I am familiar with finite element simulations and numerical schemes like the Gauss Legendre quadrature. I also know that the Gauss integration yields exact values for polynomials up to degree 2n-1 with n being the number of quadrature points. But I don't understand why, for example, second order terahedrons are evaluated in 4 Gauss points and second order hexahedrons in 27 Gauss points (following the Abaqus element definitions). I can't see any connection to the 2n-1 approach here, since the shape functions have to be of degree 2 (or am I wrong?) Maybe someone could explain, how the "right" number of integration points is derived in these elements.
Dear Jonathan Stollberg,
Comparing interpolation functions for second-order hex and tet elements
https://abaqus-docs.mit.edu/2017/English/SIMACAETHERefMap/simathe-c-solidisoquadhex.htm
https://abaqus-docs.mit.edu/2017/English/SIMACAETHERefMap/simathe-c-tritetwedge.htm
one can notice that hex element has such higher-order terms as g^2*h*r,…, r^2*g*h, whereas tet has only g^2, h^2, r^2, g*h, g*r, h*r. Therefore, integrands in stiffness matrix for hex element contain such multyplies as g^4, h^4, r^4, whereas for hex element only g^2, h^2, r^2. I think Abaqus creators took this fact into account.
You must avoid to compare tetrahedra with hexahedra because they don't have the same shape and they have different weights-locations.
You can compare two tetrahedra or two hexahedra from two meshes with different gauss integration points. Two different comparisons gives you a clear view of the convergence.
The precision depends also on the selected mesh and its refinement.
You can see:
web.mit.edu › www › 16.810...PDF
Finite Element Method - MIT
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