Cauchy's equation by Extended divergence theorem

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Start from linear momentum equation with Reynolds Transport Theorem (p.252)

 

$$\int_{\rm{CV}}{\rho \vec{g} dV} + \int_{\rm{CS}}{\sigma_{ij}\cdot\vec{n}dA} = \int_{\rm{CV}}{\frac{\partial}{\partial t}\left(\rho \vec{v}\right)dV} + \int_{\rm{CS}}{\rho \vec{v} \vec{v} \cdot \vec{n} dA}\,\cdots (1)$$

 

For who want to know how to derive equation \((1)\)

Link:

 

Extended Divergence Theorem:

$$\int_{\rm{CS}}{G_{ij}\cdot \vec{n}dA} = \int_{\rm{CV}}{\vec{\nabla}\cdot G_{ij}dV}$$

 

Let's apply extended divergence theorem in second term of left side of the equation \((1)\).

$$\int_{\rm{CS}}{\sigma_{ij}\cdot \vec{n}dA} = \int_{\rm{CV}}{\vec{\nabla}\cdot(\sigma_{ij})dV}$$

 

We can apply also on second term of right side of the equation \((1)\)

$$\int_{\rm{CS}}{(\rho \vec{v}\vec{v})\cdot\vec{n}dA} = \int_{\rm{CV}}{\vec{\nabla}\cdot(\rho \vec{v}\vec{v})dV}$$

 

Then we can re-arrange equation \((1)\) like below:

$$\begin{aligned}\int_{\rm{CV}}{\rho \vec{g}dV} + \int_{\rm{CV}}{\vec{\nabla}\cdot(\sigma_{ij})dV} & = \int_{\rm{CV}}{\dfrac{\partial}{\partial t}(\rho \vec{v})dV} + \int_{\rm{CV}}{\vec{\nabla}\cdot(\rho \vec{v}\vec{v})dV} \\ \\ 0 & = \int_{\rm{CV}}{\dfrac{\partial}{\partial t}(\rho \vec{v})dV} + \int_{\rm{CV}}{\vec{\nabla}\cdot\left(\rho \vec{v}\vec{v}\right)dV} - \int_{\rm{CV}}{\rho \vec{g}dV} - \int_{\rm{CV}}{\vec{\nabla}\cdot(\sigma_{ij})dV}\end{aligned}$$

 

$$\int_{\rm{CV}}{\left[\dfrac{\partial}{\partial t}(\rho \vec{v}) + \vec{\nabla}\cdot(\rho \vec{v}\vec{v}) - \rho \vec{g} - \vec{\nabla}\cdot\sigma_{ij}dV\right]} = 0$$

 

$$\dfrac{\partial}{\partial t}(\rho \vec{v}) + \vec{\nabla}\cdot(\rho \vec{v}\vec{v}) - \rho \vec{g} - \vec{\nabla}\cdot\sigma_{ij} = 0$$

 

Then we can derive Cauchy's equation from linear momentum equation with Reynolds Transport Theorem

$$\dfrac{\partial}{\partial t}(\rho \vec{v}) + \vec{\nabla}\cdot(\rho\vec{v}\vec{v}) = \rho \vec{g} + \vec{\nabla}\cdot\sigma_{ij}$$

 

Reference

Advanced Engineering Mathematics