7.5 极坐标中的平面问题

liblaf10/26/22About 2 min

7.5 极坐标中的平面问题

平衡方程

\begin{aligned} \pdv σ_{r} r + \frac{1}{r} \pdv τ_{r θ} θ + \frac{σ_{r} - σ_{θ}}{r} + f_{r} = 0 \\ \pdv τ_{r θ} r + \frac{1}{r} \pdv σ_{θ} θ + 2 \frac{τ_{r θ}}{r} + f_{θ} = 0 \end{aligned}

几何方程

\begin{aligned} ε_{r} & = \pdv u_{r} r \\ ε_{θ} & = \frac{1}{r} \pdv v_{θ} θ + \frac{u_{r}}{r} \\ γ_{r θ} & = \frac{1}{r} \pdv u_{r} θ + \pdv v_{θ} r - \frac{v_{θ}}{r} \end{aligned}

$u_{r}$ — 径向位移; $v_{θ}$ — 环向位移

本构方程

应变 - 应力

\begin{aligned} ε_{r} & = \frac{1}{E} \pqty σ_{r} - ν σ_{θ} \\ ε_{θ} & = \frac{1}{E} \pqty σ_{θ} - ν σ_{r} \\ γ_{r θ} & = \frac{1}{G} τ_{r θ} \end{aligned}

应力 - 应变

\begin{aligned} σ_{r} & = \frac{2 G}{1 - ν} \pqty ε_{r} + ν ε_{θ} \\ σ_{θ} & = \frac{2 G}{1 - ν} \pqty ε_{θ} + ν ε_{r} \\ τ_{r θ} & = G γ_{r θ} \end{aligned}

协调方程

\pdv [2] ε_{θ} r + \frac{1}{r^{2}} \pdv [2] ε_{r} θ - \frac{1}{r} \pdv γ_{r θ} r θ + \frac{2}{r} \pdv ε_{θ} r - \frac{1}{r} \pdv ε_{r} r - \frac{1}{r^{2}} \pdv γ_{r θ} θ = 0

应力函数解法基本方程

\pqty \pdv [2] r + \frac{1}{r} \pdv r + \frac{1}{r^{2}} \pdv [2] θ \pqty \pdv [2] r + \frac{1}{r} \pdv r + \frac{1}{r^{2}} \pdv [2] θ ϕ = 0

极坐标应力公式

\begin{aligned} σ_{r} & = \frac{1}{r^{2}} \pdv [2] ϕ θ + \frac{1}{r} \pdv ϕ r \\ σ_{θ} & = \pdv [2] ϕ r \\ τ_{r θ} & = \frac{1}{r^{2}} \pdv ϕ θ - \frac{1}{r} \pdv ϕ θ r = - \pdv r (\frac{1}{r} \pdv ϕ θ) \end{aligned}

边界条件

位移边界 $Γ_{u}$

u_{r} = {\overset{―}{u}}_{r} \qc v_{θ} = {\overset{―}{v}}_{θ}

力边界 $Γ_{σ}$

\begin{matrix} σ_{r} \cos (v, r) + τ_{r θ} \cos (v, θ) = \overset{―}{R} \\ τ_{r θ} \cos (v, r) + σ_{θ} \cos (v, θ) = \overset{―}{Θ} \end{matrix}

在环向边界 ( $r = c o n s t$ ) 上:

σ_{r} = {\overset{―}{N}}_{r} (θ) \qc τ_{r θ} = {\overset{―}{T}}_{r} (θ)

在径向边界 ( $θ = c o n s t$ ) 上:

σ_{θ} = {\overset{―}{N}}_{θ} (r) \qc τ_{θ r} = {\overset{―}{T}}_{θ} (r)

7.6 轴对称问题

应力函数解法

应力公式

σ_{r} = \frac{1}{r} \pdv ϕ r \qc σ_{θ} = \pdv [2] ϕ r \qc τ_{r θ} = 0

通解

ϕ = A \ln r + B r^{2} \ln r + C r^{2} + D

应力分量

\begin{aligned} σ_{r} & = \frac{A}{r^{2}} + B \pqty 1 + 2 \ln r + 2 C \\ σ_{θ} & = - \frac{A}{r^{2}} + B \pqty 3 + 2 \ln r + 2 C \\ τ_{r θ} & = 0 \end{aligned}

应变分量

\begin{aligned} ε_{r} & = \frac{1}{E} \bqty \pqty 1 + ν \frac{A}{r^{2}} + \pqty 1 - 3 ν B + 2 \pqty 1 - ν \times B \ln r + 2 \pqty 1 - ν C \\ ε_{θ} & = \frac{1}{E} \bqty - \pqty 1 + ν \frac{A}{r^{2}} + \pqty 3 - ν B + 2 \pqty 1 - ν \times B \ln r + 2 \pqty 1 - ν C \\ τ_{r θ} & = 0 \end{aligned}

位移分量

\begin{aligned} u_{r} & = \frac{1}{E} \bqty - \pqty 1 + ν \frac{A}{r} + 2 \pqty 1 - ν B r \ln r - \pqty 1 + ν B r + 2 \pqty 1 - ν C r + I \cos θ + K \sin θ \\ v_{θ} & = \frac{4 B r θ}{E} + H r + K \cos θ - I \sin θ \end{aligned}

限制原点的刚体位移 — $I = K = 0$ ; 位移单值 — $B = 0$ ; 限制刚体转动 — $H = 0$

位移解法

限制原点的刚体位移和转动, 通解为

u_{r} = C_{1} r + \frac{C_{2}}{r} \qc v_{θ} = 0