Shape function in Finite Element Method
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Why is it that the choice of polynomial for 6-nodes rectangular element(linear in sides 1 and 3, quadratic in sides 2 and 4) in FEM does not follow normal pascal triangle regular arrangement? i.e $u=c_1+c_2x+c_3y+c_4xy+c_5xy^2+c_6x^2y$ .
finite-element-method
edited Sep 8 at 17:23
Han de Bruijn
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asked Sep 5 at 10:32
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Why is it that the choice of polynomial for 6-nodes rectangular element(linear in sides 1 and 3, quadratic in sides 2 and 4) in FEM does not follow normal pascal triangle regular arrangement? i.e $u=c_1+c_2x+c_3y+c_4xy+c_5xy^2+c_6x^2y$ .
finite-element-method
edited Sep 8 at 17:23
Han de Bruijn
12k22260
asked Sep 5 at 10:32
COLLINS AKEREMALE
1
2
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 5 at 10:37
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Why is it that the choice of polynomial for 6-nodes rectangular element(linear in sides 1 and 3, quadratic in sides 2 and 4) in FEM does not follow normal pascal triangle regular arrangement? i.e $u=c_1+c_2x+c_3y+c_4xy+c_5xy^2+c_6x^2y$ .
finite-element-method
edited Sep 8 at 17:23
Han de Bruijn
12k22260
asked Sep 5 at 10:32
COLLINS AKEREMALE
1
Why is it that the choice of polynomial for 6-nodes rectangular element(linear in sides 1 and 3, quadratic in sides 2 and 4) in FEM does not follow normal pascal triangle regular arrangement? i.e $u=c_1+c_2x+c_3y+c_4xy+c_5xy^2+c_6x^2y$ .
finite-element-method
finite-element-method
edited Sep 8 at 17:23
Han de Bruijn
12k22260
asked Sep 5 at 10:32
COLLINS AKEREMALE
1
edited Sep 8 at 17:23
Han de Bruijn
12k22260
asked Sep 5 at 10:32
COLLINS AKEREMALE
1
edited Sep 8 at 17:23
Han de Bruijn
12k22260
edited Sep 8 at 17:23
Han de Bruijn
12k22260
edited Sep 8 at 17:23
Han de Bruijn
12k22260
12k22260
asked Sep 5 at 10:32
COLLINS AKEREMALE
1
asked Sep 5 at 10:32
COLLINS AKEREMALE
1
asked Sep 5 at 10:32
COLLINS AKEREMALE
1
1
2
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 5 at 10:37
add a comment |Â
2
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 5 at 10:37
2
2
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 5 at 10:37
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 5 at 10:37
add a comment |Â
1 Answer
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See above picture: there are only (linear) line segments in vertical direction.
And because of the three nodal points, there are two quadratic elements in horizontal direction. A reference for the latter:
- Understand 1D FEM solution using quadratics elements
Therefore the interpolation must be linear in $,y,$ and quadratic in $,x,$. According to a Cartesian product:
$$
(1,y) times (1,x,x^2) = (1,x,x^2,y,yx,yx^2)
$$
Mind that your term with $y^2$ is absent now and two terms with $x^2$ are present.
So the correct interpolation may be formulated as:
$$
u=c_1+c_2x+c_3y+c_4xy+colorredc_5x^2+c_6x^2y
$$
answered Sep 8 at 18:20
Han de Bruijn
12k22260
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
See above picture: there are only (linear) line segments in vertical direction.
And because of the three nodal points, there are two quadratic elements in horizontal direction. A reference for the latter:
- Understand 1D FEM solution using quadratics elements
Therefore the interpolation must be linear in $,y,$ and quadratic in $,x,$. According to a Cartesian product:
$$
(1,y) times (1,x,x^2) = (1,x,x^2,y,yx,yx^2)
$$
Mind that your term with $y^2$ is absent now and two terms with $x^2$ are present.
So the correct interpolation may be formulated as:
$$
u=c_1+c_2x+c_3y+c_4xy+colorredc_5x^2+c_6x^2y
$$
answered Sep 8 at 18:20
Han de Bruijn
12k22260
add a comment |Â
up vote
0
down vote
See above picture: there are only (linear) line segments in vertical direction.
And because of the three nodal points, there are two quadratic elements in horizontal direction. A reference for the latter:
- Understand 1D FEM solution using quadratics elements
Therefore the interpolation must be linear in $,y,$ and quadratic in $,x,$. According to a Cartesian product:
$$
(1,y) times (1,x,x^2) = (1,x,x^2,y,yx,yx^2)
$$
Mind that your term with $y^2$ is absent now and two terms with $x^2$ are present.
So the correct interpolation may be formulated as:
$$
u=c_1+c_2x+c_3y+c_4xy+colorredc_5x^2+c_6x^2y
$$
answered Sep 8 at 18:20
Han de Bruijn
12k22260
add a comment |Â
up vote
0
down vote
up vote
0
down vote
See above picture: there are only (linear) line segments in vertical direction.
And because of the three nodal points, there are two quadratic elements in horizontal direction. A reference for the latter:
- Understand 1D FEM solution using quadratics elements
Therefore the interpolation must be linear in $,y,$ and quadratic in $,x,$. According to a Cartesian product:
$$
(1,y) times (1,x,x^2) = (1,x,x^2,y,yx,yx^2)
$$
Mind that your term with $y^2$ is absent now and two terms with $x^2$ are present.
So the correct interpolation may be formulated as:
$$
u=c_1+c_2x+c_3y+c_4xy+colorredc_5x^2+c_6x^2y
$$
answered Sep 8 at 18:20
Han de Bruijn
12k22260
See above picture: there are only (linear) line segments in vertical direction.
And because of the three nodal points, there are two quadratic elements in horizontal direction. A reference for the latter:
- Understand 1D FEM solution using quadratics elements
Therefore the interpolation must be linear in $,y,$ and quadratic in $,x,$. According to a Cartesian product:
$$
(1,y) times (1,x,x^2) = (1,x,x^2,y,yx,yx^2)
$$
Mind that your term with $y^2$ is absent now and two terms with $x^2$ are present.
So the correct interpolation may be formulated as:
$$
u=c_1+c_2x+c_3y+c_4xy+colorredc_5x^2+c_6x^2y
$$
answered Sep 8 at 18:20
Han de Bruijn
12k22260
answered Sep 8 at 18:20
Han de Bruijn
12k22260
answered Sep 8 at 18:20
Han de Bruijn
12k22260
answered Sep 8 at 18:20
Han de Bruijn
12k22260
12k22260
add a comment |Â
add a comment |Â
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 5 at 10:37