Micromechanics of composites/Balance of angular momentum

Statement of the balance of angular momentum
The balance of angular momentum in an inertial frame can be expressed as: $$ \boldsymbol{\sigma}= \boldsymbol{\sigma}^T $$

Proof
We assume that there are no surface couples on $$\partial{\Omega}$$ or body couples in $$\Omega$$. Recall the general balance equation

\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~. $$ In this case, the physical quantity to be conserved the angular momentum density, i.e., $$f = \mathbf{x}\times(\rho~\mathbf{v})$$. The angular momentum source at the surface is then $$g = \mathbf{x}\times\mathbf{t}$$ and the angular momentum source inside the body is $$h = \mathbf{x}\times(\rho~\mathbf{b})$$. The angular momentum and moments are calculated with respect to a fixed origin. Hence we have

\cfrac{d}{dt}\left[\int_{\Omega} \mathbf{x}\times(\rho~\mathbf{v})~\text{dV}\right] = \int_{\partial{\Omega}} [\mathbf{x}\times(\rho~\mathbf{v})] [u_n - \mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \mathbf{x}\times\mathbf{t}~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ Assuming that $$\Omega$$ is a control volume, we have

\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\partial{\Omega}} [\mathbf{x}\times(\rho~\mathbf{v})][\mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \mathbf{x}\times\mathbf{t}~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ Using the definition of a tensor product we can write

[\mathbf{x}\times(\rho~\mathbf{v})][\mathbf{v}\cdot\mathbf{n}] = [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]\cdot\mathbf{n} ~. $$ Also, $$\mathbf{t} = \boldsymbol{\sigma}\cdot\mathbf{n}$$. Therefore we have

\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\partial{\Omega}} [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]\cdot\mathbf{n} ~\text{dA} + \int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ Using the divergence theorem, we get

\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]~\text{dV} + \int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ To convert the surface integral in the above equation into a volume integral, it is convenient to use index notation. Thus,

\left[\int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA}\right]_i = \int_{\partial{\Omega}} e_{ijk}~x_j~\sigma_{kl}~n_l~\text{dA}= \int_{\partial{\Omega}} A_{il}~n_l~\text{dA}= \int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} $$ where $$[~]_i$$ represents the $$i$$-th component of the vector. Using the divergence theorem

\int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \boldsymbol{\nabla} \bullet \boldsymbol{A}~\text{dV} = \int_{\Omega} \frac{\partial A_{il}}{\partial x_l}~\text{dV} = \int_{\Omega} \frac{\partial }{\partial x_l}(e_{ijk}~x_j~\sigma_{kl})~\text{dV}~. $$ Differentiating,

\int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \left[ e_{ijk}~\delta_{jl}~\sigma_{kl} + e_{ijk}~x_j~\frac{\partial \sigma_{kl}}{\partial x_l}\right]~\text{dV} = \int_{\Omega} \left[ e_{ijk}~\sigma_{kj} + e_{ijk}~x_j~\frac{\partial \sigma_{kl}}{\partial x_l}\right]~\text{dV} = \int_{\Omega} \left[ e_{ijk}~\sigma_{kj} + e_{ijk}~x_j~[\boldsymbol{\nabla} \bullet \boldsymbol{\sigma}]_k\right]~\text{dV} ~. $$ Expressed in direct tensor notation,

\int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \left[ [\mathcal{E}:\boldsymbol{\sigma}^T]_i + [\mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})]_i\right]~\text{dV} $$ where $$\mathcal{E}$$ is the third-order permutation tensor. Therefore,

\left[\int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA}\right]_i = = \int_{\Omega} \left[ [\mathcal{E}:\boldsymbol{\sigma}^T]_i + [\mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})]_i\right]~\text{dV} $$ or,

\int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA} = = \int_{\Omega} \left[ \mathcal{E}:\boldsymbol{\sigma}^T + \mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})\right]~\text{dV} ~. $$ The balance of angular momentum can then be written as

\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]~\text{dV} + \int_{\Omega} \left[ \mathcal{E}:\boldsymbol{\sigma}^T + \mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})\right]~\text{dV} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ Since $$\Omega$$ is an arbitrary volume, we have

\mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right] = - \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] + \mathcal{E}:\boldsymbol{\sigma}^T + \mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma}) + \mathbf{x}\times(\rho~\mathbf{b}) $$ or,

{\mathbf{x}}\times {\left[\frac{\partial }{\partial t}(\rho~\mathbf{v}) - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] + \mathcal{E}:\boldsymbol{\sigma}^T ~. $$ Using the identity,

\boldsymbol{\nabla} \bullet (\mathbf{u}\otimes\mathbf{v}) = (\boldsymbol{\nabla} \bullet \mathbf{v})\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})\cdot\mathbf{v} $$ we get

\boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] = (\boldsymbol{\nabla} \bullet \mathbf{v})[\mathbf{x}\times(\rho~\mathbf{v})] + (\boldsymbol{\nabla} [\mathbf{x}\times(\rho~\mathbf{v})])\cdot\mathbf{v} ~. $$ The second term on the right can be further simplified using index notation as follows.

\begin{align} \left[(\boldsymbol{\nabla} [\mathbf{x}\times(\rho~\mathbf{v})])\cdot\mathbf{v}\right]_i = \left[(\boldsymbol{\nabla} [\rho~(\mathbf{x}\times\mathbf{v})])\cdot\mathbf{v}\right]_i & = \frac{\partial }{\partial x_l}(\rho~e_{ijk}~x_j~v_k)~v_l \\ & = e_{ijk}\left[ \frac{\partial \rho}{\partial x_l}~x_j~v_k~v_l+ \rho~\frac{\partial x_j}{\partial x_l}~v_k~v_l + \rho~x_j~\frac{\partial v_k}{\partial x_l}~v_l\right] \\ & = (e_{ijk}~x_j~v_k)~\left(\frac{\partial \rho}{\partial x_l}~v_l\right)+ \rho~(e_{ijk}~\delta_{jl}~v_k~v_l) + e_{ijk}~x_j~\left(\rho~\frac{\partial v_k}{\partial x_l}~v_l\right) \\ & = [(\mathbf{x}\times\mathbf{v})(\boldsymbol{\nabla} \rho\cdot\mathbf{v}) + \rho~\mathbf{v}\times\mathbf{v} + \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})]_i \\ & = [(\mathbf{x}\times\mathbf{v})(\boldsymbol{\nabla} \rho\cdot\mathbf{v}) + \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})]_i ~. \end{align} $$ Therefore we can write

\boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] = (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) + (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})] ~. $$ The balance of angular momentum then takes the form

{\mathbf{x}}\times {\left[\frac{\partial }{\partial t}(\rho~\mathbf{v}) - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) - \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T $$ or,

{\mathbf{x}}\times {\left[\frac{\partial }{\partial t}(\rho~\mathbf{v}) + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T $$ or,

{\mathbf{x}}\times {\left[\rho\frac{\partial \mathbf{v}}{\partial t} + \frac{\partial \rho}{\partial t}~\mathbf{v} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T $$ The material time derivative of $$\mathbf{v}$$ is defined as

\dot{\mathbf{v}} = \frac{\partial \mathbf{v}}{\partial t} + \boldsymbol{\nabla} \mathbf{v}\cdot\mathbf{v} ~. $$ Therefore,

{\mathbf{x}}\times {\left[\rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - \mathbf{x}\times\cfrac{\partial \rho}{\partial t}~\mathbf{v} + - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T ~. $$ Also, from the conservation of linear momentum

\rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. $$ Hence,

\begin{align} 0 & = \mathbf{x}\times\cfrac{\partial\rho}{\partial t}~\mathbf{v} + (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) + (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) - \mathcal{E}:\boldsymbol{\sigma}^T \\ & = \left(\frac{\partial \rho}{\partial t} + \rho\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} \right)(\mathbf{x}\times\mathbf{v}) - \mathcal{E}:\boldsymbol{\sigma}^T ~. \end{align} $$ The material time derivative of $$\rho$$ is defined as

\dot{\rho} = \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} ~. $$ Hence,

(\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times\mathbf{v}) - \mathcal{E}:\boldsymbol{\sigma}^T = 0 ~. $$ From the balance of mass

\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v} = 0 ~. $$ Therefore,

\mathcal{E}:\boldsymbol{\sigma}^T = 0 ~. $$ In index notation,

e_{ijk}~\sigma_{kj} = 0 ~. $$ Expanding out, we get

\sigma_{12} - \sigma_{21} = 0 ~; \sigma_{23} - \sigma_{32} = 0 ~; \sigma_{31} - \sigma_{13} = 0 ~. $$ Hence,

{ \boldsymbol{\sigma} = \boldsymbol{\sigma}^T } $$