Energy-Momentum Integrators for Elastic Cosserat Points, Rigid Bodies, and Multibody Systems

Abstract

The goal of this chapter is to present the development of energy-momentum (EM) schemes in the framework of discrete (or finite-dimensional) mechanical systems. EM integrators belong to the class of structure-preserving numerical methods and have been originally developed in the field of nonlinear solid and structural mechanics. EM schemes and energy dissipating variants thereof typically exhibit improved numerical stability and robustness when compared to standard integrators. Due to their superior numerical properties, EM schemes have soon been extended to more involved applications such as flexible multibody dynamics and coupled thermomechanical problems. In this chapter, we start the development of second-order EM schemes in the context of the Cosserat point (or pseudo-rigid body). The theory of a Cosserat point shares main structural properties with semi-discrete formulations of elastodynamics. Indeed, the Cosserat point can be directly linked to the 4-node tetrahedral finite element. Besides its usefulness in explaining main ingredients of EM schemes such as the algorithmic stress formula, the Cosserat point is ideally suited to perform the transition to rigid body dynamics. In particular, in the present work, the rigid body formulation is obtained by imposing the zero strain condition on the Cosserat point. This way the rigid body is treated as constrained mechanical system. Moreover, we show that the EM discretization of constrained mechanical systems can be derived in a straightforward way from the EM scheme for the Cosserat point. The resulting rigid body formulation is closely connected to natural coordinates. Eventually, we deal with the extension to multibody systems which can be done in a straightforward way due to the presence of holonomic constraints in the present rigid body formulation.

Funding for this work has been provided by the German Science Foundation under Grant BE 2285/10-1.

A Appendix

1.1 A.1 Balance Laws

For comparison, the balance laws are directly derived from the variational equations (19) governing the motion of the pseudo-rigid body. To this end, we recast (19) in the form

$$\begin{aligned} \delta \overline{{\varvec{x}}} \cdot \left( M \ddot{\overline{{\varvec{x}}}} - {\varvec{f}}_\mathrm {ext} \right)&= 0 \end{aligned}$$

(130)

$$\begin{aligned} \text{ tr }\left( \delta {\varvec{F}}^T \big (\ddot{{\varvec{F}}}{\varvec{E}}_0 + 2{\varvec{F}} DU({\varvec{C}}) - {\varvec{M}}_\mathrm {ext}\big ) \right)&= 0 \end{aligned}$$

(131)

Applying the polar decomposition theorem to the deformation gradient, we get

$$\begin{aligned} {\varvec{F}} = {\varvec{R}}{\varvec{U}} \qquad \text{ and }\qquad \delta {\varvec{F}} = \delta {\varvec{R}}{\varvec{U}} + {\varvec{R}}\delta {\varvec{U}} \end{aligned}$$

(132)

Since ${\varvec{R}}{\varvec{R}}^T = {\varvec{I}}$, $\delta {\varvec{R}}{\varvec{R}}^T + {\varvec{R}}\delta {\varvec{R}}^T = {\varvec{0}}$, and consequently

$$\begin{aligned} \widehat{{\varvec{\omega }}}_\delta = \delta {\varvec{R}}{\varvec{R}}^T \end{aligned}$$

(133)

is skew-symmetric. A straightforward calculation shows that (131) can be rewritten as

$$\begin{aligned} \text{ tr }\left( \delta {\varvec{U}} {\varvec{U}} \big ( {\varvec{F}}^{-1}\ddot{{\varvec{F}}}{\varvec{E}}_0 + 2 DU({\varvec{C}}) - {\varvec{F}}^{-1}{\varvec{M}}_\mathrm {ext}\big ) \right)&= 0 \end{aligned}$$

(134)

$$\begin{aligned} {\varvec{\omega }}_\delta \cdot \left( 2\text{ vect }\big (\ddot{{\varvec{F}}}{\varvec{E}}_0{\varvec{F}}^T\big ) - 2\text{ vect }\big ({\varvec{M}}_\mathrm {ext}{\varvec{F}}^T\big )\right)&= 0 \end{aligned}$$

(135)

Accordingly, the nine independent equations emanating from (131) have been converted to six independent equations (134) plus three independent equations (135).

In (135), $\text{ vect }(\bullet )$ denotes the vector invariant of a second-order tensor defined by

$$\begin{aligned} \text{ vect }({\varvec{a}}\otimes {\varvec{b}})\times {\varvec{c}} = \text{ skew }({\varvec{a}}\otimes {\varvec{b}}){\varvec{c}} \end{aligned}$$

Since

$$\begin{aligned} \text{ skew }({\varvec{a}}\otimes {\varvec{b}}){\varvec{c}}&=\frac{1}{2}({\varvec{a}}\otimes {\varvec{b}} - {\varvec{b}}\otimes {\varvec{a}}) {\varvec{c}} \\&=\frac{1}{2}\left( ({\varvec{b}}\cdot {\varvec{c}}){\varvec{a}} - ({\varvec{a}}\cdot {\varvec{c}}){\varvec{b}}\right) \\&=\frac{1}{2}({\varvec{b}}\times {\varvec{a}})\times {\varvec{c}} \end{aligned}$$

we have

$$\begin{aligned} \text{ vect }({\varvec{a}}\otimes {\varvec{b}}) = \frac{1}{2}({\varvec{b}}\times {\varvec{a}}) \end{aligned}$$

(136)

Accordingly,

$$\begin{aligned} 2\text{ vect }\big (\ddot{{\varvec{F}}}{\varvec{E}}_0{\varvec{F}}^T\big )&= 2\text{ vect }\big (E_0^{ij}\ddot{{\varvec{d}}}_i\otimes {\varvec{d}}_j\big )= E_0^{ij}{\varvec{d}}_j\times \ddot{{\varvec{d}}}_i \end{aligned}$$

(137)

$$\begin{aligned} 2\text{ vect }\big ({\varvec{M}}_\mathrm {ext}{\varvec{F}}^T\big )&= 2\text{ vect }\big ({\varvec{f}}^i_\mathrm {ext}\otimes {\varvec{d}}_i\big )= {\varvec{d}}_i\times {\varvec{f}}^i_\mathrm {ext} = \overline{{\varvec{m}}}_\mathrm {ext} \end{aligned}$$

(138)

1.1.1 A.1.1 Balance of Angular Momentum

To get the balance law for angular momentum, substitute $\delta {{\varvec{U}}} = {{\varvec{0}}}$ into (134), ${\varvec{\omega }}_\delta ={\varvec{\xi }}$ into (135), and $\delta \overline{{\varvec{x}}}={\varvec{\xi }}\times \overline{{\varvec{x}}}$ into (130). Subsequent summation of the resulting equations yields

$$\begin{aligned} {\varvec{\xi }}\cdot \left( M\overline{{\varvec{x}}}\times \ddot{\overline{{\varvec{x}}}} + 2\text{ vect }\big (\ddot{{\varvec{F}}}{\varvec{E}}_0{\varvec{F}}^T\big ) -\overline{{\varvec{x}}}\times {\varvec{f}}_\mathrm {ext} - 2\text{ vect }\big ({\varvec{M}}_\mathrm {ext}{\varvec{F}}^T\big ) \right) = 0 \end{aligned}$$

or

$$\begin{aligned} {\varvec{\xi }}\cdot \left( \frac{d}{dt} {\varvec{j}} - {\varvec{m}}_\mathrm {ext} \right) = 0 \end{aligned}$$

The last equation has to hold for arbitrary ${\varvec{\xi }}\in \mathbb {R}^3$. Accordingly, one obtains $d{{\varvec{j}}}/dt={\varvec{m}}_\mathrm {ext}$, where

$$\begin{aligned} {{\varvec{j}}}&= M\overline{{\varvec{x}}}\times \dot{\overline{{\varvec{x}}}} + 2\text{ vect }\big (\dot{{\varvec{F}}}{\varvec{E}}_0{\varvec{F}}^T\big ) \\ {{\varvec{m}}}_\mathrm {ext}&= \overline{{\varvec{x}}}\times {\varvec{f}}_\mathrm {ext} + 2\text{ vect }\big ({\varvec{M}}_\mathrm {ext}{\varvec{F}}^T\big ) \end{aligned}$$

denote, respectively, the total angular momentum and the resultant external torque with respect to the origin of the inertial frame of reference. Note that the same conclusions can be drawn by substituting $\delta \overline{{\varvec{x}}}={\varvec{\xi }}\times \overline{{\varvec{x}}}$ into (130), and $\delta {\varvec{F}} = \widehat{{\varvec{\xi }}}{\varvec{F}}$ into (131).

1.1.2 A.1.2 Balance of Energy

Suppose that an external force ${\varvec{f}}_\mathrm {ext}\in \mathbb {R}^3$ along with external director forces ${\varvec{f}}^i_\mathrm {ext}\in \mathbb {R}^3$, $i=1,2,3$, are acting on the body under consideration. Recall that the external director forces ${\varvec{f}}^i_\mathrm {ext}$ can be linked to the second-order tensor ${\varvec{M}}_\mathrm {ext}$ via ${\varvec{M}}_\mathrm {ext} = {\varvec{f}}^i_\mathrm {ext} \otimes {\varvec{D}}_i$ (see Eq. (38) in Sect. 2.2). To define the external director forces we introduce 9 independent quantities $\mathcal {M}^{ij}$ such that

$$\begin{aligned} {\varvec{\mathcal {M}}} = \mathcal {M}^{ij} {\varvec{D}}_i \otimes {\varvec{D}}_j \end{aligned}$$

(139)

and

$$\begin{aligned} {\varvec{M}}_\mathrm {ext} = {\varvec{F}}{\varvec{\mathcal {M}}} = \mathcal {M}^{ij} {\varvec{d}}_i \otimes {\varvec{D}}_j \end{aligned}$$

(140)

Note that the last equation implies

$$\begin{aligned} {\varvec{f}}^i_\mathrm {ext} = \mathcal {M}^{ki} {\varvec{d}}_k \end{aligned}$$

(141)

Now substitute $\dot{\overline{{\varvec{x}}}}$ for $\delta \overline{{\varvec{x}}}$ into (130) and $\dot{{\varvec{F}}}$ for $\delta {\varvec{F}}$ into (131). Subsequent summation of both equations yields

$$\begin{aligned} \dot{\overline{{\varvec{x}}}} \cdot \left( M \ddot{\overline{{\varvec{x}}}} - {\varvec{f}}_\mathrm {ext} \right) + \text{ tr }\left( \dot{{\varvec{F}}}^T \big (\ddot{{\varvec{F}}}{\varvec{E}}_0 + 2{\varvec{F}} DU({\varvec{C}}) - {\varvec{M}}_\mathrm {ext}\big ) \right) = 0 \end{aligned}$$

(142)

Taking into account the relationships

$$\begin{aligned} \dot{\overline{{\varvec{x}}}} \cdot M \ddot{\overline{{\varvec{x}}}}&= \frac{d}{dt} \left( \frac{1}{2} M\dot{\overline{{\varvec{x}}}}\cdot \dot{\overline{{\varvec{x}}}} \right) \\ \text{ tr }\left( \dot{{\varvec{F}}}^T\ddot{{\varvec{F}}}{\varvec{E}}_0 \right)&= \frac{d}{dt}\left( \frac{1}{2} \text{ tr }\left( \dot{{\varvec{F}}}^T\dot{{\varvec{F}}}{\varvec{E}}_0 \right) \right) \end{aligned}$$

we define the kinetic energy

$$\begin{aligned} T = \frac{1}{2} M \dot{\overline{{\varvec{x}}}}\cdot \dot{\overline{{\varvec{x}}}}+\frac{1}{2}\text{ tr }\left( \dot{{\varvec{F}}}{\varvec{E}}_0\dot{{\varvec{F}}}^T\right) \end{aligned}$$

(143)

Moreover,

$$\begin{aligned} \text{ tr }\left( \dot{{\varvec{F}}}^T{\varvec{F}} 2DU({\varvec{C}})\right)&= \text{ tr }\left( 2DU({\varvec{C}}) \text{ sym }(\dot{{\varvec{F}}}^T{\varvec{F}})\right) \\&= \text{ tr }\left( 2DU({\varvec{C}}) \frac{1}{2}(\dot{{\varvec{F}}}^T{\varvec{F}} + {\varvec{F}}^T\dot{{\varvec{F}}}) \right) \\&= \text{ tr }\left( 2DU({\varvec{C}}) \frac{1}{2} \dot{{\varvec{C}}} \right) \\&= \frac{d}{dt} U({\varvec{C}}) \end{aligned}$$

Now (142) can be recast in the form

$$\begin{aligned} \frac{d}{dt} E = P_{\mathrm {ext}} \end{aligned}$$

(144)

Here, E is the total mechanical energy given by

$$\begin{aligned} E = T + U \end{aligned}$$

where U denotes the total strain energy defined in (13). On the right hand side of balance equation (144)

$$\begin{aligned} P_{\mathrm {ext}} = {\varvec{f}}_\mathrm {ext}\cdot \dot{\overline{{\varvec{x}}}} + \text{ tr }\left( \dot{{\varvec{F}}}^T {\varvec{M}}_\mathrm {ext} \right) \end{aligned}$$

denotes the power of the external forces acting on the pseudo-rigid body. We next focus on the power of the director forces given by

$$\begin{aligned} \overline{P}_{\mathrm {ext}} = \text{ tr }\left( \dot{{\varvec{F}}}^T {\varvec{M}}_\mathrm {ext} \right) \end{aligned}$$

Taking into account (140), the last equation can be rewritten as

$$\begin{aligned} \overline{P}_{\mathrm {ext}}&= \text{ tr }\left( \dot{{\varvec{F}}}^T{\varvec{F}}{\varvec{\mathcal {M}}} \right) \\&= \text{ tr }\left( \big (\overline{{\varvec{\mathcal {M}}}} + \widetilde{{\varvec{\mathcal {M}}}}\big )\dot{{\varvec{F}}}^T{\varvec{F}} \right) \end{aligned}$$

In the last equation

$$\begin{aligned} \overline{{\varvec{\mathcal {M}}}}&= \text{ sym }\left( {\varvec{\mathcal {M}}} \right) \\ \widetilde{{\varvec{\mathcal {M}}}}&= \text{ skew }\left( {\varvec{\mathcal {M}}} \right) \end{aligned}$$

have been introduced. Now

$$\begin{aligned} \text{ tr }\left( \overline{{\varvec{\mathcal {M}}}}\dot{{\varvec{F}}}^T{\varvec{F}}\right) =\text{ tr }\left( \overline{{\varvec{\mathcal {M}}}}\text{ sym }\left( \dot{{\varvec{F}}}^T{\varvec{F}} \right) \right) =\frac{1}{2}\text{ tr }\left( \overline{{\varvec{\mathcal {M}}}}\dot{{\varvec{C}}}\right) \end{aligned}$$

Furthermore,

$$\begin{aligned} \text{ tr }\left( \widetilde{{\varvec{\mathcal {M}}}}\dot{{\varvec{F}}}^T{\varvec{F}}\right) =\text{ tr }\left( \widetilde{{\varvec{\mathcal {M}}}}\text{ skew }\left( \dot{{\varvec{F}}}^T{\varvec{F}} \right) \right) \end{aligned}$$

Applying the polar decomposition ${\varvec{F}} = {\varvec{R}}{\varvec{U}}$ along with $\dot{{\varvec{F}}} = \dot{{\varvec{R}}}{\varvec{U}} + {\varvec{R}}\dot{{\varvec{U}}}$ and $\widehat{{\varvec{\omega }}} = \dot{{\varvec{R}}}{\varvec{R}}^T$ (cf. (132) and (133) on p. 47), we get

$$\begin{aligned} \text{ tr }\left( \widetilde{{\varvec{\mathcal {M}}}}\text{ skew }\left( \dot{{\varvec{F}}}^T{\varvec{F}} \right) \right) = {\varvec{\omega }}\cdot 2\text{ vect }\left( {\varvec{F}}\widetilde{{\varvec{\mathcal {M}}}}{\varvec{F}}^T\right) +\text{ tr }\left( \widetilde{{\varvec{\mathcal {M}}}}\dot{{\varvec{U}}}^T{\varvec{U}} \right) \end{aligned}$$

Altogether the power of the external forces can be written in the form

$$\begin{aligned} P_{\mathrm {ext}}&= {\varvec{f}}_\mathrm {ext}\cdot \dot{\overline{{\varvec{x}}}} + \frac{1}{2}\text{ tr }\left( \overline{{\varvec{\mathcal {M}}}}\dot{{\varvec{C}}}\right) + {\varvec{\omega }}\cdot 2\text{ vect }\left( {\varvec{F}}\widetilde{{\varvec{\mathcal {M}}}}{\varvec{F}}^T\right) + \text{ tr }\left( \widetilde{{\varvec{\mathcal {M}}}}\dot{{\varvec{U}}}^T{\varvec{U}} \right) \\&= {\varvec{f}}_\mathrm {ext}\cdot \dot{\overline{{\varvec{x}}}} + \frac{1}{2}\overline{\mathcal {M}}^{ij}\dot{d}_{ij} + {\varvec{\omega }}\cdot \overline{{\varvec{m}}}_\mathrm {ext} + \text{ tr }\left( \widetilde{{\varvec{\mathcal {M}}}}\dot{{\varvec{U}}}^T{\varvec{U}} \right) \end{aligned}$$

Here, $\overline{{\varvec{m}}}_\mathrm {ext}$ can be identified as the resultant external torque relative to the center of mass that has been introduced in (71). In particular, we have

$$\begin{aligned} \overline{{\varvec{m}}}_\mathrm {ext}&= 2\text{ vect }\left( {\varvec{F}}\widetilde{{\varvec{\mathcal {M}}}}{\varvec{F}}^T\right) \\&= \widetilde{\mathcal {M}}^{ji}{\varvec{d}}_i\times {\varvec{d}}_j \\&= \varepsilon _{ijk}\widetilde{\mathcal {M}}^{ji} {\varvec{d}}^k \\&= {\varvec{d}}_i\times {\varvec{f}}^i_\mathrm {ext} \end{aligned}$$

In the last equation use has been made of (141). Moreover,

$$\begin{aligned} \varepsilon _{ijk} = ({\varvec{d}}_i\times {\varvec{d}}_j)\cdot {\varvec{d}}_k = e_{ijk}\sqrt{d} \end{aligned}$$

where $d=\det (d_{ij})$ (or $\sqrt{d}=({\varvec{d}}_1\times {\varvec{d}}_2)\cdot {\varvec{d}}_3$) and $e_{ijk}$ denotes the alternating symbol.

1.2 A.2 Application of External Torques

It can be observed from the above treatment that the application of external torques $\overline{{\varvec{m}}}_\mathrm {ext}$ relative to the center of mass is linked to the skew-symmetric tensor $\widetilde{{\varvec{\mathcal {M}}}}=\widetilde{\mathcal {M}}^{ij}{\varvec{D}}_i\otimes {\varvec{D}}_j$. In particular, given the covariant components of the external torque, $m_k={\varvec{d}}_k\cdot \overline{{\varvec{m}}}_\mathrm {ext}$, we obtain

$$\begin{aligned} m_k = \varepsilon _{ijk}\widetilde{\mathcal {M}}^{ji} \end{aligned}$$

from which it follows that

$$\begin{aligned} \widetilde{\mathcal {M}}^{23}&=-\widetilde{\mathcal {M}}^{32}=-\frac{m_1}{2\sqrt{d}} \\ \widetilde{\mathcal {M}}^{31}&=-\widetilde{\mathcal {M}}^{13}=-\frac{m_2}{2\sqrt{d}} \\ \widetilde{\mathcal {M}}^{12}&=-\widetilde{\mathcal {M}}^{21}=-\frac{m_3}{2\sqrt{d}} \end{aligned}$$

or

$$\begin{aligned} \widetilde{\mathcal {M}}^{ij} = \frac{e^{jik} m_k}{2\sqrt{d}} \end{aligned}$$

(145)

where $e^{ijk}=e_{ijk}$ again denotes the alternating symbol. Accordingly, using the above formulas for $\widetilde{\mathcal {M}}^{ij}$ in terms of the torque components $m_k$, the corresponding director forces can be calculated via

$$\begin{aligned} {\varvec{f}}^j_\mathrm {ext} = \widetilde{\mathcal {M}}^{ij} {\varvec{d}}_i \end{aligned}$$

(146)

or

$$\begin{aligned} {\varvec{f}}^j_\mathrm {ext} = \frac{e^{ijk}}{2\sqrt{d}}\left( {\varvec{d}}_j\otimes {\varvec{d}}_k\right) \overline{{\varvec{m}}}_\mathrm {ext} \end{aligned}$$

(147)

To summarize, the action of an external torque $\overline{{\varvec{m}}}_\mathrm {ext}$ relative to the center of mass can be realized by applying external director forces of the form

$$\begin{aligned} \begin{bmatrix} {\varvec{f}}^1_\mathrm {ext} \\ {\varvec{f}}^2_\mathrm {ext} \\ {\varvec{f}}^3_\mathrm {ext} \end{bmatrix} = \frac{1}{2\sqrt{d}} \begin{bmatrix} {\varvec{d}}_2\otimes {\varvec{d}}_3 - {\varvec{d}}_3\otimes {\varvec{d}}_2 \\ {\varvec{d}}_3\otimes {\varvec{d}}_1 - {\varvec{d}}_1\otimes {\varvec{d}}_3 \\ {\varvec{d}}_1\otimes {\varvec{d}}_2 - {\varvec{d}}_2\otimes {\varvec{d}}_1 \end{bmatrix} \overline{{\varvec{m}}}_\mathrm {ext} \end{aligned}$$

(148)

Remark A.1

Formula (148) can be viewed as an extension to flexible Cosserat points of the method proposed in Betsch and Sänger (2013). In this work the consistent application of torques has been dealt with in the context of rigid body dynamics formulated in terms of directors (or direction cosines). The formula proposed in Betsch and Sänger (2013) is given by

$$\begin{aligned} {\varvec{f}}^j_\mathrm {ext} = \frac{1}{2} \overline{{\varvec{m}}}_\mathrm {ext}\times {\varvec{d}}^j \end{aligned}$$

(149)

The equivalence of (149) to (148) can be shown by a direct calculation:

$$\begin{aligned} {\varvec{f}}^j_\mathrm {ext}&= \frac{1}{2} \overline{{\varvec{m}}}_\mathrm {ext}\times {\varvec{d}}^j \\&= \frac{1}{2} m_k {\varvec{d}}^k\times {\varvec{d}}^j \\&= \frac{1}{2} m_k d^{-\frac{1}{2}} e^{kji}{\varvec{d}}_i \\&= \widetilde{\mathcal {M}}^{ij} {\varvec{d}}_i \end{aligned}$$

where (145) has been used.

1.2.1 A.2.1 Fully Actuated Cosserat Point

If the Cosserat point shall be fully actuated, the 9 independent quantities $\mathcal {M}^{ij}$ in (139) can be employed as control inputs. According to (141) this approach determines the external director forces

$$\begin{aligned} {\varvec{f}}^i_\mathrm {ext} = \mathcal {M}^{ji} {\varvec{d}}_j \end{aligned}$$

If required the external torque associated with the control inputs can be extracted via

$$\begin{aligned} \overline{{\varvec{m}}}_\mathrm {ext} = {\mathcal {M}}^{ji}{\varvec{d}}_i\times {\varvec{d}}_j \end{aligned}$$

Note that due to the presence of the cross product the skew-symmetric part of $\mathcal {M}^{ji}$, that is, $\widetilde{\mathcal {M}}^{ji}=(\mathcal {M}^{ji}-\mathcal {M}^{ij})/2$, is automatically extracted. The above result coincides with

$$\begin{aligned} \overline{{\varvec{m}}}_\mathrm {ext} = m_k{\varvec{d}}^k \quad \text{ where }\quad m_k = \sqrt{d}\,e_{ijk}{\mathcal {M}}^{ji} \end{aligned}$$

Again the skew-symmetric part of $\mathcal {M}^{ji}$ is extracted due to the presence of the alternating symbol.

1.3 A.3 Iteration Matrix of the EM Integrator

Consider St. Venant-Kirchhoff material with strain energy density

$$\begin{aligned} W({\varvec{G}}) = \frac{\lambda }{2}\left( \text{ tr } {\varvec{G}}\right) ^2 + \mu \text{ tr }\left( {\varvec{G}}^2\right) \end{aligned}$$

where the Green–Lagrangean strain tensor is given by

$$\begin{aligned} {\varvec{G}} = \frac{1}{2}\left( {\varvec{C}} - {\varvec{I}}\right) = \gamma _{ij} {\varvec{D}}^i\otimes {\varvec{D}}^j \end{aligned}$$

Note that the components $\gamma _{ij}=\frac{1}{2}(d_{ij}-\delta _{ij})$ have been introduced in (66). According to (11), the second Piola-Kirchhoff stress tensor is given by

$$\begin{aligned} \begin{array}{rcl} {\varvec{S}} &{}=&{} 2 DW({\varvec{C}}) \\ &{}=&{} DW({\varvec{G}}) \\ &{}=&{} \lambda \left( \text{ tr }{\varvec{G}}\right) {\varvec{I}} + 2\mu {\varvec{G}} \end{array} \end{aligned}$$

(150)

Moreover, the fourth-order elasticity tensor assumes the form

$$\begin{aligned} \begin{array}{rcl} {\varvec{\mathsf {C}}} &{}=&{} 4 D^2W({\varvec{C}}) \\ &{}=&{} D^2W({\varvec{G}}) \\ &{}=&{} \lambda {\varvec{I}}\otimes {\varvec{I}} + 2\mu {\varvec{\mathsf {I}}} \end{array} \end{aligned}$$

(151)

Since we have assumed that the director triad $\{{\varvec{D}}_i\}$ in the reference configuration is orthonormal, it suffices to consider the Cartesian components of ${\varvec{S}}$ and ${\varvec{\mathsf {C}}}$. Accordingly, we have

$$\begin{aligned} S_{ij} = \lambda \gamma _{kk}\delta _{ij} + 2\mu \gamma _{ij} \end{aligned}$$

(152)

and

$$\begin{aligned} \mathsf {C}_{ijkl} = \lambda \delta _{ij}\delta _{kl} + \mu \left( \delta _{ik}\delta _{jl} + \delta _{il}\delta _{jk}\right) \end{aligned}$$

We next deal with the linearization of the internal director forces ${\varvec{f}}^i_\mathrm {int}=S^{ij}{\varvec{d}}_j$. First consider the time-continuous case where, according to the product rule of differentiation, we get

$$\begin{aligned} \Delta {\varvec{f}}^i_\mathrm {int}=\Delta S^{ij}{\varvec{d}}_j + S^{ij}\Delta {\varvec{d}}_j \end{aligned}$$

(153)

With regard to (152)

$$\begin{aligned} \begin{array}{rcl} \Delta S_{ij} &{}=&{} \lambda \Delta \gamma _{kk}\delta _{ij} + 2\mu \Delta \gamma _{ij} \\ &{}=&{} \lambda ({\varvec{d}}_k\cdot \Delta {\varvec{d}}_k) \delta _{ij} + \mu ({\varvec{d}}_i\cdot \Delta {\varvec{d}}_j + {\varvec{d}}_j\cdot \Delta {\varvec{d}}_i) \end{array} \end{aligned}$$

(154)

Now, a straightforward calculation yields

$$\begin{aligned} \Delta {\varvec{f}}^i_\mathrm {int}=\left( {\varvec{K}}^{ij}_\mathrm {mat} + {\varvec{K}}^{ij}_\mathrm {geo}\right) \Delta {\varvec{d}}_j \end{aligned}$$

(155)

where the contributions to the iteration matrix have been split into a material part ${\varvec{K}}^{ij}_\mathrm {mat}$ and a geometric part ${\varvec{K}}^{ij}_\mathrm {geo}$. The material part is given by

$$\begin{aligned} {\varvec{K}}^{ij}_\mathrm {mat} = \lambda {\varvec{d}}_i\otimes {\varvec{d}}_j + \mu {\varvec{d}}_j\otimes {\varvec{d}}_i + \mu {\varvec{d}}_k\otimes {\varvec{d}}_k \, \delta _{ij} \end{aligned}$$

and the geometric part assumes the form

$$\begin{aligned} {\varvec{K}}^{ij}_\mathrm {geo} = S_{ij}{\varvec{I}} \end{aligned}$$

The symmetry of the iteration matrix follows from the properties ${\varvec{K}}^{ij}_\mathrm {mat}=({\varvec{K}}^{ji}_\mathrm {mat})^T$ and ${\varvec{K}}^{ij}_\mathrm {geo}=({\varvec{K}}^{ji}_\mathrm {geo})^T$. Similar to (153), in the discrete case the EM scheme (63) leads to

$$\begin{aligned} \Delta \left. ({\varvec{f}}^i_\mathrm {int})\right| _{n+\frac{1}{2}}=\Delta {S}^{ij}_A{{\varvec{d}}}_{j_{n+\frac{1}{2}}} + {S}^{ij}_A\Delta {{\varvec{d}}}_{j_{n+\frac{1}{2}}} \end{aligned}$$

Similar to (154), the algorithmic stress formula (65) leads to

$$\begin{aligned} \begin{array}{rcl} \Delta {S}^{ij}_A &{}=&{} \lambda \Delta \gamma _{kk_{n+\frac{1}{2}}}\delta _{ij} + 2\mu \Delta \gamma _{ij_{n+\frac{1}{2}}} \\ &{}=&{} \frac{\lambda }{2}({\varvec{d}}_{k_{n+1}}\cdot \Delta {\varvec{d}}_{k_{n+1}}) \delta _{ij} + \frac{\mu }{2}({\varvec{d}}_{i_{n+1}}\cdot \Delta {\varvec{d}}_{j_{n+1}} + {\varvec{d}}_{j_{n+1}}\cdot \Delta {\varvec{d}}_{i_{n+1}}) \end{array} \end{aligned}$$

(156)

Altogether the discrete counterpart of (155) is given by the consistent linearization

$$\begin{aligned} \Delta \left. ({\varvec{f}}^i_\mathrm {int})\right| _{n+\frac{1}{2}}=\left( \left. {\varvec{K}}^{ij}_\mathrm {mat}\right| _{n+\frac{1}{2}} + \left. {\varvec{K}}^{ij}_\mathrm {geo}\right| _{n+\frac{1}{2}}\right) \Delta {\varvec{d}}_{j_{n+1}} \end{aligned}$$

where the material part is given by

$$\begin{aligned} \left. {\varvec{K}}^{ij}_\mathrm {mat}\right| _{n+\frac{1}{2}} = \frac{1}{2}\left( \lambda {\varvec{d}}_{i_{n+\frac{1}{2}}}\otimes {\varvec{d}}_{j_{n+1}} + \mu {\varvec{d}}_{j_{n+\frac{1}{2}}}\otimes {\varvec{d}}_{i_{n+1}} + \mu {\varvec{d}}_{k_{n+\frac{1}{2}}}\otimes {\varvec{d}}_{k_{n+1}} \, \delta _{ij}\right) \end{aligned}$$

and the geometric part assumes the form

$$\begin{aligned} \left. {\varvec{K}}^{ij}_\mathrm {geo}\right| _{n+\frac{1}{2}} = \frac{1}{2}{S}^{ij}_A{\varvec{I}} \end{aligned}$$

It is obvious that in the discrete setting the material part destroys the symmetry of the iteration matrix, for

$$\begin{aligned} \left. \left( {\varvec{K}}^{ij}_\mathrm {mat}\right) \right| _{n+\frac{1}{2}}\ne \left. \left( {\varvec{K}}^{ji}_\mathrm {mat}\right) ^T\right| _{n+\frac{1}{2}} \end{aligned}$$

We finally remark that due to definition (13) of the total strain energy of the Cosserat point, namely $U({\varvec{C}}) = V_0 W({\varvec{C}})$, the above stress components $S^{ij}$ should be replaced by $\overline{S}^{ij}=V_0 S^{ij}$.