
Recommended