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| 弹性力学:应变分析 [2025/12/05 13:08] – [1. 预备知识：几何方程] 张叶安 | 弹性力学:应变分析 [2025/12/05 13:12] (当前版本) – [3. 第二组方程的证明（空间混合协调）] 张叶安 | ||
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| $$ \gamma_{xy} = \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}, \quad \gamma_{yz} = \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}, \quad \gamma_{zx} = \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} $$ | $$ \gamma_{xy} = \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}, \quad \gamma_{yz} = \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}, \quad \gamma_{zx} = \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} $$ | ||
| - | < | + | |
| **关键数学性质：** 证明过程中利用了混合偏导数与求导顺序无关的性质（假设位移函数连续且二阶可导），即： | **关键数学性质：** 证明过程中利用了混合偏导数与求导顺序无关的性质（假设位移函数连续且二阶可导），即： | ||
| $$ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} $$ | $$ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} $$ | ||
| - | </ | ||
| - | ==== 2. 第一组方程的证明（平面内协调） ==== | + | |
| + | ==== 5.2. 第一组方程的证明（平面内协调） ==== | ||
| 这组方程主要联系平面内的正应变与切应变。我们以 $xOy$ 平面为例进行推导。 | 这组方程主要联系平面内的正应变与切应变。我们以 $xOy$ 平面为例进行推导。 | ||
| - | === 2.1 推导过程 === | + | === 5.2.1 推导过程 === |
| 1. **对 $\varepsilon_x$ 关于 $y$ 求两次偏导：** | 1. **对 $\varepsilon_x$ 关于 $y$ 求两次偏导：** | ||
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| + | $$ \frac{\partial^2 \varepsilon_x}{\partial y^2} = \frac{\partial^2}{\partial y^2} \left( \frac{\partial u}{\partial x} \right) = \frac{\partial^3 u}{\partial x \partial y^2} \quad \dots (1) $$ | ||
| 2. **对 $\varepsilon_y$ 关于 $x$ 求两次偏导：** | 2. **对 $\varepsilon_y$ 关于 $x$ 求两次偏导：** | ||
| - | | + | |
| + | $$ \frac{\partial^2 \varepsilon_y}{\partial x^2} = \frac{\partial^2}{\partial x^2} \left( \frac{\partial v}{\partial y} \right) = \frac{\partial^3 v}{\partial y \partial x^2} \quad \dots (2) $$ | ||
| 3. **对 $\gamma_{xy}$ 关于 $x$ 和 $y$ 各求一次偏导：** | 3. **对 $\gamma_{xy}$ 关于 $x$ 和 $y$ 各求一次偏导：** | ||
| - | | + | |
| + | $$ \frac{\partial^2 \gamma_{xy}}{\partial x \partial y} = \frac{\partial^2}{\partial x \partial y} \left( \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \right) = \frac{\partial^3 v}{\partial y \partial x^2} + \frac{\partial^3 u}{\partial x \partial y^2} \quad \dots (3) $$ | ||
| 4. **联立方程：** | 4. **联立方程：** | ||
| - | 观察 (1)、(2)、(3) 式右端，显然有 (1) + (2) = (3)。 | ||
| - | === 2.2 结论 === | + | 观察 (1)、(2)、(3) 式右端，显然有 (1) + (2) = (3)。 |
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| + | === 5.2.2 结论 === | ||
| 即得第一个协调方程： | 即得第一个协调方程： | ||
| 行 140: | 行 144: | ||
| 利用轮换下标法（$x \to y \to z \to x$），可直接写出另外两个方程： | 利用轮换下标法（$x \to y \to z \to x$），可直接写出另外两个方程： | ||
| $$ \frac{\partial^2 \varepsilon_y}{\partial z^2} + \frac{\partial^2 \varepsilon_z}{\partial y^2} = \frac{\partial^2 \gamma_{yz}}{\partial y \partial z} $$ | $$ \frac{\partial^2 \varepsilon_y}{\partial z^2} + \frac{\partial^2 \varepsilon_z}{\partial y^2} = \frac{\partial^2 \gamma_{yz}}{\partial y \partial z} $$ | ||
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| $$ \frac{\partial^2 \varepsilon_z}{\partial x^2} + \frac{\partial^2 \varepsilon_x}{\partial z^2} = \frac{\partial^2 \gamma_{zx}}{\partial z \partial x} $$ | $$ \frac{\partial^2 \varepsilon_z}{\partial x^2} + \frac{\partial^2 \varepsilon_x}{\partial z^2} = \frac{\partial^2 \gamma_{zx}}{\partial z \partial x} $$ | ||
| - | ==== 3. 第二组方程的证明（空间混合协调） ==== | + | ==== 5.3. 第二组方程的证明（空间混合协调） ==== |
| 这组方程联系一个正应变和三个切应变。我们以 $\varepsilon_x$ 为例进行推导。 | 这组方程联系一个正应变和三个切应变。我们以 $\varepsilon_x$ 为例进行推导。 | ||
| - | === 3.1 构造目标项 === | + | === 5.3.1 构造目标项 === |
| 首先，对 $\varepsilon_x$ 关于 $y$ 和 $z$ 求混合偏导，并乘以 2（为了凑系数）： | 首先，对 $\varepsilon_x$ 关于 $y$ 和 $z$ 求混合偏导，并乘以 2（为了凑系数）： | ||
| $$ 2 \frac{\partial^2 \varepsilon_x}{\partial y \partial z} = 2 \frac{\partial^2}{\partial y \partial z} \left( \frac{\partial u}{\partial x} \right) = 2 \frac{\partial^3 u}{\partial x \partial y \partial z} \quad \dots (4) $$ | $$ 2 \frac{\partial^2 \varepsilon_x}{\partial y \partial z} = 2 \frac{\partial^2}{\partial y \partial z} \left( \frac{\partial u}{\partial x} \right) = 2 \frac{\partial^3 u}{\partial x \partial y \partial z} \quad \dots (4) $$ | ||
| - | === 3.2 构造切应变组合 === | + | === 5.3.2 构造切应变组合 === |
| 我们需要寻找切应变的某种导数组合，使其结果等于 (4) 式。考察以下三项导数： | 我们需要寻找切应变的某种导数组合，使其结果等于 (4) 式。考察以下三项导数： | ||
| - | * $$ \frac{\partial \gamma_{xy}}{\partial z} = \frac{\partial^2 v}{\partial x \partial z} + \frac{\partial^2 u}{\partial y \partial z} $$ | + | $$ \frac{\partial \gamma_{xy}}{\partial z} = \frac{\partial^2 v}{\partial x \partial z} + \frac{\partial^2 u}{\partial y \partial z} $$ |
| - | * $$ \frac{\partial \gamma_{zx}}{\partial y} = \frac{\partial^2 u}{\partial z \partial y} + \frac{\partial^2 w}{\partial x \partial y} $$ | + | |
| - | * $$ -\frac{\partial \gamma_{yz}}{\partial x} = -\frac{\partial^2 w}{\partial y \partial x} - \frac{\partial^2 v}{\partial z \partial x} $$ | + | |
| - | === 3.3 组合与消元 === | + | $$ \frac{\partial \gamma_{zx}}{\partial y} = \frac{\partial^2 u}{\partial z \partial y} + \frac{\partial^2 w}{\partial x \partial y} $$ |
| + | |||
| + | $$ -\frac{\partial \gamma_{yz}}{\partial x} = -\frac{\partial^2 w}{\partial y \partial x} - \frac{\partial^2 v}{\partial z \partial x} $$ | ||
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| + | === 5.3.3 组合与消元 === | ||
| 将上述三式相加： | 将上述三式相加： | ||
| 行 167: | 行 174: | ||
| 由于混合偏导数相等，含 $v$ 和 $w$ 的项相互抵消，只剩下 $u$ 的项： | 由于混合偏导数相等，含 $v$ 和 $w$ 的项相互抵消，只剩下 $u$ 的项： | ||
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| $$ \frac{\partial \gamma_{xy}}{\partial z} + \frac{\partial \gamma_{zx}}{\partial y} - \frac{\partial \gamma_{yz}}{\partial x} = 2 \frac{\partial^2 u}{\partial y \partial z} $$ | $$ \frac{\partial \gamma_{xy}}{\partial z} + \frac{\partial \gamma_{zx}}{\partial y} - \frac{\partial \gamma_{yz}}{\partial x} = 2 \frac{\partial^2 u}{\partial y \partial z} $$ | ||
| 行 172: | 行 180: | ||
| 对上式两边同时关于 $x$ 求导： | 对上式两边同时关于 $x$ 求导： | ||
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| $$ \frac{\partial}{\partial x} \left( -\frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) = 2 \frac{\partial^3 u}{\partial x \partial y \partial z} $$ | $$ \frac{\partial}{\partial x} \left( -\frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) = 2 \frac{\partial^3 u}{\partial x \partial y \partial z} $$ | ||
| 对比 (4) 式，得证： | 对比 (4) 式，得证： | ||
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| $$ 2 \frac{\partial^2 \varepsilon_x}{\partial y \partial z} = \frac{\partial}{\partial x} \left( -\frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) $$ | $$ 2 \frac{\partial^2 \varepsilon_x}{\partial y \partial z} = \frac{\partial}{\partial x} \left( -\frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) $$ | ||
| 利用轮换下标法，可得另外两个方程： | 利用轮换下标法，可得另外两个方程： | ||
| + | |||
| $$ 2 \frac{\partial^2 \varepsilon_y}{\partial x \partial z} = \frac{\partial}{\partial y} \left( \frac{\partial \gamma_{yz}}{\partial x} - \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) $$ | $$ 2 \frac{\partial^2 \varepsilon_y}{\partial x \partial z} = \frac{\partial}{\partial y} \left( \frac{\partial \gamma_{yz}}{\partial x} - \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) $$ | ||
| + | |||
| $$ 2 \frac{\partial^2 \varepsilon_z}{\partial x \partial y} = \frac{\partial}{\partial z} \left( \frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} - \frac{\partial \gamma_{xy}}{\partial z} \right) $$ | $$ 2 \frac{\partial^2 \varepsilon_z}{\partial x \partial y} = \frac{\partial}{\partial z} \left( \frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} - \frac{\partial \gamma_{xy}}{\partial z} \right) $$ | ||
张叶安