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This section gives data that is no longer used but is important enough not to be deleted.
These elements support isotropic and 2-D anisotropic materials declared with a material entry described in m_elastic. Element property declarations are p_solid subtype 2 entries
[ProId fe_mat('p_solid','SI',2) f N 0]
Where
f | Formulation : 0 plane stress, 1 plane strain, 2 axisymmetric. |
N | Fourier coefficient for axisymmetric formulations |
Integ | set to zero to select this family of elements. |
The xy plane is used with displacement DOFs .01 and .02 given at each node. Element matrix calls are implemented using .c files called by of_mk_subs.c and handled by the element function itself, while load computations are handled by fe_load. For integration rules, see section 7.19.2. The following elements are supported
There are five nodes for this incompressible quadrilateral element, four nodes at the vertices and one at the intersection of the two diagonals.
The displacement (u,v) are assumed to be linear functions of (x,y) (Linear Triangular Element), thus the strain are constant (Constant Strain Triangle).
The hexa8 and hexa20 elements are the standard 8 node 24 DOF and 20 node 60 DOF brick elements.
The hexa8 element uses the et*3q1d routines.
hexa8 volumes are integrated at 8 Gauss points
ωi = 1 /8 for i=1,4
bi for i=1,4 as below, with z=α1
bi for i=4,8 as below, with z=α2
hexa8 surfaces are integrated using a 4 point rule
ωi = 1 /4 for i=1,4
b1= ( α1 , α1 ) , b2= ( α2 , α1 ) , b3= ( α2 , α2 ) and b4= ( α1 , α2 )
with α1= 1 /2−1 /2 √3=0.2113249 and α2= 1 /2+1 /2 √3=0.7886751.
The hexa20 element uses the et*3q2c routines.
hexa20 volumes are integrated at 27 Gauss points ωl = wi wj wk for i,j,k=1,3
with
w1 = w3 = 5/18 and w2 = 8/18 bl = (αi, αj, αk) for i,j,k=1,3
with
α1 = 1 − √3/5 /2 , α2 = 0.5 and α3 = 1 + √3/5 /2
α1 = 1 − √3/5 /2 , α2 = 0.5 and
hexa20 surfaces are integrated at 9 Gauss points ωk = wi wj for i,j=1,3 with
wi as above and bk = (αi, αj) for i,j=1,3
with α1 = 1 − √3/5 /2 , α2 = 0.5 and α3 = 1 + √3/5 /2 .
The penta6 and penta15 elements are the standard 6 node 18 DOF and 15 node 45 DOF pentahedral elements. A derivation of these elements can be found in [51].
The penta6 element uses the et*3r1d routines.
penta6 volumes are integrated at 6 Gauss points
Points bk | x | y | z |
1 | a | a | c |
2 | b | a | c |
3 | a | b | c |
4 | a | a | d |
5 | b | a | d |
6 | a | b | d |
with a=1 /6=.16667, b=4/6=.66667, c=1 /2−1 /2 √3=.21132, d= 1 /2+1 /2 √3=.78868
penta6 surfaces are integrated at 3 Gauss points for a triangular face (see tetra4) and 4 Gauss points for a quadrangular face (see hexa8).
penta15 volumes are integrated at 21 Gauss points with the 21 points formula
a= 9 − 2 √15 /21 , b= 9 + 2 √15 /21 ,
c= 6 + √15 /21 , d= 6 − √15 /21 ,
e= 0.5 ( 1 − √ 3 /5 ),
f= 0.5 and g= 0.5 ( 1 + √ 3 /5 )
α = 155 − √15 /2400 , β = 5 /18 ,
γ = 155 + √15 /2400 , δ = 9 /80 and є = 8 /18 .
Positions and weights of the 21 Gauss point are
Points bk | x | y | z | weight ωk |
1 | d | d | e | α.β |
2 | b | d | e | α.β |
3 | d | b | e | α.β |
4 | c | a | e | γ.β |
5 | c | c | e | γ.β |
6 | a | c | e | γ.β |
7 | 1 /3 | 1 /3 | e | δ.β |
8 | d | d | f | α.є |
9 | b | d | f | α.є |
10 | d | b | f | α.є |
11 | c | a | f | γ.є |
12 | c | c | f | γ.є |
13 | a | c | f | γ.є |
14 | 1 /3 | 1 /3 | f | δ.є |
15 | d | d | g | α.β |
16 | b | d | g | α.β |
17 | d | b | g | α.β |
18 | c | a | g | γ.β |
19 | c | c | g | γ.β |
20 | a | c | g | γ.β |
21 | 1 /3 | 1 /3 | g | δ.β |
penta15 surfaces are integrated at 7 Gauss points for a triangular face (see tetra10) and 9 Gauss points for a quadrangular face (see hexa20).
The tetra4 element is the standard 4 node 12 DOF trilinear isoparametric solid element. tetra10 is the corresponding second order element.
You should be aware that this element can perform very badly (for poor aspect ratio, particular loading conditions, etc.) and that higher order elements should be used instead.
The tetra4 element uses the et*3p1d routines.
tetra4 volumes are integrated at the 4 vertices ωi = 1 /4 for i=1,4 and bi=Si the i-th element vertex.
tetra4 surfaces are integrated at the 3 vertices with ωi = 1 /3 for i=1,3 and bi=Si the i-th vertex of the actual face
The tetra10 element is second order and uses the et*3p2c routines.
tetra10 volumes are integrated at 15 Gauss points
Points bk | λ1 | λ2 | λ3 | λ4 | weight ωk |
1 | 1 /4 | 1 /4 | 1 /4 | 1 /4 | 8 /405 |
2 | b | a | a | a | α |
3 | a | b | a | a | α |
4 | a | a | b | a | α |
5 | a | a | a | b | α |
6 | d | c | c | c | β |
7 | c | d | c | c | β |
8 | c | c | d | c | β |
9 | c | c | c | d | β |
10 | e | e | f | f | γ |
11 | f | e | e | f | γ |
12 | f | f | e | e | γ |
13 | e | f | f | e | γ |
14 | e | f | e | f | γ |
15 | f | e | f | e | γ |
with a = 7 − √15 /34 = 0.0919711 , b = 13 + 3 √15 /34 = 0.7240868 ,
c = 7 + √15 /34 = 0.3197936 ,
d = 13 − 3 √15 /34 = 0.0406191 ,
e = 10 − 2 √15 /40 = 0.0563508 , f = 10 + 2 √15 /40 = 0.4436492
and α = 2665 + 14 √15 /226800 , β = 2665 − 14 √15 /226800 et γ = 5 /567
λj for j=1,4 are barycentric coefficients for each vertex Sj :
bk=∑j=1,4λj Sj for k=1,15
tetra10 surfaces are integrated using a 7 point rule
Points bk | λ1 | λ2 | λ3 | weight ωk |
1 | c | d | c | α |
2 | d | c | c | α |
3 | c | c | d | α |
4 | b | b | a | β |
5 | a | b | b | β |
6 | b | a | b | β |
7 | 1 /3 | 1 /3 | 1 /3 | γ |
with γ = 9 /80 = 0.11250 , α = 155 − √15 /2400 = 0.06296959 , β = 155 + √15 /2400 = 0.066197075 and a = 9 − 2 √15 /21 = 0.05961587 , b = 6 + √15 /21 = 0.47014206 , c = 6 − √15 /21 = 0.10128651 , d = 9 + 2 √15 /21 = 0.797427
λj for j=1,3 are barycentric coefficients for each surface vertex Sj :
bk=∑j=1,3λj Sj for k=1,7
The displacement (u,v) are bilinear functions over the element.
For surfaces, q4p uses numerical integration at the corner nodes with ωi=1/4 and bi=Si for i=1,4.
For edges, q4p uses numerical integration at each corner node with ωi=1/2 and bi=Si for i=1,2.
For surfaces, q4p uses a 4 point rule with
For edges, q4p uses a 2 point rule with
For surfaces, q5p uses a 5 point rule with bi=Si for i=1,4 the corner nodes and b5 the node 5.
For edges, q5p uses a 1 point rule with ω = 1 /2 and b the midside node.
For surfaces, q8p uses a 9 point rule with
For edges, q8p uses a 3 point rule with
For surfaces, q8p uses a 9 point rule with
For edges, q8p uses a 3 point rule with
For surfaces, t3p uses a 3 point rule at the vertices with ωi = 1 /3 and bi=Si.
For edges, t3p uses a 2 point rule at the vertices with ωi = 1 /2 and bi=Si.
For surfaces, t3p uses a 1 point rule at the barycenter (b1=G) with ω1 = 1 /2 .
For edges, t3p uses a 2 point rule at the vertices with ωi = 1 /2 and b1 = 1 /2 − 2 /2 √3 and b2 =1 /2 + 2 /2 √3 .
For surfaces, t6p uses a 3 point rule with
For edges, t6p uses a 3 point rule
For surfaces, t6p uses a 7 point rule
Points bk | λ1 | λ2 | λ3 | weight ωk |
1 | 1 /3 | 1 /3 | 1 /3 | a |
2 | α | β | β | b |
3 | β | β | α | b |
4 | β | α | β | b |
5 | γ | γ | δ | c |
6 | δ | γ | γ | c |
7 | γ | δ | γ | c |
with :
a = 9 / 80 = 0.11250 ,
b = 155 + √15/2400 = 0.066197075 and
c = 155 − √15/2400 = 0.06296959
α = 9 − 2 √15/ 21 = 0.05961587 ,
β = 6 + √15/ 21 = 0.47014206
γ = 6 − √15/ 21 = 0.10128651 ,
δ = 9 + 2 √15/ 21 = 0.797427
λj for j=1,3 are barycentric coefficients for each vertex Sj :
bk=∑j=1,3λj Sj for k=1,7
For edges, t6p uses a 3 point rule with ω1 = ω3 = 5/18 , ω2 = 8/18
b1 = 1− √3/5/2 = 0.1127015 , b2 = 0.5 and b3 = 1 + √ 3/5 / 2 = 0.8872985