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Navier-Stokes equation supplementary (1) 본문

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Navier-Stokes equation supplementary (1)

빛나는 전구 2023. 10. 18. 10:00

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$$\begin{aligned}\sigma & = \left(\lambda + \frac{2}{3}\mu\right)\left(\vec{\nabla}\cdot \vec{u}\right) \pmb{I}+ \mu\left\{ \vec{\nabla}\vec{u} + \left(\vec{\nabla}\vec{u}\right)^{\rm{T}} - \frac{2}{3}(\vec{\nabla}\cdot\vec{u}) \pmb{I} \right\}\\ \\ & = \left(\lambda + \frac{2}{3}\mu\right) \left(\vec{\nabla}\vec{u}\right) \pmb{I} - \frac{2}{3}\mu\left(\vec{\nabla}\vec{u}\right)\pmb{I} + \mu \vec{\nabla}\vec{u} + \mu \left(\vec{\nabla}\vec{u}\right)^{\rm{T}} \\ \\ & = \lambda \left(\vec{\nabla}\vec{u} \right)\pmb{I} + \frac{2}{3}\mu \left(\vec{\nabla}\vec{u}\right) \pmb{I} - \frac{2}{3} \left(\vec{\nabla}\vec{u}\right) \pmb{I} + \mu\vec{\nabla}\vec{u} + \mu \left(\vec{\nabla}\vec{u}\right)^{\rm{T}}\\ \\ & = \lambda \left(\vec{\nabla}\vec{u}\right) \pmb{I} + \mu \left(\vec{\nabla}\vec{u} + \left(\vec{\nabla}\vec{u}\right)\right)^{\rm{T}} \end{aligned} $$

 

이때 아래 식을 대입한다.

 

$$\varepsilon_{ij} = \begin{pmatrix} \dfrac{1}{2}\dfrac{\partial u}{\partial x} & \dfrac{1}{2}\dfrac{\partial u}{\partial y} & \dfrac{1}{2}\dfrac{\partial u}{\partial z} \\ \dfrac{1}{2}\dfrac{\partial v}{\partial x} & \dfrac{1}{2}\dfrac{\partial v}{\partial y} & \dfrac{1}{2}\dfrac{\partial v}{\partial z} \\ \dfrac{1}{2}\dfrac{\partial w}{\partial x} & \dfrac{1}{2}\dfrac{\partial w}{\partial y} & \dfrac{1}{2}\dfrac{\partial w}{\partial z} \end{pmatrix} + \begin{pmatrix} \dfrac{1}{2}\dfrac{\partial u}{\partial x} & \dfrac{1}{2}\dfrac{\partial v}{\partial x} & \dfrac{1}{2} \dfrac{\partial w}{\partial x} \\ \dfrac{1}{2} \dfrac{\partial u}{\partial y} & \dfrac{1}{2} \dfrac{\partial v}{\partial y} & \dfrac{1}{2} \dfrac{\partial w}{\partial y} \\ \dfrac{1}{2}\dfrac{\partial w}{\partial z} & \dfrac{1}{2} \dfrac{\partial v}{\partial z} & \dfrac{1}{2} \dfrac{\partial w}{\partial z} \end{pmatrix}$$

 

$$\varepsilon_{ij} = \frac{1}{2}\left(\vec{\nabla}\vec{u}\right) + \frac{1}{2}\left(\vec{\nabla}\vec{u}\right)^{\rm{T}}$$

 

따라서

$$\begin{aligned}\sigma & = \lambda (\vec{\nabla}\vec{u}) \pmb{I} + 2 \mu \left\{\frac{1}{2}(\vec{\nabla}\vec{u}) + \frac{1}{2}(\vec{\nabla}\vec{u})^{\rm{T}}\right\} \\ \\ & = \lambda (\vec{\nabla}\vec{u}) \pmb{I} + \mu (\vec{\nabla}\vec{u} + (\vec{\nabla}\vec{u})^{\rm{T}})\end{aligned}$$

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