有限元笔记6-柱坐标系下的平衡方程

柱坐标下的拉梅方程推导其实就用到一点坐标变换就可以了。

笛卡尔坐标系下的拉梅方程为：

(λ + 2 μ) \nabla (\nabla \cdot u) - μ \nabla \nabla u + F = 0

$(\lambda+2\mu)\nabla(\nabla\cdot u)-\mu\nabla\nabla u+F=0$

笛卡尔坐标到柱坐标的转换为：

x 1 = x = r cos θ, x 2 = y = r sin θ, x 3 = z = z

$x_{1}=x=r\cos\theta,x_{2}=y=r\sin\theta,x_{3}=z=z$
对应的相关分量为：

u = (u r, u θ, u z), e = (e i j), τ = (τ i j) i, j = r, θ, z

$u=(u_{r},u_{\theta},u_{z}),e=(e_{ij}),\tau=(\tau_{ij})\quad\quad i,j=r,\theta,z$

对应的算符运算法则为:

\nabla f = \partial f \partial r r ⃗ + 1 r \partial f \partial θ θ ⃗ + \partial f \partial z z ⃗

$\nabla f=\frac{\partial f}{\partial r}\vec{r}+\frac{1}{r}\frac{\partial f}{\partial\theta}\vec{\theta}+\frac{\partial f}{\partial z}\vec{z}$

\nabla \cdot u = 1 r \partial ( r u r ) \partial r + 1 r \partial u θ \partial θ + \partial u z \partial z

$\nabla\cdot u=\frac{1}{r}\frac{\partial (ru_{r})}{\partial r}+\frac{1}{r}\frac{\partial u_{\theta}}{\partial\theta}+\frac{\partial u_{z}}{\partial z}$

c u r l (u) = (1 r \partial u z \partial θ - \partial u θ \partial z) r ⃗ + (\partial u r \partial z - \partial u z \partial r) θ ⃗ + (1 r \partial ( r u θ ) \partial r - 1 r \partial u r \partial θ) z ⃗

$curl(u)=(\frac{1}{r}\frac{\partial u_{z}}{\partial\theta}-\frac{\partial u_{\theta}}{\partial z})\vec{r}+(\frac{\partial u_{r}}{\partial z}-\frac{\partial u_{z}}{\partial r})\vec{\theta}+(\frac{1}{r}\frac{\partial(ru_{\theta})}{\partial r}-\frac{1}{r}\frac{\partial u_{r}}{\partial\theta})\vec{z}$

笛卡尔坐标下的本构关系为：

τ i j = λ δ i j \nabla \cdot u + 2 μ e i j

$\tau_{ij}=\lambda\delta_{ij}\nabla\cdot u+2\mu e_{ij}$

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那么对应的柱坐标下的应力应变关系为：

τ r r = λ \nabla \cdot u + 2 μ \partial u r \partial r

$\tau_{rr}=\lambda\nabla\cdot u+2\mu\frac{\partial u_{r}}{\partial r}$

τ θ θ = λ \nabla \cdot u + 2 μ (1 r \partial u θ \partial θ + u r r)

$\tau_{\theta\theta}=\lambda\nabla\cdot u+2\mu(\frac{1}{r}\frac{\partial u_{\theta}}{\partial\theta}+\frac{u_{r}}{r})$

τ z z = λ \nabla u + 2 μ \partial u z \partial z

$\tau_{zz}=\lambda\nabla u+2\mu\frac{\partial u_{z}}{\partial z}$

τ r θ μ = τ θ r μ = \partial u θ \partial r - u θ r + 1 r \partial u r \partial θ

$\frac{\tau_{r\theta}}{\mu}=\frac{\tau_{\theta r}}{\mu}=\frac{\partial u_{\theta}}{\partial r}-\frac{u_{\theta}}{r}+\frac{1}{r}\frac{\partial u_{r}}{\partial\theta}$

τ r z μ = τ z r μ = \partial u z \partial r + \partial u r \partial z

$\frac{\tau_{rz}}{\mu}=\frac{\tau_{zr}}{\mu}=\frac{\partial u_{z}}{\partial r}+\frac{\partial u_{r}}{\partial z}$

τ θ z μ = τ z θ μ = 1 r \partial u z \partial θ + \partial u θ \partial z

$\frac{\tau_{\theta z}}{\mu}=\frac{\tau_{z\theta}}{\mu}=\frac{1}{r}\frac{\partial u_{z}}{\partial\theta}+\frac{\partial u_{\theta}}{\partial z}$

而位移和应变的关系为：

e r r = \partial u r \partial r, e θ θ = 1 r \partial u θ \partial θ + u r r, e z z = \partial u z \partial z

$e_{rr}=\frac{\partial u_{r}}{\partial r},e_{\theta\theta}=\frac{1}{r}\frac{\partial u_{\theta}}{\partial\theta}+\frac{u_{r}}{r},e_{zz}=\frac{\partial u_{z}}{\partial z}$

2 e r θ = 2 e θ r = \partial u θ \partial r - \partial u θ r + 1 r \partial u r \partial θ

$2e_{r\theta}=2e_{\theta r}=\frac{\partial u_{\theta}}{\partial r}-\frac{\partial u_{\theta}}{r}+\frac{1}{r}\frac{\partial u_{r}}{\partial\theta}$

2 e r z = 2 e z r = \partial u r \partial z + \partial u z \partial r

$2e_{rz}=2e_{zr}=\frac{\partial u_{r}}{\partial z}+\frac{\partial u_{z}}{\partial r}$

2 e z θ = 2 e θ z = 1 r \partial u z \partial θ + \partial u θ \partial z

$2e_{z\theta}=2e_{\theta z}=\frac{1}{r}\frac{\partial u_{z}}{\partial\theta}+\frac{\partial u_{\theta}}{\partial z}$

对应的应力平衡方程就可以写为：

\partial τ r r \partial r + 1 r \partial τ r θ \partial θ + \partial τ r z \partial z + τ r r - τ θ θ r + F r = 0

$\frac{\partial\tau_{rr}}{\partial r}+\frac{1}{r}\frac{\partial\tau_{r\theta}}{\partial\theta}+\frac{\partial\tau_{rz}}{\partial z}+\frac{\tau_{rr}-\tau_{\theta\theta}}{r}+F_{r}=0$

\partial τ r θ \partial r + 1 r \partial τ θ θ \partial θ + \partial τ θ z \partial z + 2 r τ r θ + F θ = 0

$\frac{\partial\tau_{r\theta}}{\partial r}+\frac{1}{r}\frac{\partial\tau_{\theta\theta}}{\partial\theta}+\frac{\partial\tau_{\theta z}}{\partial z}+\frac{2}{r}\tau_{r\theta}+F_{\theta}=0$

\partial τ r z \partial r + 1 r \partial τ θ z \partial θ + \partial τ z z \partial z + τ r z r + F z = 0

$\frac{\partial\tau_{rz}}{\partial r}+\frac{1}{r}\frac{\partial\tau_{\theta z}}{\partial\theta}+\frac{\partial\tau_{zz}}{\partial z}+\frac{\tau_{rz}}{r}+F_{z}=0$

有限元笔记6-柱坐标系下的平衡方程

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