振动近似计算方法

邓克利法

\[ \boldsymbol{M} \ddot{\boldsymbol{x}} + \boldsymbol{K} \boldsymbol{x} = \boldsymbol{0} \implies \underset{结构矩阵 \, \boldsymbol{D}}{\underbrace{(\boldsymbol{K}^{-1} \boldsymbol{M})}} \ddot{\boldsymbol{x}} + \boldsymbol{x} = \boldsymbol{0} \]

代入 \(x = A \sin(\omega t + \theta)\)，

\[ (-\omega^2 \boldsymbol{D} + \boldsymbol{I}) \boldsymbol{A} = \boldsymbol{0} \]

令 \(\bar{\lambda} = 1 / \omega^2\)，得到特征方程

\[ \det (\boldsymbol{D} - \bar{\lambda} \boldsymbol{I}) = 0 \iff \det \begin{pmatrix} D_{11} - \bar{\lambda} & & & & \\ & \ddots & & D_{ij} & \\ & & \ddots & & \\ & D_{ji} & & \ddots & \\ & & & & D_{nn} - \bar{\lambda} \end{pmatrix} = 0 \]

展开即

\[ \begin{aligned} \bar{\lambda}^n - (D_{11} + D_{22} + \ldots + D_{nn}) \bar{\lambda}^{n-1} + \cdots + (-1)^n \det(\boldsymbol{D}) &= 0 \\ \iff (\bar{\lambda} - \bar{\lambda}_1)(\bar{\lambda} - \bar{\lambda}_2) \cdots (\bar{\lambda} - \bar{\lambda}_n) &= 0 \end{aligned} \]

得到

\[ D_{11} + D_{22} + \ldots + D_{nn} = \bar{\lambda}_1 + \bar{\lambda}_2 + \ldots + \bar{\lambda}_n > \bar{\lambda}_1 = \frac{1}{\omega_1^2} \]

这个式子可以用来估计最小固有频率 \(\omega_1\)：

\[ \omega_1^2 > \frac{1}{D_{11} + D_{22} + \ldots + D_{nn}} \]

\(\boldsymbol{D}_{ii}\) 的物理意义

由结构矩阵 \(\boldsymbol{D}\) 的定义，\(D_{ii} = F_{ii} m_i\). 柔度 \(F_{ii}\) 表示单位力作用在 \(m_i\) 上第 \(i\) 个坐标的位移。下面的例子可以看出，\(\boldsymbol{D}_{ii}\) 相当于系统“抹掉”除了第 \(i\) 个自由度以外的所有自由度后，第 \(i\) 个自由度的固有频率的倒数平方。

参考例 4.2.3 的图，

\[F_{11} = \frac{1}{k_1} \implies \]

瑞利法

第 I/II 类瑞利商：

\[ \begin{align} R_{\mathrm{I}} (\boldsymbol{X}) &= \frac{\boldsymbol{X}^{\mathrm{T}} \boldsymbol{K} \boldsymbol{X}}{\boldsymbol{X}^{\mathrm{T}} \boldsymbol{M} \boldsymbol{X}} \tag{I} \\ R_{\mathrm{II}} (\boldsymbol{X}) &= \frac{\boldsymbol{X}^{\mathrm{T}} \boldsymbol{M} \boldsymbol{X}}{\boldsymbol{X}^{\mathrm{T}} \boldsymbol{M} \boldsymbol{K}^{-1} \boldsymbol{M} \boldsymbol{X}} \tag{II} \end{align} \]

其中 \(\boldsymbol{X}\) 是任意向量，可用互相正交的振型线性表示：

\[ \begin{equation} \tag{4.2.1} \label{eq:linear-combination} \boldsymbol{X} = \sum_{i=1}^n c_i \boldsymbol{\varphi}_i \end{equation} \]

重要结论：

\[ \begin{equation} \tag{4.2.2} \label{eq:rayleigh-quotient-inequality} R_{\mathrm{I}} (\boldsymbol{X}) \geq R_{\mathrm{II}} (\boldsymbol{X}) \geq \omega_1^2 \end{equation} \]

证明：

\[ \begin{aligned} \boldsymbol{X}^{\mathrm{T}} \boldsymbol{M} \boldsymbol{X} &= \left(\sum_{i=1}^n c_i \boldsymbol{\varphi}_i \right)^{\mathrm{T}} \boldsymbol{M} \left(\sum_{j=1}^n c_j \boldsymbol{\varphi}_j \right) = \sum_{i=1}^n \sum_{j=1}^n c_i c_j \cdot \boldsymbol{\varphi}_i^{\mathrm{T}} \boldsymbol{M} \boldsymbol{\varphi}_j \\ &= \sum_{i=1}^n c_i^2 \\ \boldsymbol{X}^{\mathrm{T}} \boldsymbol{K} \boldsymbol{X} &= \sum_{i=1}^n \sum_{j=1}^n c_i c_j \cdot \boldsymbol{\varphi}_i^{\mathrm{T}} \boldsymbol{K} \boldsymbol{\varphi}_j = \sum_{i=1}^n c_i^2 \omega_i^2 \end{aligned} \]

故

\[ R_{\mathrm{I}} (\boldsymbol{X}) = \frac{\boldsymbol{X}^{\mathrm{T}} \boldsymbol{K} \boldsymbol{X}}{\boldsymbol{X}^{\mathrm{T}} \boldsymbol{M} \boldsymbol{X}} = \frac{\sum_{i=1}^n c_i^2 \omega_i^2}{\sum_{i=1}^n c_i^2} \geq \omega_1^2 \]

现在考虑 \(R_{\mathrm{II}} (\boldsymbol{X})\) 的分母. 首先重写式 \eqref{eq:linear-combination} 为

\[ \boldsymbol{X} = \sum_{i=1}^n c_i \boldsymbol{\varphi}_i = [\boldsymbol{\varphi}_1, \boldsymbol{\varphi}_2, \ldots, \boldsymbol{\varphi}_n] \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix} = \boldsymbol{\Phi}_{n \times n} \boldsymbol{c}_{n \times 1} \]

由 \(\boldsymbol{\varphi}_i^{\mathrm{T}} \boldsymbol{M} \boldsymbol{\varphi}_j = \delta_{ij}\) 的正交性，有 \(\boldsymbol{\Phi}^{\mathrm{T}} \boldsymbol{M} \boldsymbol{\Phi} = \boldsymbol{I}\).

\[ \boldsymbol{X}^{\mathrm{T}} \boldsymbol{M} \boldsymbol{K}^{-1} \boldsymbol{M} \boldsymbol{X} = \boldsymbol{c}^{\mathrm{T}} \boldsymbol{\Phi}^{\mathrm{T}} \boldsymbol{M} \boldsymbol{K}^{-1} \boldsymbol{M} \boldsymbol{\Phi} \boldsymbol{c} \]

Rayleigh-Ritz 法

只选取 \(m\) 个振型近似表示系统的位移：

\[ \begin{equation} \tag{4.3.1} \label{eq:rayleigh-ritz-approximation} \boldsymbol{X} \approx \sum_{i=1}^m c_i \boldsymbol{\varphi}_i = \boldsymbol{\Phi}_{n \times m} \boldsymbol{c}_{m \times 1} \end{equation} \]

第一类瑞利商变为

\[ R_{\mathrm{I}} (\boldsymbol{X}) = \frac{\boldsymbol{c}^{\mathrm{T}} \boldsymbol{\Phi}^{\mathrm{T}} \boldsymbol{K} \boldsymbol{\Phi} \boldsymbol{c}}{\boldsymbol{c}^{\mathrm{T}} \boldsymbol{\Phi}^{\mathrm{T}} \boldsymbol{M} \boldsymbol{\Phi} \boldsymbol{c}} = \frac{\boldsymbol{c}^{\mathrm{T}} \bar{\boldsymbol{K}} \boldsymbol{c}}{\boldsymbol{c}^{\mathrm{T}} \bar{\boldsymbol{M}} \boldsymbol{c}} \]

子空间迭代法

广义本征方程

\[ \boldsymbol{K} \boldsymbol{x} = \lambda \boldsymbol{M} \boldsymbol{x} \]

1. 给定刚度矩阵 \(\boldsymbol{K}\)，质量矩阵 \(\boldsymbol{M}\) 和各阶固有频率误差向量 \(\boldsymbol{\varepsilon} = [\varepsilon_1 \, \cdots \, \varepsilon_i \, \cdots \, \varepsilon_p]^{\mathrm{T}}\)，\(p\) 为要求解的本征频率个数.
2. 选取

对角元占优，\(\frac{k_{ii}}{m_{ii}}\) 作为初始近似频率 \(\tilde{\omega}_i^2\). 再对频率排序，取前 \(p\) 个，初始振型向量除了第 \(i\) 个分量为 1 外，其余均为 0.

\(\boldsymbol{K}\) 为半正定矩阵时，\(\boldsymbol{K}^{-1}\) 不存在，采用修正刚度矩阵 \(\boldsymbol{K} + \beta \boldsymbol{M}\)，其中 \(\beta > 0\).

\[ \bar{\boldsymbol{K}} \boldsymbol{X} = \tilde{\lambda} \boldsymbol{M} \boldsymbol{X} \]

\(\tilde{\lambda} = \lambda + \beta\)