6. TDNNS

Tangential-Displacement and Normal-Normal-Stress. Thesis by Astrid Pechstein

6.1. Idea of degrees of freedom

The tangential dofs control sliding along the edge, the linear normal component of the normal stress vector compensates force and angular momentum.

The tangential odfs control sliding and rotations along the face, the linear normal component of the normal stress vector compensates force and angular momentum.

6.1.1. Dofs for a (stretched) quadrilateral

Even part of longitudinal and odd part of transversal component of displacements, and even part of longitudinal stresses prescribe stretching:

Odd part of longitudinal and even part of transversal component of displacements, and odd part of longitudinal stresses prescribe bending:

6.2. Hellinger Reissner mixed methods for elasticity

Primal mixed method: Find \(\sigma \in L_2^{SYM}\) and \(u \in [H^1]^d\) such that

\[\begin{split} \begin{array}{ccccll} \int A \sigma : \tau & - & \int \tau : \varepsilon(u) & = & 0 & \forall \, \tau \\ -\int \sigma : \varepsilon(v) & & & = & -\int f \cdot v \qquad \qquad & \forall \, v \end{array} \end{split}\]

Dual mixed method: Find \(\sigma \in H(\operatorname{div}, {\mathbb S})\) and \(u \in [L_2]^d\) such that

\[\begin{split} \begin{array}{ccccll} \int A \sigma : \tau & + & \int \operatorname{div} \tau \cdot u & = & 0 & \forall \, \tau \\ \int \operatorname{div} \sigma \cdot v & & & = & -\int f \cdot v \qquad \qquad & \forall \, v \end{array} \end{split}\]

6.3. Reduced Symmetry mixed methods

Decompose

\[\begin{split} \varepsilon(u) = \nabla u + \frac{1}{2} \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \operatorname{curl} u = \nabla u + \omega \end{split}\]

Impose symmetry of the strain tensor by an additional Lagrange parameter:

Find \(\sigma \in [H(\operatorname{div})]^2\), \(u \in [L_2]^2\), and \(\omega \in L_2^{skew}\) such that

\[\begin{split} \begin{array}{ccccll} \int A \sigma : \tau & + & \int u \operatorname{div} \tau + \int \tau : \omega & = & 0 & \forall \, \tau \\ \int v \operatorname{div} \sigma & & & = & -\int f v \qquad & \forall \, v \\ \int \sigma : \gamma & & & = & 0 & \forall \, \gamma \end{array} \end{split}\]

The solution satisfies \(u \in L_2\) and \(\omega \in L_2\), i.e.,

\[ u \in H(\operatorname{curl}) \]

Arnold+Brezzi, Stenberg,…

6.4. The Balkenwaage

\[ \int \operatorname{div} \sigma \cdot u \]

is understood as

\(\left< \operatorname{div} \sigma, u \right>_{H^{-1} \times H^1} \quad \)	\(\left< \operatorname{div} \sigma, u \right>_{H(\operatorname{curl})^* \times H(\operatorname{curl}} \quad \)	\((\operatorname{div} \sigma, u )_{L_2}\)
	displacement
\(u \in [H^1]^d\)	\(u \in H(\operatorname{curl})\)	\(u \in L_2\)
continuous f.e.	tangential continuous	discontinuous f.e.
	stress
\(\sigma \in L_2(\mathbb{S})\)	\(\sigma \in H(\operatorname{div} \operatorname{div})\)	\(\sigma \in H(\operatorname{div}, {\mathbb S})\)
discontinuous f.e.	\(\sigma_{nn}\) continuous	\(\sigma_n\) continuous

6.5. Explicit integration by parts formulas

The elasticity problem is equivalent to the mixed problem: Find \(\sigma \in H(\operatorname{div} \operatorname{div})\) and \(u \in H(\operatorname{curl})\) such that for tangentially continuous \(v\) and normal-normal continuous \(\tau\):

\[\begin{split} \begin{array}{ccccll} \int A \sigma : \tau & + & \sum_T \Big\{ \int_T \operatorname{div} \tau \cdot u - \int_{\partial T} \tau_{n\tau} u_\tau \Big\} & = & 0 & \forall \, \tau \\ \sum_T \Big\{ \int_T \operatorname{div} \sigma \cdot v - \int_{\partial T} \sigma_{n\tau} v_\tau \Big\} & & & = & -\int f \cdot v \qquad & \forall \, v \end{array} \end{split}\]

Proof: The second line is equilibrium, plus tangential continuity of the normal stress vector:

\[ \sum_T \int_T (\operatorname{div} \sigma + f) v + \sum_E \int_E [\sigma_{n\tau}] v_\tau = 0 \qquad \forall \, v \]

Since the space requires continuity of \(\sigma_{nn}\), the normal stress vector is continuous.

Element-wise integration by parts in the first line gives

\[ \sum_T \int_T (A \sigma - \varepsilon(u)) : \tau + \sum_E \int_E \tau_{nn} [u_n] = 0 \qquad \forall \, \tau \]

This is the constitutive relation, plus normal-continuity of the displacement. Tangential continuity of the displacement is implied by the space \(H(\operatorname{curl})\).

6.6. Distributional divergence

\(\DeclareMathOperator{\opdiv}{div}\) \(\DeclareMathOperator{\opcurl}{curl}\)

Let \(\sigma \in V_{dd}^k\) (normal-normal continuous). Then the distributional divergence \(f := \opdiv \sigma\) is

\[\begin{eqnarray*} \left<f, \varphi \right> & = &-\int_\Omega \sigma : \nabla \varphi = -\sum_T \int_T \sigma : \nabla \varphi = \sum_T \int_T \opdiv \sigma \, \varphi - \int_{\partial T} \sigma_n \varphi \\ & = & \sum_T \int_T \opdiv \sigma \varphi - \sum_E \int_E [\sigma_n] \varphi = \sum_T \int_T \underbrace{\opdiv \sigma}_{f_T} \varphi - \sum_E \int_E \underbrace{ [\sigma_{nt}]}_{f_E} \varphi_t \end{eqnarray*}\]

\(f = \opdiv \sigma \) consists of element-terms and facet-terms:

\[\begin{eqnarray*} f_T & = & \opdiv_T \sigma \\ f_E & = & [\sigma_{nt}] \qquad \text{vector in tangential space} \end{eqnarray*}\]

It can be applied to \(v_h \in {\mathcal N} \subset H(\opcurl)\).

Write duality pairing as $\( \left< \opdiv \sigma, v \right> \qquad \text{for} \; \sigma \in V_{dd}^k, \; v \in {\mathcal N}^k \)$