diff --git a/.obsidian/plugins/copilot/data.json b/.obsidian/plugins/copilot/data.json
index bea7b0e..faf85cb 100644
--- a/.obsidian/plugins/copilot/data.json
+++ b/.obsidian/plugins/copilot/data.json
@@ -117,9 +117,9 @@
       "apiKey": "",
       "isEmbeddingModel": false,
       "capabilities": [
-        "websearch",
+        "reasoning",
         "vision",
-        "reasoning"
+        "websearch"
       ],
       "stream": true,
       "displayName": "ngrok_gemma3:12b",
@@ -318,9 +318,9 @@
     },
     {
       "name": "check formula",
-      "prompt": "<instruct>check formula in latex \nreturn only corrected formula in latex  </instruct>\n\n<text>{copilot-selection}</text>",
+      "prompt": "<instruct>This formula is wrong, cannot display in latex. check formula \nreturn only corrected formula in latex  </instruct>\n\n<text>{copilot-selection}</text>",
       "showInContextMenu": true,
-      "modelKey": "gemma3:12b|ollama"
+      "modelKey": "phi4:latest|ollama"
     }
   ]
 }
\ No newline at end of file
diff --git a/力学书籍/力学/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library)/auto/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library).md b/力学书籍/力学/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library)/auto/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library).md
index a26ddf9..df25408 100644
--- a/力学书籍/力学/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library)/auto/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library).md	
+++ b/力学书籍/力学/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library)/auto/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library).md	
@@ -7009,54 +7009,64 @@ depend on the system generalized coordinates, and as a result they are implicit
 
 
 $$
-\begin{array}{r l}&{T^{i}=\cfrac{1}{2}\big(\dot{\mathbf{R}}^{i^{\mathrm{T}}}\mathbf{m}_{R R}^{i}\dot{\mathbf{R}}^{i}+2\dot{\mathbf{R}}^{i^{\mathrm{T}}}\mathbf{m}_{R\theta}^{i}\dot{\mathbf{\Gamma}}\dot{\mathbf{\Gamma}}+2\dot{\mathbf{R}}^{i^{\mathrm{T}}}\mathbf{m}_{R f}^{i}\dot{\mathbf{q}}_{f}^{i}+\dot{\mathbf{\Gamma}}\dot{\mathbf{\Gamma}}^{\mathrm{\dagger}}\mathbf{m}_{\theta\theta}^{i}\dot{\mathbf{\Gamma}}}\\ &{\qquad+\,2\dot{\mathbf{\Theta}}^{i^{\mathrm{T}}}\mathbf{m}_{\theta f}^{i}\dot{\mathbf{q}}_{f}^{i}+\dot{\mathbf{q}}_{f}^{i^{\mathrm{T}}}\mathbf{m}_{f f}^{i}\dot{\mathbf{q}}_{f}^{i}\big)}\end{array}
+\begin{array}{r l}&{T^{i}=\cfrac{1}{2}\big(\dot{\mathbf{R}}^{i^{\mathrm{T}}}\mathbf{m}_{R R}^{i}\dot{\mathbf{R}}^{i}+2\dot{\mathbf{R}}^{i^{\mathrm{T}}}\mathbf{m}_{R\theta}^{i}\dot{\mathbf{\theta}}^i+2\dot{\mathbf{R}}^{i^{\mathrm{T}}}\mathbf{m}_{R f}^{i}\dot{\mathbf{q}}_{f}^{i}+{\dot{\mathbf{\theta}}^i}^T\mathbf{m}_{\theta\theta}^{i}\dot{\mathbf{\theta}}^i}\\ &{\qquad+\,2{\dot{\mathbf{\theta}}^i}^T\mathbf{m}_{\theta f}^{i}\dot{\mathbf{q}}_{f}^{i}+\dot{\mathbf{q}}_{f}^{i^{\mathrm{T}}}\mathbf{m}_{f f}^{i}\dot{\mathbf{q}}_{f}^{i}\big)}\end{array}
 $$  
 
-Special Cases If the body is rigid, the vector $\dot{\mathbf{q}}_{f}^{i}$ of elastic coordinates of body $i$ vanishes and the kinetic energy reduces to  
+Special Cases If the body is rigid, the vector $\dot{\mathbf{q}}_{f}^{i}$ of elastic coordinates of body $i$ vanishes and the kinetic energy reduces to 
+特殊情况 如果物体是刚体，则物体的弹性坐标向量 $\dot{\mathbf{q}}_{f}^{i}$ 对于物体 $i$ 为零，动能随之降低为 
 
 $$
-T^{i}=\frac{1}{2}\big[\dot{\mathbf{R}}^{i}^{\mathrm{T}}\mathbf{m}_{R R}^{i}\dot{\mathbf{R}}^{i}+2\dot{\mathbf{R}}^{i}^{\mathrm{T}}\mathbf{m}_{R\theta}^{i}\dot{\mathbf{\theta}}^{i}+\dot{\mathbf{\theta}}^{i}^{\mathrm{T}}\mathbf{m}_{\theta\theta}^{i}\dot{\mathbf{\theta}}^{i}\big]
+T^{i} = \frac{1}{2} \left[ \dot{\mathbf{R}}^{i^{\mathrm{T}}} \mathbf{m}_{RR}^{i} \dot{\mathbf{R}}^{i} + 2 \dot{\mathbf{R}}^{i^{\mathrm{T}}} \mathbf{m}_{R\theta}^{i} \dot{\mathbf{\theta}}^{i} + \dot{\mathbf{\theta}}^{i^{\mathrm{T}}} \mathbf{m}_{\theta\theta}^{i} \dot{\mathbf{\theta}}^{i} \right]
 $$  
 
-which can be written in a partitioned form as  
+which can be written in a partitioned form as 可以写成分段形式如下： 
 
 $$
-T^{i}=\frac{1}{2}[\dot{\mathbf{R}}^{i^{\mathrm{\tiny{T}}}}\quad\dot{\mathbf{\theta}}^{i^{\mathrm{\tiny{T}}}}]\left[\mathbf{m}_{R R}^{i}\quad\mathbf{m}_{R\theta}^{i}\right]\left[\dot{\mathbf{R}}^{i}\right]
+T^{i}=\frac{1}{2}[\dot{\mathbf{R}}^{i^{\mathrm{\tiny{T}}}}\quad\dot{\mathbf{\theta}}^{i^{\mathrm{\tiny{T}}}}]\left[\begin{array}{l l}{\mathbf{m}_{R R}^{i}}&{\mathbf{m}_{R\theta}^{i}}\\ {\mathbf{m}_{\theta R}^{i}}&{\mathbf{m}_{\theta\theta}^{i}}\end{array}\right]\left[\begin{array}{l l}\dot{\mathbf{R}}^{i} \\ \dot{\mathbf{\theta}}^{i}\end{array}\right]
 $$  
 
-where the mass matrix in the case of a rigid body motion can be recognized as  
+where the mass matrix in the case of a rigid body motion can be recognized as 
+其中，刚体运动中的质量矩阵可以被识别为 
 
 $$
 \mathbf{M}^{i}=\left[\begin{array}{l l}{\mathbf{m}_{R R}^{i}}&{\mathbf{m}_{R\theta}^{i}}\\ {\mathbf{m}_{\theta R}^{i}}&{\mathbf{m}_{\theta\theta}^{i}}\end{array}\right]
 $$  
 
 The matrix $\mathbf{m}_{R\theta}^{i}$ and its transpose $\mathbf{m}_{\theta R}^{i}$ represent the inertia coupling between the rigid body translation and the rigid body rotation. It is shown in Chapter 3 that this coupling can be eliminated, in rigid body dynamics, if the origin of the body reference is rigidly attached to the mass center of the body. The term ${\bf{m}}_{R R}^{i}$ represents the mass matrix associated with the translational coordinates of the body reference. This matrix is diagonal, and the diagonal elements are equal to the total mass of the body. The matrix $\mathbf{m}_{\theta\theta}^{i}$ is associated with the rotational coordinates of the body reference.  
+**矩阵 $\mathbf{m}_{R\theta}^{i}$ 及其转置 $\mathbf{m}_{\theta R}^{i}$ 代表刚体平动与刚体转动之间的惯性耦合**。正如第三章所示，如果在刚体动力学中，刚体参考系的坐标原点 rigidly 地固定在刚体的质心上，则可以消除这种耦合。项 ${\bf{m}}_{R R}^{i}$ 代表与刚体参考系的平动坐标相关的质量矩阵。该矩阵是对称的，其对角元素等于刚体的总质量。矩阵 $\mathbf{m}_{\theta\theta}^{i}$ 与刚体参考系的转动坐标相关联。
 
 In the case of structural systems, the reference coordinates remain constant with respect to time, that is, $\dot{\mathbf{R}}^{i}=\mathbf{0},\ \dot{\theta}^{i}=\mathbf{0}.$ . The kinetic energy of Eq. 47 reduces to  
-
+对于结构系统而言，参考坐标随时间保持恒定，即 $\dot{\mathbf{R}}^{i}=\mathbf{0},\ \dot{\theta}^{i}=\mathbf{0}.$ 。 方程式 47 中的动能简化为
+$$
+T^{i}=\frac{1}{2}{\dot{\mathbf{q}}}^{i}{\mathbf{m}_{ff}^{i}}{\dot{\mathbf{q}}}_{f}^{i}
 $$
-T^{i}=\frac{1}{2}{\dot{\mathbf{q}}}^{i}^{\mathrm{T}}\mathbf{m}_{f f}^{i}{\dot{\mathbf{q}}}_{f}^{i}
-$$  
-
 and the mass matrix of the body can be recognized in this case as the constant matrix $\mathbf{m}_{f f}^{i}$ , which appears in the dynamic formulation of linear structural systems.  
+并且在本例中，可以识别出该物体的质量矩阵为常数矩阵 $\mathbf{m}_{f f}^{i}$，该矩阵出现在线性结构体系的动力学公式中。
 
 When a deformable body undergoes rigid body motion, the mass matrix is defined by Eq. 45 and the submatrices ${\bf{m}}_{R f}^{i}$ and $\mathbf{m}_{\theta f}^{i}$ and their transpose represent the coupling between the reference motion and elastic deformation. In this case these matrices as well as the matrices ${\bf{m}}_{R\theta}^{i}$ and ${\bf{m}}_{\theta\theta}^{i}$ depend on both the rotational reference coordinates and the elastic coordinates of the body $i$ .  
+当一个可变形体发生刚体运动时，质量矩阵由公式 45 定义，子矩阵 ${\bf{m}}_{R f}^{i}$ 和 $\mathbf{m}_{\theta f}^{i}$ 以及它们的转置，表示参考运动与弹性变形之间的耦合。 在这种情况下，这些矩阵以及矩阵 ${\bf{m}}_{R\theta}^{i}$ 和 ${\bf{m}}_{\theta\theta}^{i}$ 都取决于旋转参考坐标和体 $i$ 的弹性坐标。
 
 Spatial Motion To develop in detail the form of the components of the mass matrix in the general case of a spatial deformable body that undergoes rigid body motion, we use the general definition of the matrix $\mathbf{B}^{i}$ given by Eq. 33. We start with the submatrix ${\bf{m}}_{R R}^{i}$ associated with the translations of the origin of the body reference given by Eq. 46. This matrix can be determined as  
+空间运动
+为了详细阐述在空间变形体经历刚体运动时，质量矩阵各分量的形式，我们采用方程33给出的矩阵$\mathbf{B}^{i}$的通用定义。 我们从与刚体参考系原点平动相关的子矩阵${\bf{m}}_{R R}^{i}$开始，该子矩阵由方程46给出。 该矩阵可以确定为：
+
 
 $$
 \mathbf{m}_{R R}^{i}=\int_{V^{i}}\rho^{i}\mathbf{I}\,d V^{i}=\int_{V^{i}}\rho^{i}\left[\begin{array}{l l l}{1}&{0}&{0}\\ {0}&{1}&{0}\\ {0}&{0}&{1}\end{array}\right]d V^{i}=\left[\begin{array}{l l l}{m^{i}}&{0}&{0}\\ {0}&{m^{i}}&{0}\\ {0}&{0}&{m^{i}}\end{array}\right]
 $$  
 
 where $m^{i}$ is the total mass of body $i$ . Because of the conservation of mass, this matrix is the same for both cases of rigid and deformable bodies.  
+其中，$m^{i}$ 为第 $i$ 个体的总质量。由于质量守恒，该矩阵对于刚体和变形体两种情况均相同。
 
 Using Eq. 33, one can determine the submatrix $\mathbf{m}_{R\theta}^{i}$ of Eq. 46 as  
+使用公式33，可以确定公式46中的子矩阵 $\mathbf{m}_{R\theta}^{i}$ 为：
 
 $$
 \mathbf{m}_{R\theta}^{i}=\mathbf{m}_{\theta R}^{i^{\mathrm{T}}}\!=\int_{V^{i}}\rho^{i}\mathbf{B}^{i}d V^{i}=-\int_{V^{i}}\rho^{i}\mathbf{A}^{i}\tilde{\bar{\mathbf{u}}}^{i}\bar{\mathbf{G}}^{i}d V^{i}
 $$  
 
 Because the matrices $\mathbf{A}^{i}$ and $\bar{\mathbf{G}}^{i}$ are not space-dependent, one can write the integral of Eq. 53 in the form  
+由于矩阵 $\mathbf{A}^{i}$ 和 $\bar{\mathbf{G}}^{i}$ 与空间无关，可以将公式 53 的积分写成如下形式：
 
 $$
 \mathbf{m}_{R\theta}^{i}=\mathbf{m}_{\theta R}^{i^{\mathrm{T}}}\!=-\mathbf{A}^{i}\left[\int_{V^{i}}\rho^{i}\tilde{\bar{\mathbf{u}}}^{i}d V^{i}\right]\bar{\mathbf{G}}^{i}
@@ -7093,92 +7103,107 @@ $$
 $$  
 
 The vector $\bar{\mathbf{S}}_{t}^{i}$ represents the components of the moment of mass of the body $i$ about the axes of the body coordinate system. In the general case of a deformable body this vector can be written as  
+向量 $\bar{\mathbf{S}}_{t}^{i}$ 表示刚体 $i$ 关于其自身坐标系轴的质量矩的分量。对于一般可变形体而言，该向量可以表示为：
 
 $$
 \bar{\bf S}_{t}^{i}=\int_{V^{i}}\rho^{i}\left[\bar{\bf u}_{o}^{i}+\bar{\bf u}_{f}^{i}\right]d V^{i}
 $$  
 
 If the body is rigid, $\bar{\mathbf{S}}_{t}^{i}$ is a constant vector. Furthermore, if the origin of the body reference is attached to the center of mass, one has  
+如果body是刚性的，$\bar{\mathbf{S}}_{t}^{i}$ 是一个常数向量。此外，如果body参考系的坐标原点附着于质心，则有
 
 $$
 \mathbf{I}_{1}^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}_{o}^{i}\,d V^{i}=\int_{V^{i}}\rho^{i}\left[x_{1}^{i}\quad x_{2}^{i}\quad x_{3}^{i}\right]^{\mathrm{T}}d V^{i}=\mathbf{0}
 $$  
 
 which shows that the matrix ${\bf{m}}_{R\theta}^{i}$ in the case of rigid body dynamics is the null matrix. In this case the translation and rotation of the rigid body are dynamically decoupled. This, however, is not the case when deformable bodies are considered. This fact can be demonstrated by writing Eq. 60 in a more explicit form as  
-
+这表明在刚体动力学的情况下，矩阵 ${\bf{m}}_{R\theta}^{i}$ 为零矩阵。在这种情况下，刚体的平移和旋转是动力学上解耦的。然而，当考虑可变形体时，情况并非如此。这个事实可以通过将公式 60 以更明确的形式写出进行证明，如下所示：
 $$
 \bar{\mathbf{S}}_{t}^{i}=\int_{V^{i}}\rho^{i}\left[\bar{\mathbf{u}}_{o}^{i}+\mathbf{S}^{i}\mathbf{q}_{f}^{i}\right]d V^{i}
 $$  
 
 One can see that in the special case in which the origin of the body reference is rigidly attached to the center of mass in the undeformed state, which is the case in which Eq. 61 is satisfied, there is no guarantee that $\bar{\mathbf{S}}_{t}^{i}$ is the null matrix because of the deformation of the body. Therefore, $\bar{\mathbf{S}}_{t}^{i}$ and the submatrix $\mathbf{m}_{R\theta}^{i}$ must be iteratively updated. It is also clear from Eq. 62 that, in addition to evaluating the moment of mass in the undeformed state, one needs to evaluate the following inertia shape integrals:  
+可以观察到，在特殊情况下，即当刚体参考系的坐标原点严格地固定在未变形状态下的质心，也就是满足公式 61 的情况时，由于物体的变形，无法保证 $\bar{\mathbf{S}}_{t}^{i}$ 是零矩阵。因此，$\bar{\mathbf{S}}_{t}^{i}$ 和子矩阵 $\mathbf{m}_{R\theta}^{i}$ 必须迭代更新。 此外，从公式 62 可以看出，除了评估未变形状态下的质量矩之外，还需要评估以下惯性形状积分：
 
 $$
 \bar{\mathbf{S}}^{i}=\int_{V^{i}}\rho^{i}\mathbf{S}^{i}d V^{i}
 $$  
 
-This matrix is also required for the evaluation of the matrix ${\bf{m}}_{R f}^{i}$ of Eq. 46, since  
+This matrix is also required for the evaluation of the matrix ${\bf{m}}_{R f}^{i}$ of Eq. 46, since
+此矩阵也需要用于评估公式46中的矩阵 ${\bf{m}}_{R f}^{i}$，因为  
 
 $$
 \mathbf{m}_{R f}^{i}=\mathbf{A}^{i}\int_{V^{i}}\rho^{i}\mathbf{S}^{i}d V^{i}=\mathbf{A}^{i}\bar{\mathbf{S}}^{i}
 $$  
 
 The submatrix $\mathbf{m}_{\theta\theta}^{i}$ , of Eq. 46, associated with the rotation of the body reference can be defined by using the matrix $\mathbf{B}^{i}$ of Eq. 33, as  
+子矩阵 $\mathbf{m}_{\theta\theta}^{i}$ ，与公式 46 相关，它与身体参考系的旋转有关，可以使用公式 33 中的矩阵 $\mathbf{B}^{i}$ 来定义，如下所示：
 
 $$
 \mathbf{m}_{\theta\theta}^{i}=\int_{V^{i}}\rho^{i}\mathbf{B}^{i\mathrm{T}}\mathbf{B}^{i}d V^{i}=\int_{V^{i}}\rho^{i}(\mathbf{A}^{i}\tilde{\bar{\mathbf{u}}}^{i}\bar{\mathbf{G}}^{i})^{\mathrm{T}}(\mathbf{A}^{i}\tilde{\bar{\mathbf{u}}}^{i}\bar{\mathbf{G}}^{i})d V^{i}
 $$  
 
-Using the orthogonality of the transformation matrix $\mathbf{A}^{i}$ , that is, $\mathbf{A}^{i^{\mathrm{T}}}\mathbf{A}^{i}=\mathbf{I}$ , one can write $\mathbf{m}_{\theta\theta}^{i}$ as  
+Using the orthogonality of the transformation matrix $\mathbf{A}^{i}$ , that is, $\mathbf{A}^{i^{\mathrm{T}}}\mathbf{A}^{i}=\mathbf{I}$ , one can write $\mathbf{m}_{\theta\theta}^{i}$ as
+利用变换矩阵 $\mathbf{A}^{i}$ 的正交性，即 $\mathbf{A}^{i^{\mathrm{T}}}\mathbf{A}^{i}=\mathbf{I}$，可以将 $\mathbf{m}_{\theta\theta}^{i}$ 写为：  
 
 $$
 \mathbf{m}_{\theta\theta}^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{G}}^{i^{\mathrm{T}}}\tilde{\bar{\mathbf{u}}}^{i^{\mathrm{T}}}\tilde{\bar{\mathbf{u}}}^{i}\bar{\mathbf{G}}^{i}d V^{i}
 $$  
 
-By factoring out terms that are not space-dependent, one gets  
+By factoring out terms that are not space-dependent, one gets
+通过剔除与空间无关的项，可以得到  
 
 $$
 \mathbf{m}_{\theta\theta}^{i}=\bar{\mathbf{G}}^{i^{\mathrm{T}}}\left[\int_{V^{i}}\rho^{i}\tilde{\bar{\mathbf{u}}}^{i^{\mathrm{T}}}\tilde{\bar{\mathbf{u}}}^{i}d V^{i}\right]\bar{\mathbf{G}}^{i}
 $$  
 
-Thus $\mathbf{m}_{\theta\theta}^{i}$ depends on both rotation of the body reference and the elastic deformation. The matrix $\mathbf{m}_{\theta\theta}^{i}$ can be written as  
+Thus $\mathbf{m}_{\theta\theta}^{i}$ depends on both rotation of the body reference and the elastic deformation. The matrix $\mathbf{m}_{\theta\theta}^{i}$ can be written as 
+因此，$\mathbf{m}_{\theta\theta}^{i}$ 同时取决于刚体的旋转和弹性变形。矩阵 $\mathbf{m}_{\theta\theta}^{i}$ 可以表示为
 
 $$
 \mathbf{m}_{\theta\theta}^{i}=\bar{\mathbf{G}}^{i\mathrm{T}}\bar{\mathbf{I}}_{\theta\theta}^{i}\bar{\mathbf{G}}^{i}
 $$  
 
 where $\bar{\mathbf{I}}_{\theta\theta}^{i}$ is called the inertia tensor of the deformable body $i$ and is defined as  
+其中 $\bar{\mathbf{I}}_{\theta\theta}^{i}$ 被称为可变形体 $i$ 的惯性张量，其定义如下：
 
 $$
 \bar{\mathbf{I}}_{\theta\theta}^{i}=\int_{V^{i}}\rho^{i}\tilde{\bar{\mathbf{u}}}^{i\mathrm{T}}\tilde{\bar{\mathbf{u}}}^{i}d V^{i}
 $$  
 
 Using Eq. 35 and noting that $\widetilde{\bar{\mathbf{u}}}^{i\mathrm{T}}=-\widetilde{\bar{\mathbf{u}}}^{i}$ , we note that Eq. 68 yields  
+利用公式35，并注意到 $\widetilde{\bar{\mathbf{u}}}^{i\mathrm{T}}=-\widetilde{\bar{\mathbf{u}}}^{i}$ ，我们注意到公式68得出
 
 $$
-\bar{\mathbf{I}}_{\theta\theta}^{i}=\int_{V^{i}}\rho^{i}\left[\begin{array}{l l l}{\left(\bar{u}_{2}^{i}\right)^{2}+\left(\bar{u}_{3}^{i}\right)^{2}}&{-\bar{u}_{2}^{i}\bar{u}_{1}^{i}}&{-\bar{u}_{3}^{i}\bar{u}_{1}^{i}}\\ &{\left(\bar{u}_{1}^{i}\right)^{2}+\left(\bar{u}_{3}^{i}\right)^{2}}&{-\bar{u}_{3}^{i}\bar{u}_{2}^{i}}\\ {\mathrm{symmetric}}&{\left(\bar{u}_{1}^{i}\right)^{2}+\left(\bar{u}_{2}^{i}\right)^{2}\right]d V^{i}}\end{array}\right]
+\bar{\mathbf{I}}_{\theta\theta}^{i} = \int_{V^{i}} \rho^{i} \left[ \begin{array}{ccc} \left(\bar{u}_{2}^{i}\right)^{2} + \left(\bar{u}_{3}^{i}\right)^{2} & -\bar{u}_{2}^{i}\bar{u}_{1}^{i} & -\bar{u}_{3}^{i}\bar{u}_{1}^{i} \\ & \left(\bar{u}_{1}^{i}\right)^{2} + \left(\bar{u}_{3}^{i}\right)^{2} & -\bar{u}_{3}^{i}\bar{u}_{2}^{i} \\ symmetric &  & \left(\bar{u}_{1}^{i}\right)^{2} + \left(\bar{u}_{2}^{i}\right)^{2} \end{array} \right] dV^{i}
 $$  
 
 It can be shown that, in order to evaluate the inertia tensor, the following distinctive inertia shape integrals are required:  
+可以证明，为了评估惯性张量，需要以下具有区分性的惯性形状积分：
 
 $$
-\begin{array}{l l}{{I_{k l}^{i}=\displaystyle\int_{V^{i}}\rho^{i}x_{k}^{i}x_{l}^{i}\,d V^{i},}}&{{\bar{\mathbf{I}}_{k l}^{i}=\displaystyle\int_{V^{i}}\rho^{i}x_{k}^{i}\mathbf{S}_{l}^{i}\,d V^{i},}}\\ {{\bar{\mathbf{S}}_{k l}^{i}=\displaystyle\int_{V^{i}}\rho^{i}\mathbf{S}_{k}^{i}\mathbf{S}_{l}^{i}d V^{i},}}&{{k,l=1,2,3}}\end{array}
+\begin{array}{l l}{{I_{k l}^{i}=\displaystyle\int_{V^{i}}\rho^{i}x_{k}^{i}x_{l}^{i}\,d V^{i},}}&{{\bar{\mathbf{I}}_{k l}^{i}=\displaystyle\int_{V^{i}}\rho^{i}x_{k}^{i}\mathbf{S}_{l}^{i}\,d V^{i},}}\\ {{\bar{\mathbf{S}}_{k l}^{i}=\displaystyle\int_{V^{i}}\rho^{i}{\mathbf{S}_{k}^{i}}^T\mathbf{S}_{l}^{i}d V^{i},}}&{{k,l=1,2,3}}\end{array}
 $$  
 
 where $\mathbf{S}_{k}^{i}$ is the kth row of the body shape function $\mathbf{S}^{i}$ . In the case of rigid body analysis the inertia tensor $\bar{\mathbf{I}}_{\theta\theta}^{i}$ is a constant matrix. In deformable body dynamics, however, the inertia tensor depends on the elastic coordinates of the body. This is clear since the vector $\bar{\mathbf{u}}^{i}$ is the sum of two vectors; the first is the undeformed position vector of the arbitrary point denoted as $\bar{\mathbf{u}}_{o}^{i}$ , while the second is the deformation vector $\mathbf{S}^{i}\mathbf{q}_{f}^{i}$ .  
+其中 $\mathbf{S}_{k}^{i}$ 是体形函数 $\mathbf{S}^{i}$ 的第 k 行。在刚体分析中，惯性张量 $\bar{\mathbf{I}}_{\theta\theta}^{i}$ 是一个常数矩阵。但在变形体动力学中，惯性张量却取决于体的弹性坐标。这显而易见，因为向量 $\bar{\mathbf{u}}^{i}$ 是两个向量的和；第一个是任意点未变形的位置向量，记为 $\bar{\mathbf{u}}_{o}^{i}$，而第二个是形变向量 $\mathbf{S}^{i}\mathbf{q}_{f}^{i}$。
 
 The inertia shape integrals of Eq. 70 are also required for the evaluation of the matrix $\mathbf{m}_{\theta f}^{i}$ of Eq. 46. Using Eq. 33 and the orthogonality of the transformation matrix, we may write $\mathbf{m}_{\theta f}^{i}$ as  
+公式 70 中的惯性形状积分也需要用于评估公式 46 中的矩阵 $\mathbf{m}_{\theta f}^{i}$。利用公式 33 和变换矩阵的正交性，我们可以将 $\mathbf{m}_{\theta f}^{i}$ 写成
 
 $$
 \mathbf{m}_{\theta f}^{i}=-\int_{V^{i}}\rho^{i}\bar{\mathbf{G}}^{i^{\mathrm{T}}}\tilde{\bar{\mathbf{u}}}^{i^{\mathrm{T}}}\mathbf{A}^{i^{\mathrm{T}}}\mathbf{A}^{i}\mathbf{S}^{i}d V^{i}=\bar{\mathbf{G}}^{i^{\mathrm{T}}}\int_{V^{i}}\rho^{i}\tilde{\bar{\mathbf{u}}}^{i}\mathbf{S}^{i}d V^{i}
 $$  
 
 where the fact that $\widetilde{\bar{\mathbf{u}}}^{i^{\mathrm{T}}}=-\widetilde{\bar{\mathbf{u}}}^{i}$ is used. Equation 71 can be written in an abbreviated form as  
+其中利用了 $\widetilde{\bar{\mathbf{u}}}^{i^{\mathrm{T}}}= - \widetilde{\bar{\mathbf{u}}}^{i}$ 这个事实。 方程 71 可以写成简略形式如下：
 
 $$
 \mathbf{m}_{\theta f}^{i}=\bar{\mathbf{G}}^{i^{\mathrm{T}}}\bar{\mathbf{I}}_{\theta f}^{i}
 $$  
 
-where $\bar{\mathbf{I}}_{\theta f}^{i}$ upon the use of Eq. 35 can be written as  
+where $\bar{\mathbf{I}}_{\theta f}^{i}$ upon the use of Eq. 35 can be written as
+其中，利用公式35，$\bar{\mathbf{I}}_{\theta f}^{i}$ 可以表示为  
 
 $$
 \bar{\mathbf{I}}_{\theta f}^{i}=\int_{V^{i}}\rho^{i}\tilde{\bar{\mathbf{u}}}^{i}\mathbf{S}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\left[\begin{array}{l}{\bar{u}_{2}^{i}\mathbf{S}_{3}^{i}-\bar{u}_{3}^{i}\mathbf{S}_{2}^{i}}\\ {\bar{u}_{3}^{i}\mathbf{S}_{1}^{i}-\bar{u}_{1}^{i}\mathbf{S}_{3}^{i}}\\ {\bar{u}_{1}^{i}\mathbf{S}_{2}^{i}-\bar{u}_{2}^{i}\mathbf{S}_{1}^{i}}\end{array}\right]d V^{i}
@@ -7191,6 +7216,7 @@ $$
 $$  
 
 where $\bar{\mathbf{u}}_{o}^{i}=[x_{1}^{i}\quad x_{2}^{i}\quad x_{3}^{i}]^{\mathrm{T}}$ is the undeformed position of the arbitrary point. Using Eq. 70, one can verify the following:  
+其中 $\bar{\mathbf{u}}_{o}^{i}=[x_{1}^{i}\quad x_{2}^{i}\quad x_{3}^{i}]^{\mathrm{T}}$ 是任意点的未变形位置。利用公式 70，可以验证以下内容：
 
 $$
 \begin{array}{r}{\left[\begin{array}{l}{\mathbf{q}_{f}^{i^{\mathrm{T}}}(\bar{\mathbf{S}}_{23}^{i}-\bar{\mathbf{S}}_{23}^{i^{\mathrm{T}}})}\\ {\mathbf{q}_{f}^{i^{\mathrm{T}}}(\bar{\mathbf{S}}_{31}^{i}-\bar{\mathbf{S}}_{31}^{i^{\mathrm{T}}})}\\ {\mathbf{q}_{f}^{i^{\mathrm{T}}}(\bar{\mathbf{S}}_{12}^{i}-\bar{\mathbf{S}}_{12}^{i^{\mathrm{T}}})}\end{array}\right]=\left[\begin{array}{l}{\mathbf{q}_{f}^{i^{\mathrm{T}}}\tilde{\mathbf{S}}_{23}^{i}}\\ {\mathbf{q}_{f}^{i^{\mathrm{T}}}\tilde{\mathbf{S}}_{31}^{i}}\\ {\mathbf{q}_{f}^{i^{\mathrm{T}}}\tilde{\mathbf{S}}_{12}^{i}}\end{array}\right]}\end{array}
@@ -7199,50 +7225,60 @@ $$
 where $\tilde{\bar{{\bf S}}}_{12}^{i},\tilde{\bar{{\bf S}}}_{23}^{i}$ , and $\tilde{\bar{{\bf S}}}_{31}^{i}$ are the skew symmetric matrices defined as  
 
 $$
-\left.\begin{array}{l}{\left.\bar{\mathbf{S}}_{12}^{i}=\bar{\mathbf{S}}_{12}^{i}-\bar{\mathbf{S}}_{12}^{i^{\mathrm{T}}}\right]}\\ {\left.\tilde{\bar{\mathbf{S}}}_{23}^{i}=\bar{\mathbf{S}}_{23}^{i}-\bar{\mathbf{S}}_{23}^{i^{\mathrm{T}}}\right\}}\\ {\left.\tilde{\bar{\mathbf{S}}}_{31}^{i}=\bar{\mathbf{S}}_{31}^{i}-\bar{\mathbf{S}}_{31}^{i^{\mathrm{T}}}\right)}\end{array}\right\}
+\begin{align} \tilde{\bar{\mathbf{S}}}_{12}^{i} &= \bar{\mathbf{S}}_{12}^{i} - (\bar{\mathbf{S}}_{12}^{i})^{\mathrm{T}} \\ \tilde{\bar{\mathbf{S}}}_{23}^{i} &= \bar{\mathbf{S}}_{23}^{i} - (\bar{\mathbf{S}}_{23}^{i})^{\mathrm{T}} \\ \tilde{\bar{\mathbf{S}}}_{31}^{i} &= \bar{\mathbf{S}}_{31}^{i} - (\bar{\mathbf{S}}_{31}^{i})^{\mathrm{T}} \end{align}
 $$  
 
 and the matrices $\bar{\mathbf{S}}_{k l}^{i}$ are given by Eq. 70.  
 
-Finally, the submatrix $\mathbf{m}_{f f}^{i}$ of Eq. 46 is independent of the generalized coordinates of the body and, therefore, is constant. This matrix can be written in terms of the inertia shape integrals of Eq. 70 as  
+Finally, the submatrix $\mathbf{m}_{f f}^{i}$ of Eq. 46 is independent of the generalized coordinates of the body and, therefore, is constant. This matrix can be written in terms of the inertia shape integrals of Eq. 70 as 
+最后，公式46中的子矩阵 $\mathbf{m}_{f f}^{i}$ 与物体的广义坐标无关，因此是一个常数。该矩阵可以根据公式70中的惯性形状积分表示为：
+ 
 
 $$
 \mathbf{m}_{f f}^{i}=\int_{V^{i}}\rho^{i}\mathbf{S}^{i^{\mathrm{T}}}\mathbf{S}^{i}d V^{i}=\bar{\mathbf{S}}_{11}^{i}+\bar{\mathbf{S}}_{22}^{i}+\bar{\mathbf{S}}_{33}^{i}
 $$  
 
 This completes the formulation of the components of the mass matrix of the deformable body in the spatial analysis.  
+这完成了对在空间分析中，变形体质量矩阵的组成部分的构建。
 
 Planar Motion A special case of three-dimensional motion is the planar motion of deformable bodies. In this special case, the reference coordinates are ${\bf q}_{r}^{i}=[{\bf R}^{i^{\textup T}}\quad\theta^{i}]^{\textup T}$ , where $\mathbf{R}^{i}=[R_{1}^{i}~R_{2}^{i}]^{\mathrm{T}}$ is the vector of Cartesian coordinates that define the location of the body reference. In the case of planar motion, the matrix ${\bf{m}}_{R R}^{i}$ can be defined as  
+平面运动
 
+三维运动的一个特例是可变形体平面运动。在这种特例中，参考坐标为 ${\bf q}_{r}^{i}=[{\bf R}^{i^{\textup T}}\quad\theta^{i}]^{\textup T}$ ，其中 $\mathbf{R}^{i}=[R_{1}^{i}~R_{2}^{i}]^{\mathrm{T}}$ 是定义物体参考位置的笛卡尔坐标向量。在平面运动的情况下，矩阵 ${\bf{m}}_{R R}^{i}$ 可以定义为
 $$
 \mathbf{m}_{R R}^{i}=\int_{V^{i}}\rho^{i}\mathbf{I}\,d V^{i}={\left[\begin{array}{l l}{m^{i}}&{0}\\ {0}&{m^{i}}\end{array}\right]}
 $$  
 
-where $\mathbf{I}$ is a $2\times2$ identity matrix and $m^{i}$ is the mass of the deformable body $i$ in the multibody system. Using the planar transformation of Eq. 5, one can define the matrix $\mathbf{B}^{i}$ as  
+where $\mathbf{I}$ is a $2\times2$ identity matrix and $m^{i}$ is the mass of the deformable body $i$ in the multibody system. Using the planar transformation of Eq. 5, one can define the matrix $\mathbf{B}^{i}$ as
+其中 $\mathbf{I}$ 是一个 $2\times2$ 单位矩阵， $m^{i}$ 是多体系统中可变形体 $i$ 的质量。利用公式 5 中的平面变换，可以定义矩阵 $\mathbf{B}^{i}$ 为  
 
 $$
 \mathbf{B}^{i}=\mathbf{A}_{\theta}^{i}\bar{\mathbf{u}}^{i}
 $$  
 
 where ${\bf A}_{\theta}^{i}$ is the partial derivative of the transformation matrix $\mathbf{A}^{i}$ with respect to the rotational coordinate $\theta^{i}$ , that is,  
+其中 ${\bf A}_{\theta}^{i}$ 是变换矩阵 $\mathbf{A}^{i}$ 对旋转坐标 $\theta^{i}$ 的偏导数，即，
 
 $$
 \mathbf{A}_{\theta}^{i}=\left[\begin{array}{l l}{-\sin\theta^{i}}&{-\cos\theta^{i}}\\ {\cos\theta^{i}}&{-\sin\theta^{i}}\end{array}\right]
 $$  
 
 Using Eq. 79, we can write the submatrix ${\bf{m}}_{R\theta}^{i}$ of Eq. 46 as  
+使用公式 79，我们可以将公式 46 中的子矩阵 ${\bf{m}}_{R\theta}^{i}$ 写为
 
 $$
 \begin{array}{l}{{\displaystyle{\bf m}_{R\theta}^{i}=\int_{V^{i}}\rho^{i}{\bf B}^{i}d V^{i}={\bf A}_{\theta}^{i}\int_{V^{i}}\rho^{i}\bar{\bf u}^{i}d V^{i}={\bf A}_{\theta}^{i}\int_{V^{i}}\rho^{i}\left[\bar{\bf u}_{o}^{i}+\bar{\bf u}_{f}^{i}\right]d V^{i}}\ ~}\\ {{\displaystyle~~~~~~={\bf A}_{\theta}^{i}\left[{\bf I}_{1}^{i}+\bar{\bf S}^{i}{\bf q}_{f}^{i}\right]}}\end{array}
 $$  
 
-where the matrices $\mathbf{I}_{1}^{i}$ and $\bar{\mathbf{S}}^{i}$ are defined as  
+where the matrices $\mathbf{I}_{1}^{i}$ and $\bar{\mathbf{S}}^{i}$ are defined as
+其中，矩阵 $\mathbf{I}_{1}^{i}$ 和 $\bar{\mathbf{S}}^{i}$ 被定义如下：  
 
 $$
 \mathbf{I}_{1}^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}_{o}^{i}d V^{i},\quad\bar{\mathbf{S}}^{i}=\int_{V^{i}}\rho^{i}\mathbf{S}^{i}d V^{i}
 $$  
 
 The vector $\mathbf{I}_{1}^{i}$ is the moment of mass of the body about the axes of the body reference in the undeformed state. Therefore, if the origin of the body reference is initially attached to the body center of mass, the vector $\mathbf{I}_{1}^{i}$ vanishes. The vector $\bar{\mathbf{S}}^{i}\mathbf{q}_{f}^{i}$ represents the change in the moment of mass due to the deformation.  
+向量 $\mathbf{I}_{1}^{i}$ 表示刚体在未变形状态下，关于其自身坐标系的转动惯量。因此，如果刚体坐标系的原点最初与质量中心重合，则向量 $\mathbf{I}_{1}^{i}$ 为零。向量 $\bar{\mathbf{S}}^{i}\mathbf{q}_{f}^{i}$ 代表由于变形引起的转动惯量的变化。
 
 Using Eq. 46, one can verify that  
 
@@ -7253,22 +7289,42 @@ $$
 Because ${\bf A}_{\theta}^{i}$ of Eq. 80 is an orthogonal matrix, that is, ${\bf A}_{\theta}^{i^{\mathrm{T}}}{\bf A}_{\theta}^{i}={\bf I}.$ , Eq. 46 yields  
 
 $$
-\mathbf{m}_{\theta\theta}^{i}=\int_{V^{i}}\rho^{i}\mathbf{B}^{i}\mathbf{^{T}}\mathbf{B}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}^{i}\mathbf{^{T}}\mathbf{A}_{\theta}^{i}\bar{\mathbf{u}}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}^{i}\mathbf{^{T}}\bar{\mathbf{u}}^{i}d V^{i}
+\mathbf{m}_{\theta\theta}^{i}=\int_{V^{i}}\rho^{i}\mathbf{B}^{i}\mathbf{^{T}}\mathbf{B}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}^{i}\mathbf{^{T}}{\bf A}_{\theta}^{i^{\mathrm{T}}}{\bf A}_{\theta}^{i}\bar{\mathbf{u}}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}^{i}\mathbf{^{T}}\bar{\mathbf{u}}^{i}d V^{i}
 $$  
 
 Since $\bar{\mathbf{u}}^{i}=\bar{\mathbf{u}}_{o}^{i}+\bar{\mathbf{u}}_{f}^{i}$ , where $\bar{\mathbf{u}}_{f}^{i}=\mathbf{S}^{i}\mathbf{q}_{f}^{i}$ , one can write Eq. 84 as  
 
 $$
-\begin{array}{l}{{{\bf{m}}_{\theta\theta}^{i}=\displaystyle\int_{V^{i}}\rho^{i}\left[\bar{\bf{u}}_{o}^{i}+\bar{\bf{u}}_{f}^{i}\right]^{\mathrm{T}}\left[\bar{\bf{u}}_{o}^{i}+\bar{\bf{u}}_{f}^{i}\right]d V^{i}}\\ {\quad}\\ {{\bf{\sigma}}=\displaystyle\int_{V^{i}}\rho^{i}\left[\bar{\bf{u}}_{o}^{i^{\mathrm{T}}}\bar{\bf{u}}_{o}^{i}+2\bar{\bf{u}}_{o}^{i^{\mathrm{T}}}\bar{\bf{u}}_{f}^{i}+\bar{\bf{u}}_{f}^{i^{\mathrm{T}}}\bar{\bf{u}}_{f}^{i}\right]d V^{i}}\\ {\quad}\\ {{\bf{\sigma}}=\left(m_{\theta\theta}^{i}\right)_{r r}+\left(m_{\theta\theta}^{i}\right)_{r f}+\left(m_{\theta\theta}^{i}\right)_{f f}}\end{array}
+\begin{align}
+    \mathbf{m}_{\theta\theta}^{i} &= \int_{V^{i}} \rho^{i} \left[\bar{\mathbf{u}}_{o}^{i} + \bar{\mathbf{u}}_{f}^{i}\right]^{\mathrm{T}} \left[\bar{\mathbf{u}}_{o}^{i} + \bar{\mathbf{u}}_{f}^{i}\right] \, dV^{i}, \\
+     &= \int_{V^{i}} \rho^{i} \left(\bar{\mathbf{u}}_{o}^{i\mathrm{T}} \bar{\mathbf{u}}_{o}^{i} + 2 \bar{\mathbf{u}}_{o}^{i\mathrm{T}} \bar{\mathbf{u}}_{f}^{i} + \bar{\mathbf{u}}_{f}^{i\mathrm{T}} \bar{\mathbf{u}}_{f}^{i}\right) \, dV^{i}, \\
+     &= \left(m_{\theta\theta}^{i}\right)_{rr} + \left(m_{\theta\theta}^{i}\right)_{rf} + \left(m_{\theta\theta}^{i}\right)_{ff}.
+\end{align}
+
 $$  
 
-in which the submatrix $\mathbf{m}_{\theta\theta}^{i}$ reduces to a scalar that can be written as the sum of three components. The first component, $(m_{\theta\theta}^{i})_{r r}$ , can be recognized as the mass moment of inertia, in the undeformed state, of the body about the origin of the body reference. This scalar component can be written as  
+in which the submatrix $\mathbf{m}_{\theta\theta}^{i}$ reduces to a scalar that can be written as the sum of three components. The first component, $(m_{\theta\theta}^{i})_{r r}$ , can be recognized as the mass moment of inertia, in the undeformed state, of the body about the origin of the body reference. This scalar component can be written as 
+其中，子矩阵 $\mathbf{m}_{\theta\theta}^{i}$ 简化为一个标量，它可以写成三个分量的和。第一个分量，$(m_{\theta\theta}^{i})_{r r}$，可以被识别为在未变形状态下，关于物体的原点，物体的转动惯量。这个标量分量可以表示为：
+ 
 
 $$
-\begin{array}{l}{{\displaystyle\left(m_{\theta\theta}^{i}\right)_{r r}=\int_{V^{i}}\rho^{i}\bar{\bf u}_{o}^{i^{\mathrm{T}}}\bar{\bf u}_{o}^{i}\,d V^{i}=\int_{V^{i}}\rho^{i}\left[x_{1}^{i}\quad x_{2}^{i}\right]\left[x_{1}^{i}\right]d V^{i}}\ ~}\\ {{\displaystyle~~~~~~~~=\int_{V^{i}}\rho^{i}\left[\left(x_{1}^{i}\right)^{2}+\left(x_{2}^{i}\right)^{2}\right]d V^{i}}}\end{array}
+\begin{align}
+\left(m_{\theta\theta}^{i}\right)_{rr} &= \int_{V^{i}} \rho^{i} \bar{\mathbf{u}}_{o}^{i^{\mathrm{T}}} \bar{\mathbf{u}}_{o}^{i} \, dV^{i} \\
+&= \int_{V^{i}} \rho^{i} 
+\begin{bmatrix}
+x_{1}^{i} & x_{2}^{i}
+\end{bmatrix}
+\begin{bmatrix}
+x_{1}^{i} \\ 
+x_{2}^{i}
+\end{bmatrix}
+dV^{i} \\
+&= \int_{V^{i}} \rho^{i} \left[ (x_{1}^{i})^2 + (x_{2}^{i})^2 \right] dV^{i}
+\end{align}
 $$  
 
 Clearly, this integral has a constant value and does not depend on the body deformation. The last two scalar components, $(m_{\theta\theta}^{i})_{r f}$ and $(m_{\theta\theta}^{i})_{f f}$ , of Eq. 85 represent the change in the mass moment of inertia of the body due to deformation. These two components are evaluated according to  
+显然，这个积分具有常数值，且不依赖于物体的变形。公式85中的最后两个标量分量，$(m_{\theta\theta}^{i})_{r f}$ 和 $(m_{\theta\theta}^{i})_{f f}$，代表了由于变形引起的物体惯性矩的变化。这两个分量的计算方法如下：
 
 $$
 \left(m_{\theta\theta}^{i}\right)_{r f}=2\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}_{o}^{i\mathrm{T}}\bar{\mathbf{u}}_{f}^{i}\,d V^{i}=2\left[\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}_{o}^{i\mathrm{T}}\mathbf{S}^{i}\,d V^{i}\right]\mathbf{q}_{f}^{i}
@@ -7277,42 +7333,50 @@ $$
 and  
 
 $$
-\begin{array}{r l}&{\left(m_{\theta\theta}^{i}\right)_{f f}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}_{f}^{i}\mathbf{\bar{u}}_{f}^{i}\,d V^{i}=\int_{V^{i}}\rho^{i}\mathbf{q}_{f}^{i\mathrm{T}}\mathbf{S}^{i\mathrm{T}}\mathbf{S}^{i}\mathbf{q}_{f}^{i}\,d V^{i}}\\ &{\qquad\qquad=\mathbf{q}_{f}^{i\mathrm{T}}\left[\displaystyle\int_{V^{i}}\rho^{i}\mathbf{S}^{i}\mathbf{\bar{S}}^{i}d V^{i}\right]\mathbf{q}_{f}^{i}}\end{array}
+\begin{array}{r l}&{\left(m_{\theta\theta}^{i}\right)_{f f}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}_{f}^{i}\mathbf{\bar{u}}_{f}^{i}\,d V^{i}=\int_{V^{i}}\rho^{i}\mathbf{q}_{f}^{i\mathrm{T}}\mathbf{S}^{i\mathrm{T}}\mathbf{S}^{i}\mathbf{q}_{f}^{i}\,d V^{i}}\\ &{\qquad\qquad=\mathbf{q}_{f}^{i\mathrm{T}}\left[\displaystyle\int_{V^{i}}\rho^{i}\mathbf{S}^{i\mathrm{T}}\mathbf{S}^{i}d V^{i}\right]\mathbf{q}_{f}^{i}}\end{array}
 $$  
 
-If we use the definition of Eq. 46, Eq. 88 can be written in the following form:  
+If we use the definition of Eq. 46, Eq. 88 can be written in the following form:  如果使用公式46的定义，公式88可以写成如下形式：
 
 $$
 \left(m_{\theta\theta}^{i}\right)_{f f}=\mathbf{q}_{f}^{i^{\mathrm{T}}}\mathbf{m}_{f f}^{i}\mathbf{q}_{f}^{i}
 $$  
 
 Note that the two scalars $(m_{\theta\theta}^{i})_{r f}$ and $(m_{\theta\theta}^{i})_{f f}$ depend on the elastic deformation of the body. Finally, we write  
+请注意，两个标量 $(m_{\theta\theta}^{i})_{r f}$ 和 $(m_{\theta\theta}^{i})_{f f}$ 取决于物体的弹性变形。最后，我们写出：
+
 
 $$
 \mathbf{m}_{\theta f}^{i}=\int_{V^{i}}\rho^{i}\mathbf{B}^{i^{\mathrm{T}}}\mathbf{A}^{i}\mathbf{S}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}^{i^{\mathrm{T}}}\mathbf{A}_{\theta}^{i^{\mathrm{T}}}\mathbf{A}^{i}\mathbf{S}^{i}d V^{i}
 $$  
 
 It can be shown that the product $\mathbf{A}_{\theta}^{i^{\mathrm{T}}}\mathbf{A}^{i}$ is a skew symmetric matrix defined as  
+可以证明，矩阵 $\mathbf{A}_{\theta}^{i^{\mathrm{T}}}\mathbf{A}^{i}$ 是一个斜对称矩阵，其定义为：
 
 $$
 \tilde{\mathbf{I}}=\mathbf{A}_{\theta}^{i^{\mathrm{T}}}\mathbf{A}^{i}=\left[\begin{array}{l l}{0}&{1}\\ {-1}&{0}\end{array}\right]
 $$  
 
 Substituting Eq. 91 into Eq. 90 and writing $\bar{\mathbf{u}}^{i}$ in a more explicit form yields  
+将公式(91)代入公式(90)，并将$\bar{\mathbf{u}}^{i}$写成更明确的形式，得到
 
 $$
-\mathbf{m}_{\theta f}^{i}=\int_{V^{i}}\rho^{i}\left[\bar{\mathbf{u}}_{o}^{i}+\bar{\mathbf{u}}_{f}^{i}\right]^{\mathrm{T}}\mathbf{\widetilde{IS}}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}_{o}^{i\mathrm{T}}\mathbf{\widetilde{IS}}^{i}d V^{i}+\mathbf{q}_{f}^{i\mathrm{T}}\mathbf{\widetilde{S}}^{i}
+\mathbf{m}_{\theta f}^{i}=\int_{V^{i}}\rho^{i}\left[\bar{\mathbf{u}}_{o}^{i}+\bar{\mathbf{u}}_{f}^{i}\right]^{\mathrm{T}}\mathbf{\widetilde{I}S}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\bar{\mathbf{u}}_{o}^{i\mathrm{T}}\mathbf{\widetilde{I}S}^{i}d V^{i}+\mathbf{q}_{f}^{i\mathrm{T}}\mathbf{\widetilde{S}}^{i}
 $$  
 
 where the constant skew symmetric matrix $\tilde{\mathbf{S}}^{i}$ is defined as  
+其中，常数斜对称矩阵 $\tilde{\mathbf{S}}^{i}$ 定义如下：
 
 $$
 \widetilde{\mathbf{S}}^{i}=\int_{V^{i}}\rho^{i}\mathbf{S}^{i^{\mathrm{T}}}\widetilde{\mathbf{I}}\mathbf{S}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\left[\mathbf{S}_{1}^{i^{\mathrm{T}}}\mathbf{S}_{2}^{i}-\mathbf{S}_{2}^{i^{\mathrm{T}}}\mathbf{S}_{1}^{i}\right]d V^{i}
 $$  
 
 in which $\mathbf{S}_{1}^{i}$ and $\mathbf{S}_{2}^{i}$ are the rows of the shape function $\mathbf{S}^{i}$ . One may also observe that the submatrix $\mathbf{m}_{\theta f}^{i}$ consists of two parts; the first part is constant, while the second part depends on the elastic coordinates of the body.  
+其中，$\mathbf{S}_{1}^{i}$ 和 $\mathbf{S}_{2}^{i}$ 分别是形函数 $\mathbf{S}^{i}$ 的行向量。 值得注意的是，子矩阵 $\mathbf{m}_{\theta f}^{i}$ 由两部分组成；第一部分是常数，而第二部分则取决于结构的弹性坐标。
 
-We conclude that, to completely describe the inertia properties of the deformable body in plane motion, a set of inertia shape integrals is required. These integrals, which depend on the assumed displacement field, are  
+We conclude that, to completely describe the inertia properties of the deformable body in plane motion, a set of inertia shape integrals is required. These integrals, which depend on the assumed displacement field, are 
+我们得出结论，为了完全描述平面运动中可变形体的惯性特性，需要一组惯性形状积分。这些积分取决于所假设的位移场，并且是……
+ 
 
 $$
 \mathbf{I}_{1}^{i}=\int_{V^{i}}\rho^{i}\left[x_{1}^{i}\quad\boldsymbol{x}_{2}^{i}\right]^{\mathrm{T}}d V^{i},\ \quad I_{k l}^{i}=\int_{V^{i}}\rho^{i}x_{k}^{i}x_{l}^{i}\,d V^{i},\quad k,l=1,2
@@ -7321,36 +7385,42 @@ $$
 and  
 
 $$
-\begin{array}{r}{\bar{\mathbf{I}}_{k l}^{i}=\int_{V^{i}}\rho^{i}x_{k}^{i}\mathbf{S}_{l}^{i}d V^{i},\quad\bar{\mathbf{S}}^{i}=\int_{V^{i}}\rho^{i}\mathbf{S}^{i}d V^{i},\quad\bar{\mathbf{S}}_{k l}^{i}=\int_{V^{i}}\rho^{i}\mathbf{S}_{k}^{i}\mathbf{S}_{l}^{i}\,d V^{i},}\\ {k,l=1,2}\end{array}
+\begin{array}{r}{\bar{\mathbf{I}}_{k l}^{i}=\int_{V^{i}}\rho^{i}x_{k}^{i}\mathbf{S}_{l}^{i}d V^{i},\quad\bar{\mathbf{S}}^{i}=\int_{V^{i}}\rho^{i}\mathbf{S}^{i}d V^{i},\quad\bar{\mathbf{S}}_{k l}^{i}=\int_{V^{i}}\rho^{i}{\mathbf{S}_{k}^{i}}^T\mathbf{S}_{l}^{i}\,d V^{i},}\\ {k,l=1,2}\end{array}
 $$  
 
 where the constant matrix $\mathbf{m}_{f f}^{i}$ can be written in terms of these integrals as  
+其中常数矩阵 $\mathbf{m}_{ff}^{i}$ 可以表示为这些积分的函数：
 
 $$
-\mathbf{m}_{f f}^{i}=\int_{V^{i}}\rho^{i}\mathbf{S}^{i}\mathbf{\bar{S}}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\left[\mathbf{S}_{1}^{i^{\mathrm{T}}}\mathbf{S}_{1}^{i}+\mathbf{S}_{2}^{i^{\mathrm{T}}}\mathbf{S}_{2}^{i}\right]d V^{i}=\bar{\mathbf{S}}_{11}^{i}+\bar{\mathbf{S}}_{22}^{i}
+\mathbf{m}_{f f}^{i}=\int_{V^{i}}\rho^{i}{\mathbf{S}^{i}}^T\mathbf{\bar{S}}^{i}d V^{i}=\int_{V^{i}}\rho^{i}\left[\mathbf{S}_{1}^{i^{\mathrm{T}}}\mathbf{S}_{1}^{i}+\mathbf{S}_{2}^{i^{\mathrm{T}}}\mathbf{S}_{2}^{i}\right]d V^{i}=\bar{\mathbf{S}}_{11}^{i}+\bar{\mathbf{S}}_{22}^{i}
 $$  
 
-If the body is rigid, only the integrals of Eq. 94 are required. On the other hand, if the large rigid body displacement is not permitted, which is the case in linear structural systems, only the constant matrix of Eq. 96 is required.  
+If the body is rigid, only the integrals of Eq. 94 are required. On the other hand, if the large rigid body displacement is not permitted, which is the case in linear structural systems, only the constant matrix of Eq. 96 is required. 
+如果刚体保持固定，则只需用到公式 94 中的积分项。另一方面，如果不允许大刚体位移，这在一些线性结构系统中是这样，则只需用到公式 96 中的常数矩阵。
 
 Example 5.3 For the deformable beam given in Example 1, the inertia shape integral $\bar{\mathbf{S}}^{i}$ of Eq. 95 is given by (since we have only one body, the superscript $i$ is omitted for simplicity)  
+例 5.3 对于例 1 中给定的挠曲梁，方程 95 中的惯性形状积分 $\bar{\mathbf{S}}^{i}$ 可表示为（由于只有一个刚体，为了简化，省略了上标 $i$）
 
 $$
-\bar{\bf S}=\int_{V}\rho{\bf S}\,d V=\int_{V}\rho\left[\xi\begin{array}{c c}{{\displaystyle0}}&{{\displaystyle0}}\\ {{\displaystyle0}}&{{\displaystyle3(\xi)^{2}-2(\xi)^{3}}}\end{array}\right]d V
+\bar{\bf S}=\int_{V}\rho{\bf S}\,d V=\int_{V}\rho\left[\begin{array}{c c}{{\xi}}&{{\displaystyle0}}\\ {{\displaystyle0}}&{{\displaystyle3(\xi)^{2}-2(\xi)^{3}}}\end{array}\right]d V
 $$  
 
 where $\rho$ is the mass density of the beam material; $\xi=(x/l)$ ; $l$ is the length of the beam; and $V$ is the volume. If the beam is assumed to have a constant cross-sectional area $a$ , then  
+其中，$\rho$ 为梁材料的质量密度；$\xi=(x/l)$；$l$ 为梁的长度；且 $V$ 为体积。如果假设梁具有恒定的横截面积 $a$，那么
 
 $$
 d V=a\;d x=a l{\frac{d x}{l}}=V d\xi
 $$  
 
-Assuming that the mass density $\rho$ is constant, we have $\rho V d\xi=m\,d\xi$ , where $m$ is the total mass of the beam. The matrix $\bar{\bf S}$ can then be written as  
+Assuming that the mass density $\rho$ is constant, we have $\rho V d\xi=m\,d\xi$ , where $m$ is the total mass of the beam. The matrix $\bar{\bf S}$ can then be written as 
+假设质量密度 $\rho$ 为常数，则有 $\rho V d\xi=m\,d\xi$，其中 $m$ 为梁的总质量。 矩阵 $\bar{\bf S}$ 因而可以写成： 
 
 $$
-\bar{\bf S}=\int_{0}^{1}m\left[\xi\begin{array}{r r}{{}}&{{0}}\\ {{0}}&{{3(\xi)^{2}-2(\xi)^{3}}}\end{array}\right]d\xi=\frac{m}{2}\left[\!\!\begin{array}{r r}{{1}}&{{0}}\\ {{0}}&{{1}}\end{array}\right]
+\bar{\bf S}=\int_{0}^{1}m\left[\begin{array}{r r}{\xi}&{{0}}\\ {{0}}&{{3(\xi)^{2}-2(\xi)^{3}}}\end{array}\right]d\xi=\frac{m}{2}\left[\begin{array}{c c}{{1}}&{{0}}\\ {{0}}&{{1}}\end{array}\right]
 $$  
 
 The skew symmetric matrix $\tilde{\bf S}$ of Eq. 93 is  
+式 (93) 中的斜对称矩阵 $\tilde{\bf S}$ 为
 
 $$
 \tilde{\mathbf{S}}=\int_{V}\rho\mathbf{S}^{\mathrm{T}}\tilde{\mathbf{I}}\mathbf{S}\,d V=\int_{0}^{1}m\mathbf{S}^{\mathrm{T}}\tilde{\mathbf{I}}\mathbf{S}\,d\xi=m\left[\begin{array}{c c}{0}&{\frac{7}{20}}\\ {-\frac{7}{20}}&{0}\end{array}\right]
@@ -7371,22 +7441,25 @@ $$
 one has the following:  
 
 $$
-\begin{array}{l}{{{\bf{I}}_{1}=\displaystyle\int_{V}\rho{\bar{\bf{u}}}_{o}\,d V=m l\int_{0}^{1}{[\xi{\bf{\Lambda}}~{\bf{0}}]^{\mathrm{T}}}\,d\xi=\frac{m l}{2}\,[1{\mathrm{\boldmath~\Psi~}}0]^{\mathrm{T}}}}\\ {{\displaystyle\int_{V}\rho{\bar{\bf{u}}}_{o}^{\mathrm{T}}{\bf{S}}\,d V=\frac{m l}{3}\,[1{\mathrm{\boldmath~\Psi~}}0]}}\\ {{\displaystyle\int_{V}\rho{\bar{\bf{u}}}_{o}^{\mathrm{T}}{\bf{\tilde{I}}}{\bf{\tilde{S}}}\,d V=m l\biggl[0{\mathrm{\boldmath~\Psi~}}\frac{7}{20}\biggr]^{\mathrm{T}}}}\end{array}
+\begin{array}{l}{{{\bf{I}}_{1}=\displaystyle\int_{V}\rho{\bar{\bf{u}}}_{o}\,d V=m l\int_{0}^{1}{[\xi\quad0]^{\mathrm{T}}}\,d\xi=\frac{m l}{2}\,[1\quad0]^{\mathrm{T}}}}\\ {{\displaystyle\int_{V}\rho{\bar{\bf{u}}}_{o}^{\mathrm{T}}{\bf{S}}\,d V=\frac{m l}{3}\,[1\quad0]}}\\ {{\displaystyle\int_{V}\rho{\bar{\bf{u}}}_{o}^{\mathrm{T}}{\bf{\tilde{I}}}{\bf{{S}}}\,d V=m l\biggl[0\quad\frac{7}{20}\biggr]^{\mathrm{T}}}}\end{array}
 $$  
 
 $$
-(m_{\theta\theta})_{r r}=\int_{V}\rho\bar{\bf u}_{o}^{\mathrm{T}}\bar{\bf u}_{o}\,d V=m(l)^{2}\int_{0}^{1}[\xi\quad0]\left[\underline{{{\xi}}}\right]d\xi=\frac{m(l)^{2}}{3}
+(m_{\theta\theta})_{r r}=\int_{V}\rho\bar{\bf u}_{o}^{\mathrm{T}}\bar{\bf u}_{o}\,d V=m(l)^{2}\int_{0}^{1}[\xi\quad0]\left[\begin{array}{c}\xi\\ 0\end{array}\right]d\xi=\frac{m(l)^{2}}{3}
 $$  
 
-which is the mass moment of inertia of the beam, in the undeformed state, about the $\mathbf{X}_{3}$ axis.  
+which is the mass moment of inertia of the beam, in the undeformed state, about the $\mathbf{X}_{3}$ axis. 
+即为梁在未变形状态下，关于 $\mathbf{X}_{3}$ 轴的转动惯量。 
 
 Let the mass of the beam be $1.236\,\mathrm{kg}$ and the length $0.5\;\mathrm{m}$ , and let at a given instant of time the vector of beam coordinates be given by  
+设梁的质量为 $1.236\,\mathrm{kg}$，长度为 $0.5\;\mathrm{m}$，并且在某一时刻，梁的坐标矢量由
 
 $$
 \begin{array}{l}{{\bf{q}}=\left[{\bf{q}}_{r}^{\mathrm{T}}\quad{\bf{q}}_{f}^{\mathrm{T}}\right]^{\mathrm{T}}=[R_{1}\quad R_{2}\quad\theta\quad q_{f1}\quad q_{f2}]^{\mathrm{T}}}\\ {=[1.0\quad0.5\quad30^{\circ}\quad0.001\quad0.01]^{\mathrm{T}}}\end{array}
 $$  
 
 The components of the mass matrix of Eq. 45 can be evaluated as follows. By use of Eq. 78, the matrix $\mathbf{m}_{R R}$ associated with the translational coordinates is given by  
+公式45的质量矩阵各分量可按以下方式计算。利用公式78，与平动坐标相关的矩阵 $\mathbf{m}_{R R}$ 可表示为：
 
 $$
 \mathbf{m}_{R R}={\left[\begin{array}{l l}{m}&{0}\\ {0}&{m}\end{array}\right]}={\left[\begin{array}{l l}{1.236}&{0}\\ {\ 0}&{1.236}\end{array}\right]}
@@ -7401,13 +7474,13 @@ $$
 and  
 
 $$
-\bar{\bf S}\mathbf{q}_{f}=\frac{m}{2}\left[1\begin{array}{c c}{0}\\ {0}\end{array}\right]\left[\begin{array}{c}{q_{f1}}\\ {q_{f2}}\end{array}\right]=\left[\begin{array}{c}{0.000618}\\ {0.00618}\end{array}\right]
+\bar{\bf S}\mathbf{q}_{f}=\frac{m}{2}{\left[\begin{array}{l l}{1}&{0}\\ {\ 0}&{1}\end{array}\right]}\left[\begin{array}{c}{q_{f1}}\\ {q_{f2}}\end{array}\right]=\left[\begin{array}{c}{0.000618}\\ {0.00618}\end{array}\right]
 $$  
 
 that is,  
 
 $$
-\mathbf{I}_{1}+\bar{\mathbf{S}}\mathbf{q}_{f}=\left[\begin{array}{c}{0.309}\\ {0}\end{array}\right]+\left[\begin{array}{c}{0.000618}\\ {0.00618}\end{array}\right]=\left[\boldsymbol{0.30962}^{0.30962}\right]
+\mathbf{I}_{1}+\bar{\mathbf{S}}\mathbf{q}_{f}=\left[\begin{array}{c}{0.309}\\ {0}\end{array}\right]+\left[\begin{array}{c}{0.000618}\\ {0.00618}\end{array}\right]=\left[\begin{array}{c}{0.30962}\\ {0.00618}\end{array}\right]
 $$  
 
 Since the matrix $\mathbf{A}_{\theta}$ in this case is given by  
@@ -7419,7 +7492,7 @@ $$
 the matrix $\mathbf{m}_{R\theta}$ of Eq. 81 is given by  
 
 $$
-\mathbf{m}_{R\theta}=\mathbf{A}_{\theta}\left[\mathbf{I}_{1}+\bar{\mathbf{S}}\mathbf{q}_{f}\right]=\left[{}-0.5\quad-0.866\,\right]\left[\mathbf{0.30962}\,\right]=\left[{\overset{-0.1602}{0.2650}}\,\right]
+\mathbf{m}_{R\theta}=\mathbf{A}_{\theta}\left[\mathbf{I}_{1}+\bar{\mathbf{S}}\mathbf{q}_{f}\right]={\left[\begin{array}{l l}{-0.5}&{-0.866}\\ {0.866}&{-0.5}\end{array}\right]}\left[\begin{array}{c}{0.30962}\\ {0.00618}\end{array}\right]=\left[\begin{array}{c}{-0.1602}\\ {0.2650}\end{array}\right]
 $$  
 
 For the value of $\theta=30^{\circ}$ , the transformation matrix $\mathbf{A}$ is  
@@ -7431,40 +7504,61 @@ $$
 and the matrix $\mathbf{m}_{R f}$ of Eq. 83 is given by  
 
 $$
-\begin{array}{r l}&{\mathbf{m}_{R f}=\mathbf{A}\bar{\mathbf{S}}=\left[0.866\quad-0.5\right]\left[\begin{array}{l l}{1}&{0}\\ {0}&{1}\end{array}\right]\left(\frac{1.236}{2}\right)}\\ &{\qquad=\left[0.5352\quad-0.309\right]}\end{array}
+\begin{array}{r l}&{\mathbf{m}_{R f}=\mathbf{A}\bar{\mathbf{S}}={\left[\begin{array}{l l}{0.866}&{-0.5}\\ {0.500}&{0.866}\end{array}\right]}\left[\begin{array}{l l}{1}&{0}\\ {0}&{1}\end{array}\right]\left(\frac{1.236}{2}\right)}\\ &{\qquad={\left[\begin{array}{l l}{0.5352}&{−0.309}\\ {0.309}&{0.5352}\end{array}\right]}}\end{array}
 $$  
 
-The scalars $(m_{\theta\theta})_{r f}$ and $(m_{\theta\theta})_{f f}$ of Eqs. 87 and 88 are given by  
+The scalars $(m_{\theta\theta})_{r f}$ and $(m_{\theta\theta})_{f f}$ of Eqs. 87 and 88 are given by 
+标量 $(m_{\theta\theta})_{r f}$ 和 $(m_{\theta\theta})_{f f}$，如公式 87 和 88 所示，由以下给出： 
 
 $$
-(m_{\theta\theta})_{r f}=2\left[\int_{V}\rho\bar{\mathbf{u}}_{o}^{\mathrm{T}}\mathbf{S}\;d V\right]\mathbf{q}_{f}=2\frac{m l}{3}\left[1\quad0\right]\left[q_{f2}\right]
+\begin{array}{c}{{(m_{\theta\theta})_{r f}=2\left[\int_{V}\rho\bar{\mathbf{u}}_{o}^{\mathrm{T}}\mathbf{S}\;d V\right]\mathbf{q}_{f}=2\frac{m l}{3}\left[1\quad0\right]\left[\begin{array}{c}{q_{f1}}\\ {q_{f2}}\end{array}\right]\\
+
+}}\end{array}
 $$  
 
 $$
-\begin{array}{c}{{=\displaystyle\frac{2m l}{3}q_{f1}=\frac{2(1.236)(0.5)}{3}(0.001)=4.12\times10^{-4}}}\\ {{(m_{\theta\theta})_{f f}=\displaystyle\mathbf{q}_{f}^{\mathrm{T}}\mathbf{m}_{f f}\mathbf{q}_{f}=\frac{1.236}{2}\left[0.001\quad0.01\right]\left[\frac{1}{0}\quad0\right]\left[\stackrel{0.001}{0.01}\right]}}\\ {{=2.316\times10^{-5}}}\end{array}
+\begin{array}{c}{{=\displaystyle\frac{2m l}{3}q_{f1}=\frac{2(1.236)(0.5)}{3}(0.001)=4.12\times10^{-4}}}\\ {{(m_{\theta\theta})_{f f}=\displaystyle\mathbf{q}_{f}^{\mathrm{T}}\mathbf{m}_{f f}\mathbf{q}_{f}=\frac{1.236}{2}\left[0.001\quad0.01\right]\left[\begin{array}{c c}{\frac{1}{3}}&{0}\\ {0}&{\frac{13}{35}}\end{array}\right]\left[\begin{array}{c}{0.001}\\ {0.01}\end{array}\right]
+\\ {= 2.316\times10^{-5}}
+}}\end{array}
 $$  
 
 Therefore, $\mathbf{m}_{\theta\theta}$ in Eq. 45 is given by  
 
 $$
-\begin{array}{l}{{{\bf{m}}_{\theta\theta}=(m_{\theta\theta})_{r r}+(m_{\theta\theta})_{r f}+(m_{\theta\theta})_{f f}\mathrm{~}}}\\ {{\mathrm{~}}}\\ {{\bf{\zeta}}=\frac{m(l)^{2}}{3}+(4.12)\times10^{-4}+2.316\times10^{-5}\mathrm{~}}\\ {{\mathrm{~}}}\\ {{\bf{\zeta}}=\frac{1.236(0.5)^{2}}{3}+(4.12)\times10^{-4}+2.316\times10^{-5}=0.10306}\end{array}
+\begin{array}{l}{{{\bf{m}}_{\theta\theta}=(m_{\theta\theta})_{r r}+(m_{\theta\theta})_{r f}+(m_{\theta\theta})_{f f}\mathrm{~}}}\\ {{\mathrm{~}}}\\ {=\frac{m(l)^{2}}{3}+(4.12)\times10^{-4}+2.316\times10^{-5}\mathrm{~}}\\ {{\mathrm{~}}}\\ {=\frac{1.236(0.5)^{2}}{3}+(4.12)\times10^{-4}+2.316\times10^{-5}=0.10306}\end{array}
 $$  
 
 Using Eq. 92, we can evaluate the matrix $\mathbf{m}_{\theta f}$ as  
 
 $$
-\begin{array}{r l}&{\mathbf{m}_{\theta f}=\displaystyle\int_{V}\rho\bar{\mathbf{u}}_{0}^{\top}\bar{\mathbf{B}}\,d V+\mathbf{q}_{f}^{\top}\bar{\mathbf{S}}}\\ &{\phantom{\sum{\Biggl(}}=m l\left[0\quad\frac{7}{20}\right]+\left[q_{f1}\quad q_{f2}\right](m)\left[\begin{array}{c c}{0}&{\frac{7}{20}}\\ {-\frac{7}{20}}&{0}\end{array}\right]}\\ &{\phantom{\sum{\Biggl(}}=(1.236)(0.5)\left[0\quad\frac{7}{20}\right]+1.236\left[0.001\quad0.01\right]\left[\begin{array}{c c}{0}&{\frac{7}{20}}\\ {-\frac{7}{20}}&{0}\end{array}\right]}\\ &{\phantom{\sum{\Biggl(}}=\left[-4.326\times10^{-3}\quad0.2167\right]}\end{array}
+\begin{array}{r l}&{\mathbf{m}_{\theta f}=\displaystyle\int_{V}\rho\bar{\mathbf{u}}_{0}^{T}\widetilde{\mathbf{I}}\mathbf{S}\,d V+\mathbf{q}_{f}^{T}\widetilde{\mathbf{S}}}\\ &{\phantom{\sum{\Biggl(}}=m l\left[0\quad\frac{7}{20}\right]+\left[q_{f1}\quad q_{f2}\right](m)\left[\begin{array}{c c}{0}&{\frac{7}{20}}\\ {-\frac{7}{20}}&{0}\end{array}\right]}\\ &{\phantom{\sum{\Biggl(}}=(1.236)(0.5)\left[0\quad\frac{7}{20}\right]+1.236\left[0.001\quad0.01\right]\left[\begin{array}{c c}{0}&{\frac{7}{20}}\\ {-\frac{7}{20}}&{0}\end{array}\right]}\\ &{\phantom{\sum{\Biggl(}}=\left[-4.326\times10^{-3}\quad0.2167\right]}\end{array}
 $$  
 
 Finally, the matrix $\mathbf{m}_{f f}$ is, in this example, the $2\times2$ matrix given by  
+最后，在本例中，矩阵 $\mathbf{m}_{f f}$ 是一个 $2\times2$ 矩阵，具体如下：
 
 $$
-{\bf m}_{f f}=1.236\left[\frac{1}{3}\quad0\atop\frac{13}{35}\right]=\left[0.412\begin{array}{c c}{{0}}\\ {{0}}\end{array}\right]
+{\bf m}_{f f}=1.236\left[\frac{1}{3}\quad0\atop 0\quad\frac{13}{35}\right]=\left[\begin{array}{c c}{0.412 \quad{0}}\\ {{0}\quad 0.4591}\end{array}\right]
 $$  
 
-Therefore, the mass matrix of the beam at this instant of time is given by  
+Therefore, the mass matrix of the beam at this instant of time is given by 
+因此，在此时刻，梁的质量矩阵可表示为 
+$$
+\mathbf{M}=\left[\begin{array}{c c c}{\mathbf{m}_{R R}}&{\mathbf{m}_{R\theta}}&{\mathbf{m}_{R f}}\\ &{\mathbf{m}_{\theta\theta}}&{\mathbf{m}_{\theta f}}\\ {\mathrm{symmetric}} &&{\mathbf{m}_{f f}}\end{array}\right]
+\\ {=\begin{bmatrix}
+1.236 & 0 & -0.1602 & 0.5352 & -0.309 \\
+{} &1.236 & 0.2650 & 0.309 & 0.5352 \\
+{} &{} &0.1031 & -4.326 \times 10^{-3} & 0.2167 \\
+{} &{} &{} &0.412 & 0 \\
+\mathrm{symmetric} &{} &{} &{} &0.4591
+\end{bmatrix}
+}
+$$
 
 Lumped Masses In this section, the kinetic energy of the deformable body was developed in terms of a finite set of coordinates. This was achieved by assuming the deformation shape using the body shape functions that depend on the spatial coordinates. Therefore, the deformation at any point on the body can be obtained by specifying the coordinates of this point in the body shape function. This approach leads to what is called consistent mass formulation. Another approach that is also used to formulate the dynamic equations of deformable bodies is based on using lumped mass techniques. In the lumped mass formulation the interest is focused on the displacement of selected grid points on the deformable body. Instead of using shape functions, a set of shape vectors are used to describe the relative motion between these grid points. These shape vectors can be assumed or can be determined experimentally. They can also be the mode shapes of vibration of the deformable body. In the lumped mass formulation the total mass of the body is distributed among the grid points. By increasing the number of the grid points more accurate results can be obtained.  
+**块状质量**
+
+在本节中，我们根据有限坐标集推导了变形体的动能。这通过假设依赖于空间坐标的形函数来描述变形形状来实现。因此，通过在形函数中指定该点的坐标，可以获得体上任意一点的变形。这种方法导出了所谓的“一致质量公式”。另一种用于建立变形体动力学方程的方法是基于块状质量技术。在块状质量公式中，重点关注变形体上选定网格点的位移。代替使用形函数，使用一组形向量来描述这些网格点之间的相对运动。这些形向量可以是假设的，也可以是实验确定的。它们也可以是变形体的振动模式。在块状质量公式中，变形体的总质量被分配到网格点上。通过增加网格点的数量，可以获得更准确的结果。
 
 In the remainder of this section we develop the inertia properties of deformable bodies that undergo finite rotations using a lumped mass technique. This development leads to a set of inertia shape matrices similar to the ones that appeared in the consistent mass formulation. As pointed out earlier, in the lumped mass formulation, the motion of the deformable body is identified by a set of shape vectors that describe the displacement of selected grid points. The shape vectors should be linearly independent and should contain the low-frequency modes of vibration of the body. In this section, grid point displacements are expressed in terms of the elastic generalized coordinates of the deformable body. The deformation vector of a grid point $j$ on body $i$ can be written as