有限元方法问答 08：等参单元

上一篇：有限元方法问答 07：建模实践 · 系列总目录 · 下一篇：有限元方法问答 09：数值积分

L9 — Isoparametric Formulation（等参单元）

Q61 ⭐ "等参(isoparametric)"的含义

EN: “Isoparametric” means that the same shape functions are used to interpolate both the geometry (coordinates) AND the displacement field:

Geometry: $x = \sum N_i(s,t) x_i$ , $y = \sum N_i(s,t) y_i$
Displacement: $u = \sum N_i(s,t) u_i$ , $v = \sum N_i(s,t) v_i$

This is the defining characteristic: geometry and displacement use identical interpolation. If geometry uses fewer nodes than displacement → “subparametric”; if more → “superparametric.”

中文： “等参” = 几何坐标插值和位移插值使用同一组形函数。几何： $x=\sum N_i x_i$ , $y=\sum N_i y_i$ ；位移： $u=\sum N_i u_i$ , $v=\sum N_i v_i$ 。这是等参单元的定义性特征。

Q62 ⭐ 4节点四边形形函数

EN: The 4-node quadrilateral shape functions in natural coordinates $(s,t) \in [-1,1] \times [-1,1]$ :

$\boxed{N_i = \frac{1}{4}(1 + s_i s)(1 + t_i t)}$

where $(s_i, t_i)$ are the natural coordinates of node $i$ :

$N_1$ at $(-1,-1)$ : $N_1 = \frac{1}{4}(1-s)(1-t)$
$N_2$ at $(+1,-1)$ : $N_2 = \frac{1}{4}(1+s)(1-t)$
$N_3$ at $(+1,+1)$ : $N_3 = \frac{1}{4}(1+s)(1+t)$
$N_4$ at $(-1,+1)$ : $N_4 = \frac{1}{4}(1-s)(1+t)$

Properties: $N_i = 1$ at node $i$ , $N_i = 0$ at all other nodes; $\sum N_i = 1$ .

中文： 四节点四边形形函数为 $N_i=\frac{1}{4}(1+s_i s)(1+t_i t)$ 。自然坐标范围 $[-1,1]\times[-1,1]$ 。 $N_i$ 在节点i为1、其他节点为0、和为1。

Q63 ⭐ Jacobian矩阵

EN: The Jacobian matrix $\mathbf{J}$ maps derivatives from natural coordinates $(s,t)$ to global coordinates $(x,y)$ :

$\boxed{\mathbf{J} = \begin{bmatrix} \frac{\partial x}{\partial s} & \frac{\partial y}{\partial s} \\ \frac{\partial x}{\partial t} & \frac{\partial y}{\partial t} \end{bmatrix} = \begin{bmatrix} \sum \frac{\partial N_i}{\partial s}x_i & \sum \frac{\partial N_i}{\partial s}y_i \\ \sum \frac{\partial N_i}{\partial t}x_i & \sum \frac{\partial N_i}{\partial t}y_i \end{bmatrix}}$

The derivative transformation is:

$\begin{Bmatrix} \partial N_i / \partial s \\ \partial N_i / \partial t \end{Bmatrix} = \mathbf{J} \begin{Bmatrix} \partial N_i / \partial x \\ \partial N_i / \partial y \end{Bmatrix} \quad \Rightarrow \quad \begin{Bmatrix} \partial N_i / \partial x \\ \partial N_i / \partial y \end{Bmatrix} = \mathbf{J}^{-1} \begin{Bmatrix} \partial N_i / \partial s \\ \partial N_i / \partial t \end{Bmatrix}$

Area transformation: $dx\,dy = |\mathbf{J}|\,ds\,dt$ .

中文： Jacobian矩阵 $\mathbf{J}$ 将自然坐标导数映射为全局坐标导数。 $\mathbf{J}$ 的每个元素 = $\sum\frac{\partial N_i}{\partial s}x_i$ 等。导数变换：全局导数 = $\mathbf{J}^{-1}$ × 自然导数。面积变换： $dxdy=|\mathbf{J}|dsdt$ 。

Q64 ⭐ $|\mathbf{J}| > 0$ 的有效性条件

EN: For a valid isoparametric element, the Jacobian determinant must be positive everywhere:

$|\mathbf{J}| > 0 \quad \text{at all points } (s,t) \in [-1,1]^2$

$|\mathbf{J}| \leq 0$ indicates:

The element is excessively distorted or concave
One-to-one mapping between $(s,t)$ and $(x,y)$ is destroyed
Interior angles exceed 180°
Node ordering is wrong (not counterclockwise)

For convex quadrilaterals with proper node ordering, $|\mathbf{J}| > 0$ is guaranteed.

中文： ⚠️ $|\mathbf{J}|>0$ 是等参单元有效的充要条件。 $|\mathbf{J}|\leq0$ 意味着单元过度扭曲、凹四边形、节点顺序错误。对于凸四边形和正确的逆时针顺序， $|\mathbf{J}|>0$ 自动满足。

Q65 🔴 $|\mathbf{J}| < 0$ 的物理含义

EN: When $|\mathbf{J}| < 0$ , the natural-to-global coordinate mapping is no longer one-to-one. This means:

A single point in natural coordinates maps to multiple points in global coordinates (or vice versa)
The element has “folded over” onto itself
Interior angles exceed 180° (concave quadrilateral)
The element stiffness matrix may contain negative or erroneous entries
Solution results from such elements are meaningless

Prevention: Maintain convex element shapes, use proper node ordering (counterclockwise), avoid excessive distortion.

中文： $|\mathbf{J}|<0$ 意味着映射不再是一对一的——单元"折叠"了，内角超过180°（凹四边形），刚度矩阵可能含负值或错误项，结果毫无意义。预防：保持凸形状、正确节点顺序（逆时针）、避免过度扭曲。

Q66 🔴 8节点四边形角节点形函数的构造方法

EN: For 8-node quadrilateral (Q8, serendipity element), the corner shape functions are constructed through a correction procedure:

Mid-edge shape functions (for nodes 5-8):
- $N_5 = \frac{1}{2}(1-s^2)(1-t)$ (between nodes 1 and 2)
- $N_6 = \frac{1}{2}(1+s)(1-t^2)$ (between nodes 2 and 3)
- $N_7 = \frac{1}{2}(1-s^2)(1+t)$ (between nodes 3 and 4)
- $N_8 = \frac{1}{2}(1-s)(1-t^2)$ (between nodes 4 and 1)
Start with the standard 4-node bilinear shape functions: $\hat{N}_i = \frac{1}{4}(1+s_i s)(1+t_i t)$
Correct each corner function by subtracting half of the adjacent mid-edge functions:

$N_1 = \hat{N}_1 - \frac{1}{2}N_5 - \frac{1}{2}N_8$

and similarly for $N_2, N_3, N_4$ .

This ensures corner nodes still satisfy $N_i = 1$ at node $i$ and $N_i = 0$ at all other nodes.

中文： Q8角节点形函数不能直接用双线性形函数，需要修正：角节点函数 = 4节点双线性形函数 − 1/2×(相邻两边中点形函数)。这保证了角节点处 $N_i=1$ 、其他节点处 $N_i=0$ 的性质。