3. Mixed finite element discretization#

3.1. Choice of discretization#

When using a mixed formulation, it is necessary to discretize both the space of travel, the Lagrange multiplier p and the « swelling » \(g\). The experience gained on mixed elements, especially with 2 fields for incompressible elements, allows you to know that the discretization of these fields cannot be any, under penalty of obtaining oscillations phenomena (especially at the level of pressures) or blocking phenomena (elements that cannot be deformed or are too rigid). So it is necessary to have a sufficiently large number of pressure Gauss points to verify the incompressibility condition almost everywhere and a number of pressure Gauss points low enough to have more degrees of freedom at calculate as many constraints to check. One of the conditions necessary to obtain satisfactory results is the verification by the finite element in question of the condition LBB (LADYJENSKAIA, BREZZI, BABUSKA). Examples of elements satisfying condition LBB can be found in [bib5] _ and [biB6] _.

Here the problem is a bit different when the 3-field formulation is used.

As it currently stands, the discretizations used are not the same in version HPP and in the large deformations version.

3.1.1. Small deformations#

For small deformations, we were inspired by the classic uses of mixed formulations (e.g. [bib7] _), using a \(\mathit{P2}\mathrm{/}\mathit{P1}\mathrm{/}\mathit{P1}\) element for the 3-field formulation. In other words, the displacement is quadratic, the pressure and the swelling are both linear. Finite elements used for the 3-field formulation are therefore the following:

in 2D:	\(u\)	6-knot triangle	quadrilateral with 8 knots
in 2D:	\(p,g\)	triangle with 3 knots	quadrilateral with 4 knots
in 3D:	\(u\)	10-knot tetrahedron	20-knot hexahedron	pentahedron with 15 knots
in 3D:	\(p,g\)	4-knot tetrahedron	hexahedron with 8 knots	6-knot pentahedron

For each type of element, a single family of Gauss points is used:

3 points for the triangles
4 points for quadrilaterals
4 points for tetrahedra
8 points for hexahedra
21 points for pentahedra

For the two-field formulation, an element of the type \(\mathit{P2}\mathrm{/}\mathit{P1}\) has been introduced. The displacement therefore has a quadratic interpolation while the pressure is interpolated linearly. In the case of using discretization in triangles or linear tetrahedra, two stabilization methods have was introduced. The first corresponds to the stabilized finite element \(\mathit{P1}\text{+/}\mathit{P1}\). The \(\text{+}\) corresponds to the introduction of an additional degree of freedom in the center of the element in the interpolation of movements. This additional degree is commonly referred to as a « bubble. » This stabilization method only works on simplex elements (triangle in 2D and tetrahedron in 3D). It has the advantage of using very few degrees of freedom. The second stabilization method corresponds to the Orthogonal Sub-Grid Scale method (OSGS) [bib11] _. The advantage of this method is that it works for all element topologies. Its main disadvantage is that it introduces a third unknown (and therefore additional degrees of freedom) corresponding to the pressure field projected \(\pi\) onto the space orthogonal to the displacement fields.

The finite elements used for the 2-field formulation are therefore as follows:

Interpolation		P1+/P1	P1/P1 OSGS	P2/P1	P1/P1 OSGS	P2/P1	P1/P1 OSGS	P2/P1
in 2D:	\(u\)	triangle with 3 knots	triangle with 3 knots	6-knot triangle	quadrilateral with 4 knots	quadrilateral with 8 knots
	\(p\)	triangle with 3 knots	triangle with 3 knots	triangle with 3 knots	quadrilateral with 4 knots	quadrilateral with 4 knots
	\(\pi\)		triangle with 3 knots		quadrilateral with 4 knots
in 3D:	\(u\)	tetrahedron with 4 knots + bubble	4-knot tetrahedron	10-knot tetrahedron	8-knot cube	20-knot cube	6-knot pentahedron	pentahedron with 15 knots
	\(p\)	4-knot tetrahedron	4-knot tetrahedron	4-knot tetrahedron	8-knot cube	8-knot cube	6-knot pentahedron	6-knot pentahedron
	\(\pi\)		4-knot tetrahedron		8-knot cube		6-knot pentahedron

The families of Gauss points used are the same as those in the 3-field formulation. Note that for For elements \(\mathit{P1}\text{+/}\mathit{P1}\), only one Gauss point is used for integration.

3.1.2. Large deformations#

Starting with version 11, the choice of interpolations in large deformations is identical to that of small deformations. The elements are \(\mathit{P2}/\mathit{P1}/\mathit{P1}\) for 3-field formulations and \(\mathit{P2}/\mathit{P1}\) for 2-field formulations. The finite elements used for the 3-field formulation are therefore as follows:

in 2D:	\(u\)	6-knot triangle	quadrilateral with 8 knots
in 2D:	\(p,g\)	triangle with 3 knots	quadrilateral with 4 knots
in 3D:	\(u\)	10-knot tetrahedron	20-knot hexahedron	pentahedron with 15 knots
in 3D:	\(p,g\)	4-knot tetrahedron	hexahedron with 8 knots	6-knot pentahedron

The same families of Gauss points as those for the small deformations were used.

3.2. Writing the discrete problem#

First, we discuss the writing of the discrete problem in the context of the 3-field formulation. Let \({u}^{e}\), \({p}^{e}\) and \({g}^{e}\) be the vectors of the elementary nodal unknowns (respectively displacement, pressure and swelling). If \({N}^{u}\), \({N}^{p}\) and \({N}^{g}\) are the shape functions (respectively interpolations of displacement, pressure and swelling) associated with the finite element in question:

\[\]

begin {array} {c} u= {N} u= {N} ^ {u} ^ {u} {u} ^ {e}\ p= {n} {e}\ g= {N}g= {N} ^ {g} {g} {g} {g} ^ {e}end {array}end {array}

3.2.1. Writing in small deformations#

\(B\) is the classic derivation matrix that allows you to go from \({u}^{e}\) to \(\varepsilon\):

\[\]

varepsilon =B {u} ^ {e}

In the formulation, we distinguish \({e}_{\text{dev}}\) and \({e}_{\text{dil}}\), which leads us to define the operators \({B}_{\text{dev}}\) and \({B}_{\text{dil}}\) such as: \({\epsilon }^{D}={B}_{\text{dev}}{U}^{e}\) and \(\frac{\text{tr}\epsilon }{3}={B}_{\text{dil}}{U}^{e}\)

The discretized form of the equations of the 3-field problem [eq] is written as:

\[\]

begin {array} {l}: {F} _ {u} =underset {omega} {omega} {omega} {omega} {omega} + pi) domega = {F} _ {mathrm {ext}}\ {F} _ {p} =underset {omega} {omega} {omega} {omega} {int} {int}}} ^ {T} ({B} _ {mathrm {dil}} u-g) domega =0\ {F} _ {g} =underset {Omega} {omega} {int} {int} {int} {({N} ^ {g})} ^ {T} (frac {1} {3}text {tr} (tr} (tr} (tr}) (sigma) -p) domega =0end {array}

The tangent matrix of the problem is symmetric and is based on the following terms:

\[\]

begin {array} {l}: {K} _ {mathrm {uu}} =frac {partial {F}} _ {u}} {partial {u} ^ {e}} =underset {omega} {int} {int} {B} {B}} _ {mathrm {dev}} {B}} =underset {omega} {int} {B}} =underset {omega} {int} {B} {B}} =underset {omega} {int} {B} {B}} _ {mathrm {dev}} {B}} domega\ {K} _ {mathrm {up}}} =frac {partial {F} _ {u}} {partial {p} ^ {e}} =underset {omega} {int} {int} {B} {B} {B}} _ {mathrm {dil}}} _ {mathrm {dil}}} ^ {N} ^ {p} domega\ {K} _ {mathrm {ug}} =frac {partial {f}} _ {u}} {partial {g} ^ {e}} =frac {1} {3}underset {omega} {omega} {omega} {omega} {omega} {omega} {omega} {omega} {int} {int}}int}text {int}\ int}\ text {tr}} ({B}} _ {mathrm {dev}}} ^ {T} D) {N} ^ {g} d) {N} ^ {g} dOmega\ {K} _ {mathrm {pp}} =frac {partial {F} _ {p}} {partial {p} ^ {e}} =0\ {K} _ {mathrm {pg}} =frac {partial {F} _ {p}} {partial {g} ^ {e}} =-underset {Omega} {omega} {int} {int} {int} {int} {int} {int} {int} {int} {int} {int} {int} {{int} {{int} {{int} {{int}} {(N} ^ {p})}} ^ {T} {N} ^ {g} domega\ {K} _ {gg} =frac {partial {F} _ {F} _ {g}}} {partial {g} ^ {e}} =frac {1} {9}underset {Omega} {omega} {int} {int} {int} {int} {int} {int} {{int} {} {int} {{int} {{int} {{}} {{int} {{}} {{int} {{}} {{int} {{}} {{int} {{}} {{int} {{}} {{int} {{}} {{int} {{}} {{int} {{}} {{int} {{}} {{int} {{}} {

As far as the formulation with 2 fields is concerned, it is easily deduced from the previous one. The discretized forms of the equations give us:

\[\]

begin {array} {l}: {F} _ {u}mathrm {=}underset {omega} {mathrm {int}} {B} ^ {T} ({sigma} ^ {D} +pi) domegamathrm {=} {F}} _ {mathit {ext}}\ {F} _ {p} =underset {Omega} {omega} {omega} {omega} {omega} {omega}} u-frac {{N}} {frac {{N}} {int} {int} {int} {int} {int} {(n}} {p})} ^ {p})} ^ {p})} ^ {T} ({B} _ {mathrm {dil}}} u-frac {{N}} {N} ^ {p}}) domega =0end {array}} u-frac {{N}} {

The tangent matrix of the problem is symmetric and is based on the following terms:

\[\]

begin {array} {l}: {K} _ {mathrm {uu}} =frac {partial {F}} _ {u}} {partial {u} ^ {e}} =underset {omega} {int} {int} {B} {B}} _ {mathrm {dev}} {B}} =underset {omega} {int} {B}} =underset {omega} {int} {B} {B}} =underset {omega} {int} {B} {B}} _ {mathrm {dev}} {B}} domega\ {K} _ {mathrm {up}}} =frac {partial {F} _ {u}} {partial {p} ^ {e}} =underset {omega} {int} {int} {B} {B} {B}} _ {mathrm {dil}}} _ {mathrm {dil}}} ^ {N} ^ {p} domega\ {K} _ {mathit {pp}}mathrm {=}frac {mathrm {partial} {F} _ {p}} {mathrm {partial} {p} {p} ^ {e}}mathrm {=}}mathrm {=}mathrm {=}mathrm {-}frac {1} {kappa}underset {omega} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {p} {int}} {({N} ^ {p})}} ^ {T} {N} ^ {p} dOmegaend {array}

3.2.2. Writing in large transformations#

Since writing is a bit tedious, the reader can refer to reading [bib8] _ for more information.