上一篇：有限元方法问答 02：预备知识 · 系列总目录 · 下一篇：有限元方法问答 04：杆与桁架单元

L3 — Direct Stiffness Method（直接刚度法）

Q14 🟡 弹簧单元刚度矩阵

EN: For a spring element with stiffness kk connecting nodes ii and jj:

ke=[k−k−kk],{fifj}=ke{uiuj}\mathbf{k}_e = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix}, \quad \begin{Bmatrix} f_i \\ f_j \end{Bmatrix} = \mathbf{k}_e \begin{Bmatrix} u_i \\ u_j \end{Bmatrix}

Note: a single unconstrained spring has a singular stiffness matrix (rigid-body mode).

中文： 弹簧单元刚度矩阵为 k[1−1−11]k\begin{bmatrix}1&-1\\-1&1\end{bmatrix}。单个未约束弹簧的刚度矩阵不可逆（刚体位移）。

Q15 🟡 直接刚度法的装配思想

EN: The Direct Stiffness Method assembles the global stiffness matrix by:

  1. Writing each element’s local stiffness matrix in terms of its nodal DOFs
  2. Mapping local DOF numbers to global DOF numbers
  3. Adding (superposing) stiffness contributions at shared DOF positions

For example, two springs in series: element 1 (DOFs 1,2) and element 2 (DOFs 2,3). The global K\mathbf{K} is 3×33 \times 3 with K22=k1+k2K_{22} = k_1 + k_2 (both elements contribute to the shared node).

中文： 直接刚度法按"对号入座"原则装配：将每个单元刚度矩阵按节点自由度编号叠加到全局刚度矩阵中。共享节点的刚度贡献相加。这是FEM装配的核心操作。

Q16 🔴 总刚度矩阵的性质（4条）

EN: Properties of the global stiffness matrix K\mathbf{K}:

  1. Symmetry: KT=K\mathbf{K}^T = \mathbf{K} (from Betti’s reciprocal theorem / symmetric material matrix)
  2. Singularity (before BCs): det⁡(K)=0\det(\mathbf{K}) = 0 due to rigid-body modes
  3. Sparsity: Most entries are zero — each node only connects to neighboring nodes
  4. Banded structure: Non-zero entries cluster near the diagonal; bandwidth depends on node numbering

中文： 四大性质——(1)对称性（由本构对称性和Betti互等定理保证）；(2)奇异性（约束前，存在刚体位移）；(3)稀疏性（节点只与相邻节点连接，大量零元素）；(4)带状结构（非零元素集中在主对角线附近，带宽取决于节点编号）。

Q17 🔴 减小刚度矩阵带宽的方法

EN: The bandwidth of K\mathbf{K} can be minimized by optimizing the node numbering scheme. Nodes should be numbered such that the maximum difference between connected node numbers is as small as possible. For elongated structures, number nodes across the shorter dimension first. This reduces computational cost (solution time ∝N⋅B2\propto N \cdot B^2, where BB is the bandwidth).

中文： 优化节点编号顺序可减小带宽。应使相连节点的编号差尽可能小。对于细长结构，沿短方向优先编号。减小带宽可大幅降低求解时间（求解时间正比于 N×B2N \times B^2）。

Q18 🟡 施加位移边界条件的方法

EN: Two common methods for applying displacement BCs:

  1. Partitioning method (划行划列法): Partition K\mathbf{K} into free (ff) and constrained (cc) DOFs. Solve KffUf=Ff−KfcUc\mathbf{K}_{ff}\mathbf{U}_f = \mathbf{F}_f - \mathbf{K}_{fc}\mathbf{U}_c. Then reactions: Fc=KcfUf+KccUc\mathbf{F}_c = \mathbf{K}_{cf}\mathbf{U}_f + \mathbf{K}_{cc}\mathbf{U}_c.

  2. Penalty method (置大数法): Add a very large number CC to the diagonal term KiiK_{ii} and replace FiF_i with C⋅uˉiC \cdot \bar{u}_i for the constrained DOF. This approximately enforces ui≈uˉiu_i \approx \bar{u}_i.

中文： 两种方法——(1)划行划列法：将K按自由度和约束自由度分块，求解约化方程组，再回代求反力；(2)置大数法（罚函数法）：在约束自由度对应的对角线元素上加一个大数C，并修改右端项，近似强制位移条件。

Q19 🔴 施加约束后K为何正定？

EN: After applying sufficient boundary conditions to eliminate all rigid-body modes, the reduced stiffness matrix Kff\mathbf{K}_{ff} becomes positive definite (xTKffx>0\mathbf{x}^T\mathbf{K}_{ff}\mathbf{x} > 0 for any non-zero x\mathbf{x}). This means: (1) the matrix is invertible; (2) the strain energy U=12UTKU>0U = \frac{1}{2}\mathbf{U}^T\mathbf{KU} > 0 for any non-rigid-body deformation; (3) Cholesky decomposition and other efficient solvers can be used.

中文： 消除刚体模态后，Kff\mathbf{K}_{ff} 正定——任何非零位移都产生正应变能（U>0U>0），矩阵可逆，可用Cholesky分解等方法高效求解。正定性也是解的唯一性的保证。

Q20 🟡 支座反力的计算

EN: After solving for unknown displacements Uf\mathbf{U}_f, the reaction forces at constrained DOFs are:

Fc=KcfUf+KccUc\mathbf{F}_c = \mathbf{K}_{cf}\mathbf{U}_f + \mathbf{K}_{cc}\mathbf{U}_c

For zero prescribed displacement (Uc=0\mathbf{U}_c = \mathbf{0}): Fc=KcfUf\mathbf{F}_c = \mathbf{K}_{cf}\mathbf{U}_f.

中文： 反力 = 约束自由度对应的刚度行 × 所有位移。若约束位移为零，则 Fc=KcfUf\mathbf{F}_c = \mathbf{K}_{cf}\mathbf{U}_f（已知位移为零时简化）。

上一篇：有限元方法问答 02：预备知识 · 系列总目录 · 下一篇：有限元方法问答 04：杆与桁架单元