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:

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$\mathit{\sigma }={\mathit{\sigma }}^T$
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Proof

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

$\frac{d}{dt}[\underset{\mathrm{\Omega }}{\int }f(𝐱,t)\text{dV}]=\underset{\partial \mathrm{\Omega }}{\int }f(𝐱,t)[u_n(𝐱,t)-𝐯(𝐱,t)\cdot 𝐧(𝐱,t)]\text{dA}+\underset{\partial \mathrm{\Omega }}{\int }g(𝐱,t)\text{dA}+\underset{\mathrm{\Omega }}{\int }h(𝐱,t)\text{dV}.$

In this case, the physical quantity to be conserved the angular momentum density, i.e., $f=𝐱\times (\rho 𝐯)$. The angular momentum source at the surface is then $g=𝐱\times 𝐭$ and the angular momentum source inside the body is $h=𝐱\times (\rho 𝐛)$. The angular momentum and moments are calculated with respect to a fixed origin. Hence we have

$\frac{d}{dt}[\underset{\mathrm{\Omega }}{\int }𝐱\times (\rho 𝐯)\text{dV}]=\underset{\partial \mathrm{\Omega }}{\int }[𝐱\times (\rho 𝐯)][u_n-𝐯\cdot 𝐧]\text{dA}+\underset{\partial \mathrm{\Omega }}{\int }𝐱\times 𝐭\text{dA}+\underset{\mathrm{\Omega }}{\int }𝐱\times (\rho 𝐛)\text{dV}.$

Assuming that $\mathrm{\Omega }$ is a control volume, we have

$\underset{\mathrm{\Omega }}{\int }𝐱\times [\frac{\partial }{\partial t}(\rho 𝐯)]\text{dV}=-\underset{\partial \mathrm{\Omega }}{\int }[𝐱\times (\rho 𝐯)][𝐯\cdot 𝐧]\text{dA}+\underset{\partial \mathrm{\Omega }}{\int }𝐱\times 𝐭\text{dA}+\underset{\mathrm{\Omega }}{\int }𝐱\times (\rho 𝐛)\text{dV}.$

Using the definition of a tensor product we can write

$[𝐱\times (\rho 𝐯)][𝐯\cdot 𝐧]=[[𝐱\times (\rho 𝐯)]\otimes 𝐯]\cdot 𝐧.$

Also, $𝐭=\mathit{\sigma }\cdot 𝐧$. Therefore we have

$\underset{\mathrm{\Omega }}{\int }𝐱\times [\frac{\partial }{\partial t}(\rho 𝐯)]\text{dV}=-\underset{\partial \mathrm{\Omega }}{\int }[[𝐱\times (\rho 𝐯)]\otimes 𝐯]\cdot 𝐧\text{dA}+\underset{\partial \mathrm{\Omega }}{\int }𝐱\times (\mathit{\sigma }\cdot 𝐧)\text{dA}+\underset{\mathrm{\Omega }}{\int }𝐱\times (\rho 𝐛)\text{dV}.$

Using the divergence theorem, we get

$\underset{\mathrm{\Omega }}{\int }𝐱\times [\frac{\partial }{\partial t}(\rho 𝐯)]\text{dV}=-\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet [[𝐱\times (\rho 𝐯)]\otimes 𝐯]\text{dV}+\underset{\partial \mathrm{\Omega }}{\int }𝐱\times (\mathit{\sigma }\cdot 𝐧)\text{dA}+\underset{\mathrm{\Omega }}{\int }𝐱\times (\rho 𝐛)\text{dV}.$

To convert the surface integral in the above equation into a volume integral, it is convenient to use index notation. Thus,

${[\underset{\partial \mathrm{\Omega }}{\int }𝐱\times (\mathit{\sigma }\cdot 𝐧)\text{dA}]}_i=\underset{\partial \mathrm{\Omega }}{\int }{e}_{ijk}{x}_j{\sigma }_{kl}{n}_l\text{dA}=\underset{\partial \mathrm{\Omega }}{\int }{A}_{il}{n}_l\text{dA}=\underset{\partial \mathrm{\Omega }}{\int }𝑨\cdot 𝐧\text{dA}$

where $[{]}_i$ represents the $i$-th component of the vector. Using the divergence theorem

$\underset{\partial \mathrm{\Omega }}{\int }𝑨\cdot 𝐧\text{dA}=\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet 𝑨\text{dV}=\underset{\mathrm{\Omega }}{\int }\frac{\partial A_{il}}{\partial x_l}\text{dV}=\underset{\mathrm{\Omega }}{\int }\frac{\partial }{\partial x_l}(e_{ijk}{x}_j{\sigma }_{kl})\text{dV}.$

Differentiating,

$\underset{\partial \mathrm{\Omega }}{\int }𝑨\cdot 𝐧\text{dA}=\underset{\mathrm{\Omega }}{\int }[e_{ijk}{\delta }_{jl}{\sigma }_{kl}+e_{ijk}{x}_j\frac{\partial {\sigma }_{kl}}{\partial x_l}]\text{dV}=\underset{\mathrm{\Omega }}{\int }[e_{ijk}{\sigma }_{kj}+e_{ijk}{x}_j\frac{\partial {\sigma }_{kl}}{\partial x_l}]\text{dV}=\underset{\mathrm{\Omega }}{\int }[e_{ijk}{\sigma }_{kj}+e_{ijk}{x}_j[\mathrm{\nabla }\bullet \mathit{\sigma }{]}_k]\text{dV}.$

Expressed in direct tensor notation,

$\underset{\partial \mathrm{\Omega }}{\int }𝑨\cdot 𝐧\text{dA}=\underset{\mathrm{\Omega }}{\int }[[ℰ:{\mathit{\sigma }}^T{]}_i+[𝐱\times (\mathrm{\nabla }\bullet \mathit{\sigma }){]}_i]\text{dV}$

where $ℰ$ is the third-order permutation tensor. Therefore,

${[\underset{\partial \mathrm{\Omega }}{\int }𝐱\times (\mathit{\sigma }\cdot 𝐧)\text{dA}]}_i==\underset{\mathrm{\Omega }}{\int }[[ℰ:{\mathit{\sigma }}^T{]}_i+[𝐱\times (\mathrm{\nabla }\bullet \mathit{\sigma }){]}_i]\text{dV}$

or,

$\underset{\partial \mathrm{\Omega }}{\int }𝐱\times (\mathit{\sigma }\cdot 𝐧)\text{dA}==\underset{\mathrm{\Omega }}{\int }[ℰ:{\mathit{\sigma }}^T+𝐱\times (\mathrm{\nabla }\bullet \mathit{\sigma })]\text{dV}.$

The balance of angular momentum can then be written as

$\underset{\mathrm{\Omega }}{\int }𝐱\times [\frac{\partial }{\partial t}(\rho 𝐯)]\text{dV}=-\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet [[𝐱\times (\rho 𝐯)]\otimes 𝐯]\text{dV}+\underset{\mathrm{\Omega }}{\int }[ℰ:{\mathit{\sigma }}^T+𝐱\times (\mathrm{\nabla }\bullet \mathit{\sigma })]\text{dV}+\underset{\mathrm{\Omega }}{\int }𝐱\times (\rho 𝐛)\text{dV}.$

Since $\mathrm{\Omega }$ is an arbitrary volume, we have

$𝐱\times [\frac{\partial }{\partial t}(\rho 𝐯)]=-\mathrm{\nabla }\bullet [[𝐱\times (\rho 𝐯)]\otimes 𝐯]+ℰ:{\mathit{\sigma }}^T+𝐱\times (\mathrm{\nabla }\bullet \mathit{\sigma })+𝐱\times (\rho 𝐛)$

or,

$𝐱\times [\frac{\partial }{\partial t}(\rho 𝐯)-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛]=-\mathrm{\nabla }\bullet [[𝐱\times (\rho 𝐯)]\otimes 𝐯]+ℰ:{\mathit{\sigma }}^T.$

Using the identity,

$\mathrm{\nabla }\bullet (𝐮\otimes 𝐯)=(\mathrm{\nabla }\bullet 𝐯)𝐮+(\mathrm{\nabla }𝐮)\cdot 𝐯$

we get

$\mathrm{\nabla }\bullet [[𝐱\times (\rho 𝐯)]\otimes 𝐯]=(\mathrm{\nabla }\bullet 𝐯)[𝐱\times (\rho 𝐯)]+(\mathrm{\nabla }[𝐱\times (\rho 𝐯)])\cdot 𝐯.$

The second term on the right can be further simplified using index notation as follows.

$\begin{array}{cc}{[(\mathrm{\nabla }[𝐱\times (\rho 𝐯)])\cdot 𝐯]}_i={[(\mathrm{\nabla }[\rho (𝐱\times 𝐯)])\cdot 𝐯]}_i& =\frac{\partial }{\partial x_l}(\rho e_{ijk}{x}_j{v}_k)v_l\\ & =e_{ijk}[\frac{\partial \rho }{\partial x_l}{x}_j{v}_k{v}_l+\rho \frac{\partial x_j}{\partial x_l}{v}_k{v}_l+\rho x_j\frac{\partial v_k}{\partial x_l}{v}_l]\\ & =(e_{ijk}{x}_j{v}_k)\left(\frac{\partial \rho }{\partial x_l}{v}_l\right)+\rho (e_{ijk}{\delta }_{jl}{v}_k{v}_l)+e_{ijk}{x}_j\left(\rho \frac{\partial v_k}{\partial x_l}{v}_l\right)\\ & =[(𝐱\times 𝐯)(\mathrm{\nabla }\rho \cdot 𝐯)+\rho 𝐯\times 𝐯+𝐱\times (\rho \mathrm{\nabla }𝐯\cdot 𝐯){]}_i\\ & =[(𝐱\times 𝐯)(\mathrm{\nabla }\rho \cdot 𝐯)+𝐱\times (\rho \mathrm{\nabla }𝐯\cdot 𝐯){]}_i.\end{array}$

Therefore we can write

$\mathrm{\nabla }\bullet [[𝐱\times (\rho 𝐯)]\otimes 𝐯]=(\rho \mathrm{\nabla }\bullet 𝐯)(𝐱\times 𝐯)+(\mathrm{\nabla }\rho \cdot 𝐯)(𝐱\times 𝐯)+𝐱\times (\rho \mathrm{\nabla }𝐯\cdot 𝐯)].$

The balance of angular momentum then takes the form

$𝐱\times [\frac{\partial }{\partial t}(\rho 𝐯)-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛]=-(\rho \mathrm{\nabla }\bullet 𝐯)(𝐱\times 𝐯)-(\mathrm{\nabla }\rho \cdot 𝐯)(𝐱\times 𝐯)-𝐱\times (\rho \mathrm{\nabla }𝐯\cdot 𝐯)+ℰ:{\mathit{\sigma }}^T$

or,

$𝐱\times [\frac{\partial }{\partial t}(\rho 𝐯)+\rho \mathrm{\nabla }𝐯\cdot 𝐯-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛]=-(\rho \mathrm{\nabla }\bullet 𝐯)(𝐱\times 𝐯)-(\mathrm{\nabla }\rho \cdot 𝐯)(𝐱\times 𝐯)+ℰ:{\mathit{\sigma }}^T$

or,

$𝐱\times [\rho \frac{\partial 𝐯}{\partial t}+\frac{\partial \rho }{\partial t}𝐯+\rho \mathrm{\nabla }𝐯\cdot 𝐯-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛]=-(\rho \mathrm{\nabla }\bullet 𝐯)(𝐱\times 𝐯)-(\mathrm{\nabla }\rho \cdot 𝐯)(𝐱\times 𝐯)+ℰ:{\mathit{\sigma }}^T$

The material time derivative of $𝐯$ is defined as

$\dot{𝐯}=\frac{\partial 𝐯}{\partial t}+\mathrm{\nabla }𝐯\cdot 𝐯.$

Therefore,

$𝐱\times [\rho \dot{𝐯}-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛]=-𝐱\times \frac{\partial \rho }{\partial t}𝐯+-(\rho \mathrm{\nabla }\bullet 𝐯)(𝐱\times 𝐯)-(\mathrm{\nabla }\rho \cdot 𝐯)(𝐱\times 𝐯)+ℰ:{\mathit{\sigma }}^T.$

Also, from the conservation of linear momentum

$\rho \dot{𝐯}-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛=0.$

Hence,

$\begin{array}{cc}0& =𝐱\times \frac{\partial \rho }{\partial t}𝐯+(\rho \mathrm{\nabla }\bullet 𝐯)(𝐱\times 𝐯)+(\mathrm{\nabla }\rho \cdot 𝐯)(𝐱\times 𝐯)-ℰ:{\mathit{\sigma }}^T\\ & =(\frac{\partial \rho }{\partial t}+\rho \mathrm{\nabla }\bullet 𝐯+\mathrm{\nabla }\rho \cdot 𝐯)(𝐱\times 𝐯)-ℰ:{\mathit{\sigma }}^T.\end{array}$

The material time derivative of $\rho$ is defined as

$\dot{\rho }=\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\rho \cdot 𝐯.$

Hence,

$(\dot{\rho }+\rho \mathrm{\nabla }\bullet 𝐯)(𝐱\times 𝐯)-ℰ:{\mathit{\sigma }}^T=0.$

From the balance of mass

$\dot{\rho }+\rho \mathrm{\nabla }\bullet 𝐯=0.$

Therefore,

$ℰ:{\mathit{\sigma }}^T=0.$

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,

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

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