diff --git a/.obsidian/plugins/copilot/data.json b/.obsidian/plugins/copilot/data.json
index 2404752..07fdde0 100644
--- a/.obsidian/plugins/copilot/data.json
+++ b/.obsidian/plugins/copilot/data.json
@@ -299,7 +299,7 @@
     },
     {
       "name": "Translate to Chinese",
-      "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>",
+      "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. **$ $ 符号包裹的内容保持与原文内容一致**\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>",
       "showInContextMenu": true,
       "modelKey": "ministral-3:14b|ollama"
     },
diff --git a/书籍/力学书籍/力学/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library)/auto/Chap 12 ANALYSIS OF DYNAMIC RESPONSE USING SUPERPOSITION.md b/书籍/力学书籍/力学/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library)/auto/Chap 12 ANALYSIS OF DYNAMIC RESPONSE USING SUPERPOSITION.md
index cce0bbe..3005978 100644
--- a/书籍/力学书籍/力学/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library)/auto/Chap 12 ANALYSIS OF DYNAMIC RESPONSE USING SUPERPOSITION.md	
+++ b/书籍/力学书籍/力学/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library)/auto/Chap 12 ANALYSIS OF DYNAMIC RESPONSE USING SUPERPOSITION.md	
@@ -11,49 +11,49 @@ FIGURE 12-1 Representing deflections as sum of modal components.
 Consider, for example, the cantilever column shown in Fig. 12-1, for which the deflected shape is expressed in terms of translational displacements at three levels. Any displacement vector $\mathbf{v}$ (static or dynamic) for this structure can be developed by superposing suitable amplitudes of the normal modes as shown. For any modal component ${\bf v}_{n}$ , the displacements are given by the product of the mode-shape vector $\phi_{n}$ and the modal amplitude $Y_{n}$ ; thus  
 以图 12-1 所示的悬臂柱为例，其变形形状可通过三个高度处的平移位移来描述。该结构的任意位移向量 **v**（静态或动态）均可通过叠加适当振幅的简正模态来表示。对于任一模态分量 **vₙ**，其位移由模态形状向量 **φₙ** 与模态振幅 **Yₙ** 的乘积给出，即：
 $$
-\mathbf{v}_{n}=\phi_{n}\ Y_{n}
+\mathbf{v}_{n}=\phi_{n}\ Y_{n} \tag{12-1}
 $$  
 
 The total displacement vector $\mathbf{v}$ is then obtained by summing the modal vectors as expressed by  
 总位移向量 **v** 则通过对各模态向量求和得到，具体表达式为：
 $$
-\mathbf{v}=\phi_{1}\,Y_{1}+\phi_{2}\,Y_{2}+\cdot\cdot\cdot+\phi_{N}\,Y_{N}=\sum_{n=1}^{N}\phi_{n}\,Y_{n}
+\mathbf{v}=\phi_{1}\,Y_{1}+\phi_{2}\,Y_{2}+\cdot\cdot\cdot+\phi_{N}\,Y_{N}=\sum_{n=1}^{N}\phi_{n}\,Y_{n}\tag{12-2}
 $$  
 
 or, in matrix notation,  
 
 $$
-\mathbf{v}=\Phi\,Y
+\mathbf{v}=\Phi\,Y\tag{12-3}
 $$  
 
 In this equation, it is apparent that the $N\times N$ mode-shape matrix $\Phi$ serves to transform the generalized coordinate vector $Y$ to the geometric coordinate vector $\mathbf{v}$ . The generalized components in vector $Y$ are called the normal coordinates of the structure.  
 
 Because the mode-shape matrix consists of $N$ independent modal vectors, $\pmb{\Phi}\,=\,\left[\pmb{\phi}_{1}\,\,\,\,\pmb{\phi}_{2}\,\,\,\,\cdot\,\cdot\,\,\,\pmb{\phi}_{N}\right]$ , it is nonsingular and can be inverted. Thus, it is always possible to solve Eq. (12-3) directly for the normal-coordinate amplitudes in $Y$ which are associated with any given displacement vector $\mathbf{v}$ . In doing so, however, it is unnecessary to solve a set of simultaneous equations, due to the orthogonality property of the mode shapes. To evaluate any arbitrary normal coordinate, $Y_{n}$ for example, premultiply Eq. (12-2) by $\phi_{n}^{T}\,\mathbf{m}$ to obtain  
 
-在此方程中，显然大小为 $N \times N$ 的模态形状矩阵 $\Phi$ 用于将广义坐标向量 \( Y \) 转换为几何坐标向量$\mathbf{v}$。向量$Y$中的广义分量被称为结构的**正则坐标**（或**法向坐标**）。
+在此方程中，显然大小为 $N \times N$ 的模态形状矩阵 $\Phi$ 用于将广义坐标向量 $Y$ 转换为几何坐标向量$\mathbf{v}$。向量$Y$中的广义分量被称为结构的**正则坐标**（或**法向坐标**）。
 
-由于模态形状矩阵由 \( N \) 个独立的模态向量组成，即
-$\pmb{\Phi} = \left[\pmb{\phi}_{1}\,\,\,\,\pmb{\phi}_{2}\,\,\,\,\cdot\,\cdot\,\,\,\pmb{\phi}_{N}\right]$，该矩阵是非奇异的，因此可以求逆。因此，对于任意给定的位移向量 \(\mathbf{v}\)，总是可以直接求解方程（12-3），得到与之对应的正则坐标幅值 \( Y \)。不过，由于模态形状具有**正交性**，无需解联立方程组。为了求取任意正则坐标（例如 \( Y_n \)），可将方程（12-2）左乘以 \(\phi_n^T \mathbf{m}\)，得到：
+由于模态形状矩阵由 $N$ 个独立的模态向量组成，即
+$\pmb{\Phi} = \left[\pmb{\phi}_{1}\,\,\,\,\pmb{\phi}_{2}\,\,\,\,\cdot\,\cdot\,\,\,\pmb{\phi}_{N}\right]$，该矩阵是非奇异的，因此可以求逆。因此，对于任意给定的位移向量 $\mathbf{v}$，总是可以直接求解方程（12-3），得到与之对应的正则坐标幅值 $Y$。不过，由于模态形状具有**正交性**，无需解联立方程组。为了求取任意正则坐标（例如 $Y_n$），可将方程（12-2）左乘以 $\phi_n^T \mathbf{m}$，得到：
 $$
-\phi_{n}^{T}\,\mathbf{m}\,\mathbf{v}=\phi_{n}^{T}\,\mathbf{m}\,\phi_{1}\,Y_{1}+\phi_{n}^{T}\,\mathbf{m}\,\phi_{2}\,Y_{2}+\hdots+\phi_{n}^{T}\,\mathbf{m}\,\phi_{N}\,Y_{N}
+\phi_{n}^{T}\,\mathbf{m}\,\mathbf{v}=\phi_{n}^{T}\,\mathbf{m}\,\phi_{1}\,Y_{1}+\phi_{n}^{T}\,\mathbf{m}\,\phi_{2}\,Y_{2}+\dots+\phi_{n}^{T}\,\mathbf{m}\,\phi_{N}\,Y_{N}\tag{12-4}
 $$  
 
 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  
-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。
+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。
 $$
-\phi_{n}^{T}\,\mathbf{m}\,\mathbf{v}=\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}\,Y_{n}
+\phi_{n}^{T}\,\mathbf{m}\,\mathbf{v}=\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}\,Y_{n}\tag{12-5}
 $$  
 
 from which  
 
 $$
-Y_{n}=\frac{\phi_{n}^{T}\,\mathbf{m}\,\mathbf{v}}{\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}}\qquad\qquad n=1,2,\cdots,N
+Y_{n}=\frac{\phi_{n}^{T}\,\mathbf{m}\,\mathbf{v}}{\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}}\qquad\qquad n=1,2,\cdots,N\tag{12-6}
 $$  
 
 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  
 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
 $$
-\dot{Y}_{n}(t)=\frac{\phi_{n}^{T}\,\mathbf{m}\,\dot{\mathbf{v}}(t)}{\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}}
+\dot{Y}_{n}(t)=\frac{\phi_{n}^{T}\,\mathbf{m}\,\dot{\mathbf{v}}(t)}{\phi_{n}^{T}\,\mathbf{m}\,\phi_{n}}\tag{12-7}
 $$  
 
 Note that the above procedure is equivalent to that used to evaluate the coefficients in the Fourier series given by Eqs. (4-3).  
@@ -64,38 +64,38 @@ The orthogonality properties of the normal modes will now be used to simplify th
 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：
 
 $$
-\mathbf{m}\;\ddot{\mathbf{v}}(t)+\mathbf{k}\;\mathbf{v}(t)=\mathbf{p}(t)
+\mathbf{m}\;\ddot{\mathbf{v}}(t)+\mathbf{k}\;\mathbf{v}(t)=\mathbf{p}(t)\tag{12-8}
 $$  
 
 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  
 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
 $$
-\mathbf{m}\oplus{\ddot{Y}}(t)+\mathbf{k}\;\Phi\;Y(t)=\mathbf{p}(t)
+\mathbf{m}\Phi{\ddot{Y}}(t)+\mathbf{k}\;\Phi\;Y(t)=\mathbf{p}(t)\tag{12-9}
 $$  
 
 If Eq. (12-9) is premultiplied by the transpose of the $n$ th mode-shape vector $\phi_{n}^{T}$ , it becomes  
 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
 $$
-\pmb{\phi}_{n}^{T}\pmb{\mathrm{m}}\,\pmb{\Phi}\,\ddot{\pmb{Y}}(t)+\pmb{\phi}_{n}^{T}\,\mathbf{k}\,\pmb{\Phi}\,\pmb{Y}(t)=\pmb{\phi}_{n}^{T}\,\mathbf{p}(t)
+\pmb{\phi}_{n}^{T}\pmb{\mathrm{m}}\,\pmb{\Phi}\,\ddot{\pmb{Y}}(t)+\pmb{\phi}_{n}^{T}\,\mathbf{k}\,\pmb{\Phi}\,\pmb{Y}(t)=\pmb{\phi}_{n}^{T}\,\mathbf{p}(t)\tag{12-10}
 $$  
 
 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  
 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：
 $$
-\pmb{\phi}_{n}^{T}\mathbf{m}\pmb{\phi}_{n}~\ddot{Y}_{n}(t)+\pmb{\phi}_{n}^{T}\mathbf{k}\pmb{\phi}_{n}~Y_{n}(t)=\pmb{\phi}_{n}^{T}\mathbf{p}(t)
+\pmb{\phi}_{n}^{T}\mathbf{m}\pmb{\phi}_{n}~\ddot{Y}_{n}(t)+\pmb{\phi}_{n}^{T}\mathbf{k}\pmb{\phi}_{n}~Y_{n}(t)=\pmb{\phi}_{n}^{T}\mathbf{p}(t)\tag{12-11}
 $$  
 
 Now new symbols will be defined as follows:  
 以下新符号定义如下：
 
 $$
-\begin{array}{r l}&{M_{n}\equiv\boldsymbol{\phi}_{n}^{T}\mathbf{m}\boldsymbol{\phi}_{n}}\\ &{K_{n}\equiv\boldsymbol{\phi}_{n}^{T}\mathbf{k}\boldsymbol{\phi}_{n}}\\ &{P_{n}(t)\equiv\boldsymbol{\phi}_{n}^{T}\mathbf{p}(t)}\end{array}
+\begin{array}{r l}&{M_{n}\equiv\boldsymbol{\phi}_{n}^{T}\mathbf{m}\boldsymbol{\phi}_{n}}\\ &{K_{n}\equiv\boldsymbol{\phi}_{n}^{T}\mathbf{k}\boldsymbol{\phi}_{n}}\\ &{P_{n}(t)\equiv\boldsymbol{\phi}_{n}^{T}\mathbf{p}(t)}\end{array}\tag{12-12 a b c}
 $$  
 
 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  
 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：
 $$
-M_{n}\,\ddot{Y}_{n}(t)+K_{n}\,Y_{n}(t)=P_{n}(t)
+M_{n}\,\ddot{Y}_{n}(t)+K_{n}\,Y_{n}(t)=P_{n}(t)\tag{12-13}
 $$  
 
 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  
@@ -104,7 +104,7 @@ which is a SDOF equation of motion for mode $n$ . If Eq. (11-39), ${\bf k}\phi_{
 or  
 
 $$
-\begin{array}{r l}&{\phi_{n}^{T}\,\mathbf k\,\phi_{n}=\omega_{n}^{2}\phi_{n}^{T}\,\mathbf m\,\phi_{n}}\\ &{\qquad\qquad\qquad\qquad\qquad\qquad}\\ &{K_{n}=\omega_{n}^{2}M_{n}}\end{array}
+\begin{array}{r l}&{\phi_{n}^{T}\,\mathbf k\,\phi_{n}=\omega_{n}^{2}\phi_{n}^{T}\,\mathbf m\,\phi_{n}}\\ &{\qquad\qquad\qquad\qquad\qquad\qquad}\\ &{K_{n}=\omega_{n}^{2}M_{n}}\end{array}\tag{12-12d}
 $$  
 
 (Capital letters are used to denote all normal-coordinate properties.)  
@@ -122,9 +122,9 @@ $$
 $$  
 
 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  
-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：
+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：
 $$
-\phi_{n}^{T}\mathbf{m}\Phi\ddot{Y}(t)+\phi_{n}^{T}\mathbf{c}\Phi\dot{Y}(t)+\phi_{n}^{T}\mathbf{k}\Phi Y(t)=\phi_{n}^{T}\mathbf{p}(t)
+\phi_{n}^{T}\mathbf{m}\Phi\ddot{Y}(t)+\phi_{n}^{T}\mathbf{c}\Phi\dot{Y}(t)+\phi_{n}^{T}\mathbf{k}\Phi Y(t)=\phi_{n}^{T}\mathbf{p}(t)\tag{12-14}
 $$  
 
 It was noted above that the orthogonality conditions  
@@ -137,31 +137,31 @@ cause all components except the nth-mode term in the mass and stiffness expressi
 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：
 
 $$
-\phi_{m}^{T}\,\mathbf{c}\,\phi_{n}=0\qquad m\neq n
+\phi_{m}^{T}\,\mathbf{c}\,\phi_{n}=0\qquad m\neq n\tag{12-15}
 $$  
 
 In this case Eq. (12-14) may be written  
 在这种情况下，式（12-14）可表示为：
 $$
-M_{n}\ \ddot{Y}_{n}(t)+C_{n}\ \dot{Y}_{n}(t)+K_{n}\ Y_{n}(t)=P_{n}(t)
+M_{n}\ \ddot{Y}_{n}(t)+C_{n}\ \dot{Y}_{n}(t)+K_{n}\ Y_{n}(t)=P_{n}(t)\tag{12-14a}
 $$  
 
 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  
 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。
 $$
-C_{n}=\pmb{\phi}_{n}^{T}\pmb{\mathrm{c}}\,\pmb{\phi}_{n}
+C_{n}=\pmb{\phi}_{n}^{T}\pmb{\mathrm{c}}\,\pmb{\phi}_{n}\tag{12-15a}
 $$  
 
 If Eq. (12-14a) is divided by the generalized mass, this modal equation of motion may be expressed in alternative form:  
 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：
 $$
-\ddot{Y}_{n}(t)+2\,\xi_{n}\,\omega_{n}\,\dot{Y}_{n}(t)+\omega_{n}^{2}\,Y_{n}(t)=\frac{P_{n}(t)}{M_{n}}
+\ddot{Y}_{n}(t)+2\,\xi_{n}\,\omega_{n}\,\dot{Y}_{n}(t)+\omega_{n}^{2}\,Y_{n}(t)=\frac{P_{n}(t)}{M_{n}}\tag{12-14b}
 $$  
 
 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  
 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。
 $$
-\xi_{n}={\frac{C_{n}}{2\,\omega_{n}\,M_{n}}}
+\xi_{n}={\frac{C_{n}}{2\,\omega_{n}\,M_{n}}}\tag{12-15b}
 $$  
 
 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.  
@@ -169,52 +169,52 @@ As was noted earlier, it generally is more convenient and physically reasonable
 # 12-4 RESPONSE ANALYSIS BY MODE DISPLACEMENT SUPERPOSITION **模态位移叠加法响应分析**
 
 
-# Viscous Damping **粘滞阻尼**  
+## Viscous Damping **粘滞阻尼**  
 
 The normal coordinate transformation was used in Section 12-3 to convert the $N$ coupled linear damped equations of motion  
 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。
 $$
-\mathbf{m}\,\ddot{\mathbf{v}}(t)+\mathbf{c}\,\dot{\mathbf{v}}(t)+\mathbf{k}\,\mathbf{v}(t)=\mathbf{p}(t)
+\mathbf{m}\,\ddot{\mathbf{v}}(t)+\mathbf{c}\,\dot{\mathbf{v}}(t)+\mathbf{k}\,\mathbf{v}(t)=\mathbf{p}(t)\tag{12-16}
 $$  
 
 to a set of $N$ uncoupled equations given by 成由以下 $N$ 个 **解耦方程** 组成的集合：
   
 
 $$
-\ddot{Y}_{n}(t)+2\,\xi_{n}\,\omega_{n}\,\dot{Y}_{n}(t)+\omega_{n}^{2}\,Y_{n}(t)=\frac{P_{n}(t)}{M_{n}}\qquad n=1,2,\cdots,N
+\ddot{Y}_{n}(t)+2\,\xi_{n}\,\omega_{n}\,\dot{Y}_{n}(t)+\omega_{n}^{2}\,Y_{n}(t)=\frac{P_{n}(t)}{M_{n}}\qquad n=1,2,\cdots,N\tag{12-17}
 $$  
 
 in which  
 
 $$
-M_{n}=\phi_{n}^{T}\textbf{m}\phi_{n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;P_{n}(t)=\phi_{n}^{T}\textbf{p}(t)
+M_{n}=\phi_{n}^{T}\textbf{m}\phi_{n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;P_{n}(t)=\phi_{n}^{T}\textbf{p}(t)\tag{12-18}
 $$  
 
 To proceed with the solution of these uncoupled equations of motion, one must first solve the eigenvalue problem  
 要解决这些**解耦**的运动方程，首先必须求解**特征值问题**。
 $$
-\left[\mathbf{k}-\omega^{2}\mathbf{m}\right]\,\hat{\mathbf{v}}=\mathbf{0}
+\left[\mathbf{k}-\omega^{2}\mathbf{m}\right]\,\hat{\mathbf{v}}=\mathbf{0}\tag{12-19}
 $$  
 
 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.  
 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 $\xi_{n}$。
 
 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)]  
-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。
+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Duhamel积分**（见式（6-7））表示。
 $$
-Y_{n}(t)=\frac{1}{M_{n}\omega_{n}}\,\int_{0}^{t}P_{n}(\tau)\;\exp\big[-\xi_{n}\omega_{n}\left(t-\tau\right)\big]\;\sin\omega_{D n}(t-\tau)\;d\tau
+Y_{n}(t)=\frac{1}{M_{n}\omega_{n}}\,\int_{0}^{t}P_{n}(\tau)\;\exp\big[-\xi_{n}\omega_{n}\left(t-\tau\right)\big]\;\sin\omega_{D n}(t-\tau)\;d\tau\tag{12-20}
 $$  
 
 which also may be written in standard convolution integral form  
 也可表示为标准卷积积分形式。
 $$
-Y_{n}(t)=\int_{0}^{t}P_{n}(\tau)\;h_{n}(t-\tau)\;d\tau
+Y_{n}(t)=\int_{0}^{t}P_{n}(\tau)\;h_{n}(t-\tau)\;d\tau\tag{12-21}
 $$  
 
 in which  
 
 $$
-h_{n}(t-\tau)=\frac{1}{M_{n}\omega_{D n}}\ \sin\omega_{D n}(t-\tau)\ \exp\big[-\xi_{n}\omega_{n}\left(t-\tau\right)\big]\ \ \ 0<\xi_{n}<1
+h_{n}(t-\tau)=\frac{1}{M_{n}\omega_{D n}}\ \sin\omega_{D n}(t-\tau)\ \exp\big[-\xi_{n}\omega_{n}\left(t-\tau\right)\big]\ \ \ 0<\xi_{n}<1\tag{12-22}
 $$  
 
 is the unit-impulse response function, similar to Eq. (6-8).  
@@ -222,81 +222,89 @@ is the unit-impulse response function, similar to Eq. (6-8).
 In the frequency domain, the response is obtained similarly to Eq. (6-24) from  
 在频域中，响应的获得方式与方程（6-24）类似。
 $$
-Y_{n}(t)=\frac{1}{2\pi}\;\int_{-\infty}^{\infty}\mathrm{H}_{n}(i\overline{{{\omega}}})\;\mathrm{P}_{n}(i\overline{{{\omega}}})\;\exp i\overline{{{\omega}}}t\;d\overline{{{\omega}}}
+Y_{n}(t)=\frac{1}{2\pi}\;\int_{-\infty}^{\infty}\mathrm{H}_{n}(i\overline{{{\omega}}})\;\mathrm{P}_{n}(i\overline{{{\omega}}})\;\exp i\overline{{{\omega}}}t\;d\overline{{{\omega}}}\tag{12-23}
 $$  
 
 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  
 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
 $$
-\mathbf{P}_{n}(i\overline{{\omega}})=\int_{-\infty}^{\infty}P_{n}(t)\,\exp(-i\overline{{\omega}}t)\;d t
+\mathbf{P}_{n}(i\overline{{\omega}})=\int_{-\infty}^{\infty}P_{n}(t)\,\exp(-i\overline{{\omega}}t)\;d t\tag{12-24}
 $$  
 
 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:  
 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：
 $$
-\begin{array}{r l r}{\mathrm{H}_{n}(i\overline{{\omega}})=\frac{1}{\omega_{n}^{2}M_{n}}\left[\frac{1}{\left(1-\beta_{n}^{2}\right)+i(2\xi_{n}\beta_{n})}\right]}&{}&\\ {=\frac{1}{\omega_{n}^{2}M_{n}}\left[\frac{\left(1-\beta_{n}^{2}\right)-i(2\xi_{n}\beta_{n})}{(1-\beta_{n}^{2})^{2}+(2\xi_{n}\beta_{n})^{2}}\right]}&{}&{\xi_{n}\ge0}\end{array}
+\begin{array}{r l r}{\mathrm{H}_{n}(i\overline{{\omega}})=\frac{1}{\omega_{n}^{2}M_{n}}\left[\frac{1}{\left(1-\beta_{n}^{2}\right)+i(2\xi_{n}\beta_{n})}\right]}&{}&\\ {=\frac{1}{\omega_{n}^{2}M_{n}}\left[\frac{\left(1-\beta_{n}^{2}\right)-i(2\xi_{n}\beta_{n})}{(1-\beta_{n}^{2})^{2}+(2\xi_{n}\beta_{n})^{2}}\right]}&{}&{\xi_{n}\ge0}\end{array}\tag{12-25}
 $$  
 
 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)]  
+
 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））。
 $$
-Y_{n}(t)=\left[Y_{n}(0)\,\cos{\omega_{D n}t}+\left(\frac{\dot{Y}_{n}(0)+Y_{n}(0)\xi_{n}\omega_{n}}{\omega_{D n}}\right)\,\sin{\omega_{D n}t}\right]\,\exp(-\xi_{n}\omega_{n}t)
+Y_{n}(t)=\left[Y_{n}(0)\,\cos{\omega_{D n}t}+\left(\frac{\dot{Y}_{n}(0)+Y_{n}(0)\xi_{n}\omega_{n}}{\omega_{D n}}\right)\,\sin{\omega_{D n}t}\right]\,\exp(-\xi_{n}\omega_{n}t)\tag{12-26}
 $$  
 
 must be added to the forced-vibration response given by Eqs. (12-20) or (12-23). The initial conditions $Y_{n}(0)$ and $\dot{Y}_{n}(0)$ in this equation are determined from $\mathbf{v}(0)$ and $\dot{\mathbf{v}}(0)$ using Eqs. (12-6) and (12-7) in the forms  
-
+必须将其添加到方程（12-20）或（12-23）给出的强迫振动响应中。该方程中的初始条件 $Y_{n}(0)$ 和 $\dot{Y}_{n}(0)$ 可通过方程（12-6）和（12-7）的形式，利用 $\mathbf{v}(0)$ 和 $\dot{\mathbf{v}}(0)$ 确定。
 $$
-\begin{array}{l}{\displaystyle Y_{n}(0)=\frac{\phi_{n}^{T}\textbf{m v}(0)}{\phi_{n}^{T}\textbf{m}\phi_{n}}}\\ {\displaystyle\dot{Y}_{n}(0)=\frac{\phi_{n}^{T}\textbf{m}\dot{\mathbf{v}}(0)}{\phi_{n}^{T}\textbf{m}\phi_{n}}}\end{array}
+{\displaystyle Y_{n}(0)=\frac{\phi_{n}^{T}\textbf{m v}(0)}{\phi_{n}^{T}\textbf{m}\phi_{n}}}\tag{12-27}
+$$
+$$
+{\displaystyle\dot{Y}_{n}(0)=\frac{\phi_{n}^{T}\textbf{m}\dot{\mathbf{v}}(0)}{\phi_{n}^{T}\textbf{m}\phi_{n}}}\tag{12-28}
 $$  
 
 Having generated the total response for each mode $Y_{n}(t)$ using either Eq. (12- 20) or Eq. (12-23) and Eq. (12-26), the displacements expressed in the geometric coordinates can be obtained using Eq. (12-2), i.e.,  
-
+在使用公式（12-20）或公式（12-23）和（12-26）生成各模态 $Y_{n}(t)$ 的总响应后，可通过几何坐标下的位移表达式（即公式（12-2））得到各模态的位移，即：
 $$
-\mathbf{v}(t)=\phi_{1}\,\,Y_{1}(t)+\phi_{2}\,\,Y_{2}(t)+\cdot\,\cdot\,\cdot+\phi_{N}\,\,Y_{N}(t)
+\mathbf{v}(t)=\phi_{1}\,\,Y_{1}(t)+\phi_{2}\,\,Y_{2}(t)+\cdot\,\cdot\,\cdot+\phi_{N}\,\,Y_{N}(t)\tag{12-29}
 $$  
 
 which superposes the separate modal displacement contributions; hence, the commonly referred to name mode superposition method. It should be noted that for most types of loadings the displacement contributions generally are greatest for the lower modes and tend to decrease for the higher modes. Consequently, it usually is not necessary to include all the higher modes of vibration in the superposition process; the series can be truncated when the response has been obtained to any desired degree of accuracy. Moreover, it should be kept in mind that the mathematical idealization of any complex structural system also tends to be less reliable in predicting the higher modes of vibration; for this reason, too, it is well to limit the number of modes considered in a dynamic-response analysis.  
+该方法将各独立模态位移贡献进行叠加，因此常被称为**模态叠加法**。需要注意的是，对于大多数类型的荷载作用，位移贡献通常在低阶模态中最大，而随着模态阶数升高而逐渐减小。因此，在叠加过程中通常不需要包含所有高阶模态；当响应达到所需精度时，级数可以截断。此外，还应考虑到，任何复杂结构系统的数学理想化在预测高阶模态时可靠性较低；基于此，在动力响应分析中也应适当限制所考虑的模态数量。
 
 The displacement time-histories in vector $\mathbf{v}(t)$ may be considered to be the basic measure of a structure’s overall response to dynamic loading. In general, other response parameters such as stresses or forces developed in various structural components can be evaluated directly from the displacements. For example, the elastic forces $\mathbf{f}_{S}$ which resist the deformation of the structure are given directly by  
-
+结构在动载作用下的整体响应可通过向量形式的位移时程 **$\mathbf{v}(t)$** 来衡量。通常，其他响应参数（如各结构构件中的应力或作用力）均可直接由位移推导得出。例如，抵抗结构变形的弹性力 **$\mathbf{f}_{S}$** 可直接表示为：
 $$
-\mathbf{f}_{S}(t)=\mathbf{k}\;\mathbf{v}(t)=\mathbf{k}\;\Phi\;Y(t)
+\mathbf{f}_{S}(t)=\mathbf{k}\;\mathbf{v}(t)=\mathbf{k}\;\Phi\;Y(t)\tag{12-30}
 $$  
 
 An alternative expression for the elastic forces may be useful in cases where the frequencies and mode shapes have been determined from the flexibility form of the eigenvalue equation [Eq. (11-17)]. Writing Eq. (12-30) in terms of the modal contributions  
-
+在某些情况下，当频率和模态形状是通过柔度形式的特征值方程（式（11-17））确定的，弹性力的另一种表达方式可能更为有用。将式（12-30）用模态贡献表示。
 $$
-{\mathbf{f}}_{S}(t)={\mathbf{k}}\;\phi_{1}\;Y_{1}(t)+{\mathbf{k}}\;\phi_{2}\;Y_{2}(t)+{\mathbf{k}}\;\phi_{3}\;Y_{3}(t)+\cdot\cdot\cdot
+{\mathbf{f}}_{S}(t)={\mathbf{k}}\;\phi_{1}\;Y_{1}(t)+{\mathbf{k}}\;\phi_{2}\;Y_{2}(t)+{\mathbf{k}}\;\phi_{3}\;Y_{3}(t)+\cdot\cdot\cdot\tag{12-31}
 $$  
 
 and substituting Eq. (11-39) leads to  
 
 $$
-\mathbf{f}_{S}(t)=\omega_{1}^{2}\,\mathbf{m}\,\phi_{1}\,Y_{1}(t)+\omega_{2}^{2}\,\mathbf{m}\,\phi_{2}\,Y_{2}(t)+\omega_{3}^{2}\,\mathbf{m}\,\phi_{3}\,Y_{3}(t)+\cdot\cdot\cdot
+\mathbf{f}_{S}(t)=\omega_{1}^{2}\,\mathbf{m}\,\phi_{1}\,Y_{1}(t)+\omega_{2}^{2}\,\mathbf{m}\,\phi_{2}\,Y_{2}(t)+\omega_{3}^{2}\,\mathbf{m}\,\phi_{3}\,Y_{3}(t)+\cdot\cdot\cdot\tag{12-32}
 $$  
 
 Writing this series in matrix form gives  
+将该系列以矩阵形式表示，得到：
 
 $$
-\mathbf{f}_{S}(t)=\mathbf{m}\boldsymbol{\Phi}\left\{\omega_{n}^{2}\,Y_{n}(t)\right\}
+\mathbf{f}_{S}(t)=\mathbf{m}\boldsymbol{\Phi}\left\{\omega_{n}^{2}\,Y_{n}(t)\right\}\tag{12-33}
 $$  
 
 where $\{\omega_{n}^{2}\,Y_{n}(t)\}$ represents a vector of modal amplitudes each multiplied by the square of its modal frequency.  
+其中 $\{\omega_{n}^{2}\,Y_{n}(t)\}$ 表示一个向量，包含每个模态振幅乘以其模态频率的平方。
 
 In Eq. (12-33), the elastic force associated with each modal component has been replaced by an equivalent modal inertial-force expression. The equivalence of these expressions was demonstrated from the equations of free-vibration equilibrium [Eq. (11-39)]; however, it should be noted that this substitution is valid at any time, even for a static analysis.  
+在式（12-33）中，每个模态分量相关的弹性力已被等效的模态惯性力表达式所替代。虽然这些表达式的等效性是通过自由振动平衡方程 [式（11-39）] 进行了证明，但应注意的是，这种替换在任何时刻都成立，即使是在静力分析中。
 
 Because each modal contribution is multiplied by the square of the modal frequency in Eq. (12-33), it is evident that the higher modes are of greater significance in defining the forces in the structure than they are in the displacements. Consequently, it will be necessary to include more modal components to define the forces to any desired degree of accuracy than to define the displacements.  
+由于方程（12-33）中每个模态贡献均乘以模态频率的平方，显然高阶模态在定义结构中的**力**方面比在**位移**方面具有更大的重要性。因此，为了以所需的精确度定义**力**，需要包含更多的模态分量，而定义**位移**时则不需如此。
 
-Example E12-1. Various aspects of the mode-superposition procedure will be illustrated by reference to the three-story frame structure of Example E11-1 (Fig. E11-1). For convenience, the physical and vibration properties of  
-
-the structure are summarized here:  
+Example E12-1. Various aspects of the mode-superposition procedure will be illustrated by reference to the three-story frame structure of Example E11-1 (Fig. E11-1). For convenience, the physical and vibration properties of the structure are summarized here:  
 
+各种模态叠加方法的具体应用将通过参考示例 E11-1（图 E11-1）中的三层框架结构进行说明。为了方便起见，该结构的物理和振动特性在此总结如下：
 $$
-\mathbf{m}={\left[\begin{array}{l l l}{1.0}&{0}&{0}\\ {0}&{1.5}&{0}\\ {0}&{0}&{2.0}\end{array}\right]}{\begin{array}{l}{k i p s\cdot s e c^{2}/i n}\\ {k i p s\cdot s e c^{2}/i n}\\ {0}\end{array}}
+\mathbf{m}=\begin{array}{array}{\left[\begin{array}{l l l}{1.0}&{0}&{0}\\ {0}&{1.5}&{0}\\ {0}&{0}&{2.0}\end{array}\right]}{k i p s\cdot s e c^{2}/i n}\end{array}
 $$  
 
 $$
-\mathbf{k}=600\begin{array}{array}{r}{\left[\begin{array}{r r r r}{1}&{-1}&{0}\\ {-1}&{3}&{-2}\\ {0}&{-2}&{5}\end{array}\right]\,\,k i p s/i n}\end{array}
+\mathbf{k}=600\begin{array}{array}{\left[\begin{array}{r r r r}{1}&{-1}&{0}\\ {-1}&{3}&{-2}\\ {0}&{-2}&{5}\end{array}\right]\,\,k i p s/i n}\end{array}
 $$  
 
 $$
@@ -413,31 +421,34 @@ $$
 
 That the higher-mode contributions are more significant with respect to the force response than for the displacements is quite evident from a comparison of expressions (f) and (g).  
 
-# Complex-Stiffness Damping  
+## Complex-Stiffness Damping**复杂刚度阻尼**
 
 As pointed out in Section 3-7, damping of the linear viscous form represented in Eqs. (12-17) has a serious deficiency because the energy loss per cycle at a fixed displacement amplitude is dependent upon the response frequency; see Eq. (3-61). Since this dependency is at variance with a great deal of test evidence which indicates that the energy loss per cycle is essentially independent of the frequency, it would be better to solve the uncoupled normal mode equations of motion in the frequency domain using complex-stiffness damping rather than viscous damping; in that case the energy loss per cycle would be independent of frequency; see Eq. (3-84).  
+如第 3-7 节所述，方程（12-17）中所示的线性黏性阻尼形式存在严重不足，因为其在固定位移振幅下每周期的能量损失依赖于响应频率（见方程（3-61））。由于这一依赖关系与大量实验证据相矛盾——实验表明每周期的能量损失在本质上与频率无关，因此更合理的做法是采用复数刚度阻尼（而非黏性阻尼）在频域内求解解耦的简正模态运动方程；此时，每周期的能量损失将不依赖于频率（见方程（3-84））。
 
 Making this change in type of damping by using a complex-generalized-stiffness of the form given by Eq. (3-79), that is, using  
-
+将阻尼类型的变化通过采用 Eq. (3-79) 中给出的复数广义刚度形式实现，即采用
 $$
-\hat{K}_{n}=K_{n}\,\left[1+i\,2\,\xi_{n}\right]
+\hat{K}_{n}=K_{n}\,\left[1+i\,2\,\xi_{n}\right]\tag{12-34}
 $$  
 
 in which  
 
 $$
-K_{n}=\omega_{n}^{2}\:M_{n}
+K_{n}=\omega_{n}^{2}\:M_{n}\tag{12-35}
 $$  
 
 the response will be given by Eq. (12-23) using the complex-frequency-response transfer function  
-
+响应将由式（12-23）给出，采用复频域响应传递函数。
 $$
-\mathrm{H}_{n}(i\overline{{\omega}})=\frac{1}{\omega_{n}^{2}\,M_{n}}\left[\frac{1}{\left(1-\beta_{n}^{2}\right)+i\left(2\xi_{n}\right)}\right]=\frac{1}{\omega_{n}^{2}\,M_{n}}\left[\frac{(1-\beta_{n}^{2})-i\left(2\xi_{n}\right)}{(1-\beta_{n}^{2})^{2}+(2\xi_{n})^{2}}\right]
+\mathrm{H}_{n}(i\overline{{\omega}})=\frac{1}{\omega_{n}^{2}\,M_{n}}\left[\frac{1}{\left(1-\beta_{n}^{2}\right)+i\left(2\xi_{n}\right)}\right]=\frac{1}{\omega_{n}^{2}\,M_{n}}\left[\frac{(1-\beta_{n}^{2})-i\left(2\xi_{n}\right)}{(1-\beta_{n}^{2})^{2}+(2\xi_{n})^{2}}\right]\tag{12-36}
 $$  
 
 rather than the corresponding transfer function given by Eq. (12-25) for viscous damping; see Eq. (6-46). All quantities in this transfer function are defined the same as those in the transfer function of Eq. (12-25).  
+与粘滞阻尼对应的传递函数（见方程（12-25））不同，此处采用方程（6-46）所示的传递函数。该传递函数中所有量的定义与方程（12-25）中的传递函数一致。
 
 Having obtained the forced-vibration response $Y_{n}(t)$ for each normal mode of interest (a limited number of the lower modes) using Eqs. (12-23), (12-24), and (12-36), the free-vibration response of Eq. (12-26) can be added to it giving the total response. One can then proceed to obtain the displacement vector $\mathbf{v}(t)$ by superposition using Eq. (12-29) and the elastic force vector $\mathbf{f}_{S}(t)$ using either Eq. (12-30) or Eq. (12-33).  
+通过使用方程（12-23）、（12-24）和（12-36）获得各感兴趣简正模态（仅限少数低阶模态）的强迫振动响应 $Y_{n}(t)$ 后，可将其与方程（12-26）中的自由振动响应相加，得到总响应。接下来，可通过方程（12-29）进行叠加得到位移向量 $\mathbf{v}(t)$，并利用方程（12-30）或（12-33）求得弹性力向量 $\mathbf{f}_{S}(t)$。
 
 Example E12-3. A mechanical exciter placed on the top mass of the frame shown in Fig. E11-1 subjects the structure to a harmonic lateral loading of amplitude $p_{0}$ at frequency $\overline{{\omega}}$ , i.e., it produces the force  
 
@@ -503,77 +514,87 @@ $$
 
 In the above Example E12-3, the loading was of a simple harmonic form which allowed an easy solution using the appropriate transfer functions in the frequency domain. However, if each component in vector $\mathbf p(t)$ had been nonperiodic of arbitrary form giving corresponding nonperiodic normal-coordinate generalized loads $P_{n}(t)$ $(n=1,2,3)$ ) in accordance with Eq. (12-12c), it would be necessary to Fourier transform each of these generalized loads as indicated by Eq. (12-24) using the FFT procedure described in Chapter 6, thus obtaining $N-1$ discrete harmonics in accordance with $N=2^{\gamma}$ where $\gamma$ is an integer selected appropriately; see discussion of solutions in Fig. 6-4. Assuming zero initial conditions on each normal coordinate $Y_{n}(t)$ $(n=1,2,3)$ ), its time-history of response following $t=0$ would be obtained upon multiplying each discrete harmonic in $P_{n}(t)$ by the corresponding complex frequency response transfer function as illustrated in Example E12-3. The $N\!-\!1$ products would then be summed giving $Y_{n}(t)$ . Carrying out this procedure for all values of $n=1,2,3$ , the time-histories of response $\mathbf{v}(t)$ would be obtained by superposition as in Example E12-3.  
 
-# 12-5 CONSTRUCTION OF PROPORTIONAL VISCOUS DAMPING MATRICES  
+# 12-5 CONSTRUCTION OF PROPORTIONAL VISCOUS DAMPING MATRICES **比例粘滞阻尼矩阵的构建**  
 
-# Rayleigh Damping  
+## Rayleigh Damping  
 
 As was stated above, generally there is no need to express the damping of a typical viscously damped MDOF system by means of the damping matrix because it is represented more conveniently in terms of the modal damping ratios $\xi_{n}$ $\left[n\right.=$ $1,2,\cdots,N)$ . However, in at least two dynamic analysis situations the response is not obtained by superposition of the uncoupled modal responses, so the damping cannot be expressed by the damping ratios — instead an explicit damping matrix is needed. These two situations are: (1) nonlinear responses, for which the mode shapes are not fixed but are changing with changes of stiffness, and (2) analysis of a linear system having nonproportional damping. In both of these circumstances, the most effective way to determine the required damping matrix is to first evaluate one or more proportional damping matrices. In performing a nonlinear analysis, it is appropriate to define the proportional damping matrix for the initial elastic state of the system (before nonlinear deformations have occurred) and to assume that this damping property remains constant during the response even though the stiffness may be changing and causing hysteretic energy losses in addition to the viscous damping losses. In cases where the damping is considered to be nonproportional, an appropriate damping matrix can be constructed by assembling a set of suitably derived proportional damping matrices, as explained later in this section. Thus for these two situations, it is necessary to be able to derive appropriate proportional damping matrices.  
 
+如前所述，通常无需通过阻尼矩阵来表达典型黏滞阻尼多自由度（MDOF）系统的阻尼，因为其更方便地以**模态阻尼比** $\xi_{n}$（$n=1,2,\cdots,N$）表示。然而，在至少两种动力学分析场景中，响应并非通过解耦模态响应的叠加获得，因此无法用阻尼比表示阻尼——此时需明确给出一个**阻尼矩阵**。这两种情况包括：
+1. **非线性响应**，此时模态形状不固定，且随刚度变化而变化；
+2. **具有非比例阻尼的线性系统**分析。
+在这两种情况下，**确定所需阻尼矩阵的最有效方法**是先评估一个或多个**比例阻尼矩阵**。在进行非线性分析时，应定义系统初始弹性状态（未发生非线性变形前）的比例阻尼矩阵，并假设该阻尼特性在响应过程中保持不变，即使刚度可能变化并导致额外的**滞回能量损失**（除黏滞阻尼损失外）。对于非比例阻尼情况，可通过组合一组合理推导的比例阻尼矩阵构建适当的阻尼矩阵（后文将详细说明）。因此，**在这两种情况下，能够推导出合适的比例阻尼矩阵是必要的**。
+
 Clearly the simplest way to formulate a proportional damping matrix is to make it proportional to either the mass or the stiffness matrix because the undamped mode shapes are orthogonal with respect to each of these. Thus the damping matrix might be given by  
+显然，构造一个比例阻尼矩阵最简单的方法是将其与质量矩阵或刚度矩阵成比例关系，因为无阻尼模态形状在这两种矩阵下都是正交的。因此，阻尼矩阵可以表示为：
 
 $$
-\mathbf{c}=a_{0}\;\mathbf{m}\qquad{\mathrm{or}}\qquad\mathbf{c}=a_{1}\;\mathbf{k}
+\mathbf{c}=a_{0}\;\mathbf{m}\qquad{\mathrm{or}}\qquad\mathbf{c}=a_{1}\;\mathbf{k}\tag{12-37a}
 $$  
 
 in which the proportionality constants $a_{0}$ and $a_{1}$ have units of $s e c^{-1}$ and $s e c$ , respectively. These are called mass proportional and stiffness proportional damping, and the damping behavior associated with them may be recognized by evaluating the generalized modal damping value for each [see Eq. (12-15a)],  
-
+其中，比例常数$a_{0}$ 和$a_{1}$ 的单位分别为$\mathrm{sec}^{-1}$ 和$\mathrm{sec}$。它们被称为质量比例阻尼和刚度比例阻尼，其对应的阻尼行为可通过计算各自的广义模态阻尼值来识别（见式（12-15a））。
 $$
-\begin{array}{r}{C_{n}=\phi_{n}^{T}\,c\,\phi_{n}=a_{0}\,\phi_{n}^{T}\,{\bf m}\,\phi_{n}\qquad\mathrm{or}\qquad a_{1}\,\phi_{n}^{T}\,{\bf k}\,\phi_{n}}\end{array}
+\begin{array}{r}{C_{n}=\phi_{n}^{T}\,c\,\phi_{n}=a_{0}\,\phi_{n}^{T}\,{\bf m}\,\phi_{n}\qquad\mathrm{or}\qquad a_{1}\,\phi_{n}^{T}\,{\bf k}\,\phi_{n}}\end{array}\tag{12-37b}
 $$  
 
 or combining with Eq. (12-15b)  
 
 $$
-2\omega_{n}\,M_{n}\,\xi_{n}=a_{0}\;M_{n}\quad\mathrm{or}\quad a_{1}\,K_{n}\qquad\mathrm{(where}\quad K_{n}=\omega_{n}^{2}\,M_{n})
+2\omega_{n}\,M_{n}\,\xi_{n}=a_{0}\;M_{n}\quad\mathrm{or}\quad a_{1}\,K_{n}\qquad\mathrm{(where}\quad K_{n}=\omega_{n}^{2}\,M_{n})\tag{12-37c}
 $$  
 
 from which  
 
 $$
-\xi_{n}=\frac{a_{0}}{2\omega_{n}}\qquad\mathrm{or}\qquad\xi_{n}=\frac{a_{1}\omega_{n}}{2}
+\xi_{n}=\frac{a_{0}}{2\omega_{n}}\qquad\mathrm{or}\qquad\xi_{n}=\frac{a_{1}\omega_{n}}{2}\tag{12-37d}
 $$  
 
 These expressions show that for mass proportional damping, the damping ratio is inversely proportional to the frequency while for stiffness proportional damping it is directly in proportion with the frequency. In this regard it is important to note that the dynamic response generally will include contributions from all $N$ modes even though only a limited number of modes are included in the uncoupled equations of motion. Thus, neither of these types of damping matrix is suitable for use with an MDOF system in which the frequencies of the significant modes span a wide range because the relative amplitudes of the different modes will be seriously distorted by inappropriate damping ratios.  
+这些表达式表明，对于质量比例阻尼，阻尼比与频率成反比；而对于刚度比例阻尼，阻尼比则与频率成正比。在此情况下，需要注意的是，动态响应通常会包含来自所有 $N$ 个模态的贡献，尽管在解耦的运动方程中仅包含有限数量的模态。因此，这两种类型的阻尼矩阵都不适用于频率跨度较大的多自由度（MDOF）系统，因为不同模态的相对振幅将因不恰当的阻尼比而严重扭曲。
 
 An obvious improvement results if the damping is assumed to be proportional to a combination of the mass and the stiffness matrices as given by the sum of the two alternative expressions shown in Eq. (12-37a):  
-
+假设阻尼与质量矩阵和刚度矩阵的组合成正比，如方程（12-37a）中所示的两种表达式之和，则显然可以获得显著的改进。
 $$
-\mathbf{c}=a_{0}\ \mathbf{m}+a_{1}\ \mathbf{k}
+\mathbf{c}=a_{0}\ \mathbf{m}+a_{1}\ \mathbf{k}\tag{12-38a}
 $$  
 
 This is called Rayleigh damping, after Lord Rayleigh, who first suggested its use. By analogy with the development in Eqs. (12-37b) to (12-37d), it is evident that Rayleigh damping leads to the following relation between damping ratio and frequency  
+此方法称为 **瑞利阻尼**（Rayleigh damping），以瑞利勋爵（Lord Rayleigh）命名，他首次提出了这一概念。通过类比方程（12-37b）至（12-37d）的推导过程，可以得出瑞利阻尼下阻尼比与频率之间的关系如下：
 
 $$
-\xi_{n}=\frac{a_{0}}{2\omega_{n}}+\frac{a_{1}\omega_{n}}{2}
+\xi_{n}=\frac{a_{0}}{2\omega_{n}}+\frac{a_{1}\omega_{n}}{2}\tag{12-38b}
 $$  
 
 The relationships between damping ratio and frequency expressed by Eqs. (12-37d) and (12-38b) are shown graphically in Fig. 12-2.  
-
-Now it is apparent that the two Rayleigh damping factors, $a_{0}$ and $a_{1}$ , can be evaluated by the solution of a pair of simultaneous equations if the damping ratios $\xi_{m}$ and $\xi_{n}$ associated with two specific frequencies (modes) $\omega_{m},\omega_{n}$ are known. Writing Eq. (12-38b) for each of these two cases and expressing the two equations in matrix  
+方程（12-37d）和（12-38b）所描述的阻尼比与频率之间的关系在图 12-2 中以图形形式展示。
+Now it is apparent that the two Rayleigh damping factors, $a_{0}$ and $a_{1}$ , can be evaluated by the solution of a pair of simultaneous equations if the damping ratios $\xi_{m}$ and $\xi_{n}$ associated with two specific frequencies (modes) $\omega_{m},\omega_{n}$ are known. Writing Eq. (12-38b) for each of these two cases and expressing the two equations in matrix form leads to  
+现在可以明显看出，如果已知与两个特定频率（模态）$\omega_{m}、\omega_{n}$ 对应的阻尼比 $\xi_{m}$ 和 $\xi_{n}$，则两个瑞利阻尼系数 $a_{0}$ 和 $a_{1}$ 可以通过解一对联立方程来求得。将式（12-38b）分别应用于这两种情况，并以矩阵形式表示这两个方程，可得到
 
 ![](c07eae58b24a73c0ac680d621c20ece8a5227136570d53582a9a6aa950090fcd.jpg)  
 
-form leads to  
 
 $$
-\left\{\begin{array}{c}{\xi_{m}}\\ {\xi_{n}}\end{array}\right\}=\frac{1}{2}\begin{array}{c}{\left[1/\omega_{m}\quad\omega_{m}\right]}\\ {\left[1/\omega_{n}\quad\omega_{n}\right]}\end{array}\left\{\begin{array}{c}{a_{0}}\\ {a_{1}}\end{array}\right\}
+\left\{\begin{array}{c}{\xi_{m}}\\ {\xi_{n}}\end{array}\right\}=\frac{1}{2}\begin{array}{c}{\left[1/\omega_{m}\quad\omega_{m}\right]}\\ {\left[1/\omega_{n}\quad\omega_{n}\right]}\end{array}\left\{\begin{array}{c}{a_{0}}\\ {a_{1}}\end{array}\right\}\tag{12-39}
 $$  
 
 and the factors resulting from the simultaneous solution are  
-
+同时求解得到的各因子为：
 $$
-\left\{{\begin{array}{c}{a_{0}}\\ {a_{1}}\end{array}}\right\}=2\;{\frac{\omega_{m}\omega_{n}}{\omega_{n}^{2}-\omega_{m}^{2}}}\;\left[{\begin{array}{c c}{\omega_{n}}&{-\omega_{m}}\\ {-1/\omega_{n}}&{1/\omega_{m}}\end{array}}\right]\;{\left\{{\begin{array}{c}{\xi_{m}}\\ {\xi_{n}}\end{array}}\right\}}
+\left\{{\begin{array}{c}{a_{0}}\\ {a_{1}}\end{array}}\right\}=2\;{\frac{\omega_{m}\omega_{n}}{\omega_{n}^{2}-\omega_{m}^{2}}}\;\left[{\begin{array}{c c}{\omega_{n}}&{-\omega_{m}}\\ {-1/\omega_{n}}&{1/\omega_{m}}\end{array}}\right]\;{\left\{{\begin{array}{c}{\xi_{m}}\\ {\xi_{n}}\end{array}}\right\}}\tag{12-40}
 $$  
 
 When these factors have been evaluated, the proportional damping matrix that will give the required values of damping ratio at the specified frequencies is given by the Rayleigh damping expression, Eq. (12-38a), as shown by Fig. 12-2.  
+当对这些因素进行评估后，能够在指定频率下提供所需阻尼比的比例阻尼矩阵，可通过瑞利阻尼表达式（式（12-38a））给出，如图 12-2 所示。
 
 Because detailed information about the variation of damping ratio with frequency seldom is available, it usually is assumed that the same damping ratio applies to both control frequencies; i.e., $\xi_{m}=\xi_{n}\equiv\xi$ . In this case, the proportionality factors are given by a simplified version of Eq. (12-40):  
-
+由于关于阻尼比随频率变化的详细信息鲜有可用，通常假设相同的阻尼比适用于两种控制频率，即：$\xi_{m} = \xi_{n} \equiv \xi$。此时，比例系数由 Eq. (12-40) 的简化形式给出：
 $$
-\left\{{\begin{array}{c}{a_{0}}\\ {a_{1}}\end{array}}\right\}={\frac{2\xi}{\omega_{m}+\omega_{n}}}\,\left\{{\begin{array}{c}{\omega_{m}\omega_{n}}\\ {1}\end{array}}\right\}
+\left\{{\begin{array}{c}{a_{0}}\\ {a_{1}}\end{array}}\right\}={\frac{2\xi}{\omega_{m}+\omega_{n}}}\,\left\{{\begin{array}{c}{\omega_{m}\omega_{n}}\\ {1}\end{array}}\right\}\tag{12-41}
 $$  
 
 In applying this proportional damping matrix derivation procedure in practice, it is recommended that $\omega_{m}$ generally be taken as the fundamental frequency of the MDOF system and that $\omega_{n}$ be set among the higher frequencies of the modes that contribute significantly to the dynamic response. The derivation ensures that the desired damping ratio is obtained for these two modes (i.e., $\xi_{1}=\xi_{n}=\xi)$ ; then as shown by Fig. 12-2, modes with frequencies between these two specified frequencies will have somewhat lower values of damping ratio, while all modes with frequencies greater than $\omega_{n}$ will have damping ratios that increases above $\xi_{n}$ monotonically with frequency. The end result of this situation is that the responses of very high frequency modes are effectively eliminated by their high damping ratios.  
+在实际应用该比例阻尼矩阵推导程序时，建议通常将 $\omega_{m}$ 取为多自由度（MDOF）系统的基频，而将 $\omega_{n}$ 设定为对动力响应有显著贡献的高阶模态频率之一。该推导确保这两个模态（即第1模态和第 $n$ 模态）能达到所需的阻尼比（即 $\xi_{1} = \xi_{n} = \xi$）；如图 12-2 所示，频率位于这两者之间的模态将具有略低的阻尼比，而所有频率高于 $\omega_{n}$ 的模态其阻尼比则会随着频率单调递增超过 $\xi_{n}$。这种情况的最终结果是，高频模态的响应因其较高的阻尼比而被有效抑制。
 
 Example E12-4. For the structure of Example E11-1, an explicit damping matrix is to be defined such that the damping ratio in the first and third modes will be 5 percent of critical. Assuming Rayleigh damping, the proportionality factors $a_{0}$ and $a_{1}$ can be evaluated from Eq. (12-39), using the frequency data listed in Example E12-1, as follows:  
 
@@ -607,48 +628,49 @@ $$
 
 Hence, even though only the first and third damping ratios were specified, the resulting damping ratio for the second mode is a reasonable value.  
 
-# Extended Rayleigh Damping  
+## Extended Rayleigh Damping  
 
 The mass and stiffness matrices used to formulate Rayleigh damping are not the only matrices to which the free-vibration mode-shape orthogonality conditions apply; in fact, it was shown earlier in Eq. (11-44) that an infinite number of matrices have this property. Therefore a proportional damping matrix can be made up of any combination of these matrices, as follows:  
-
+以下矩阵和刚度矩阵用于雷利阻尼（Rayleigh damping）的建模，并非唯一满足自由振动模态正交性条件的矩阵；实际上，如前所述（见式（11-44）），无限多个矩阵具有这一性质。因此，比例阻尼矩阵可以由这些矩阵的任意组合构成，具体如下：
 $$
-\mathbf{c}=m\,\sum_{b}a_{b}[m^{-1}\,k]^{b}\equiv\sum_{b}\,c_{b}
+\mathbf{c}=m\,\sum_{b}a_{b}[m^{-1}\,k]^{b}\equiv\sum_{b}\,c_{b}\tag{12-42}
 $$  
 
 in which the coefficients $a_{b}$ are arbitrary. It is evident that Rayleigh damping is given by Eq. (12-42) if only the terms $b=0$ and $b=1$ are retained in the series. By retaining additional terms of the series a proportional damping matrix can be constructed that gives any desired damping ratio $\xi_{n}$ at a specified frequency $\omega_{n}$ for as many frequencies as there are terms in the series of Eq. (12-42).  
+在其中，$a_{b}$ 为任意系数。显然，若仅保留级数中 $b=0$ 和 $b=1$ 的项，则瑞利阻尼（Rayleigh damping）即由式（12-42）给出。通过保留级数中的额外项，可以构造出一个比例阻尼矩阵，该矩阵能够在指定频率 $\omega_{n}$ 下为任意阻尼比 $\xi_{n}$ 提供所需的阻尼效果，且可支持的频率数量与式（12-42）级数中的项数相同。
 
-To understand the procedure, consider the generalized damping value $C_{n}$ for any normal mode $\cdot\cdot$ [see Eqs. (12-37b) and (12-37c)]:  
-
+To understand the procedure, consider the generalized damping value $C_{n}$ for any normal mode "n" see Eqs. (12-37b) and (12-37c):  
+要理解该程序，请考虑任意简正模态“n”的广义阻尼系数 $C_{n}$（见方程（12-37b）和（12-37c））：
 $$
-C_{n}=\pmb\phi_{n}^{T}\,\mathbf c\,\pmb\phi_{n}=2\xi_{n}\,\omega_{n}\,M_{n}
+C_{n}=\pmb\phi_{n}^{T}\,\mathbf c\,\pmb\phi_{n}=2\xi_{n}\,\omega_{n}\,M_{n}\tag{12-43}
 $$  
 
 If c in this expression is given by Eq. (12-42), the contribution of term $b$ to the generalized damping value is  
-
+如果表达式中的 $c$ 由式（12-42）给出，则项 $b$ 对广义阻尼值的贡献为：
 $$
-C_{n b}=\phi_{n}^{T}\,\mathbf{c}_{b}\,\phi_{n}=a_{b}\,\mathbf{m}\,[\mathbf{m}^{-1}\,\mathbf{k}]^{b}\,\phi_{n}
+C_{n b}=\phi_{n}^{T}\,\mathbf{c}_{b}\,\phi_{n}=a_{b}\,\mathbf{m}\,[\mathbf{m}^{-1}\,\mathbf{k}]^{b}\,\phi_{n}\tag{12-44a}
 $$  
 
 Now if Eq. (11-39) $(k\,\phi_{n}=\omega_{n}^{2}\,m\,\phi_{n})$ is premultiplied on both sides by $\phi_{n}^{T}\,\mathbf{k}\,\mathbf{m}^{-1}$ , the result is  
-
+现在，如果将方程（11-39）$(k\,\phi_{n}=\omega_{n}^{2}\,m\,\phi_{n})$ 两边左乘以 $\phi_{n}^{T}\,\mathbf{k}\,\mathbf{m}^{-1}$，则结果为：
 $$
 \phi_{n}^{T}\,\mathbf{k\,m}^{-1}\,\mathbf{k\,}\phi_{n}=\omega_{n}^{2}\,\phi_{n}^{T}\,\mathbf{k\,}\phi_{n}\equiv\omega_{n}^{4}\,M_{n}
 $$  
 
 By operations equivalent to this it can be shown that  
-
+通过与上述操作等效的方法可以证明：
 $$
-\pmb{\phi}_{n}^{T}\,\mathbf{m}\left[\mathbf{m}^{-1}\,\mathbf{k}\right]^{b}\,\pmb{\phi}_{n}=\omega_{n}^{2b}\,M_{n}
+\pmb{\phi}_{n}^{T}\,\mathbf{m}\left[\mathbf{m}^{-1}\,\mathbf{k}\right]^{b}\,\pmb{\phi}_{n}=\omega_{n}^{2b}\,M_{n}\tag{12-45}
 $$  
 
 and consequently  
 
 $$
-C_{n b}=a_{b}\,\omega_{n}^{2b}\,M_{n}
+C_{n b}=a_{b}\,\omega_{n}^{2b}\,M_{n}\tag{12-44b}
 $$  
 
 On this basis, the generalized damping value associated with any mode $n$ is  
-
+在此基础上，与任意模态 $n$ 相关的广义阻尼值为：
 $$
 C_{n}=\sum_{b}C_{n b}=\sum_{b}a_{b}\;\omega_{n}^{2b}\,M_{n}=2\xi_{n}\,\omega_{n}\,M_{n}
 $$  
@@ -656,62 +678,64 @@ $$
 from which  
 
 $$
-\xi_{n}=\frac{1}{2\omega_{n}}\sum_{b}a_{b}\;\omega_{n}^{2b}
+\xi_{n}=\frac{1}{2\omega_{n}}\sum_{b}a_{b}\;\omega_{n}^{2b}\tag{12-46}
 $$  
 
 Equation (12-46) provides the means for evaluating the constants $a_{b}$ to give the desired damping ratios at any specified number of modal frequencies. As many terms must be included in the series as there are specified modal damping ratios; then the constants are given by the solution of the set of equations, one written for each damping ratio. In principle, the values of $b$ can lie anywhere in the range $-\infty<b<\infty$ , but in practice it is desirable to select values of these exponents as close to zero as possible. For example, to evaluate the coefficients that will provide specified damping ratios in any four modes having the frequencies $\omega_{m}$ $m,\omega_{n},\omega_{o},$ $\omega_{p}$ , the equations resulting from Eq. (12-46) using the terms for $b=-1,0,+1$ , and $+2$ are  
-
+方程（12-46）提供了评估常数 $a_{b}$ 的方法，以实现任意指定模态频率下所需的阻尼比。在级数展开中，必须包含与指定阻尼比数量相等的项；然后，这些常数由一组方程的解给出，每个方程对应一个阻尼比。理论上，指数 $b$ 的取值范围为 $-\infty < b < \infty$，但在实际应用中，应尽可能选择接近零的指数值。例如，为了评估在频率分别为 $\omega_{m}$、$\omega_{n}$、$\omega_{o}$ 和 $\omega_{p}$ 的任意四个模态中提供指定阻尼比的系数，根据方程（12-46）并采用 $b = -1, 0, +1$ 和 $+2$ 的项所得到的方程为：
 $$
-\left\{\begin{array}{c}{\displaystyle\xi_{m}}\\ {\displaystyle\xi_{n}}\\ {\displaystyle\xi_{o}}\\ {\displaystyle\xi_{p}}\end{array}\right\}=\frac{1}{2}\left[\begin{array}{c c c c}{1/\omega_{m}^{2}}&{1/\omega_{m}}&{\omega_{m}}&{\omega_{m}^{3}}\\ {1/\omega_{n}^{2}}&{1/\omega_{n}}&{\omega_{n}}&{\omega_{n}^{3}}\\ {1/\omega_{o}^{2}}&{1/\omega_{o}}&{\omega_{o}}&{\omega_{o}^{3}}\\ {1/\omega_{p}^{2}}&{1/\omega_{p}}&{\omega_{p}}&{\omega_{p}^{3}}\end{array}\right]\left\{\begin{array}{c}{\displaystyle a_{-1}}\\ {\displaystyle a_{0}}\\ {\displaystyle a_{1}}\\ {\displaystyle a_{2}}\end{array}\right\}
+\left\{\begin{array}{c}{\displaystyle\xi_{m}}\\ {\displaystyle\xi_{n}}\\ {\displaystyle\xi_{o}}\\ {\displaystyle\xi_{p}}\end{array}\right\}=\frac{1}{2}\left[\begin{array}{c c c c}{1/\omega_{m}^{2}}&{1/\omega_{m}}&{\omega_{m}}&{\omega_{m}^{3}}\\ {1/\omega_{n}^{2}}&{1/\omega_{n}}&{\omega_{n}}&{\omega_{n}^{3}}\\ {1/\omega_{o}^{2}}&{1/\omega_{o}}&{\omega_{o}}&{\omega_{o}^{3}}\\ {1/\omega_{p}^{2}}&{1/\omega_{p}}&{\omega_{p}}&{\omega_{p}^{3}}\end{array}\right]\left\{\begin{array}{c}{\displaystyle a_{-1}}\\ {\displaystyle a_{0}}\\ {\displaystyle a_{1}}\\ {\displaystyle a_{2}}\end{array}\right\}\tag{12-47}
 $$  
 
 When the coefficients $a_{-1},\,a_{0},\,a_{1}$ , and $a_{2}$ have been evaluated by the simultaneous solution of Eq. (12-47), the viscous damping matrix that provides the four required damping ratios at the four specified frequencies is obtained by superposing four matrices (one for each value of $b$ ) in accordance with Eq. (12-42). Figure $12{-}3a$ illustrates the relation between damping ratio and frequency that would result from this matrix. To simplify the figure it has been assumed here that the same damping ratio, $\xi_{x}$ , was specified for all four frequencies; however, each of the damping ratios could have been specified arbitrarily. Also, $\omega_{m}$ has been taken as the fundamental mode frequency, $\omega_{1}$ , and $\omega_{p}$ is intended to approximate the frequency of the highest mode that contributes significantly to the response, while $\omega_{n}$ and $\omega_{0}$ are spaced about equally within the frequency range. It is evident in Fig. 12-3a that the damping ratio remains close to the desired value $\xi_{x}$ throughout the frequency range, being exact at the four specified frequencies and ranging slightly above or below at other frequencies in the range. It is important to note, however, that the damping increases monotonically with frequency for frequencies increasing above $\omega_{p}$ . This has the effect of excluding any significant contribution from any modes with frequencies much greater than $\omega_{p}$ , thus such modes need not be included in the response superposition.  
-
+当通过联立求解式（12-47）评估系数 $a_{-1}、a_{0}、a_{1}$ 和 $a_{2}$ 后，通过按式（12-42）叠加四个矩阵（每个 $b$ 值对应一个矩阵）即可得到满足四个指定频率下四个所需阻尼比的黏性阻尼矩阵。图 12-3a 说明了该矩阵所产生的阻尼比与频率之间的关系。为了简化图示，此处假设四个频率下均采用相同的阻尼比 $\xi_x$；然而，每个阻尼比也可以任意指定。此外，$\omega_m$ 被取为基频 $\omega_1$，而 $\omega_p$ 则旨在近似表示对响应有显著贡献的最高模态频率，$\omega_n$ 和 $\omega_0$ 在频率范围内大致均匀分布。从图 12-3a 可以看出，阻尼比在整个频率范围内保持接近期望值 $\xi_x$，在四个指定频率处完全匹配，而在其他频率处则略高或略低。然而，需要注意的是，对于高于 $\omega_p$ 的频率，阻尼比随频率单调递增。这使得任何频率远高于 $\omega_p$ 的模态对响应的贡献均可忽略不计，因此此类模态无需包含在响应叠加中。
 ![](22e11203357d49b0872b743b582c7a7f4e8d3e1804c54ff89347611e132c89d7.jpg)  
 FIGURE 12-3 Extended Rayleigh damping (damping ratio vs. frequency).  
 
-An even more important point to note is the consequence of including only three terms in the derivation of the viscous damping matrix using Eq. (12-42). In that case a set of three simultaneous equations equivalent to Eq. (12-47) would be obtained and solved for the coefficients $a_{-1}$ , $a_{0}$ , $a_{1}$ , and if these were substituted into Eq. (12-42), the resulting damping ratio-frequency relation would be as shown in Fig. $12{-}3b$ . As required by the simultaneous equation solution, the desired damping ratio is obtained exactly at the three specified frequencies, and is approximated well at intermediate frequencies. However, the serious defect of this result is that the damping decreases monotonically with frequencies increasing above $\omega_{o}$ and negative damping is indicated for all the highest modal frequencies. This is an unacceptable result because the contribution of the negatively damped modes would tend to increase without limit in the analysis but certainly would not do so in actuality.  
+An even more important point to note is the consequence of including only three terms in the derivation of the viscous damping matrix using Eq. (12-42). In that case a set of three simultaneous equations equivalent to Eq. (12-47) would be obtained and solved for the coefficients $a_{-1}$ , $a_{0}$ , $a_{1}$ , and if these were substituted into Eq. (12-42), the resulting damping ratio-frequency relation would be as shown in Fig. $12{-}3b$ . As required by the simultaneous equation solution, the desired damping ratio is obtained exactly at the three specified frequencies, and is approximated well at intermediate frequencies. However, the serious defect of this result is that the damping decreases monotonically with frequencies increasing above $\omega_{o}$ and negative damping is indicated for all the highest modal frequencies. This is an unacceptable result because the contribution of the negatively damped modes would tend to increase without limit in the analysis but certainly would not do so in actuality.  更为重要的是，在仅包含三项的黏性阻尼矩阵推导中（采用式（12-42）），将得到一组等效于式（12-47）的三个同步方程，并可解出系数 $a_{-1}$、$a_{0}$ 和 $a_{1}$。若将这些系数代入式（12-42），则得到的阻尼比-频率关系如图 12-3b 所示。根据同步方程的解，在三个指定频率处可精确获得所需的阻尼比，而在中间频率处则能较好地近似。然而，这一结果的严重缺陷在于：当频率高于 $\omega_{o}$ 时，阻尼比单调递减，且在所有最高阶模态频率下均显示出负阻尼。这是一个不可接受的结果，因为负阻尼模态在分析中会趋向于无限增大，而在实际情况下则绝不可能出现这种情况。
 
 The general implication of this observation is that extended Rayleigh damping may be used effectively only if an even number of terms is included in the series expression, Eq. (12-42). In such cases, modes with frequencies greater than the range considered in evaluating the coefficients will be effectively excluded from the mode superposition response. However, if an odd number of terms (greater than one) were included in Eq. (12-42), the modes with frequencies much greater than the controlled range would be negatively damped and would invalidate the results of the analysis.  
-
-# Alternative Formulation  
+此观察的普遍含义是，**扩展瑞利阻尼**仅在级数表达式（式（12-42））中包含**偶数项**时才能有效应用。在这种情况下，频率高于评估系数时所考虑范围的模态将被有效排除于模态叠加响应之外。然而，若式（12-42）中包含**奇数项**（大于1），则频率远高于控制范围的模态将出现**负阻尼**，从而使分析结果失效。
+## Alternative Formulation **替代表述**
 
 A second method is available for evaluating the damping matrix associated with any given set of modal damping ratios. In principle, the procedure can be explained by considering the complete diagonal matrix of generalized damping coefficients, given by pre- and postmultiplying the damping matrix by the mode-shape matrix:  
-
+第二种方法可用于评估与任意给定模态阻尼比相关的阻尼矩阵。从原理上讲，该程序可以通过考虑广义阻尼系数的完整对角矩阵来解释，该矩阵由阻尼矩阵与模态形状矩阵的**前乘**和**后乘**得到：
 $$
-C=\Phi^{T}\mathrm{\bf~c}\;\Phi=2\;\left[\begin{array}{c c c c}{\xi_{1}\omega_{1}M_{1}}&{0}&{0}&{\cdot\cdot\cdot}\\ {0}&{\xi_{2}\omega_{2}M_{2}}&{0}&{\cdot\cdot}\\ {0}&{0}&{\xi_{3}\omega_{3}M_{3}}&{\vdots}\\ {\vdots}&{\vdots}&{\vdots}&{\vdots}\end{array}\right]\;
+C=\Phi^{T}\mathrm{\bf~c}\;\Phi=2\;\left[\begin{array}{c c c c}{\xi_{1}\omega_{1}M_{1}}&{0}&{0}&{\cdot\cdot\cdot}\\ {0}&{\xi_{2}\omega_{2}M_{2}}&{0}&{\cdot\cdot}\\ {0}&{0}&{\xi_{3}\omega_{3}M_{3}}&{\vdots}\\ {\vdots}&{\vdots}&{\vdots}&{\vdots}\end{array}\right]\;\tag{12-48}
 $$  
 
 It is evident from this equation that the damping matrix can be obtained by pre- and postmultiplying matrix $c$ by the inverse of the transposed mode-shape matrix and the inverse of the mode-shape matrix, respectively, yielding  
+由此方程可以看出，阻尼矩阵可以通过将矩阵 $c$ 左乘以模态形状矩阵的转置矩阵的逆矩阵，右乘以模态形状矩阵的逆矩阵，从而得到，即：
 
 $$
-\left[\Phi^{T}\right]^{-1}C\;\Phi^{-1}=\left[\Phi^{T}\right]^{-1}\Phi^{T}\;\mathbf{c}\;\Phi\;\Phi^{-1}=\mathbf{c}
+\left[\Phi^{T}\right]^{-1}C\;\Phi^{-1}=\left[\Phi^{T}\right]^{-1}\Phi^{T}\;\mathbf{c}\;\Phi\;\Phi^{-1}=\mathbf{c}\tag{12-49}
 $$  
-
 Since for any specified set of modal damping ratios $\xi_{n}$ , the generalized damping coefficients in matrix $c$ can be evaluated, as indicated in Eq. (12-43), the damping matrix c can be evaluated using Eq. (12-49).  
+对于任意指定的模态阻尼比 $\xi_{n}$ 集合，可根据式（12-43）评估矩阵 $c$ 中的广义阻尼系数，从而利用式（12-49）求解阻尼矩阵 $c$。
 
 In practice, however, this is not a convenient procedure because inversion of the mode-shape matrix requires a large computational effort. Instead, it is useful to take advantage of the orthogonality properties of the mode shapes relative to the mass matrix. The diagonal generalized-mass matrix of the system is obtained using the relation  
+在实际应用中，该方法并不方便，因为模态形状矩阵的求逆需耗费大量计算资源。相反，更有效的方法是利用模态形状相对于质量矩阵的正交性。系统的对角广义质量矩阵可通过以下关系式获得：
 
 $$
-M=\Phi^{T}\textbf{m}\Phi
+M=\Phi^{T}\textbf{m}\Phi\tag{12-50}
 $$  
 
 Premultiplying this equation by the inverse of the generalized-mass matrix then gives  
-
+将该方程两边左乘广义质量矩阵的逆矩阵，得到：
 $$
-\mathbf{I}=M^{-1}\,M=\left[M^{-1}\,\,\Phi^{T}\,\mathbf{\Phi}\mathbf{m}\right]\,\Phi=\Phi^{-1}\,\Phi
+\mathbf{I}=M^{-1}\,M=\left[M^{-1}\,\,\Phi^{T}\,\mathbf{\Phi}\mathbf{m}\right]\,\Phi=\Phi^{-1}\,\Phi\tag{12-51}
 $$  
 
 from which it is evident that the mode-shape-matrix inverse is  
 
 $$
-\Phi^{-1}=M^{-1}\;\Phi^{T}\;{\bf m}
+\Phi^{-1}=M^{-1}\;\Phi^{T}\;{\bf m}\tag{12-52}
 $$  
 
 Operating on this expression, one can obtain  
 
 $$
-\left[\Phi^{T}\right]^{-1}=\mathbf{m}\;\Phi\;M^{-1}
+\left[\Phi^{T}\right]^{-1}=\mathbf{m}\;\Phi\;M^{-1}\tag{12-53}
 $$  
 
 Substituting Eqs. (12-52) and (12-53) into Eq. (12-49) yields  
@@ -721,7 +745,7 @@ $$
 $$  
 
 Since matrix $c$ is a diagonal matrix containing elements $C_{n}~=~2\,\xi_{n}\,\omega_{n}\,M_{n}$ , the elements of the diagonal matrix obtained as the product of the three central diagonal matrices in this equation are  
-
+由于矩阵 $c$ 是一个对角矩阵，其元素为 $C_{n} = 2\,\xi_{n}\,\omega_{n}\,M_{n}$，则该方程中三个中心对角矩阵相乘所得的对角矩阵的元素为
 $$
 d_{n}\equiv\frac{2\,\xi_{n}\,\omega_{n}}{M_{n}}
 $$  
@@ -729,214 +753,228 @@ $$
 so that Eq. (12-54) may be written  
 
 $$
-\mathbf{c}=\mathbf{m}\oplus\mathbf{d}\;\Phi^{T}\;\mathbf{m}
+\mathbf{c}=\mathbf{m}\Phi\mathbf{d}\;\Phi^{T}\;\mathbf{m}
 $$  
 
 where $\mathbf{d}$ is the diagonal matrix containing elements $d_{n}$ . In the analysis, however, it is more convenient to note that each modal damping ratio provides an independent contribution to the damping matrix, as follows:  
-
+其中 $\mathbf{d}$ 是包含元素 $d_{n}$ 的对角矩阵。然而，在分析中，**更方便的是指出每个模态阻尼比对阻尼矩阵提供独立的贡献**，具体如下：
 $$
 \mathbf{c}_{n}=\mathbf{m}\;\phi_{n}\;d_{n}\;\phi_{n}^{T}\;\mathbf{m}
 $$  
 
 Thus the total damping matrix is obtained as the sum of the modal contributions  
-
+因此，总阻尼矩阵可表示为各模态贡献的总和。
 $$
 \mathbf{c}=\sum_{n=1}^{N}\mathbf{c}_{n}=\mathbf{m}\,\Big[\sum_{n=1}^{N}\phi_{n}\,d_{n}\,\phi_{n}^{T}\Big]\,\mathbf{m}
 $$  
 
 By substituting from Eq. (12-55), this equation may be written  
-
+将方程（12-55）中的表达式代入后，该方程可表示为：
 $$
 \mathbf{c}=\mathbf{m}\,\Big[\sum_{n=1}^{N}\frac{2\,\xi_{n}\omega_{n}}{M_{n}}\,\phi_{n}\,\phi_{n}^{T}\Big]\,\mathbf{m}
 $$  
 
 In this equation, the contribution to the damping matrix from each mode is proportional to the modal damping ratio; thus any undamped mode will contribute nothing to the damping matrix. In other words, only those modes specifically included in the formation of the damping matrix will have any damping and all other modes will be undamped.  
+在这个方程中，每个模态对阻尼矩阵的贡献与该模态的阻尼比成正比；因此，任何无阻尼的模态都不会对阻尼矩阵产生任何贡献。换句话说，只有那些**明确包含**在阻尼矩阵构建中的模态才会具有阻尼，而其他所有模态则保持无阻尼状态。
 
 In order to avoid undesirable amplification of undamped modal responses, damping of the type provided by Eq. (12-56c) should be used only as a supplement to a stiffness proportional damping matrix, for which the damping ratio increases in proportion with the modal frequencies as shown by the right hand expression of Eq. (12-37d); i.e., $\xi=\frac{a_{1}\omega}{2}$ . The coefficient $a_{1}$ of this stiffness proportional damping matrix should be calculated to provide the damping ratio $\xi_{c}$ required at the frequency $\omega_{c}$ of the highest mode for which damping is specified; thus from Eq. (12-37d),  
-
+为了避免未阻尼模态响应的不良放大，仅应将如 Eq. (12-56c) 所示的阻尼作为刚度比例阻尼矩阵的补充使用。刚度比例阻尼矩阵的阻尼比随模态频率成比例增加，如 Eq. (12-37d) 右侧表达式所示，即：$\xi = \frac{a_{1}\omega}{2}$。该刚度比例阻尼矩阵的系数 $a_{1}$ 应根据最高模态频率 $\omega_{c}$（阻尼要求已明确的模态）所需的阻尼比 $\xi_{c}$ 进行计算；因此，根据 Eq. (12-37d)，
 $$
 a_{1}=\frac{2\xi_{c}}{\omega_{c}}
 $$  
 
 The stiffness proportional damping ratios at other frequencies then are given by  
-
+其他频率下的比例阻尼刚度系数由以下给出：
 $$
 \hat{\xi}_{n}=\frac{a_{1}\omega_{n}}{2}=\xi_{c}\,\left(\frac{\omega_{n}}{\omega_{c}}\right)
 $$  
 
 Hence if the total damping ratio desired in any mode $n$ is $\xi_{n}$ , it is evident that the damping of the type of Eq. (12-56c), designated $\bar{\xi}_{n}$ , required to supplement the stiffness proportional damping must be  
-
+因此，若第 $n$ 阶模态所需的总阻尼比为 $\xi_n$，则显然需要补充刚度比例阻尼的阻尼形式（如式（12-56c）所示，记为 $\bar{\xi}_n$）应为：
 $$
 \overline{{\xi}}_{n}=\xi_{n}-\xi_{c}\Big(\frac{\omega_{n}}{\omega_{c}}\Big)
 $$  
 
 The final result of this development is a proportional damping matrix c given by  
-
+最终发展结果是由比例阻尼矩阵 **c** 给出的表达式为：
 $$
 \mathbf{c}=a_{1}\,\mathbf{k}+\mathbf{m}\,\left[\sum_{n=1}^{c-1}\frac{2\overline{{\xi}}_{n}\omega_{n}}{M_{n}}\,\phi_{n}\phi_{n}^{T}\right]\mathbf{m}
 $$  
 
 which provides the desired modal damping ratios for frequencies less than or equal to $\omega_{c}$ and which has linearly increasing damping for higher frequencies.  
+该系统在频率小于或等于 $\omega_{c}$ 时提供所需的模态阻尼比，而在更高频率下则呈线性增加的阻尼特性。
+## Construction of Nonproportional Damping Matrices **非比例阻尼矩阵的构建**
 
-# Construction of Nonproportional Damping Matrices  
 
 The proportional damping matrices described in the preceding paragraphs are suitable for modeling the behavior of most structural systems, in which the damping mechanism is distributed rather uniformly throughout the structure. However, for structures made up of more than a single type of material, where the different materials provide drastically differing energy-loss mechanisms in various parts of the structure, the distribution of damping forces will not be similar to the distribution of the inertial and elastic forces; in other words, the resulting damping will be nonproportional.  
+前文中提及的比例阻尼矩阵适用于建模大多数结构系统的行为，这些系统的阻尼机制在结构中较为均匀分布。然而，对于由多种材料组成的结构，若不同材料在结构各部分提供显著不同的能量耗散机制，则阻尼力的分布将不再与惯性力和弹性力的分布相似；换句话说，所产生的阻尼将是**非比例阻尼**。
 
 A nonproportional damping matrix that will represent this situation may be constructed by applying procedures similar to those discussed above in developing proportional damping matrices, with a proportional matrix being developed for each distinct part of the structure and then the combined system matrix being formed by direct assembly. The procedure is explained with reference to Fig. 12-4, which portrays a five-story steel building frame erected on top of a five-story reinforced concrete building frame. As shown, it is assumed that the modal damping of the steel frame alone would be 5 percent of critical while that of the concrete frame alone would be 10 percent of critical.  
+一个非比例阻尼矩阵，用于描述此种情况，可通过类似上文中讨论的构建比例阻尼矩阵的方法来构造。具体步骤是：为结构中的每个独立部分分别开发一个比例阻尼矩阵，然后通过直接组装形成整体系统矩阵。以下以图 12-4 为例进行说明，该图展示了一个五层钢结构框架建筑建造在五层钢筋混凝土框架建筑之上。如图所示，假设仅钢框架的模态阻尼为临界阻尼的 5%，而仅混凝土框架的模态阻尼为临界阻尼的 10%。
 
 ![](6a950fd0db707ccb5051ac4f4c7ab12ee844dba633c5a98dbbcf98092e452511.jpg)  
-Combined frame: steel and concrete.  
+figure 12-4 Combined frame: steel and concrete.  
 
 ![](a413df76ce3e154d455ca64d6fa028f325f4f3159e57e13b58a71880f6e1a46d.jpg)  
 
 The stiffness and mass matrices of the combined system are shown qualitatively in Fig. 12-5, with the contributions from the steel frame located in the upper left corner of the combined matrices and the concrete frame contributions in the lower right corner. The contributions associated with the common degrees of freedom at the interface between the two substructures (designated areas “ $X^{\bullet}$ in the figure) include contributions from both the steel and the concrete frames. The damping matrix for the combined frame may be developed by a similar assembly procedure as shown in Fig. $12.5c$ after the damping submatrices for the steel and concrete substructures have been derived. In principle these could be evaluated by any of the procedures for developing proportional damping matrices described above, but for most cases the recommended procedure is to assume Rayleigh damping. Thus the steel and the concrete submatrices will be given respectively by [see Eq. (12-38a)]  
 
+结构系统的刚度矩阵和质量矩阵如图 12-5 所示，其中钢框架的贡献位于组合矩阵的左上角，混凝土框架的贡献位于右下角。两个子结构接口处的共同自由度（图中标记为区域“$X^{\bullet}$”）的贡献则同时包含钢框架和混凝土框架的影响。组合框架的阻尼矩阵可通过类似的组装程序（如图 12.5c 所示）构建，前提是先分别推导出钢框架和混凝土框架的子阻尼矩阵。理论上，这些子矩阵可通过上述任一构建比例阻尼矩阵的方法求得，但大多数情况下推荐采用**瑞利阻尼**假设。因此，钢框架和混凝土框架的子矩阵分别由以下公式给出（见式（12-38a））：
+
 $$
 \begin{array}{r}{\mathbf{c}_{s}=a_{0s}\;\mathbf{m}_{s}+a_{1s}\;\mathbf{k}_{s}}\\ {\mathbf{c}_{c}=a_{0c}\;\mathbf{m}_{c}+a_{1c}\;\mathbf{k}_{c}}\end{array}
 $$  
 
 in which the constants for the steel frame are evaluated as shown by Eq. (12-41):  
-
+其中，钢结构框架的常数如方程（12-41）所示进行评估：
 $$
 \left\{{\begin{array}{c}{a_{0s}}\\ {a_{1s}}\end{array}}\right\}={\frac{2\xi_{s}}{\omega_{m}+\omega_{n}}}\,\left\{{\begin{array}{c}{\omega_{m}\omega_{n}}\\ {1}\end{array}}\right\}
 $$  
 
 and the corresponding values for the concrete frame, $a_{0c}$ and $a_{1c}$ , are twice as great because $\xi_{c}=10\%$ is twice as great as $\xi_{s}=5\%$ .  
+而对于具体的框架结构，$a_{0c}$ 和 $a_{1c}$ 的值则是两倍，因为 $\xi_{c}=10\%$ 是 $\xi_{s}=5\%$ 的两倍。
 
-These values depend on the frequencies $\omega_{m}$ and $\omega_{n}$ , and the frequencies to be used must be determined by solving the eigenproblem of the combined system (i.e., using the combined stiffness and mass matrices $\mathbf{k}$ and $\mathbf{m}$ . As was mentioned above, it is recommended that $\omega_{m}$ be taken as the first mode frequency of the combined system, while for this 10-story frame it would be appropriate to use the seventh or eighth mode frequency as $\omega_{n}$ . The nonproportional damping matrix for the combined system is obtained finally by assembly as shown in Fig. $12.5c$ .  
+These values depend on the frequencies $\omega_{m}$ and $\omega_{n}$ , and the frequencies to be used must be determined by solving the eigenproblem of the combined system (i.e., using the combined stiffness and mass matrices $\mathbf{k}$ and $\mathbf{m}$ . As was mentioned above, it is recommended that $\omega_{m}$ be taken as the first mode frequency of the combined system, while for this 10-story frame it would be appropriate to use the seventh or eighth mode frequency as $\omega_{n}$ . The nonproportional damping matrix for the combined system is obtained finally by assembly as shown in Fig. 12.5c .  
+这些值取决于频率 $\omega_{m}$ 和 $\omega_{n}$，而所需使用的频率应通过联合系统的特征值问题（即利用联合系统的刚度矩阵和质量矩阵 $\mathbf{k}$ 和 $\mathbf{m}$）求解得到。如前所述，建议将 $\omega_{m}$ 取为联合系统的第一模态频率，而对于该10层框架结构，则可选择第七或第八模态频率作为 $\omega_{n}$。联合系统的非比例阻尼矩阵最终通过如图12.5c所示的组装方法获得。
 
 Using this damping matrix in the equations of motion [Eq. (9-13)] and transforming to normal coordinates by pre- and postmultiplying by the mode-shape matrix $\Phi$ for the combined system leads to the modal coordinate equations of motion  
-
+将阻尼矩阵代入运动方程 [式（9-13）]，并通过在系统耦合模态下的模态形状矩阵 $\Phi$ 左乘与右乘进行坐标变换，可得到模态坐标下的运动方程。
 $$
 M\,\ddot{\pmb{Y}}+{\pmb{C}}\,\dot{\pmb{Y}}+{\pmb{K}}\,{\pmb{Y}}={\pmb{P}}(t)
 $$  
 
 where $M$ and $\pmb{K}$ are the diagonal modal coordinate mass and stiffness matrices and $P(t)$ is the standard modal coordinate load vector. However, the modal coordinate damping matrix  
-
+其中，$M$ 和 $\pmb{K}$ 分别是对角模态坐标质量矩阵和刚度矩阵，$P(t)$ 是标准模态坐标载荷向量。然而，模态坐标阻尼矩阵
 $$
 {\cal C}=\Phi^{T}\:{\bf c}\:\Phi=\left[\begin{array}{r r r r}{{C_{11}}}&{{C_{12}}}&{{C_{13}}}&{{\cdots}}\\ {{C_{21}}}&{{C_{22}}}&{{C_{23}}}&{{\cdots}}\\ {{C_{31}}}&{{C_{32}}}&{{C_{33}}}&{{\cdots}}\\ {{\vdots}}&{{\vdots}}&{{\vdots}}&{{\vdots}}\end{array}\right]
 $$  
 
 is not diagonal but includes modal coupling coefficients $C_{i j}\left(i\neq j\right)$ because the matrix c is nonproportional.  
+该矩阵 **c** 不是对角矩阵，因为它包含非比例的模态耦合系数 $C_{ij}$（$i \neq j$）。
 
 An effective method of solving for the dynamic response using this coupled modal equation set is to merely use direct step-by-step integration, as is explained by means of an example in Chapter 15. An approximate solution may be obtained by ignoring the off-diagonal coupling coefficients of the modal damping matrix and then solving the resulting uncoupled equations as a typical mode superposition analysis.  
+要解决该耦合模态方程组的动态响应，一种有效的方法是直接采用逐步积分法，具体说明可参考第 15 章中的示例。若忽略模态阻尼矩阵中的非对角耦合系数，则可将方程简化为解耦形式，进而通过典型的模态叠加分析获得近似解。
 
 The errors resulting from this approximation are indicated in the example presented in Chapter 15; however, it must be remembered that the errors resulting in other cases from this assumed uncoupling may be larger or smaller than those found in this example.  
+此近似方法所带来的误差在第 15 章的示例中已有说明；然而，必须注意的是，在其他情况下，基于此假设的解耦所产生的误差可能会比该示例中发现的误差更大或更小。
+
+# 12-6 RESPONSE ANALYSIS USING COUPLED EQUATIONS OF MOTION  **基于耦合运动方程的响应分析**
 
-# 12-6 RESPONSE ANALYSIS USING COUPLED EQUATIONS OF MOTION  
 
 Mode superposition is a very effective means of evaluating the dynamic response of structures having many degrees of freedom because the response analysis is performed only for a series of SDOF systems. However, the computational cost in this type of calculation is transferred from the MDOF dynamic analysis to the solution of the $N$ degree of freedom undamped eigenproblem followed by the modal coordinate transformation, which must be done before the individual modal responses can be evaluated. Certainly the eigenproblem solution represents the major part of the cost of a typical mode superposition analysis, but also it must be recalled that the equations of motion will be uncoupled by the resulting undamped mode shapes only if the damping is represented by a proportional damping matrix.  
+模态叠加法是一种非常有效的手段，用于评估具有多自由度结构的动态响应，因为响应分析仅需针对一系列**单自由度（SDOF）**系统进行。然而，此类计算中的计算成本则从**多自由度（MDOF）**动力学分析转移到了求解具有 **$N$ 自由度** 的无阻尼特征值问题（eigenproblem），并随后进行模态坐标变换，这一步骤必须在各个模态响应评估之前完成。诚然，特征值问题的求解构成了典型模态叠加分析中成本的主要部分，但还应注意的是，只有当阻尼由比例阻尼矩阵表示时，运动方程才能通过所得的无阻尼模态形状实现解耦。
 
 For these reasons it is useful to examine the possibility of avoiding the modal coordinate transformation by carrying out the dynamic response analysis directly in the original geometric coordinate equations of motion; these were stated previously by Eq. (9-13) and are renumbered here for convenience:  
-
+由于以下原因，直接在原始几何坐标系下的运动方程（即动力响应分析）中避免模态坐标变换是有益的；这些方程之前已由式（9-13）给出，现为方便重新编号如下：
 $$
 \mathbf{m}\,\ddot{\mathbf{v}}(t)+\mathbf{c}\,\dot{\mathbf{v}}(t)+\mathbf{k}\,\mathbf{v}(t)=\mathbf{p}(t)
 $$  
 
 One approach to the solution of this set of coupled equations that often may be worth consideration is the step-by-step procedure, as is described in Chapter 15. However, for linear systems to which superposition is applicable, a more convenient solution may be obtained by Fourier transform (frequency-domain) procedures, as well as — at least in principle — by applying convolution integral (time-domain) methods; these MDOF procedures are analogous to the corresponding methods described previously for SDOF systems. A brief conceptual description of these techniques follows; however, the convolution integral approach is not generally suitable for practical use, and it is not discussed further after this brief description.  
-
-# Time Domain  
+解决这组耦合方程的一种方法，通常值得考虑的是逐步求解程序，如第 15 章所述。然而，对于适用叠加原理的线性系统，可以通过傅里叶变换（频域）方法获得更为方便的解，至少在理论上，还可以通过卷积积分（时域）方法求解；这些多自由度（MDOF）方法与之前针对单自由度（SDOF）系统所述的对应方法相类似。以下将简要介绍这些技术的概念；不过，卷积积分方法通常不适用于实际应用，因此在简要描述后不再进一步讨论。
+## Time Domain  
 
 First is considered the case where the MDOF structure is subjected to a unitimpulse loading in the $j$ th degree of freedom, while no other loads are applied. Thus the force vector $\mathbf p(t)$ consists only of zero components except for the $j$ th term, and that term is expressed by $p_{j}(t)=\delta(t)$ , where $\delta(t)$ is the Dirac delta function defined as  
+首先考虑多自由度（MDOF）结构在第 *j* 个自由度上受到单位脉冲载荷作用的情况，而其他自由度上不施加任何载荷。因此，力向量 **p**(*t*) 中除了第 *j* 项外，其他分量均为零，而该项由以下表达式给出：$p_{j}(t) = \delta(t)$其中，$\delta(t)$ 是狄拉克δ函数（Dirac delta function），定义为
 
 $$
 \delta(t)={\left\{\begin{array}{l l}{0}&{t\neq0}\\ {\infty}&{t=0}\end{array}\right.}\qquad\qquad\int_{-\infty}^{\infty}\delta(t)\;d t=1
 $$  
 
 Assuming now that Eq. (12-60) can be solved for the displacements caused by this loading, the ith component in the resulting displacement vector $\mathbf{r}(t)$ will then be the free-vibration response in that degree of freedom caused by a unit-impulse loading in coordinate $j$ ; therefore by definition this $i$ component motion is a unit-impulse transfer function, which will be denoted herein by $h_{i j}(t)$ .  
+假设方程（12-60）可以求解由该载荷引起的位移，则结果位移向量 $\mathbf{r}(t)$ 中的第 $i$ 个分量将对应于在坐标 $j$ 处受到单位脉冲载荷作用下该自由度的自由振动响应；因此，根据定义，该第 $i$ 个分量的运动即为单位脉冲传递函数，本文中将其表示为 $h_{ij}(t)$。
 
 If the loading in coordinate $j$ were a general time varying load $p_{j}(t)$ rather than a unit-impulse loading, the dynamic response in coordinate $i$ could be obtained by superposing the effects of a succession of impulses in the manner of the Duhamel integral, assuming zero initial conditions. The generalized expression for the response in coordinate $i$ to the load at $j$ is the convolution integral, as follows:  
-
+如果坐标系 *j* 中的载荷为一般随时间变化的载荷 $p_{j}(t)$（而非单位脉冲载荷），则坐标系 $i$ 的动态响应可通过杜哈美积分（Duhamel integral）的方式，将一系列脉冲效应叠加得到，假设初始条件为零。坐标系 $i$ 对载荷 $j$ 的响应的广义表达式为卷积积分，具体如下：
 $$
 v_{i j}(t)=\int_{0}^{t}p_{j}(\tau)\;h_{i j}(t-\tau)\;d\tau\qquad i=1,2,\cdots,N
 $$  
 
 and the total response in coordinate $i$ produced by a general loading involving all components of the load vector $p(t)$ is obtained by summing the contributions from all load components:  
-
+在坐标系 $i$ 中，由涉及载荷向量 $p(t)$ 所有分量的一般载荷作用产生的总响应可通过对所有载荷分量的贡献进行叠加得到：
 $$
 v_{i}(t)=\sum_{j=1}^{N}\left[\int_{0}^{t}p_{j}(\tau)\;h_{i j}(t-\tau)\;d\tau\right]\qquad i=1,2,\cdots,N
 $$  
 
-# Frequency Domain  
+## Frequency Domain  
 
 The frequency-domain analysis is similar to the time-domain procedure in that it involves superposition of the effects in coordinate $i$ of a unit load applied in coordinate $j$ ; however, in this case both the load and the response are harmonic. Thus the loading is an applied force vector ${\bf p}(t)$ having all zero components except for the $j$ th term which is a unit harmonic loading, $p_{j}(t)\;=\;1\exp(i\overline{{\omega}}t)$ . Assuming now that the steady-state solution of Eq. (12-60) to this loading can be obtained, the resulting steady-state response in the ith component of the displacement vector $\mathbf{v}(t)$ will be $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)~\exp(i\overline{{\omega}}t)$ in which $\mathrm{H}_{i j}\left(i{\overline{{\omega}}}\right)$ is defined as the complex-frequency-response transfer function.  
+频域分析与时域过程类似，均涉及在坐标 $i$ 中单位载荷在坐标 $j$ 作用下的效应叠加；然而，此处载荷与响应均为谐波形式。因此，载荷为一应用力向量 ${\bf p}(t)$，其所有分量均为零，仅第 $j$ 项为单位谐波载荷，即 $p_{j}(t) = 1 \exp(i\overline{{\omega}}t)$。假设方程（12-60）的稳态解可求得，则位移向量 $\mathbf{v}(t)$ 的第 $i$ 分量的稳态响应将为 $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right) \exp(i\overline{{\omega}}t)$，其中 $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)$ 被定义为**复频响应传递函数**。
 
 If the loading in coordinate $j$ were a general time-varying load $p_{j}(t)$ rather than a unit-harmonic loading, the forced-vibration response in coordinate $i$ could be obtained by superposing the effects of all the harmonics contained in $p_{j}(t)$ . For this purpose the time-domain expression of the loading is Fourier transformed to obtain  
-
+如果坐标 $j$ 处的载荷为一般随时间变化的载荷 $p_{j}(t)$ 而非单一谐波载荷，则坐标 $i$ 处的强迫振动响应可通过叠加 $p_{j}(t)$ 中所包含的所有谐波分量的效应来获得。为此，将载荷的时域表达式进行傅里叶变换，得到
 $$
 {\bf P}_{j}\left(i\overline{{\omega}}\right)=\int_{-\infty}^{\infty}p_{j}(t)\;\exp(-i\overline{{\omega}}t)\;d t
 $$  
 
-and then by inverse Fourier transformation the responses to all of these harmonics are combined to obtain the total forced-vibration response in coordinate $i$ as follows  
-
-(assuming zero initial conditions):  
-
+and then by inverse Fourier transformation the responses to all of these harmonics are combined to obtain the total forced-vibration response in coordinate $i$ as follows  (assuming zero initial conditions):  
+然后通过**逆傅里叶变换**，将所有这些谐波的响应组合起来，得到坐标 **$i$** 处的总强迫振动响应（假设初始条件为零）：
 $$
 v_{i j}(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{H}_{i j}(i\overline{{\omega}})~\mathrm{P}_{j}(i\overline{{\omega}})~\exp(i\overline{{\omega}}t)~d\overline{{\omega}}
 $$  
 
 Finally, the total response in coordinate $i$ produced by a general loading involving all components of the load vector ${\bf p}(t)$ could be obtained by superposing the contributions from all the load components:  
-
+最终，**$i$** 坐标处由涉及载荷向量 **${\bf p}(t)$** 所有分量的一般加载所产生的总响应可通过叠加所有载荷分量的贡献得到：
 $$
 v_{i}(t)=\frac{1}{2\pi}\,\sum_{j=1}^{N}\,\left[\int_{-\infty}^{\infty}\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)\mathrm{P}_{j}(i\overline{{\omega}})\,\exp(i\overline{{\omega}}t)\,d\overline{{\omega}}\right]\,\,\,\,\,i=1,2,\cdots,N
 $$  
 
 Equations (12-63) and (12-66) consistute general solutions to the coupled equations of motion (12-60), assuming zero initial conditions. Their successful implementation depends on being able to generate the transfer functions $h_{i j}(t)$ and $\mathrm{H}_{i j}(i\overline{{\omega}})$ efficiently, and it was suggested above that this is not practical for the time-domain functions, in general. However, procedures for implementing the frequency-domain formulation will be developed in this chapter after the following section.  
-
+方程（12-63）和（12-66）**构成**耦合运动方程（12-60）在零初始条件下的通解。其实现的成功取决于能否高效生成传递函数 **$h_{i j}(t)$** 和 **$\mathrm{H}_{i j}(i\overline{{\omega}})$**，而之前已指出，对于时域函数来说，这在一般情况下并不实际可行。不过，本章将在下一节之后，详细阐述频域表达式的实现方法。
 # 12-7 RELATIONSHIP BETWEEN TIME- AND FREQUENCYDOMAIN TRANSFER FUNCTIONS  
 
 To develop the interrelationships between transfer functions $h_{i j}(t)$ and $\mathrm{H}_{i j}(i\overline{{\omega}})$ , it is necessary to define a complex function $\mathrm{V}_{i j}(i\overline{{\omega}})$ as the Fourier transform of function $v_{i j}(t)$ given by Eq. (12-62); thus,  
-
+为了建立传递函数 $h_{i j}(t)$ 和 $\mathrm{H}_{i j}(i\overline{{\omega}})$ 之间的相互关系，有必要定义一个复函数 $\mathrm{V}_{i j}(i\overline{{\omega}})$ 作为函数 $v_{i j}(t)$ 的傅里叶变换，该函数由方程（12-62）给出；因此，
 $$
 {\bf V}_{i j}(i\overline{{\omega}})\equiv\int_{-\infty}^{\infty}\;\left[\int_{-\infty}^{t}\,p_{j}(\tau)\;h_{i j}(t-\tau)\;d\tau\right]\;\exp(-i\overline{{\omega}}t)\;d t
 $$  
 
-Note that because Eq. (12-62) assumes zero initial conditions, which is equivalent to assuming $p_{j}(t)=0$ for $t<0$ , one can change the lower limit of the integral in that equation from zero to $-\infty$ as shown in Eq. (12-67) without affecting the results of the integral. It will be assumed here that damping is present in the system so that the integral  
-
+Note that because Eq. (12-62) assumes zero initial conditions, which is equivalent to assuming $p_{j}(t)=0$ for $t<0$ , one can change the lower limit of the integral in that equation from zero to $-\infty$ as shown in Eq. (12-67) without affecting the results of the integral. It will be assumed here that damping is present in the system so that the integral is finite. 
+请注意，由于方程（12-62）假设初始条件为零，即等价于假设 **$p_{j}(t)=0$** 对于 **$t<0$** ，因此可以将该方程中积分的下限从零改为 **$-\infty$** ，如方程（12-67）所示，而不会影响积分结果。在此假设系统中存在阻尼，以确保积分结果为有限值。
 $$
 I_{1}\equiv\int_{-\infty}^{\infty}\left|v_{i j}(t)\right|\,d t
 $$  
 
-is finite. This is a necessary condition for the Fourier transform given by Eq. (12-67) to exist.  
+This is a necessary condition for the Fourier transform given by Eq. (12-67) to exist.  
+这是傅里叶变换（如方程（12-67）所示）存在的**$必要条件$**。
 
 Since the function $h_{i j}(t-\tau)$ equals zero for $\tau>t$ , the upper limit of the second integral in Eq. (12-67) can be changed from $t$ to $\infty$ without influencing the final result. Therefore, Eq. (12-67) can be expressed in the equivalent form  
-
+由于函数 $h_{i j}(t-\tau)$ 在 $\tau>t$ 时等于零，因此式（12-67）中第二个积分的上限可从 $t$ 扩展至 $\infty$，而不会影响最终结果。因此，式（12-67）可表示为等效形式：
 $$
 {\bf V}_{i j}(i\overline{{{\omega}}})=\operatorname*{lim}_{s\rightarrow\infty}\,\int_{-s}^{s}\,\int_{-s}^{s}\,p_{j}(\tau)\,\,h_{i j}(t-\tau)\,\,\exp(-i\overline{{{\omega}}}t)\,\,d t\,\,d\tau
 $$  
 
 When a new variable $\theta\equiv t-\tau$ is introduced, this equation becomes  
-
+当引入一个新的变量 $\theta \equiv t - \tau$ 时，该方程变为：
 $$
 \mathbf{V}_{i j}(i\overline{{\omega}})=\operatorname*{lim}_{s\to\infty}\;\int_{-s}^{s}\,p_{j}(\tau)\,\exp(-i\overline{{\omega}}t)\,d\tau\;\int_{-s-\tau}^{s-\tau}\,h_{i j}(\theta)\,\exp(-i\overline{{\omega}}\theta)\,d\theta
 $$  
 
-The expanding domain of integration given by this equation is shown in Fig. 12-6a. Since the function $\mathrm{V}_{i j}(i\overline{{\omega}})$ exists only when the integrals  
-
+The expanding domain of integration given by this equation is shown in Fig. 12-6a. Since the function $\mathrm{V}_{i j}(i\overline{{\omega}})$ exists only when the integrals are finite,  
+该方程所定义的积分域的扩展范围如图 12-6a 所示。由于函数 $\mathrm{V}_{i j}(i\overline{{\omega}})$ **仅在积分收敛时才存在**，
 $$
 I_{2}\equiv\int_{-\infty}^{\infty}\,\left|p_{j}(\tau)\right|\,d\tau\qquad\qquad I_{3}\equiv\int_{-\infty}^{\infty}\,\left|h_{i j}(\theta)\right|\,d\theta
 $$  
-
-are finite, which is always the case in practice due to the loadings being of finite duration and the unit-impulse-response function being a decayed function, it is valid to drop $\tau$ from the limits of the second integral in Eq. (12-69), resulting in  
-
+which is always the case in practice due to the loadings being of finite duration and the unit-impulse-response function being a decayed function, it is valid to drop $\tau$ from the limits of the second integral in Eq. (12-69), resulting in  
+在实际应用中，由于载荷持续时间有限且单位脉冲响应函数为衰减函数，因此可以在方程（12-69）的第二个积分限中去掉 $\tau$，从而得到
 $$
 \mathbf{V}_{i j}(i\overline{{\omega}})=\left[\operatorname*{lim}_{s\to\infty}\,\int_{-s}^{s}p_{j}(\tau)\,\exp(-i\overline{{\omega}}t)\,d\tau\right]\,\left[\operatorname*{lim}_{s\to\infty}\,\int_{-s}^{s}\,h_{i j}(\theta)\,\exp(-i\overline{{\omega}}\theta)\,d\theta\right]
 $$  
 
 which changes the expanding domain of integration to that shown in Fig. $12{-}6b$ . Variable $\theta$ can now be changed to $t$ since it is serving only as a dummy time variable. Equation (12-70) then becomes  
-
+将积分域扩展至如图 $12{-}6b$ 所示的范围。此时变量 $\theta$ 可替换为 $t$，因为它仅作为虚拟时间变量使用。于是，方程（12-70）变为：
 $$
 \mathbf{V}_{i j}(i\overline{{\omega}})=\mathbf{P}_{j}(i\overline{{\omega}})\,\int_{-\infty}^{\infty}\,h_{i j}(t)\,\exp(-i\overline{{\omega}}t)\,d t
 $$  
 
 When it is noted that Eq. (12-65) in its inverse form gives  
-
+当注意到方程（12-65）的逆形式给出
 $$
 {\bf V}_{i j}(i\overline{{\omega}})={\bf H}_{i j}(i\overline{{\omega}})\ {\bf P}_{j}(i\overline{{\omega}})
 $$  
 
 a comparison of Eqs. (12-71) and (12-72) makes it apparent that  
-
+比较方程（12-71）和（12-72）可以明显看出
 $$
 \mathrm{H}_{i j}\big(i\overline{{\omega}}\big)=\int_{-\infty}^{\infty}\,h_{i j}(t)\,\exp(-i\overline{{\omega}}t)\,d t
 $$  
@@ -952,6 +990,7 @@ $$
 FIGURE 12-6 Expanding domains of integration.  
 
 This derivation shows that any unit-impulse-response transfer function $h_{i j}(t)$ and the corresponding complex-frequency-response transfer function $\mathrm{H}_{i j}(i{\overline{{\omega}}})$ are Fourier transform pairs, provided damping is present in the system. This is a requirement for mathematical stability to exist.  
+该推导表明，在系统中存在阻尼的情况下，**$h_{i j}(t)$** 和对应的复频域响应传递函数 **$\mathrm{H}_{i j}(i{\overline{{\omega}}})$** 是傅里叶变换对。这是数学稳定性存在的必要条件。
 
 Example E12-5. Show that the complex-frequency-response function given by Eq. (6-52) and the unit-impulse-response function given by Eq. (6- 51) are Fourier transform pairs in accordance with Eqs. (6-53) and (6-54) which correspond to Eqs. (12-73).  
 
@@ -990,89 +1029,95 @@ $$
 
 in which $\omega_{D}\,=\,\omega\,\sqrt{1-\xi^{2}}$ . Note that the first of Eqs. (f) does indeed agree with Eq. (6-51), thus showing the validity of Eqs. (6-53) and (6-54) in this case. Note also that the inverse Fourier transform of $\mathrm{H}(i{\overline{{\omega}}})$ yields the unit-impulse response functions for all values of damping, i.e., for $0\leq\xi<1,\xi>1$ , and $\xi=1$ .  
 
-# 12-8 PRACTICAL PROCEDURE FOR SOLVING COUPLED EQUATIONS OF MOTION  
+# 12-8 PRACTICAL PROCEDURE FOR SOLVING COUPLED EQUATIONS OF MOTION  **解耦运动方程求解的实用步骤**
+
 
 The solution of coupled sets of equations of motion is carried out most easily in the frequency domain; therefore, this section will be devoted to developing procedures for this approach only. In doing so, consideration will be given to three different sets of equations as expressed in the frequency domain by  
+在频域中求解耦合运动方程组的解最为简便；因此，本节将仅针对该方法进行相关步骤的开发。在此过程中，将考虑频域中表达的三种不同方程组
 
 $$
 \begin{array}{r l}&{\left[(\mathbf{k}-\overline{{\omega}}^{2}\,\mathbf{m})+i\,\hat{\mathbf{k}}\right]\,\mathbf{V}(i\overline{{\omega}})=\mathbf{P}(i\overline{{\omega}})}\\ &{\left[(\mathbf{k}-\overline{{\omega}}^{2}\,\mathbf{m})+i\left(\overline{{\omega}}\,\mathbf{c}\right)\right]\,\mathbf{V}(i\overline{{\omega}})=\mathbf{P}(i\overline{{\omega}})}\\ &{\left[(\mathbf{K}-\overline{{\omega}}^{2}\,\mathbf{M})+i\left(\overline{{\omega}}\,\mathbf{C}\right)\right]\,\mathbf{Y}(i\overline{{\omega}})=\overline{{\mathbf{P}}}(i\overline{{\omega}})}\end{array}
-$$  
-
+$$
 in which the complex matrix in the bracket term on the left hand side of each equation is the impedance (or dynamic stiffness) matrix for the complete structural system being represented.  
+在每个方程左侧括号中的复杂矩阵是所表示的**完整结构系统**的**阻抗（或动态刚度）**矩阵。
 
 Equation (12-74) represents a complete $N$ -DOF system using the complexstiffness form of damping equivalent to Eq. (3-79) for the SDOF system. Matrix $\hat{\mathbf{k}}$ in this equation is a stiffness matrix for the entire system obtained by assembling individual finite-element stiffness matrices $\hat{\mathbf{k}}^{(m)}$ [superscript $(m)$ denotes element $m$ ] of the form  
-
+方程（12-74）描述了一个完整的 **$N$ 自由度系统**，采用了与单自由度系统（SDOF）中方程（3-79）类似的复刚度阻尼形式。该方程中的矩阵 **$\hat{\mathbf{k}}$** 是整个系统的刚度矩阵，由各个有限元刚度矩阵 **$\hat{\mathbf{k}}^{(m)}$** [上标 $(m)$ 表示第 $m$ 个单元] 汇总而成，其形式如下：
 $$
 \hat{\mathbf{k}}^{(m)}=2\,\xi^{(m)}\,\mathbf{k}^{(m)}
-$$  
-
+$$ 
 in which $\mathbf{k}^{\left(m\right)}$ denotes the individual elastic stiffness matrix for finite element $m$ as used in the assembly process to obtain matrix $\mathbf{k}$ for the entire system; and $\xi^{(m)}$ is a damping ratio selected to be appropriate for the material used in finite element $m$ . If the material is the same throughout the system so that the same damping ratio is used for each element, i.e., $\xi^{(1)}=\xi^{(2)}=\cdots=\xi$ , then the overall system matrix $\hat{\mathbf{k}}$ would be proportional to $\mathbf{k}$ as given by ${\hat{\mathbf{k}}}=2\xi\,\mathbf{k}$ . Matrix $\hat{\mathbf{k}}$ would then possess the same orthogonality property as $\mathbf{k}$ . However, when different materials are included in the system, e.g., soil and steel, the finite elements consisting of these materials would be assigned different values of $\xi^{(m)}$ . In this case, the assembled matrix $\hat{\mathbf{k}}$ would not satisfy the orthogonality condition, and modal coupling would be present. Vectors $\mathbf{V}(i\overline{{\omega}})$ and $\mathbf{P}(i{\overline{{\omega}}})$ in Eq. (12-74) are the Fourier transforms of vectors $\mathbf{v}(t)$ and $\mathbf p(t)$ , respectively, and all other quantities are the same as previously defined.  
+在其中，$\mathbf{k}^{(m)}$ 表示有限元单元 *m* 在组装过程中用于获得整个系统刚度矩阵 **k** 的单个弹性刚度矩阵；而 $\xi^{(m)}$ 是为有限元单元 *m* 所选的阻尼比，该比例应与所用材料相适配。若系统内材料相同，即每个单元采用相同的阻尼比（即 $\xi^{(1)} = \xi^{(2)} = \cdots = \xi$），则整体系统矩阵 $\hat{\mathbf{k}}$ 将与 **k** 成比例关系，具体表达式为 $\hat{\mathbf{k}} = 2\xi\,\mathbf{k}$。此时，矩阵 $\hat{\mathbf{k}}$ 将保持与 **k** 相同的正交性质。然而，当系统中包含不同材料（例如土壤和钢材）时，由这些材料构成的有限元单元将被分配不同的 $\xi^{(m)}$ 值。这种情况下，组装后的矩阵 $\hat{\mathbf{k}}$ 将不满足正交条件，且会出现模态耦合现象。方程（12-74）中的向量 $\mathbf{V}(i\overline{\omega})$ 和 $\mathbf{P}(i\overline{\omega})$ 分别是向量 $\mathbf{v}(t)$ 和 $\mathbf{p}(t)$ 的傅里叶变换，其余量均与之前定义一致。
 
 Equation (12-75) is the Fourier transform of Eq. (12-60) which represents an $N$ -DOF system having the viscous form of damping. Using the solution procedure developed subsequently in this section, it is not necessary for matrix c to satisfy the orthogonality condition. Therefore, the case of modal coupling through damping can be treated, whether it is of the viscous form or of the complex-stiffness form described above.  
+方程（12-75）是方程（12-60）的傅里叶变换，该方程描述的是一个具有**$N$**自由度且带有黏性阻尼形式的系统。通过本节后续发展的求解方法，矩阵 c** 不需要满足正交性条件。因此，**通过阻尼耦合的模态（无论是黏性形式还是上述提到的复刚度形式）均可处理。
 
 Equation (12-76) gives the normal mode equations of motion [Eq. (12-58)] in the frequency domain, in which $\overline{{\mathbf{P}}}(i\overline{{\omega}})$ is the Fourier transform of the generalized (modal) loading vector $P(t)$ which contains components $P_{1}(t),P_{2}(t),\cdot\cdot\cdot,P_{n}(t)$ as defined by Eq. (12-12c), $\mathbf{Y}(i{\overline{{\omega}}})$ is the Fourier transform of the normal coordinate vector $\pmb{Y}(t),\pmb{K}$ and $M$ are the diagonal normal mode stiffness and mass matrices containing elements in accordance with Eqs. (12-12b) and (12-12a), respectively, and $c$ is the normal mode damping matrix having elements as given by Eq. (12-15a). As noted carlier, if the damping matrix c possesses the orthogonality property, matrix $c$ will be of diagonal form; however, if matrix c does not possess the orthogonality property, the modal damping matrix will be full. The analysis procedure developed subsequently can treat this coupled form of matrix without difficulty, however. Note that Eqs. (12- 76) may contain all $N$ normal mode equations or only a smaller specified number representing the lower modes according to the degree of approximation considered acceptable. Reducing the number of equations to be solved does not change the analysis procedure but it does reduce the computational effort involved.  
+方程（12-76）给出了简正模态运动方程 [Eq. (12-58)] 在频域中的表达式，其中 $\overline{{\mathbf{P}}}(i\overline{{\omega}})$ 是广义（模态）载荷向量 $P(t)$ 的傅里叶变换，$P(t)$ 包含由方程（12-12c）定义的分量 $P_{1}(t), P_{2}(t), \cdots, P_{n}(t)$；$\mathbf{Y}(i{\overline{{\omega}}})$ 是简正坐标向量 $\pmb{Y}(t)$ 的傅里叶变换；$\pmb{K}$ 和 $M$ 分别是简正模态刚度矩阵和质量矩阵，其元素分别符合方程（12-12b）和（12-12a）的定义；$c$ 是简正模态阻尼矩阵，其元素由方程（12-15a）给出。如前所述，**如果阻尼矩阵 c 具有正交性，则矩阵 c 将为对角形式；否则，模态阻尼矩阵将为满矩阵。** 后续开发的分析方法能够轻松处理这种耦合形式的矩阵。需要注意的是，方程（12-76）可能包含所有 $N$ 个简正模态方程，或者仅包含根据可接受的近似程度指定的较少数量的低阶模态方程。减少需求解的方程数量不会改变分析程序，但会降低计算工作量。
 
 To develop the analysis procedure, let us consider only Eq. (12-74) since the procedure is applied to the other cases [Eqs. (12-75) and (12-76)] in exactly the same way. Equation (12-74) may be written in the abbreviated form:  
-
+在开发分析程序时，我们仅考虑 **Eq. (12-74)**，因为其他情况 [Eqs. (12-75) 和 (12-76)] 的处理方式完全相同。方程 (12-74) 可简化表示为：
 $$
 \mathbf{I}(i\overline{{\omega}})\;\mathbf{V}(i\overline{{\omega}})=\mathbf{P}(i\overline{{\omega}})
 $$  
 
-in which the impedance matrix $\mathbf{I}(i\overline{{\omega}})$ is given by the entire bracket matrix on the left hand side. Premultiplying both sides of this equation by the inverse of the impedance  
-
-matrix, response vector $\mathbf{V}(i{\overline{{\omega}}})$ can be expressed in the form  
-
+in which the impedance matrix $\mathbf{I}(i\overline{{\omega}})$ is given by the entire bracket matrix on the left hand side. Premultiplying both sides of this equation by the inverse of the impedance matrix, response vector $\mathbf{V}(i{\overline{{\omega}}})$ can be expressed in the form  
+其中，阻抗矩阵 $\mathbf{I}(i\overline{{\omega}})$ 由等式左侧的**整个括号矩阵**给出。将等式两边左乘以阻抗矩阵的逆矩阵后，**响应向量** $\mathbf{V}(i{\overline{{\omega}}})$ 可表示为如下形式：
 $$
 \mathbf{V}(i\overline{{\omega}})=\mathbf{I}(i\overline{{\omega}})^{-1}\ \mathbf{P}(i\overline{{\omega}})
 $$  
 
 which implies that multiplying a complex matrix by its inverse results in the identity matrix, similar to the case involving a real matrix. The inversion procedure is the same as that involving a real matrix with the only difference being that the coefficients involved are complex rather than real. Although computer programs are readily available for carrying out this type of inversion solution, it is impractical for direct use as it involves inverting the $N\times N$ complex impedance matrix for each of the closelyspaced discrete values of $\overline{{\omega}}$ as required in performing the fast Fourier transform (FFT) of loading vector ${\bf p}(t)$ to obtain the vector $\mathbf{P}(i{\overline{{\omega}}})$ ; this approach requires an excessive amount of computer time. The required time can be reduced to a practical level, however, by first solving for the complex-frequency-response transfer functions $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)$ at a set of widely-spaced discrete values of $\overline{{\omega}}$ , and then using an effective and efficient interpolation procedure to obtain the transfer functions at the intermediate closely-spaced discrete values of $\overline{{\omega}}$ required by the FFT procedure.  
+这意味着将一个复数矩阵乘以其逆矩阵会得到单位矩阵，这与实数矩阵的情况类似。求逆的过程与实数矩阵相同，唯一的区别在于系数为复数而非实数。虽然计算机程序可以方便地执行此类求逆操作，但直接使用这种方法并不实际，因为它需要对每一个用于执行加载向量 ${\bf p}(t)$ 的快速傅里叶变换（FFT）以获得向量 $\mathbf{P}(i{\overline{{\omega}}})$ 所需的、密集间隔的 $\overline{{\omega}}$ 的 $N \times N$ 复数阻抗矩阵进行求逆；这种方法需要消耗大量的计算时间。然而，可以通过先计算在一组间隔较宽的 $\overline{{\omega}}$ 离散值下的复频率响应传递函数 $\mathrm{H}_{ij}\left(i\overline{{\omega}}\right)$，再利用有效且高效的插值方法来获得 FFT 过程所需的、中间密集间隔的 $\overline{{\omega}}$ 离散值下的传递函数，从而将所需时间降低到实际可行的水平。
 
 The complex-frequency-response transfer functions $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)$ are obtained for the widely spaced discrete values of $\overline{{\omega}}$ using Eq. (12-79) consistent with the definition of these functions given previously; that is, using  
-
+复频域响应传递函数 $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)$ 通过 Eq. (12-79) 在广泛分布的离散 $\overline{{\omega}}$ 值上获得，与之前对这些函数的定义一致；也就是说，使用
 $$
 <\mathrm{H}_{1j}\big(i\overline{{\omega}}\big)\quad\mathrm{H}_{2j}\big(i\overline{{\omega}}\big)\quad\cdot\cdot\cdot\quad\mathrm{H}_{N j}\big(i\overline{{\omega}}\big)>^{T}=\mathbf{I}\big(i\overline{{\omega}}\big)^{-1}\,I_{j}\qquad j=1,2,\cdot\cdot\cdot,N
-$$  
-
+$$
 in which $I_{j}$ denotes an $N$ -component vector containing all zeros except for the $j$ th component which equals unity. Because these transfer functions are smooth, as indicated in Fig. 12-7, even though they peak at the natural frequencies of the system, interpolation can be used effectively to obtained their complex values at the intermediate closely-spaced discrete values of $\overline{{\omega}}$ . Note that natural frequencies can be obtained, corresponding to the frequencies at the peaks in the transfer functions, without solving the eigenvalue problem. The effective interpolation procedure required to carry out the analysis in this way will be developed in the following Section 12-9.  
+在其中，$I_{j}$ 表示一个 $N$ 分量的向量，除了第 $j$ 个分量为 1 外，其余分量均为零。由于这些传递函数如图 12-7 所示是平滑的，尽管它们在系统的固有频率处达到峰值，但仍可有效地通过插值法获取在中间紧密分布的离散 $\overline{\omega}$ 值处的复数值。需要注意的是，固有频率可以通过对应于传递函数峰值处的频率获得，而无需解特征值问题。本节所述的有效插值方法将在第 12-9 节中详细阐述。
 
 ![](fd1300cc77e1aab70e2dbc7ab5858e2cbec4812014ee073d61dc7e6fa1ca52a9.jpg)  
 FIGURE 12-7 Interpolation of transfer function.  
 
 Having obtained all transfer functions $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)$ using Eq. (12-80) and the interpolation procedure of Section 12-9, the response vector $\mathbf{V}(i\overline{{\omega}})$ is easily obtained by superposition using  
-
+获得所有传递函数 $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)$ 后，通过第 12-9 节所述的插值方法（以及 **Eq. (12-80)**）计算得到，响应向量 $\mathbf{V}(i\overline{{\omega}})$ 可通过叠加法轻松求得。
 $$
 \mathbf{V}(i\overline{{\omega}})=\mathbf{H}(i\overline{{\omega}})\ \mathbf{P}(i\overline{{\omega}})
 $$  
 
 in which $\mathbf{H}(i{\overline{{\omega}}})$ is the $N\times N$ complex-frequency-response transfer matrix  
-
+其中 $\mathbf{H}(i{\overline{{\omega}}})$ 是 $N \times N$ **复频域响应传递矩阵**
 $$
 \mathbf{H}(i\overline{{\omega}})=\left[\begin{array}{c c c c}{\mathrm{H}_{11}(i\overline{{\omega}})}&{\mathrm{H}_{12}(i\overline{{\omega}})}&{\cdots\cdot}&{\mathrm{H}_{1N}(i\overline{{\omega}})}\\ {\mathrm{H}_{21}(i\overline{{\omega}})}&{\mathrm{H}_{22}(i\overline{{\omega}})}&{\cdots\cdot}&{\mathrm{H}_{2N}(i\overline{{\omega}})}\\ {\vdots}&{\vdots}&{\vdots}&{\vdots}\\ {\mathrm{H}_{N1}(i\overline{{\omega}})}&{\mathrm{H}_{N2}(i\overline{{\omega}})}&{\cdots\cdot}&{\mathrm{H}_{N N}(i\overline{{\omega}})}\end{array}\right]
 $$  
 
 obtained for each frequency required in the response analysis. Note that once this transfer matrix has been obtained, the responses of the system to multiple sets of loadings can be obtained very easily by simply Fourier transforming each set by the FFT procedure and then multiplying the resulting vector set in each case by the transfer matrix in accordance with Eq. (12-81). Having vector $\mathbf{V}(i\overline{{\omega}})$ for each set, it can be inverse transformed by the FFT procedure to obtain the corresponding set of displacements in vector $\mathbf{v}(t)$ .  
+对于响应分析中每个频率所需的数据，一旦获得该**传递矩阵**，系统在多组载荷作用下的响应可以通过以下方式快速获得： 对每组载荷进行快速傅里叶变换（FFT），然后将变换后的向量集分别与传递矩阵相乘（按 Eq. (12-81) 执行）。获得每组的向量 **$\mathbf{V}(i\overline{{\omega}})$** 后，可通过反快速傅里叶变换（FFT）得到对应的位移向量 **$\mathbf{v}(t)$**。
 
 It is evident that by Fourier transforming each element $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)$ in Eq. (12-82), one could easily obtain the corresponding unit-impulse-response function $h_{i j}(t)$ as shown by the second of Eqs. (12-73). This is of academic interest only, however, as one would not use the convolution integral formulation given by Eq. (12-63) to evaluate the response of a complicated structural system.  
+显然，通过对方程（12-82）中每个元素 $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)$ 进行傅里叶变换，可以很容易地得到对应的单位脉冲响应函数 $h_{i j}(t)$，如方程（12-73）中的第二式所示。然而，**这在学术上仅具有理论意义，因为在评估复杂结构系统的响应时，不会使用方程（12-63）中给出的卷积积分公式。**
+
+# 12-9 INTERPOLATION PROCEDURE FOR GENERATION OF TRANSFER FUNCTIONS  **插值程序用于传递函数的生成**
 
-# 12-9 INTERPOLATION PROCEDURE FOR GENERATION OF TRANSFER FUNCTIONS  
 
 Because both the real and imaginary parts of a complex-frequency-response transfer function are smooth functions of $\overline{{\omega}}$ , interpolation of their values at equal intervals $\triangle\overline{{\omega}}$ over relatively wide frequency bands can be done effectively using an interpolation function corresponding to the forms of the complex-frequency-response transfer functions for a 2-DOF system having the complex-stiffness uncoupled-type of damping. The frequency-domain normal mode equations of motion for such a system are  
-
+由于复频域传递函数的实部和虚部都是关于 $\overline{{\omega}}$ 的平滑函数，因此在相对宽广的频率范围内，以等间隔 $\triangle\overline{{\omega}}$ 对其值进行插值可以有效地使用与具有复数刚度解耦型阻尼的二自由度系统的复频域传递函数形式相对应的插值函数。此类系统的频域简正模态运动方程为：
 $$
 \begin{array}{r}{\left[\left(K_{1}-\overline{{\omega}}^{2}\,M_{1}\right)+i\left(2\xi\,K_{1}\right)\right]\,\Upsilon_{1}(i\overline{{\omega}})=\phi_{1}^{T}\,\,{\bf P}(i\overline{{\omega}})}\\ {\left[\left(K_{2}-\overline{{\omega}}^{2}\,M_{2}\right)+i\left(2\xi\,K_{2}\right)\right]\,\Upsilon_{2}(i\overline{{\omega}})=\phi_{2}^{T}\,{\bf P}(i\overline{{\omega}})}\end{array}
 $$  
 
 in which vector $\mathbf{P}(i{\overline{{\omega}}})$ is the Fourier transform of loading vector ${\bf p}(t)$ .  
-
+在其中，向量 $\mathbf{P}(i{\overline{{\omega}}})$ 是载荷向量 ${\bf p}(t)$ 的傅里叶变换。
 Let us now generate a single complex-frequency-response transfer function, e.g., $\mathrm{H}_{11}(i\overline{{\omega}})$ , which is the transfer function between loading $p_{1}(t)$ and displacement $v_{1}(t)$ . In the frequency domain $v_{1}(t)$ is given in terms of the normal mode coordinates by  
-
+现在我们生成一个复杂频率响应传递函数，例如 $\mathrm{H}_{11}(i\overline{{\omega}})$ ，即加载 $p_{1}(t)$ 与位移 $v_{1}(t)$ 之间的传递函数。在频域中，$v_{1}(t)$ 可用简正模态坐标表示为：
 $$
 \mathbf{V}_{1}(i\overline{{\omega}})=\phi_{11}\,\mathbf{Y}_{1}(i\overline{{\omega}})+\phi_{12}\mathbf{Y}_{2}(i\overline{{\omega}})
 $$  
 
 To generate $\mathrm{H_{11}}$ , let $\mathbf{P}(i\overline{{\omega}})=<1\,\mathbf{\epsilon}\,0>^{T}$ giving  
-
+要生成 $\mathrm{H_{11}}$ ，令 $\mathbf{P}(i\overline{\omega}) = <1\,\mathbf{\epsilon}\,0>^{T}$ ，则得到
 $$
 \phi_{1}^{T}\,\mathbf{P}(i\overline{{\omega}})=<\phi_{11}\quad\phi_{21}><1\ \ 0>^{T}=\phi_{11}
 $$  
@@ -1084,29 +1129,43 @@ $$
 $$  
 
 in which case, substituting the resulting values of $\mathrm{Y}_{1}(i\overline{{\omega}})$ and $\mathrm{Y}_{2}(i\overline{{\omega}})$ given by Eqs. (12-83) and (12-84), respectively, into Eq. (12-85) gives ${\bf V}_{1}(i\overline{{\omega}})\,=\,{\bf H}_{11}(i\overline{{\omega}})$ . Taking this action, one obtains  
-
+在这种情况下，将方程（12-83）和（12-84）分别给出的 $\mathrm{Y}_{1}(i\overline{{\omega}})$ 和 $\mathrm{Y}_{2}(i\overline{{\omega}})$ 的结果值代入方程（12-85），可得 ${\bf V}_{1}(i\overline{{\omega}})\,=\,{\bf H}_{11}(i\overline{{\omega}})$。经过这一步骤后，可以得到
 $$
 \mathrm{H}_{11}(i\overline{{\omega}})=\frac{\phi_{11}^{2}}{\left[\left(K_{1}-\overline{{\omega}}^{2}\,M_{1}\right)+i\left(2\xi_{1}\,K_{1}\right)\right]}+\frac{\phi_{12}^{2}}{\left[\left(K_{2}-\overline{{\omega}}^{2}\,M_{2}\right)+i\left(2\xi_{2}\,K_{2}\right)\right]}
 $$  
 
 By operating on this equation, it can be put in the equivalent single-fraction form  
-
+通过对该方程进行运算，可以将其转化为等效的单分式形式。
 $$
 \mathrm{H}_{11}(i\overline{{\omega}})=\frac{A\,\overline{{\omega}}^{2}+B}{\overline{{\omega}}^{4}+C\,\overline{{\omega}}^{2}+D}
 $$  
 
 in which $A$ is a real constant and $B,C$ , and $D$ are complex constants, all expressed in terms of the known quantities in Eq. (12-86). The forms of these expressions are of no interest, however, as only the functional form of $\mathrm{H}_{11}(i\overline{{\omega}})$ with respect to $\overline{{\omega}}$ is needed. Repeating the above development, one finds that each of the other three transfer functions $\mathrm{H}_{12}(i\overline{{\omega}})$ , $\mathrm{H}_{\mathrm{21}}(i\overline{{\omega}})$ , and $\mathrm{H}_{22}(i\overline{{\omega}})$ has the same form as that given by Eq. (12-87).  
+在其中，$A$ 是一个实数常数，$B$、$C$ 和 $D$ 是复数常数，它们均以方程（12-86）中的已知量表示。不过，这些表达式的具体形式并不重要，因为仅需关注 $\mathrm{H}_{11}(i\overline{{\omega}})$ 关于 $\overline{{\omega}}$ 的函数形式。通过重复上述推导过程，可以发现其他三个传递函数 $\mathrm{H}_{12}(i\overline{{\omega}})$、$\mathrm{H}_{21}(i\overline{{\omega}})$ 和 $\mathrm{H}_{22}(i\overline{{\omega}})$ 的形式与方程（12-87）中给出的形式完全相同。
 
 To use Eq. (12-87) purely as an interpolation function for any transfer function $\mathbf{H}_{i j}(i\overline{{\omega}})$ of a complex $N$ -DOF system, express it in the discrete form  
-
+将复杂 $N$ 自由度系统中任一传递函数 $\mathbf{H}_{i j}(i\overline{{\omega}})$ 的Eq. (12-87) 纯粹作为插值函数使用时，应将其表示为离散形式。
 $$
 \mathbf{H}_{i j}(i\overline{{\omega}}_{m})=\frac{A_{m n}\,\overline{{\omega}}_{m}^{2}+B_{m n}}{\overline{{\omega}}_{m}^{4}+C_{m n}\,\overline{{\omega}}_{m}^{2}+D_{m n}}\qquad\bigl(n-\frac{3}{2}\,q\bigr)<m<\bigl(n+\frac{3}{2}\,q\bigr)
 $$  
 
-in which $\overline{{{\omega}}}_{m}\,=\,m\,\,\triangle\overline{{{\omega}}},\,\triangle\overline{{{\omega}}}$ being the constant frequency interval of the narrowly spaced discrete frequencies required by the FFT procedure in generating loading vector $\mathbf{P}(i{\overline{{\omega}}})$ , and $A_{m n}$ , $B_{m n}$ , $C_{m n}$ , and $D_{m n}$ are all treated as complex constants, even though coefficient $A$ in Eq. (12-87) for a 2-DOF system is real. These four constants are evaluated by applying Eq. (12-88) separately to four consecutive widely-spaced discrete values of $\overline{{\omega}}$ , as given by $m=\left(n-{\textstyle{\frac{3}{2}}}\,q\right)$ , $m=\left(n-{\textstyle{\frac{1}{2}}}\,q\right)$ , $m=\left(n+{\textstyle{\frac{1}{2}}}\,q\right)$ , and $m\,=\,{\bigl(}n+{\textstyle{\frac{3}{2}}}\,q{\bigr)}$ , as shown in Fig. 12-7, in which $q$ represents the number of closely-spaced frequency intervals within one of the widely-spaced intervals. Knowing $\mathrm{H}_{i j}\left(i\overline{{\omega}}\right)$ for the above four values of $m$ as obtained using Eq. (12-80), separate applications of Eq. (12-88) to the corresponding four values of $\overline{{\omega}}_{m}$ yields four simultaneous complex algebraic equations involving unknowns $A_{m n}$ , $B_{m n}$ , $C_{m n}$ , and $D_{m n}$ . Solving for these constants and entering their numerical values back into Eq. (12-88), this equation can be used to calculate the intermediate values of $\mathrm{H}_{i j}\big(i\overline{{\omega}}_{m}\big)$ at the closely-spaced discrete frequencies in the range $\left(n-{\textstyle{\frac{3}{2}}}\,q\right)\,<m<\,\left(n+{\textstyle{\frac{3}{2}}}\,q\right)$ . This same procedure is then repeated for $\begin{array}{r}{n\,=\,\frac{3}{2}\,q,\,\frac{9}{2}\,q,\,\frac{15}{2}\,q,\,\frac{21}{2}\,q,\,\cdot\,\cdot\,.}\end{array}$ so as to cover the entire range of frequencies of interest. Better accuracy can be obtained by this interpolation if it is applied over only the central frequency interval, i.e., over the range $\begin{array}{r}{(n-\frac{1}{2}q)<m<(n+\frac{1}{2}q)}\end{array}$ ; this is a greater computational task, however, because the set of constants in Eq. (12-88) then must be evaluated for $\begin{array}{r}{n=\frac{3}{2}q,\,\frac{5}{2}q,\,\frac{7}{2}q,\,\cdot\,\cdot\,}\end{array}$ .  
+
+在其中，$$\overline{\omega}_m = m \triangle \overline{\omega}$$，其中 $\triangle \overline{\omega}$ 是快速傅里叶变换（FFT）程序在生成载荷向量 $\mathbf{P}(i\overline{\omega})$ 时所需的**狭窄间隔离散频率**的**常数频率间隔**，而 $A_{mn}$、$B_{mn}$、$C_{mn}$ 和 $D_{mn}$ 虽然在二自由度系统的 Eq. (12-87) 中系数 $A$ 为实数，但在本文中均被视为**复数常数**。这四个常数是通过将 Eq. (12-88) 分别应用于四个**宽间隔离散频率值** $\overline{\omega}$ 的情况来求解的，具体为：
+$m = \left(n - \frac{3}{2}q\right)$、$m = \left(n - \frac{1}{2}q\right)$、$m = \left(n + \frac{1}{2}q\right)$ 以及 $m = \left(n + \frac{3}{2}q\right)$，如图 12-7 所示，$q$ 表示一个宽间隔内**紧密间隔频率区间**的数量。已知通过 Eq. (12-80) 得到的上述四种 $m$ 值对应的 $\mathrm{H}_{ij}(i\overline{\omega})$，将 Eq. (12-88) 分别应用于对应的 $\overline{\omega}_m$ 的四个值，可得到包含未知数 $A_{mn}$、$B_{mn}$、$C_{mn}$ 和 $D_{mn}$ 的**四个复数代数方程**。通过求解这些常数，并将其数值代入 Eq. (12-88)，即可计算出范围为 $\left(n - \frac{3}{2}q\right) < m < \left(n + \frac{3}{2}q\right)$ 内**紧密间隔离散频率**的中间值 $\mathrm{H}_{ij}\big(i\overline{\omega}_m\big)$。该程序将重复应用于
+$$
+n = \frac{3}{2}q, \frac{9}{2}q, \frac{15}{2}q, \frac{21}{2}q, \cdots
+$$
+以覆盖整个感兴趣的频率范围。若仅在**中心频率区间**内应用插值，即范围为
+$$
+\left(n - \frac{1}{2}q\right) < m < \left(n + \frac{1}{2}q\right)
+$$
+则可获得更高的精度，但计算量更大，因为此时 Eq. (12-88) 中的常数集需重新计算，对应的 $n$ 值为
+$$
+n = \frac{3}{2}q, \frac{5}{2}q, \frac{7}{2}q, \cdots
+$$。
 
 To set the optimum value of $q$ , considering computational effort and accuracy, requires considerable experience with the procedure. While it is difficult to provide guidelines for this purpose, one should at least be aware that the frequency interval of $3\,q\,\triangle\overline{{\omega}}$ should never include more than two natural frequencies because the form of the interpolation function is that of a transfer function for a 2-DOF system, for which only two peaks can be represented.  
-
+要确定 $q$ 的最优值，在计算效率和精度之间权衡，需依赖对该流程的丰富经验。虽然难以提供具体指导，但至少应明确：频率区间 $3\,q\,\triangle\overline{{\omega}}$ 不应包含超过两个固有频率，因为该插值函数的形式类似于二自由度系统的传递函数，后者最多只能表示两个峰值。
 # PROBLEMS  
 
 12-1. A cantilever beam supporting three equal lumped masses is shown in Fig. P12-1; also listed there are its undamped mode shapes $\Phi$ and frequencies of vibration $\pmb{\omega},$ . Write an expression for the dynamic response of mass 3 of this system after an 8-kips step function load is applied at mass 2 (i.e., 8 kips is suddenly applied at time $t\,=\,0$ and remains on the structure permanently), including all three modes and neglecting damping. Plot the history of response $v_{3}(t)$ for the time interval $0<t<T_{1}$ where $T_{1}=2\pi/\omega_{1}=2\pi/3.61$ .