有限元方法刚度矩阵K详细表达式推导

三维刚度矩阵K的详细推导

前置弹性力学矩阵知识

几何变换矩阵S

$\mathbf{S} = \begin{bmatrix} \frac{\partial}{\partial x} & 0 & 0 \\ 0 & \frac{\partial}{\partial y} & 0 \\ 0 & 0 & \frac{\partial}{\partial z} \\ \frac{\partial}{\partial y} & \frac{\partial}{\partial x} & 0 \\ 0 & \frac{\partial}{\partial z} & \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} & 0 & \frac{\partial}{\partial x} \end{bmatrix}\tag{1.1}$

位移矩阵u

$\mathbf{u} = \begin{Bmatrix} u \\ v \\ w \end{Bmatrix}\tag{1.2}$

应变矩阵$\epsilon$

由(1.1)式和(1.2)式导出：

$\boldsymbol{\varepsilon} = \left\{ \begin{array}{c} \varepsilon_x \\ \varepsilon_y \\ \varepsilon_z \\ \gamma_{xy} \\ \gamma_{yz} \\ \gamma_{zx} \end{array} \right\} = \left\{ \begin{array}{c} \dfrac{\partial u}{\partial x} \\ \dfrac{\partial v}{\partial y} \\ \dfrac{\partial w}{\partial z} \\ \dfrac{\partial u}{\partial y} + \dfrac{\partial v}{\partial x} \\ \dfrac{\partial v}{\partial z} + \dfrac{\partial w}{\partial y} \\ \dfrac{\partial w}{\partial x} + \dfrac{\partial u}{\partial z} \end{array} \right\} = \mathbf{S}\mathbf{u} \tag{1.3}$

节点位移a

$\begin{aligned} \mathbf{a}^e &= \left\{ \begin{array}{c} \mathbf{a}_i \\ \mathbf{a}_j \\ \mathbf{a}_m \\ \mathbf{a}_p \\ \end{array} \right\} \\ \mathbf{a}_i &= \left\{ \begin{array}{c} u_i \\ v_i \\ w_i \\ \end{array} \right\} \quad \text{etc.} \end{aligned}\tag{1.4}$

形函数N

设任意点位移为

$\begin{aligned} u &= \alpha_1 + \alpha_2 x + \alpha_3 y + \alpha_4 z \\ v &= \alpha_5 + \alpha_6 x + \alpha_7 y + \alpha_8 z \\ w &= \alpha_9 + \alpha_{10} x + \alpha_{11} y + \alpha_{12} z \end{aligned}\tag{1.5}$

每个节点的空间位置也已知：

$X_i=(x_i,y_i,z_i), etc\tag{1.6}$

对于$u$，可以得到

$\begin{aligned} u_i &= \alpha_1 + \alpha_2 x_i + \alpha_3 y_i + \alpha_4 z_i \\ u_j &= \alpha_1 + \alpha_2 x_j + \alpha_3 y_j + \alpha_4 z_j \\ u_m &= \alpha_1 + \alpha_2 x_m + \alpha_3 y_m + \alpha_4 z_m \\ u_p &= \alpha_1 + \alpha_2 x_p + \alpha_3 y_p + \alpha_4 z_p \end{aligned}\tag{1.7}$

写成矩阵形式

$\underbrace{ \begin{bmatrix} 1 & x_i & y_i & z_i \\ 1 & x_j & y_j & z_j \\ 1 & x_m & y_m & z_m \\ 1 & x_p & y_p & z_p \end{bmatrix}}_{\mathbf{X} \in \mathbb{R}^{4 \times 4}} \cdot \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \\ \alpha_4 \end{bmatrix} = \begin{bmatrix} u_i \\ u_j \\ u_m \\ u_p \end{bmatrix}\tag{1.8}$

可以解出

$\begin{aligned} u &= \frac{1}{6V} \Big[ (a_i + b_i x + c_i y + d_i z) u_i + (a_j + b_j x + c_j y + d_j z) u_j \\ &\quad + (a_m + b_m x + c_m y + d_m z) u_m + (a_p + b_p x + c_p y + d_p z) u_p \Big] \\ \\ v &= \frac{1}{6V} \Big[ (a_i + b_i x + c_i y + d_i z) v_i + (a_j + b_j x + c_j y + d_j z) v_j \\ &\quad + (a_m + b_m x + c_m y + d_m z) v_m + (a_p + b_p x + c_p y + d_p z) v_p \Big] \\ \\ w &= \frac{1}{6V} \Big[ (a_i + b_i x + c_i y + d_i z) w_i + (a_j + b_j x + c_j y + d_j z) w_j \\ &\quad + (a_m + b_m x + c_m y + d_m z) w_m + (a_p + b_p x + c_p y + d_p z) w_p \Big] \end{aligned}\tag{1.9}$

其中

$6V = \det \begin{vmatrix} 1 & x_i & y_i & z_i \\ 1 & x_j & y_j & z_j \\ 1 & x_m & y_m & z_m \\ 1 & x_p & y_p & z_p \\ \end{vmatrix}\tag{1.10}$

$\begin{aligned} a_i &= \det \begin{vmatrix} x_j & y_j & z_j \\ x_m & y_m & z_m \\ x_p & y_p & z_p \\ \end{vmatrix} \\ b_i &= -\det \begin{vmatrix} 1 & y_j & z_j \\ 1 & y_m & z_m \\ 1 & y_p & z_p \\ \end{vmatrix} \\ c_i &= -\det \begin{vmatrix} x_j & 1 & z_j \\ x_m & 1 & z_m \\ x_p & 1 & z_p \\ \end{vmatrix} \\ d_i &= -\det \begin{vmatrix} x_j & y_j & 1 \\ x_m & y_m & 1 \\ x_p & y_p & 1 \\ \end{vmatrix} \end{aligned}\tag{1.11}$

其他的系数由更换角标得到，按照右手准则按顺序变换。

可以得到形函数$N_i$:

$N_i=\frac{1}{6V} \begin{bmatrix} a_i + b_i x + c_i y + d_i z & 0 & 0 \\ 0 & a_i + b_i x + c_i y + d_i z & 0 \\ 0 & 0 & a_i + b_i x + c_i y + d_i z \\ \end{bmatrix}\tag{1.12}$

形函数插值表达式$u=Na^e$

$\boldsymbol{\varepsilon} = \mathbf{S} \mathbf{N} \mathbf{a}^e = \mathbf{B} \mathbf{a}^e = \begin{bmatrix} \mathbf{B}_i & \mathbf{B}_j & \mathbf{B}_m & \mathbf{B}_p \end{bmatrix} \mathbf{a}^e\tag{1.13}$

应变-位移矩阵B

由(1.13)式导出，即$B=SN$

$\mathbf{B}_i = \begin{bmatrix} \frac{\partial N_i}{\partial x} & 0 & 0 \\ 0 & \frac{\partial N_i}{\partial y} & 0 \\ 0 & 0 & \frac{\partial N_i}{\partial z} \\ \frac{\partial N_i}{\partial y} & \frac{\partial N_i}{\partial x} & 0 \\ 0 & \frac{\partial N_i}{\partial z} & \frac{\partial N_i}{\partial y} \\ \frac{\partial N_i}{\partial z} & 0 & \frac{\partial N_i}{\partial x} \end{bmatrix} = \frac{1}{6V} \begin{bmatrix} b_i & 0 & 0 \\ 0 & c_i & 0 \\ 0 & 0 & d_i \\ c_i & b_i & 0 \\ 0 & d_i & c_i \\ d_i & 0 & b_i \end{bmatrix}\tag{1.14}$

其他块直接更换角标即可。

各向同性弹性矩阵D

$\mathbf{D} = \frac{E}{(1+\nu)(1-2\nu)} \begin{bmatrix} 1-\nu & \nu & \nu & 0 & 0 & 0 \\ \nu & 1-\nu & \nu & 0 & 0 & 0 \\ \nu & \nu & 1-\nu & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1-2\nu}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1-2\nu}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1-2\nu}{2} \end{bmatrix} \tag{1.15}$

刚度矩阵K表达式推导

基于虚功原理的推导

为使节点力在静力上等效于真实边界应力与体力，最简单的方法是对节点施加一个任意的虚位移 $\delta \mathbf{a}^e$，然后使该虚位移下的外功等于内功。

1. 虚位移与应变表达式

$\delta \mathbf{u} = \mathbf{N} \, \delta \mathbf{a}^e, \quad \delta \boldsymbol{\varepsilon} = \mathbf{B} \, \delta \mathbf{a}^e \tag{2.1}$

2. 外力虚功

$\delta \mathbf{a}^{eT} \mathbf{q}^e \tag{2.2}$

3. 内力虚功（单位体积）

$\delta \boldsymbol{\varepsilon}^T \boldsymbol{\sigma} - \delta \mathbf{u}^T \mathbf{b} \tag{2.3}$

代入式 (2.1) 得：

$\delta \mathbf{a}^{eT} \left( \mathbf{B}^T \boldsymbol{\sigma} - \mathbf{N}^T \mathbf{b} \right) \tag{2.4}$

4. 平衡条件：外功 = 内功

$\delta \mathbf{a}^{eT} \mathbf{q}^e = \delta \mathbf{a}^{eT} \left( \int_{V^e} \mathbf{B}^T \boldsymbol{\sigma} \, dV - \int_{V^e} \mathbf{N}^T \mathbf{b} \, dV \right) \tag{2.5}$

因为该等式对任意虚位移成立，得：

$\boxed{\mathbf{q}^e = \int_{V^e} \mathbf{B}^T \boldsymbol{\sigma} \, dV - \int_{V^e} \mathbf{N}^T \mathbf{b} \, dV } \tag{2.6}$

5. 引入线性本构关系

若采用线性应力-应变关系：

$\boldsymbol{\sigma} = \mathbf{D}(\boldsymbol{\varepsilon} - \boldsymbol{\varepsilon}_0) + \boldsymbol{\sigma}_0 \tag{2.7}$

则有：

$\boxed{ \mathbf{q}^e = \mathbf{K}^e \mathbf{a}^e + \mathbf{f}^e } \tag{2.8}$

其中刚度矩阵为：

$\boxed{ \mathbf{K}^e = \int_{V^e} \mathbf{B}^T \mathbf{D} \mathbf{B} \, dV } \tag{2.9a}$

等效节点力为：

$\boxed{ \mathbf{f}^e = - \int_{V^e} \mathbf{N}^T \mathbf{b} \, dV - \int_{V^e} \mathbf{B}^T \mathbf{D} \boldsymbol{\varepsilon}_0 \, dV + \int_{V^e} \mathbf{B}^T \boldsymbol{\sigma}_0 \, dV } \tag{2.9b}$

总势能原理的变分推导（最小势能原理）（从虚功原理出发）

1 虚功原理变体

当虚位移 $\delta \mathbf{a}$、$\delta \mathbf{u}$、$\delta \boldsymbol{\varepsilon}$ 被视为实际量的变分（variation）时，虚功原理可改写为：

$\delta \left( \mathbf{a}^T \mathbf{r} + \int_V \mathbf{u}^T \mathbf{b} \, dV + \int_A \mathbf{u}^T \bar{\mathbf{t}} \, dA \right) = - \delta W \tag{2.10}$

其中，$W$ 是外载荷的势能（potential energy of external loads）。当 $\mathbf{r}, \mathbf{b}, \bar{\mathbf{t}}$ 与位移无关时（即保守系统），上式成立。

2 内功变分项（应变能）

对于弹性材料，虚功原理中的内力项变为：

$\delta U = \int_V \delta \boldsymbol{\varepsilon}^T \boldsymbol{\sigma} \, dV \tag{2.11}$

其中 $U$ 是应变能（strain energy）。对于线弹性材料，代入本构关系：

$\boldsymbol{\sigma} = \mathbf{D} (\boldsymbol{\varepsilon} - \boldsymbol{\varepsilon}_0) + \boldsymbol{\sigma}_0$

得应变能表达式：

$U = \frac{1}{2} \int_V \boldsymbol{\varepsilon}^T \mathbf{D} \boldsymbol{\varepsilon} \, dV - \int_V \boldsymbol{\varepsilon}^T \mathbf{D} \boldsymbol{\varepsilon}_0 \, dV + \int_V \boldsymbol{\varepsilon}^T \boldsymbol{\sigma}_0 \, dV \tag{2.12}$

当 $\mathbf{D}$ 是对称矩阵时，以上变分结果才是正确的，这是单值应变能 $U$ 存在的必要条件。

3 总势能泛函及其驻值条件

因此，虚功原理可以简洁表示为：

$\delta (U + W) = \delta (\Pi) = 0 \tag{2.13}$

其中，$\Pi$ 是总势能（total potential energy）。

对于给定位移模式下的有限元系统，令总势能对节点参数 $\mathbf{a}$ 的偏导为零，即可求解平衡方程：

$\frac{\partial \Pi}{\partial \mathbf{a}} = \begin{Bmatrix} \frac{\partial \Pi}{\partial a_1} \\ \frac{\partial \Pi}{\partial a_2} \\ \vdots \end{Bmatrix} = \mathbf{0} \tag{2.14}$

这就是有限元系统的离散平衡方程。

在稳定弹性系统中，总势能不仅是驻值点（stationary），而且是最小值。
因此，有限元法的本质是：在预设位移模式下，寻找总势能的最小值解。

可以导出和上一节(2.8)的方程

刚度矩阵K具体形式

很容易得到K矩阵的维度为$12\times12$。

随着单元尺寸减小，(2.9a)式可以简化为

$\mathbf{K}^e = \mathbf{B}^\mathrm{T} \mathbf{D} \mathbf{B} V^e$

由于B是分块矩阵，故可以写出K的分块形式：

$\mathbf{K}^e=V^e \begin{bmatrix} \mathbf{B}_i^\mathrm{T} \mathbf{D} \mathbf{B}_i & \mathbf{B}_i^\mathrm{T} \mathbf{D} \mathbf{B}_j & \mathbf{B}_i^\mathrm{T} \mathbf{D} \mathbf{B}_m & \mathbf{B}_i^\mathrm{T} \mathbf{D} \mathbf{B}_p \\ \mathbf{B}_j^\mathrm{T} \mathbf{D} \mathbf{B}_i & \mathbf{B}_j^\mathrm{T} \mathbf{D} \mathbf{B}_j & \mathbf{B}_j^\mathrm{T} \mathbf{D} \mathbf{B}_m & \mathbf{B}_j^\mathrm{T} \mathbf{D} \mathbf{B}_p \\ \mathbf{B}_m^\mathrm{T} \mathbf{D} \mathbf{B}_i & \mathbf{B}_m^\mathrm{T} \mathbf{D} \mathbf{B}_j & \mathbf{B}_m^\mathrm{T} \mathbf{D} \mathbf{B}_m & \mathbf{B}_m^\mathrm{T} \mathbf{D} \mathbf{B}_p \\ \mathbf{B}_p^\mathrm{T} \mathbf{D} \mathbf{B}_i & \mathbf{B}_p^\mathrm{T} \mathbf{D} \mathbf{B}_j & \mathbf{B}_p^\mathrm{T} \mathbf{D} \mathbf{B}_m & \mathbf{B}_p^\mathrm{T} \mathbf{D} \mathbf{B}_p \end{bmatrix} =\begin{bmatrix} \mathbf{K}_{11} & \mathbf{K}_{12} & \mathbf{K}_{13} & \mathbf{K}_{14}\\ \mathbf{K}_{21} & \mathbf{K}_{22} & \mathbf{K}_{23} & \mathbf{K}_{24}\\ \mathbf{K}_{31} & \mathbf{K}_{32} & \mathbf{K}_{33} & \mathbf{K}_{34}\\ \mathbf{K}_{41} & \mathbf{K}_{42} & \mathbf{K}_{43} & \mathbf{K}_{44}\\ \end{bmatrix} \tag{2.15}$

又由于D是对称矩阵，因此有：

$(\mathbf{B_i}^\mathrm{T} \mathbf{D} \mathbf{B_j})^\mathrm{T}=\mathbf{B_j}^\mathrm{T} \mathbf{D} \mathbf{B_i} \tag{2.16}$

$\mathbf{K}_{11}$:

$\mathbf{K}_{11} = \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2\nu (b_i^2 + c_i^2 + d_i^2) - 2 b_i^2 - c_i^2 - d_i^2 & - b_i c_i & - b_i d_i \\ - b_i c_i & 2 \nu (b_i^2 + c_i^2 + d_i^2) - b_i^2 - 2 c_i^2 - d_i^2 & - c_i d_i \\ - b_i d_i & - c_i d_i & 2 \nu (b_i^2 + c_i^2 + d_i^2) - b_i^2 - c_i^2 - 2 d_i^2 \end{bmatrix}\tag{2.17a}$

$\mathbf{K}_{22}$:

$\mathbf{K}_{22} = \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2\nu (b_j^2 + c_j^2 + d_j^2) - 2 b_j^2 - c_j^2 - d_j^2 & - b_j c_j & - b_j d_j \\ - b_j c_j & 2 \nu (b_j^2 + c_j^2 + d_j^2) - b_j^2 - 2 c_j^2 - d_j^2 & - c_j d_j \\ - b_j d_j & - c_j d_j & 2 \nu (b_j^2 + c_j^2 + d_j^2) - b_j^2 - c_j^2 - 2 d_j^2 \end{bmatrix}\tag{2.17b}$

$\mathbf{K}_{33}$:

$\mathbf{K}_{33} = \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2\nu (b_m^2 + c_m^2 + d_m^2) - 2 b_m^2 - c_m^2 - d_m^2 & - b_m c_m & - b_m d_m \\ - b_m c_m & 2 \nu (b_m^2 + c_m^2 + d_m^2) - b_m^2 - 2 c_m^2 - d_m^2 & - c_m d_m \\ - b_m d_m & - c_m d_m & 2 \nu (b_m^2 + c_m^2 + d_m^2) - b_m^2 - c_m^2 - 2 d_m^2 \end{bmatrix}\tag{2.17c}$

$\mathbf{K}_{44}$:

$\mathbf{K}_{44} = \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2\nu (b_p^2 + c_p^2 + d_p^2) - 2 b_p^2 - c_p^2 - d_p^2 & - b_p c_p & - b_p d_p \\ - b_p c_p & 2 \nu (b_p^2 + c_p^2 + d_p^2) - b_p^2 - 2 c_p^2 - d_p^2 & - c_p d_p \\ - b_p d_p & - c_p d_p & 2 \nu (b_p^2 + c_p^2 + d_p^2) - b_p^2 - c_p^2 - 2 d_p^2 \end{bmatrix}\tag{2.17d}$

$\mathbf{K}_{12}$和$\mathbf{K}_{21}$:

$\mathbf{K}_{12} = \mathbf{K}_{21} = - \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2 b_i b_j + c_i c_j + d_i d_j - 2 \nu (b_i b_j + c_i c_j + d_i d_j) & b_j c_i + 2 \nu b_i c_j - 2 \nu b_j c_i & b_j d_i + 2 \nu b_i d_j - 2 \nu b_j d_i \\[1ex] b_i c_j - 2 \nu b_i c_j + 2 \nu b_j c_i & b_i b_j + 2 c_i c_j + d_i d_j - 2 \nu (b_i b_j + c_i c_j + d_i d_j) & c_j d_i + 2 \nu c_i d_j - 2 \nu c_j d_i \\[1ex] b_i d_j - 2 \nu b_i d_j + 2 \nu b_j d_i & c_i d_j - 2 \nu c_i d_j + 2 \nu c_j d_i & b_i b_j + c_i c_j + 2 d_i d_j - 2 \nu (b_i b_j + c_i c_j + d_i d_j) \end{bmatrix}\tag{2.17e}$

$\mathbf{K}_{13}$和$\mathbf{K}_{31}$:

$\mathbf{K}_{13} = \mathbf{K}_{31} = - \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2 b_i b_m + c_i c_m + d_i d_m - 2 \nu (b_i b_m + c_i c_m + d_i d_m) & b_m c_i + 2 \nu b_i c_m - 2 \nu b_m c_i & b_m d_i + 2 \nu b_i d_m - 2 \nu b_m d_i \\[1ex] b_i c_m - 2 \nu b_i c_m + 2 \nu b_m c_i & b_i b_m + 2 c_i c_m + d_i d_m - 2 \nu (b_i b_m + c_i c_m + d_i d_m) & c_m d_i + 2 \nu c_i d_m - 2 \nu c_m d_i \\[1ex] b_i d_m - 2 \nu b_i d_m + 2 \nu b_m d_i & c_i d_m - 2 \nu c_i d_m + 2 \nu c_m d_i & b_i b_m + c_i c_m + 2 d_i d_m - 2 \nu (b_i b_m + c_i c_m + d_i d_m) \end{bmatrix}\tag{2.17f}$

$\mathbf{K}_{14}$ 和 $\mathbf{K}_{41}$:

$\mathbf{K}_{14} = \mathbf{K}_{41} = - \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2 b_i b_p + c_i c_p + d_i d_p - 2 \nu (b_i b_p + c_i c_p + d_i d_p) & b_p c_i + 2 \nu b_i c_p - 2 \nu b_p c_i & b_p d_i + 2 \nu b_i d_p - 2 \nu b_p d_i \\[1ex] b_i c_p - 2 \nu b_i c_p + 2 \nu b_p c_i & b_i b_p + 2 c_i c_p + d_i d_p - 2 \nu (b_i b_p + c_i c_p + d_i d_p) & c_p d_i + 2 \nu c_i d_p - 2 \nu c_p d_i \\[1ex] b_i d_p - 2 \nu b_i d_p + 2 \nu b_p d_i & c_i d_p - 2 \nu c_i d_p + 2 \nu c_p d_i & b_i b_p + c_i c_p + 2 d_i d_p - 2 \nu (b_i b_p + c_i c_p + d_i d_p) \end{bmatrix}\tag{2.17g}$

$\mathbf{K}_{23}$ 和 $\mathbf{K}_{32}$:

$\mathbf{K}_{23} = \mathbf{K}_{32} = - \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2 b_j b_m + c_j c_m + d_j d_m - 2 \nu (b_j b_m + c_j c_m + d_j d_m) & b_m c_j + 2 \nu b_j c_m - 2 \nu b_m c_j & b_m d_j + 2 \nu b_j d_m - 2 \nu b_m d_j \\[1ex] b_j c_m - 2 \nu b_j c_m + 2 \nu b_m c_j & b_j b_m + 2 c_j c_m + d_j d_m - 2 \nu (b_j b_m + c_j c_m + d_j d_m) & c_m d_j + 2 \nu c_j d_m - 2 \nu c_m d_j \\[1ex] b_j d_m - 2 \nu b_j d_m + 2 \nu b_m d_j & c_j d_m - 2 \nu c_j d_m + 2 \nu c_m d_j & b_j b_m + c_j c_m + 2 d_j d_m - 2 \nu (b_j b_m + c_j c_m + d_j d_m) \end{bmatrix}\tag{2.17h}$

$\mathbf{K}_{24}$ 和 $\mathbf{K}_{42}$:

$\mathbf{K}_{24} = \mathbf{K}_{42} = - \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2 b_j b_p + c_j c_p + d_j d_p - 2 \nu (b_j b_p + c_j c_p + d_j d_p) & b_p c_j + 2 \nu b_j c_p - 2 \nu b_p c_j & b_p d_j + 2 \nu b_j d_p - 2 \nu b_p d_j \\[1ex] b_j c_p - 2 \nu b_j c_p + 2 \nu b_p c_j & b_j b_p + 2 c_j c_p + d_j d_p - 2 \nu (b_j b_p + c_j c_p + d_j d_p) & c_p d_j + 2 \nu c_j d_p - 2 \nu c_p d_j \\[1ex] b_j d_p - 2 \nu b_j d_p + 2 \nu b_p d_j & c_j d_p - 2 \nu c_j d_p + 2 \nu c_p d_j & b_j b_p + c_j c_p + 2 d_j d_p - 2 \nu (b_j b_p + c_j c_p + d_j d_p) \end{bmatrix}\tag{2.17i}$

$\mathbf{K}_{34}$ 和 $\mathbf{K}_{43}$:

$\mathbf{K}_{34} = \mathbf{K}_{43} = - \frac{E}{72 V (2 \nu^2 + \nu - 1)} \begin{bmatrix} 2 b_m b_p + c_m c_p + d_m d_p - 2 \nu (b_m b_p + c_m c_p + d_m d_p) & b_p c_m + 2 \nu b_m c_p - 2 \nu b_p c_m & b_p d_m + 2 \nu b_m d_p - 2 \nu b_p d_m \\[1ex] b_m c_p - 2 \nu b_m c_p + 2 \nu b_p c_m & b_m b_p + 2 c_m c_p + d_m d_p - 2 \nu (b_m b_p + c_m c_p + d_m d_p) & c_p d_m + 2 \nu c_m d_p - 2 \nu c_p d_m \\[1ex] b_m d_p - 2 \nu b_m d_p + 2 \nu b_p d_m & c_m d_p - 2 \nu c_m d_p + 2 \nu c_p d_m & b_m b_p + c_m c_p + 2 d_m d_p - 2 \nu (b_m b_p + c_m c_p + d_m d_p) \end{bmatrix}\tag{2.17j}$

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