有限元方法问答 06：轴对称单元

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L8 — Axisymmetric Elements（轴对称单元）

Q43 🔴 轴对称的四个必要条件

EN: For a problem to be treated as axisymmetric, ALL four conditions must be satisfied:

Geometry — the structure is a solid of revolution (rotationally symmetric about an axis)
Material properties — axisymmetric (same in all circumferential directions)
Boundary conditions — axisymmetric (no variation in $\theta$ )
Loading — axisymmetric (no variation in $\theta$ )

If any condition is violated, a full 3D or other simplified model must be used.

中文： 轴对称必须同时满足几何、材料、边界条件、载荷四个方面的轴对称性。任何一个不满足都不能简化为轴对称模型。典型结构：轴、管道、压力容器、回转体。

Q44 🟡 轴对称坐标系统

EN: Axisymmetric problems use cylindrical coordinates $(r, \theta, z)$ . Due to axisymmetry:

$\frac{\partial(\cdot)}{\partial\theta} = 0$ — nothing varies with circumferential angle
$v_\theta = 0$ — no circumferential displacement
Displacements are functions of $r$ and $z$ only: $u = u(r,z)$ , $w = w(r,z)$

中文： 柱坐标 $(r,\theta,z)$ 。轴对称条件： $\partial(\cdot)/\partial\theta=0$ （无不随θ变化）， $v_\theta=0$ （无环向位移）。位移仅为r和z的函数，问题从3D降为2D（r-z平面）。

Q45 ⭐ 环向应变 $\varepsilon_\theta = u/r$

EN: The distinguishing feature of axisymmetric problems is the hoop (circumferential) strain:

$\boxed{\varepsilon_\theta = \frac{u}{r}}$

where $u$ is the radial displacement and $r$ is the radial coordinate. This strain arises because radial expansion/contraction changes the circumference. The axisymmetric strain vector has 4 components (one more than plane problems):

$\boldsymbol{\varepsilon} = \begin{Bmatrix} \varepsilon_r \\ \varepsilon_z \\ \varepsilon_\theta \\ \gamma_{rz} \end{Bmatrix}$

中文： 轴对称的标志性特征——环向应变 $\varepsilon_\theta=u/r$ 。当结构径向变形时，周长改变产生环向应变。轴对称有4个应变分量（比平面问题多了 $\varepsilon_\theta$ ）。

Q46 ⭐ 3节点轴对称单元不是常应变单元

EN: Although the 3-node axisymmetric element uses the same shape functions as CST in the $r$ - $z$ plane, it is NOT a constant strain element. The hoop strain $\varepsilon_\theta = u/r$ contains $1/r$ terms (through the shape function $N_i/r = (\alpha_i + \beta_i r + \gamma_i z)/(2Ar)$ ), which vary with radial position. Therefore, the B-matrix is not constant, and the strains vary within the element. This is a common exam trap.

中文： ⚠️ 高频易错——3节点轴对称单元虽然在r-z平面看像CST三角形，但不是常应变单元！因为 $\varepsilon_\theta=u/r$ 含有 $1/r$ 项，导致B矩阵随r变化，应变在单元内不是常数。这是与平面CST的本质区别。

Q47 🟡 轴对称D矩阵（4×4）

EN: The axisymmetric constitutive matrix (4×4):

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

Corresponding stress and strain vectors: $\{\sigma_r, \sigma_z, \sigma_\theta, \tau_{rz}\}^T$ and $\{\varepsilon_r, \varepsilon_z, \varepsilon_\theta, \gamma_{rz}\}^T$ .

中文： 轴对称D矩阵为4×4，含 $(1-\nu)/((1+\nu)(1-2\nu))$ 因子（与3D问题类似）。

Q48 🔴 轴对称刚度积分中的 $2\pi r$

EN: The volume element in axisymmetric coordinates is $dV = r\,dr\,d\theta\,dz$ . Integrating over $\theta$ from 0 to $2\pi$ gives:

$dV = 2\pi r \, dA \quad \text{(where } dA = dr\,dz \text{ is the area in the r-z plane)}$

Therefore, the element stiffness integral for axisymmetric elements is:

$\boxed{\mathbf{k}_e = \int_{A_e} 2\pi r \, \mathbf{B}^T \mathbf{D} \mathbf{B} \, dA}$

This is fundamentally different from plane problems ( $t\,dA$ ). The $r$ inside the integral makes analytical integration difficult, so numerical integration is generally required.

中文： ⚠️ 轴对称的刚度积分中有 $2\pi r$ （而不是平面问题的 $t$ ）！体积微元 $dV = 2\pi r\,dA$ 。这个额外的 $r$ 因子使得即使B矩阵近似常数，积分也不是简单的面积乘积，通常需要数值积分。

Q49 🔴 轴对称与平面CST的全面对比

EN:

Feature	Plane CST	3-Node Axisymmetric
Coordinate system	$x$ - $y$ (Cartesian)	$r$ - $z$ (Cylindrical)
Strain components	3 ( $\varepsilon_x, \varepsilon_y, \gamma_{xy}$ )	4 (adds $\varepsilon_\theta$ )
Hoop strain	None	$\varepsilon_\theta = u/r$
D-matrix	3×3	4×4
Strain variation	Constant	NOT constant ( $1/r$ terms)
Volume element	$t\,dA$	$2\pi r\,dA$
Numerical integration	Not needed for CST	Generally required

中文： 这个对比表是必背内容。最关键的区别：轴对称有环向应变，应变不是常数，积分含 $2\pi r$ 。