弹性力学:应变分析

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弹性力学:应变分析 [2025/12/05 13:08] – [5.1. 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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-==== 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:**
-    $$ \frac{\partial^2 \varepsilon_x}{\partial y^2} = \frac{\partial^2}{\partial y^2} \left( \frac{\partial u}{\partial x} \right) = \frac{\partial^3 u}{\partial x \partial y^2} \quad \dots (1) $$+ 
 +$$ \frac{\partial^2 \varepsilon_x}{\partial y^2} = \frac{\partial^2}{\partial y^2} \left( \frac{\partial u}{\partial x} \right) = \frac{\partial^3 u}{\partial x \partial y^2} \quad \dots (1) $$
  
 2.  **对 $\varepsilon_y$ 关于 $x$ 求两次偏导:** 2.  **对 $\varepsilon_y$ 关于 $x$ 求两次偏导:**
-    $$ \frac{\partial^2 \varepsilon_y}{\partial x^2} = \frac{\partial^2}{\partial x^2} \left( \frac{\partial v}{\partial y} \right) = \frac{\partial^3 v}{\partial y \partial x^2} \quad \dots (2) $$+ 
 +$$ \frac{\partial^2 \varepsilon_y}{\partial x^2} = \frac{\partial^2}{\partial x^2} \left( \frac{\partial v}{\partial y} \right) = \frac{\partial^3 v}{\partial y \partial x^2} \quad \dots (2) $$
  
 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:**
-    $$ \frac{\partial^2 \gamma_{xy}}{\partial x \partial y} = \frac{\partial^2}{\partial x \partial y} \left( \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \right) = \frac{\partial^3 v}{\partial y \partial x^2} + \frac{\partial^3 u}{\partial x \partial y^2} \quad \dots (3) $$+ 
 +$$ \frac{\partial^2 \gamma_{xy}}{\partial x \partial y} = \frac{\partial^2}{\partial x \partial y} \left( \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \right) = \frac{\partial^3 v}{\partial y \partial x^2} + \frac{\partial^3 u}{\partial x \partial y^2} \quad \dots (3) $$
  
 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)。 
 + 
 +=== 5.2.2 结论 ===
  
 即得第一个协调方程: 即得第一个协调方程:
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 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:
 $$ \frac{\partial^2 \varepsilon_y}{\partial z^2} + \frac{\partial^2 \varepsilon_z}{\partial y^2} = \frac{\partial^2 \gamma_{yz}}{\partial y \partial z} $$ $$ \frac{\partial^2 \varepsilon_y}{\partial z^2} + \frac{\partial^2 \varepsilon_z}{\partial y^2} = \frac{\partial^2 \gamma_{yz}}{\partial y \partial z} $$
 +
 $$ \frac{\partial^2 \varepsilon_z}{\partial x^2} + \frac{\partial^2 \varepsilon_x}{\partial z^2} = \frac{\partial^2 \gamma_{zx}}{\partial z \partial x} $$ $$ \frac{\partial^2 \varepsilon_z}{\partial x^2} + \frac{\partial^2 \varepsilon_x}{\partial z^2} = \frac{\partial^2 \gamma_{zx}}{\partial z \partial x} $$
  
-==== 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 ===+$$ \frac{\partial \gamma_{zx}}{\partial y} \frac{\partial^2 u}{\partial z \partial y} + \frac{\partial^2 w}{\partial x \partial y} $$ 
 + 
 +$$ -\frac{\partial \gamma_{yz}}{\partial x} -\frac{\partial^2 w}{\partial y \partial x} - \frac{\partial^2 v}{\partial z \partial x} $$ 
 + 
 +=== 5.3.3 组合与消元 ===
  
 将上述三式相加: 将上述三式相加:
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 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:
 +
 $$ \frac{\partial \gamma_{xy}}{\partial z} + \frac{\partial \gamma_{zx}}{\partial y} - \frac{\partial \gamma_{yz}}{\partial x} = 2 \frac{\partial^2 u}{\partial y \partial z} $$ $$ \frac{\partial \gamma_{xy}}{\partial z} + \frac{\partial \gamma_{zx}}{\partial y} - \frac{\partial \gamma_{yz}}{\partial x} = 2 \frac{\partial^2 u}{\partial y \partial z} $$
  
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 对上式两边同时关于 $x$ 求导: 对上式两边同时关于 $x$ 求导:
 +
 $$ \frac{\partial}{\partial x} \left( -\frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) = 2 \frac{\partial^3 u}{\partial x \partial y \partial z} $$ $$ \frac{\partial}{\partial x} \left( -\frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) = 2 \frac{\partial^3 u}{\partial x \partial y \partial z} $$
  
 对比 (4) 式,得证: 对比 (4) 式,得证:
 +
 $$ 2 \frac{\partial^2 \varepsilon_x}{\partial y \partial z} = \frac{\partial}{\partial x} \left( -\frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) $$ $$ 2 \frac{\partial^2 \varepsilon_x}{\partial y \partial z} = \frac{\partial}{\partial x} \left( -\frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) $$
  
 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:
 +
 $$ 2 \frac{\partial^2 \varepsilon_y}{\partial x \partial z} = \frac{\partial}{\partial y} \left( \frac{\partial \gamma_{yz}}{\partial x} - \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) $$ $$ 2 \frac{\partial^2 \varepsilon_y}{\partial x \partial z} = \frac{\partial}{\partial y} \left( \frac{\partial \gamma_{yz}}{\partial x} - \frac{\partial \gamma_{zx}}{\partial y} + \frac{\partial \gamma_{xy}}{\partial z} \right) $$
 +
 $$ 2 \frac{\partial^2 \varepsilon_z}{\partial x \partial y} = \frac{\partial}{\partial z} \left( \frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} - \frac{\partial \gamma_{xy}}{\partial z} \right) $$ $$ 2 \frac{\partial^2 \varepsilon_z}{\partial x \partial y} = \frac{\partial}{\partial z} \left( \frac{\partial \gamma_{yz}}{\partial x} + \frac{\partial \gamma_{zx}}{\partial y} - \frac{\partial \gamma_{xy}}{\partial z} \right) $$
  

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  • 弹性力学/应变分析.1764911333.txt.gz
  • 最后更改: 2025/12/05 13:08
  • 张叶安