Micromechanics of composites/Balance of linear momentum

Statement of the balance of linear momentum

The balance of linear momentum can be expressed as:

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$\rho \dot{𝐯}-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛=0$ 
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where $\rho (𝐱,t)$ is the mass density, $𝐯(𝐱,t)$ is the velocity, $\mathit{\sigma }(𝐱,t)$ is the Cauchy stress, and $\rho 𝐛$ is the body force density.

Proof

Recall the general equation for the balance of a physical quantity

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

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

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

The surface tractions are related to the Cauchy stress by

$𝐭=\mathit{\sigma }\cdot 𝐧.$

Therefore,

$\frac{d}{dt}\left[\underset{\mathrm{\Omega }}{\int }\rho 𝐯\text{dV}\right]=\underset{\partial \mathrm{\Omega }}{\int }\rho 𝐯[u_n-𝐯\cdot 𝐧]\text{dA}+\underset{\partial \mathrm{\Omega }}{\int }\mathit{\sigma }\cdot 𝐧\text{dA}+\underset{\mathrm{\Omega }}{\int }\rho 𝐛\text{dV}.$

Let us assume that $\mathrm{\Omega }$ is an arbitrary fixed control volume. Then,

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

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

$(𝐮\otimes 𝐯)\cdot 𝐚=(𝐚\cdot 𝐯)𝐮.$

Therefore,

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

Using the divergence theorem

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet 𝐯\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }𝐯\cdot 𝐧\text{dA}$

we have

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

or,

$\underset{\mathrm{\Omega }}{\int }[\frac{\partial }{\partial t}(\rho 𝐯)+\mathrm{\nabla }\bullet [(\rho 𝐯)\otimes 𝐯)]-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛]\text{dV}=0.$

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

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

Using the identity

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

we get

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

or,

$[\frac{\partial \rho }{\partial t}+\rho \mathrm{\nabla }\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+\mathrm{\nabla }(\rho 𝐯)\cdot 𝐯-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛=0$

Using the identity

$\mathrm{\nabla }(\phi 𝐯)=\phi \mathrm{\nabla }𝐯+𝐯\otimes (\mathrm{\nabla }\phi )$

we get

$[\frac{\partial \rho }{\partial t}+\rho \mathrm{\nabla }\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+[\rho \mathrm{\nabla }𝐯+𝐯\otimes (\mathrm{\nabla }\rho )]\cdot 𝐯-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛=0$

From the definition

$(𝐮\otimes 𝐯)\cdot 𝐚=(𝐚\cdot 𝐯)𝐮$

we have

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

Hence,

$[\frac{\partial \rho }{\partial t}+\rho \mathrm{\nabla }\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+\rho \mathrm{\nabla }𝐯\cdot 𝐯+[𝐯\cdot (\mathrm{\nabla }\rho )]𝐯-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛=0$

or,

$[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\rho \cdot 𝐯+\rho \mathrm{\nabla }\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+\rho \mathrm{\nabla }𝐯\cdot 𝐯-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛=0.$

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

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

Therefore,

$[\dot{\rho }+\rho \mathrm{\nabla }\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+\rho \mathrm{\nabla }𝐯\cdot 𝐯-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛=0.$

From the balance of mass, we have

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

Therefore,

$\rho \frac{\partial 𝐯}{\partial t}+\rho \mathrm{\nabla }𝐯\cdot 𝐯-\mathrm{\nabla }\bullet \mathit{\sigma }-\rho 𝐛=0.$

The material time derivative of $𝐯$ is defined as

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

Hence,

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

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