vault backup: 2026-03-06 10:54:58
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12
.obsidian/plugins/copilot/data.json
vendored
12
.obsidian/plugins/copilot/data.json
vendored
@ -13,7 +13,7 @@
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"googleApiKey": "",
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"openRouterAiApiKey": "",
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"defaultChainType": "llm_chain",
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"defaultModelKey": "gemini-2.5-flash|google",
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"defaultModelKey": "llama3.2:latest|ollama",
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"embeddingModelKey": "nomic-embed-text|ollama",
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"temperature": 0.1,
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"maxTokens": 1000,
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@ -165,12 +165,12 @@
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"apiKey": "",
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"isEmbeddingModel": false,
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"capabilities": [
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"websearch",
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"vision"
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"vision",
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"websearch"
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],
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"stream": true,
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"displayName": "",
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"enableCors": true
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"enableCors": true,
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"displayName": "ministral-3:14b-6005"
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}
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],
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"activeEmbeddingModels": [
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@ -299,7 +299,7 @@
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},
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{
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"name": "Translate to Chinese",
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"prompt": "<instruction>Translate the text below into Chinese:\n 1. Preserve the meaning and tone\n 2. Maintain appropriate cultural context\n 3. Keep formatting and structure\n \n </instruction>\n\n<text>{copilot-selection}</text>\n<restrictions>\n1. Blade翻译为叶片，flapwise翻译为挥舞，edgewise翻译为摆振，pitch angle翻译成变桨角度，twist angle翻译为扭角，rotor翻译为风轮，turbine、wind turbine翻译为机组、风电机组，span翻译为展向，deflection翻译为变形，mode翻译为模态，normal mode翻译为简正模态，jacket 翻译为导管架，superelement翻译为超单元，shaft翻译为主轴，azimuth、azimuth angle翻译为方位角，neutral axes 翻译为中性轴\n2. Return only the translated text.\n</restrictions>",
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"prompt": "<instruction>Translate the text below into Chinese:\n 1. Preserve the meaning and tone\n 2. Maintain appropriate cultural context\n 3. Keep formatting and structure\n 4. In line math formular use LaTex.\n </instruction>\n\n<text>{copilot-selection}</text>\n<restrictions>\n1. Blade翻译为叶片，flapwise翻译为挥舞，edgewise翻译为摆振，pitch angle翻译成变桨角度，twist angle翻译为扭角，rotor翻译为风轮，turbine、wind turbine翻译为机组、风电机组，span翻译为展向，deflection翻译为变形，mode翻译为模态，normal mode翻译为简正模态，jacket 翻译为导管架，superelement翻译为超单元，shaft翻译为主轴，azimuth、azimuth angle翻译为方位角，neutral axes 翻译为中性轴\n2. Return only the translated text.\n</restrictions>",
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"showInContextMenu": true,
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"modelKey": "ministral-3:14b|ollama"
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},
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@ -39,7 +39,7 @@ $$
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$$
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Because of the orthogonality property with respect to mass, i.e., $\phi_{n}^{T}\,\mathbf{m}\,\phi_{m}\,=\,0$ for $m\neq n$ , all terms on the right hand side of this equation vanish, except for the term containing $\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}$ , leaving
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由于质量矩阵的正交性，即当 \( m \neq n \) 时，有 \(\phi_{n}^{T}\,\mathbf{m}\,\phi_{m}\,=\,0\)，方程右侧的所有项均消失，仅剩下包含 \(\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}\) 的项。
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由于质量矩阵的正交性，即对于 \( m \neq n \)，有 \( \phi_{n}^{T}\,\mathbf{m}\,\phi_{m} = 0 \)，方程右侧的所有项均消失，仅剩下包含 \( \phi_{n}^{T}\,\mathbf{m}\,\phi_{n} \) 的项。
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$$
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\phi_{n}^{T}\,\mathbf{m}\,\mathbf{v}=\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}\,Y_{n}
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$$
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@ -51,53 +51,56 @@ Y_{n}=\frac{\phi_{n}^{T}\,\mathbf{m}\,\mathbf{v}}{\phi_{n}^{T}\,\mathbf{m}\,\phi
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$$
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If vector $\mathbf{v}$ is time dependent, the $Y_{n}$ coordinates will also be time dependent; in this case, taking the time derivative of Eq. (12-6) yields
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如果向量 $\mathbf{v}$ 随时间变化，则其 $Y_{n}$ 坐标也将随时间变化；此时，对式（12-6）取时间导数，可得
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$$
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\dot{Y}_{n}(t)=\frac{\phi_{n}^{T}\,\mathbf{m}\,\dot{\mathbf{v}}(t)}{\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}}
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$$
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Note that the above procedure is equivalent to that used to evaluate the coefficients in the Fourier series given by Eqs. (4-3).
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请注意，上述程序与用于评估傅里叶级数系数的方法（如方程（4-3）所示）等价。
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# 12-2 UNCOUPLED EQUATIONS OF MOTION: UNDAMPED
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The orthogonality properties of the normal modes will now be used to simplify the equations of motion of the MDOF system. In general form these equations are given by Eq. (9-13) [or its equivalent Eq. (9-19) if axial forces are present]; for the undamped system they become
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现在将利用简正模态的正交性质来简化多自由度（MDOF）系统的运动方程。这些方程的一般形式由式（9-13）给出（或在轴向力存在时等效的式（9-19））；对于无阻尼系统，方程则简化为：
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$$
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\mathbf{m}\;\ddot{\mathbf{v}}(t)+\mathbf{k}\;\mathbf{v}(t)=\mathbf{p}(t)
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$$
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Introducing Eq. (12-3) and its second time derivative $\ddot{\mathbf{v}}=\Phi\ \ddot{\mathbf{Y}}$ (noting that the mode shapes do not change with time) leads to
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将方程（12-3）及其二阶时间导数 $\ddot{\mathbf{v}}=\Phi\ \ddot{\mathbf{Y}}$（注意模态形状不随时间变化）代入后，得到
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$$
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\mathbf{m}\oplus{\ddot{Y}}(t)+\mathbf{k}\;\Phi\;Y(t)=\mathbf{p}(t)
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$$
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If Eq. (12-9) is premultiplied by the transpose of the $n$ th mode-shape vector $\phi_{n}^{T}$ , it becomes
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如果将方程（12-9）左乘第 $n$ 阶模态形状向量的转置 $\phi_{n}^{T}$，则可得到
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$$
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\pmb{\phi}_{n}^{T}\pmb{\mathrm{m}}\,\pmb{\Phi}\,\ddot{\pmb{Y}}(t)+\pmb{\phi}_{n}^{T}\,\mathbf{k}\,\pmb{\Phi}\,\pmb{Y}(t)=\pmb{\phi}_{n}^{T}\,\mathbf{p}(t)
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$$
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but if the two terms on the left hand side are expanded as shown in Eq. (12-4), all terms except the $n$ th will vanish because of the mode-shape orthogonality properties; hence the result is
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但如果左侧的两项按照 Eq. (12-4) 展开，由于模态形状的正交性，除了第 $n$ 项外，所有项都将消失；因此最终结果为：
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$$
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\pmb{\phi}_{n}^{T}\mathbf{m}\pmb{\phi}_{n}~\ddot{Y}_{n}(t)+\pmb{\phi}_{n}^{T}\mathbf{k}\pmb{\phi}_{n}~Y_{n}(t)=\pmb{\phi}_{n}^{T}\mathbf{p}(t)
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$$
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Now new symbols will be defined as follows:
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以下新符号定义如下：
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$$
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\begin{array}{r l}&{M_{n}\equiv\boldsymbol{\phi}_{n}^{T}\mathbf{m}\boldsymbol{\phi}_{n}}\\ &{K_{n}\equiv\boldsymbol{\phi}_{n}^{T}\mathbf{k}\boldsymbol{\phi}_{n}}\\ &{P_{n}(t)\equiv\boldsymbol{\phi}_{n}^{T}\mathbf{p}(t)}\end{array}
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$$
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which are called the normal-coordinate generalized mass, generalized stiffness, and generalized load for mode $n$ , respectively. With them Eq. (12-11) can be written
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这些分别被称为第 $n$ 阶模态的**广义质量**、**广义刚度**和**广义载荷**。利用它们，方程（12-11）可以表示为：
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$$
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M_{n}\,\ddot{Y}_{n}(t)+K_{n}\,Y_{n}(t)=P_{n}(t)
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$$
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which is a SDOF equation of motion for mode $n$ . If Eq. (11-39), ${\bf k}\phi_{n}=\omega_{n}^{2}{\bf m}\phi_{n}$ , is multiplied on both sides by $\phi_{n}^{T}$ , the generalized stiffness for mode $n$ is related to the generalized mass by the frequency of vibration
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该方程是第 $n$ 阶单自由度（SDOF）运动方程。若将式（11-39）${\bf k}\phi_{n}=\omega_{n}^{2}{\bf m}\phi_{n}$ 两边同时左乘以 $\phi_{n}^{T}$，则第 $n$ 阶模态的**广义刚度**与**广义质量**之间通过振动频率建立联系。
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or
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$$
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@ -105,72 +108,77 @@ $$
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$$
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(Capital letters are used to denote all normal-coordinate properties.)
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（用大写字母表示所有**正则坐标**的性质。）
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The procedure described above can be used to obtain an independent SDOF equation for each mode of vibration of the undamped structure. Thus the use of the normal coordinates serves to transform the equations of motion from a set of $N$ simultaneous differential equations, which are coupled by the off-diagonal terms in the mass and stiffness matrices, to a set of $N$ independent normal-coordinate equations. The dynamic response therefore can be obtained by solving separately for the response of each normal (modal) coordinate and then superposing these by Eq. (12-3) to obtain the response in the original geometric coordinates. This procedure is called the mode-superposition method, or more precisely the mode displacement superposition method.
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上述所述程序可用于为无阻尼结构的每一振动模态推导出一个独立的单自由度（SDOF）方程。因此，使用**正则坐标**（normal coordinates）将运动方程从一组包含质量矩阵和刚度矩阵中非对角耦合项的 $N$ 个**耦合微分方程**转化为 $N$ 个独立的**正则坐标方程**。动力响应因此可以通过分别求解每个正则（模态）坐标的响应，再通过 **式（12-3）** 进行叠加，从而得到原几何坐标系下的响应。该方法称为**模态叠加法**（mode-superposition method），更精确地说，称为**模态位移叠加法**（mode displacement superposition method）。
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# 12-3 UNCOUPLED EQUATIONS OF MOTION: VISCOUS DAMPING
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Now it is of interest to examine the conditions under which this normalcoordinate transformation will also serve to uncouple the damped equations of motion. These equations [Eq. (9-13)] are
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现在，我们有兴趣探讨在何种条件下，这种**正则坐标变换**也能够将**阻尼运动方程**解耦。这些方程（见式（9-13））如下：
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$$
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\mathbf{m}\,\ddot{\mathbf{v}}(t)+\mathbf{c}\,\dot{\mathbf{v}}(t)+\mathbf{k}\,\mathbf{v}(t)=\mathbf{p}(t)
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$$
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Introducing the normal-coordinate expression of Eq. (12-3) and its time derivatives and premultiplying by the transpose of the $n$ th mode-shape vector $\phi_{n}^{T}$ leads to
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将方程（12-3）的正则坐标表达式及其时间导数代入，并左乘第 \( n \) 阶模态形状向量的转置 $\phi_{n}^{T}$，可得到：
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$$
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\phi_{n}^{T}\mathbf{m}\Phi\ddot{Y}(t)+\phi_{n}^{T}\mathbf{c}\Phi\dot{Y}(t)+\phi_{n}^{T}\mathbf{k}\Phi Y(t)=\phi_{n}^{T}\mathbf{p}(t)
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$$
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It was noted above that the orthogonality conditions
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之前已提及，正交条件如下：
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$$
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\begin{array}{r l}{\phi_{m}^{T}\,\mathbf{m}\,\phi_{n}=0}&{{}}\\ {\phi_{m}^{T}\,\mathbf{k}\,\phi_{n}=0}&{{}}\end{array}\quad m\neq n
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$$
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cause all components except the nth-mode term in the mass and stiffness expressions of Eq. (12-14) to vanish. A similar reduction will apply to the damping expression if it is assumed that the corresponding orthogonality condition applies to the damping matrix; that is, assume that
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在 Eq. (12-14) 的质量和刚度表达式中，**除了第 *n* 阶模态项外**，所有其他组分均消失。若假设阻尼矩阵同样满足相应的正交性条件，则阻尼表达式也将出现类似的简化，即假设：
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$$
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\phi_{m}^{T}\,\mathbf{c}\,\phi_{n}=0\qquad m\neq n
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$$
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In this case Eq. (12-14) may be written
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在这种情况下，式（12-14）可表示为：
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$$
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M_{n}\ \ddot{Y}_{n}(t)+C_{n}\ \dot{Y}_{n}(t)+K_{n}\ Y_{n}(t)=P_{n}(t)
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$$
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where the definitions of modal coordinate mass, stiffness, and load have been introduced from Eq. (12-12) and where the modal coordinate viscous damping coefficient has been defined similarly
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其中，模态坐标下的质量、刚度和载荷的定义已在式（12-12）中引入，而模态坐标下的黏性阻尼系数则类似地进行了定义。
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$$
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C_{n}=\pmb{\phi}_{n}^{T}\pmb{\mathrm{c}}\,\pmb{\phi}_{n}
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$$
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If Eq. (12-14a) is divided by the generalized mass, this modal equation of motion may be expressed in alternative form:
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若将式（12-14a）除以广义质量，该模态运动方程可表示为另一种形式：
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$$
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\ddot{Y}_{n}(t)+2\,\xi_{n}\,\omega_{n}\,\dot{Y}_{n}(t)+\omega_{n}^{2}\,Y_{n}(t)=\frac{P_{n}(t)}{M_{n}}
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$$
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where Eq. (12-12d) has been used to rewrite the stiffness term and where the second term on the left hand side represents a definition of the modal viscous damping ratio
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其中，式（12-12d）被用于重写刚度项，而左侧第二项则定义了模态黏性阻尼比。
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$$
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\xi_{n}={\frac{C_{n}}{2\,\omega_{n}\,M_{n}}}
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$$
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As was noted earlier, it generally is more convenient and physically reasonable to define the damping of a MDOF system using the damping ratio for each mode in this way rather than to evaluate the coefficients of the damping matrix c because the modal damping ratios $\xi_{n}$ can be determined experimentally or estimated with adequate precision in many cases.
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如前所述，通常更为方便且在物理上更合理的是以这种方式为每个模态定义多自由度系统的阻尼，而不是直接评估阻尼矩阵 **c** 的系数，因为模态阻尼比 $\xi_{n}$ 在许多情况下可以通过实验确定或进行足够精确的估计。
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# 12-4 RESPONSE ANALYSIS BY MODE DISPLACEMENT SUPERPOSITION **模态位移叠加法响应分析**
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# 12-4 RESPONSE ANALYSIS BY MODE DISPLACEMENT SUPERPOSITION
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# Viscous Damping
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# Viscous Damping **粘滞阻尼**
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The normal coordinate transformation was used in Section 12-3 to convert the $N$ coupled linear damped equations of motion
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在第 12-3 节中，采用了**正常坐标变换**将由 $N$ 个耦合的线性阻尼运动方程进行转换。
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$$
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\mathbf{m}\,\ddot{\mathbf{v}}(t)+\mathbf{c}\,\dot{\mathbf{v}}(t)+\mathbf{k}\,\mathbf{v}(t)=\mathbf{p}(t)
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$$
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to a set of $N$ uncoupled equations given by
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to a set of $N$ uncoupled equations given by 成由以下 $N$ 个 **解耦方程** 组成的集合：
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$$
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\ddot{Y}_{n}(t)+2\,\xi_{n}\,\omega_{n}\,\dot{Y}_{n}(t)+\omega_{n}^{2}\,Y_{n}(t)=\frac{P_{n}(t)}{M_{n}}\qquad n=1,2,\cdots,N
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@ -183,21 +191,22 @@ M_{n}=\phi_{n}^{T}\textbf{m}\phi_{n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;P_{n}(t)=\phi_
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$$
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To proceed with the solution of these uncoupled equations of motion, one must first solve the eigenvalue problem
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要解决这些**解耦**的运动方程，首先必须求解**特征值问题**。
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$$
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\left[\mathbf{k}-\omega^{2}\mathbf{\deltam}\right]\,\hat{\mathbf{v}}=\mathbf{0}
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\left[\mathbf{k}-\omega^{2}\mathbf{m}\right]\,\hat{\mathbf{v}}=\mathbf{0}
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$$
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to obtain the required mode shapes $\phi_{n}$ $(n=1,2,\cdot\cdot\cdot)$ and corresponding frequencies $\omega_{n}$ . The modal damping ratios $\xi_{n}$ are usually assumed based on experimental evidence.
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为了获得所需的模态形状 $\phi_{n}$（$n=1,2,\cdot\cdot\cdot$）及其对应的固有频率 $\omega_{n}$。通常根据实验数据假设各阶模态阻尼比 $\xi_{n}$。
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The total response of the MDOF system now can be obtained by solving the $N$ uncoupled modal equations and superposing their effects, as indicated by Eq. (12-3). Each of Eqs. (12-17) is a standard SDOF equation of motion and can be solved in either the time domain or the frequency domain by the procedures described in Chapter 6. The time-domain solution is expressed by the Duhamel integral [see Eq. (6-7)]
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多自由度（MDOF）系统的总响应可通过求解 $N$ 个**解耦**的**模态**方程并叠加其效应来获得，如式（12-3）所示。式（12-17）中的每一个方程均为标准的单自由度（SDOF）运动方程，可通过第 6 章所述的方法在**时域**或**频域**中求解。时域解可用**杜哈美积分**（见式（6-7））表示。
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$$
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Y_{n}(t)=\frac{1}{M_{n}\omega_{n}}\,\int_{0}^{t}P_{n}(\tau)\;\exp\big[-\xi_{n}\omega_{n}\left(t-\tau\right)\big]\;\sin\omega_{D n}(t-\tau)\;d\tau
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$$
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which also may be written in standard convolution integral form
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也可表示为标准卷积积分形式。
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$$
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Y_{n}(t)=\int_{0}^{t}P_{n}(\tau)\;h_{n}(t-\tau)\;d\tau
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$$
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@ -209,27 +218,27 @@ h_{n}(t-\tau)=\frac{1}{M_{n}\omega_{D n}}\ \sin\omega_{D n}(t-\tau)\ \exp\big[-\
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$$
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is the unit-impulse response function, similar to Eq. (6-8).
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是单位脉冲响应函数，类似于式（6-8）。
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In the frequency domain, the response is obtained similarly to Eq. (6-24) from
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在频域中，响应的获得方式与方程（6-24）类似。
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$$
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Y_{n}(t)=\frac{1}{2\pi}\;\int_{-\infty}^{\infty}\mathrm{H}_{n}(i\overline{{{\omega}}})\;\mathrm{P}_{n}(i\overline{{{\omega}}})\;\exp i\overline{{{\omega}}}t\;d\overline{{{\omega}}}
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$$
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In this equation, the complex load function $\mathrm{P}_{n}(i\overline{{\omega}})$ is the Fourier transform of the modal loading $P_{n}(t)$ , and similar to Eq. (6-23) it is given by
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在此方程中，复杂载荷函数 $\mathrm{P}_{n}(i\overline{{\omega}})$ 是模态载荷 $P_{n}(t)$ 的傅里叶变换，与式（6-23）类似，其表达式为
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$$
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\mathbf{P}_{n}(i\overline{{\omega}})=\int_{-\infty}^{\infty}P_{n}(t)\,\exp(-i\overline{{\omega}}t)\;d t
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$$
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Also in Eq. (12-23), the complex frequency response function, $\mathrm{H}_{n}(i\overline{{\omega}})$ , may be expressed similarly to Eq. (6-25) as follows:
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在方程（12-23）中，复频响应函数 $\mathrm{H}_{n}(i\overline{{\omega}})$ 也可类似于方程（6-25）表示如下：
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$$
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\begin{array}{r l r}{\mathrm{H}_{n}(i\overline{{\omega}})=\frac{1}{\omega_{n}^{2}M_{n}}\left[\frac{1}{\left(1-\beta_{n}^{2}\right)+i(2\xi_{n}\beta_{n})}\right]}&{}&\\ {=\frac{1}{\omega_{n}^{2}M_{n}}\left[\frac{\left(1-\beta_{n}^{2}\right)-i(2\xi_{n}\beta_{n})}{(1-\beta_{n}^{2})^{2}+(2\xi_{n}\beta_{n})^{2}}\right]}&{}&{\xi_{n}\ge0}\end{array}
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$$
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In these functions, $\beta_{n}\equiv\,\overline{{\omega}}/\omega_{n}$ and $\omega_{D n}\,=\,\omega_{n}\sqrt{1-\xi_{n}^{2}}$ . As indicated previously by Eqs. (6-53) and (6-54), $h_{n}(t)$ and $\mathrm{H}_{n}(i\overline{{\omega}})$ are Fourier transform pairs. Solving Eq. (12-20) or (12-23) for any general modal loading yields the modal response $Y_{n}(t)$ for $t\,\geq\,0$ , assuming zero initial conditions, i.e., $Y_{n}(0)\,=\,\dot{Y}_{n}(0)\,=\,0$ . Should the initial conditions not equal zero, the damped free-vibration response [Eq. (2-49)]
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在这些函数中，$\beta_{n}\equiv\,\overline{{\omega}}/\omega_{n}$ 且 $\omega_{D n}\,=\,\omega_{n}\sqrt{1-\xi_{n}^{2}}$。如前所述（式（6-53）和（6-54）所示），$h_{n}(t)$ 和 $\mathrm{H}_{n}(i\overline{{\omega}})$ 是傅里叶变换对。对于任意一般模态载荷，通过求解方程（12-20）或（12-23）可得到模态响应 $Y_{n}(t)$（$t\,\geq\,0$），假设初始条件为零，即 $Y_{n}(0)\,=\,\dot{Y}_{n}(0)\,=\,0$。若初始条件不为零，则需考虑阻尼自由振动响应（式（2-49））。
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$$
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Y_{n}(t)=\left[Y_{n}(0)\,\cos{\omega_{D n}t}+\left(\frac{\dot{Y}_{n}(0)+Y_{n}(0)\xi_{n}\omega_{n}}{\omega_{D n}}\right)\,\sin{\omega_{D n}t}\right]\,\exp(-\xi_{n}\omega_{n}t)
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$$
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Before Width: | Height: | Size: 83 KiB After Width: | Height: | Size: 83 KiB |
@ -26,6 +26,7 @@ Craig [[Craig-Coupling of substructures for dynamic an]]
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## 动力学方程如何建立
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## 如何直接使用bladed的模态结果？
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