Micromechanics of composites/Balance of linear momentum

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[edit] Statement of the balance of linear momentum

The balance of linear momentum can be expressed as:

where $\rho(\mathbf{x},t)$ is the mass density, $\mathbf{v}(\mathbf{x},t)$ is the velocity, $\boldsymbol{\sigma}(\mathbf{x},t)$ is the Cauchy stress, and $\rho~\mathbf{b}$ is the body force density.

[edit] Proof

Recall the general equation for the balance of a physical quantity

$\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~.$

In this case the physical quantity of interest is the momentum density, i.e., $f(\mathbf{x},t) = \rho(\mathbf{x},t)~\mathbf{v}(\mathbf{x},t)$ . The source of momentum flux at the surface is the surface traction, i.e., $g(\mathbf{x},t) = \mathbf{t}$ . The source of momentum inside the body is the body force, i.e., $h(\mathbf{x},t) = \rho(\mathbf{x},t)~\mathbf{b}(\mathbf{x},t)$ . Therefore, we have

$\cfrac{d}{dt}\left[\int_{\Omega} \rho~\mathbf{v}~\text{dV}\right] = \int_{\partial{\Omega}} \rho~\mathbf{v}[u_n - \mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \mathbf{t}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~.$

The surface tractions are related to the Cauchy stress by

$\mathbf{t} = \boldsymbol{\sigma}\cdot\mathbf{n} ~.$

Therefore,

$\cfrac{d}{dt}\left[\int_{\Omega} \rho~\mathbf{v}~\text{dV}\right] = \int_{\partial{\Omega}} \rho~\mathbf{v}[u_n - \mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \boldsymbol{\sigma}\cdot\mathbf{n}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~.$

Let us assume that $Ω$ is an arbitrary fixed control volume. Then,

$\int_{\Omega} \frac{\partial }{\partial t}(\rho~\mathbf{v})~\text{dV} = - \int_{\partial{\Omega}} \rho~\mathbf{v}~(\mathbf{v}\cdot\mathbf{n})~\text{dA} + \int_{\partial{\Omega}} \boldsymbol{\sigma}\cdot\mathbf{n}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~.$

Now, from the definition of the tensor product we have (for all vectors $\mathbf{a}$ )

$(\mathbf{u}\otimes\mathbf{v})\cdot\mathbf{a} = (\mathbf{a}\cdot\mathbf{v})~\mathbf{u} ~.$

Therefore,

$\int_{\Omega} \frac{\partial }{\partial t}(\rho~\mathbf{v})~\text{dV} = - \int_{\partial{\Omega}} \rho~(\mathbf{v}\otimes\mathbf{v})\cdot\mathbf{n}~\text{dA} + \int_{\partial{\Omega}} \boldsymbol{\sigma}\cdot\mathbf{n}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~.$

Using the divergence theorem

$\int_{\Omega} \boldsymbol{\nabla} \bullet \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\cdot\mathbf{n}~\text{dA}$

we have

$\int_{\Omega} \frac{\partial }{\partial t}(\rho~\mathbf{v})~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet [\rho~(\mathbf{v}\otimes\mathbf{v})]~\text{dV} + \int_{\Omega} \boldsymbol{\nabla} \bullet \boldsymbol{\sigma}~\text{dV} + \int_{\Omega} \rho~\mathbf{b}~\text{dV}$

or,

$\int_{\Omega}\left[ \frac{\partial }{\partial t}(\rho~\mathbf{v}) + \boldsymbol{\nabla} \bullet [(\rho~\mathbf{v})\otimes\mathbf{v})] - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b}\right]~\text{dV} = 0 ~.$

Since $Ω$ is arbitrary, we have

$\frac{\partial }{\partial t}(\rho~\mathbf{v}) + \boldsymbol{\nabla} \bullet [(\rho~\mathbf{v})\otimes\mathbf{v})] - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0~.$

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

$\frac{\partial \rho}{\partial t}~\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + (\boldsymbol{\nabla} \bullet \mathbf{v})(\rho\mathbf{v}) + \boldsymbol{\nabla} (\rho~\mathbf{v})\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0$

or,

$\left[\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \boldsymbol{\nabla} (\rho~\mathbf{v})\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0$

Using the identity

$\boldsymbol{\nabla} (\varphi~\mathbf{v}) = \varphi~\boldsymbol{\nabla}\mathbf{v} + \mathbf{v}\otimes(\boldsymbol{\nabla} \varphi)$

we get

$\left[\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \left[\rho~\boldsymbol{\nabla}\mathbf{v} + \mathbf{v}\otimes(\boldsymbol{\nabla} \rho)\right]\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0$

From the definition

$(\mathbf{u}\otimes\mathbf{v})\cdot\mathbf{a} = (\mathbf{a}\cdot\mathbf{v})~\mathbf{u}$

we have

$[\mathbf{v}\otimes(\boldsymbol{\nabla} \rho)]\cdot\mathbf{v} = [\mathbf{v}\cdot(\boldsymbol{\nabla} \rho)]~\mathbf{v} ~.$

Hence,

$\left[\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} +[\mathbf{v}\cdot(\boldsymbol{\nabla} \rho)]~\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0$

or,

$\left[\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~.$

The material time derivative of $ρ$ is defined as

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

Therefore,

$\left[\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~.$

From the balance of mass, we have

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

Therefore,

$\rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~.$