diff --git a/书籍/力学书籍/力学/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library)/auto/Chap 10 EVALUATION OF STRUCTURALPROPERTY MATRICES.md b/书籍/力学书籍/力学/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library)/auto/Chap 10 EVALUATION OF STRUCTURALPROPERTY MATRICES.md
index 4588996..21ce274 100644
--- a/书籍/力学书籍/力学/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library)/auto/Chap 10 EVALUATION OF STRUCTURALPROPERTY MATRICES.md	
+++ b/书籍/力学书籍/力学/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library)/auto/Chap 10 EVALUATION OF STRUCTURALPROPERTY MATRICES.md	
@@ -191,7 +191,7 @@ The problem of defining the stiffness properties of any structure is thus reduce
 
 因此，定义任何结构刚度特性的问题基本上简化为评估典型单元的刚度。例如，考虑图10-4所示的非均匀直梁段。这种单元可以组装成结构的两个节点位于其两端，并且如果只考虑横向平面位移，每个节点有两个自由度，即垂直平移和转动。在单元左端施加每种类型的单位位移，同时约束其他三个节点位移所产生的变形形状如图10-4所示。这些位移函数可以取为满足节点和内部连续性要求的任何任意形状，但它们通常被假定为在承受这些节点位移的均匀梁中产生的形状。这些是三次Hermite多项式，可以表示为
 ![](a65f1a26eb07d38b2788ec586cf11330619f88bbfc2961ef7d974343873b9c0e.jpg)  
-FIGURE 10-4 Beam deflections due to unit nodal displacements at left end.  
+FIGURE 10-4 Beam deflections due to unit nodal displacements at left end. 由于左端单位节点位移引起的梁变形。 
 
 $$
 \begin{array}{l}{\displaystyle{\psi_{1}(x)=1-3\left(\frac{x}{L}\right)^{2}+2\left(\frac{x}{L}\right)^{3}}}\\ {\displaystyle{\psi_{3}(x)=x\left(1-\frac{x}{L}\right)^{2}}}\end{array}
@@ -212,7 +212,7 @@ $$
 where the numbered degrees of freedom are related to those shown in Fig. 10-4 as follows:  
 其中，编号的自由度与图10-4中所示的自由度对应关系如下：
 $$
-\left\{\begin{array}{c}{v_{1}}\\ {v_{2}}\\ {v_{3}}\\ {v_{4}}\end{array}\right\}\equiv\left\{\begin{array}{c}{v_{a}}\\ {v_{b}}\\ {v_{b}}\end{array}\right\}
+\left\{\begin{array}{c}{v_{1}}\\ {v_{2}}\\ {v_{3}}\\ {v_{4}}\end{array}\right\}\equiv\left\{\begin{array}{c}{v_{a}}\\ {v_{b}}\\ {\theta_{a}}\\ {\theta_{b}}\end{array}\right\}
 $$  
 
 It should be noted that both rotations and translations are represented as basic nodal degrees of freedom $v_{i}$ .  
@@ -303,7 +303,7 @@ The structure stiffness matrix finally obtained by assembling all these coeffici
 
 通过组装所有这些系数最终得到的结构刚度矩阵是
 $$
-\left\{\begin{array}{c}{{f_{S1}}}\\ {{f_{S2}}}\\ {{f_{S3}}}\end{array}\right\}=\frac{2E I}{L^{3}}\begin{array}{c}{{\left[12\quad3L\quad3L\quad3L\right]}}\\ {{\left[3L\quad6L^{2}\quad2L^{2}\right]}}\\ {{\left[3L\quad2L^{2}\quad6L^{2}\right]}}\end{array}\right\{\begin{array}{c}{{v_{1}}}\\ {{v_{2}}}\\ {{v_{3}}}\end{array}\right\}
+\begin{Bmatrix} f_{s1} \\ f_{s2} \\ f_{s3} \end{Bmatrix} = \frac{2EI}{L^3} \begin{bmatrix} 12 & 3L & 3L \\ 3L & 6L^2 & 2L^2 \\ 3L & 2L^2 & 6L^2 \end{bmatrix} \begin{Bmatrix} v_1 \\ v_2 \\ v_3 \end{Bmatrix}
 $$  
 
 # 10-2 MASS PROPERTIES  
@@ -321,7 +321,17 @@ For a system in which only translational degrees of freedom are defined, the lum
 written  
 
 $$
-\mathbf{m}={\left[\begin{array}{l l l l l l l}{m_{1}}&{0}&{0}&{\cdots}&{0}&{\cdots}&{0}\\ {}&{m_{2}}&{0}&{\cdots}&{0}&{\cdots}&{0}\\ {0}&{0}&{m_{3}}&{\cdots}&{0}&{\cdots}&{0}\\ &{\cdots\cdots\cdots\cdots\cdots\cdots\cdots}&{}&{}&{}\\ {}&{0}&{0}&{\cdots}&{m_{i}}&{\cdots}&{0}\\ &{\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}&{}&{}\\ {0}&{0}&{0}&{\cdots}&{0}&{\cdots}\end{array}\right]}
+\mathbf{m} =
+\begin{bmatrix}
+m_1 & 0 & 0 & \cdots & 0 & \cdots & 0 \\
+0 & m_2 & 0 & \cdots & 0 & \cdots & 0 \\
+0 & 0 & m_3 & \cdots & 0 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
+0 & 0 & 0 & \cdots & m_i & \cdots & 0 \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
+0 & 0 & 0 & \cdots & 0 & \cdots & m_N
+\end{bmatrix}
+\tag{10-24}
 $$  
 
 in which there are as many terms as there are degrees of freedom. The off-diagonal terms $m_{i j}$ of this matrix vanish because an acceleration of any mass point produces an inertial force at that point only. The inertial force at $i$ due to a unit acceleration of point $i$ is obviously equal to the mass concentrated at that point; thus the mass influence coefficient $m_{i i}=m_{i}$ in a lumped-mass system.