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# 10-2 MASS PROPERTIES
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# Lumped-Mass Matrix
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## Lumped-Mass Matrix集中质量矩阵
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The simplest procedure for defining the mass properties of any structure is to assume that the entire mass is concentrated at the points at which the translational displacements are defined. The usual procedure for defining the point mass to be located at each node is to assume that the structure is divided into segments, the nodes serving as connection points. Figure 10-6 illustrates the procedure for a beam-type structure. The mass of each segment is assumed to be concentrated in point masses at each of its nodes, the distribution of the segment mass to these points being determined by statics. The total mass concentrated at any node of the complete structure then is the sum of the nodal contributions from all the segments attached to that node. In the beam system of Fig. 10-6 there are two segments contributing to each node; for example, $m_{1}=m_{1a}+m_{1b}$ .
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For a system in which only translational degrees of freedom are defined, the lumped-mass matrix has a diagonal form; for the system of Fig. 10-6 it would be
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定义任何结构质量特性的最简单方法是假设整个质量集中在定义平移位移的点上。定义位于每个节点的点质量的通常方法是假设结构被分成若干段，节点作为连接点。图10-6阐述了梁式结构的方法。每段的质量被假设集中在其每个节点的点质量中，段质量到这些点的分布由静力学确定。那么，集中在完整结构任何节点的总质量是连接到该节点的所有段的节点贡献之和。在图10-6的梁系统中，有两段对每个节点有贡献；例如，$m_{1}=m_{1a}+m_{1b}$。
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对于一个只定义了平移自由度的系统，集中质量矩阵具有对角形式；对于图10-6的系统，它将是
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written
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@ -298,34 +300,36 @@ $$
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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.
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If more than one translational degree of freedom is specified at any nodal point, the same point mass will be associated with each degree of freedom. On the other hand, the mass associated with any rotational degree of freedom will be zero because of the assumption that the mass is lumped in points which have no rotational inertia. (Of course, if a rigid mass having a finite rotational inertia is associated with a rotational degree of freedom, the diagonal mass coefficient for that degree of freedom would be the rotational inertia of the mass.) Thus the lumped-mass matrix is a diagonal matrix which will include zero diagonal elements for the rotational degrees of freedom, in general.
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其中项数与自由度数相同。该矩阵的非对角项 $m_{i j}$ 为零，因为任何质点的加速度仅在该点产生惯性力。点 $i$ 处由于点 $i$ 的单位加速度产生的惯性力显然等于集中在该点的质量；因此，在集中质量系统中，质量影响系数 $m_{i i}=m_{i}$。
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# Consistent-Mass Matrix
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如果在任何节点处指定了多个平移自由度，则相同的质点质量将与每个自由度相关联。另一方面，与任何旋转自由度相关联的质量将为零，因为假设质量集中在没有转动惯量的点上。（当然，如果一个具有有限转动惯量的刚性质量与一个旋转自由度相关联，那么该自由度的对角质量系数将是该质量的转动惯量。）因此，集中质量矩阵是一个对角矩阵，通常会包含旋转自由度的零对角元素。
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## Consistent-Mass Matrix一致质量矩阵
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Making use of the finite-element concept, it is possible to evaluate mass influence coefficients for each element of a structure by a procedure similar to the analysis of element stiffness coefficients. Consider, for example, the nonuniform beam segment shown in Fig. 10-7, which may be assumed to be the same as that of Fig. 10-4. The degrees of freedom of the segment are the translation and rotation at each end, and it will be assumed that the displacements within the span are defined by the same interpolation functions $\psi_{i}(x)$ used in deriving the element stiffness.
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利用有限元概念，可以通过类似于单元刚度系数分析的程序，评估结构中每个单元的质量影响系数。例如，考虑图10-7所示的非均匀梁段，该梁段可假定与图10-4所示的梁段相同。该梁段的自由度是其两端的平移和转动，并且假定展向内的位移由推导单元刚度时使用的相同插值函数$\psi_{i}(x)$定义。
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FIGURE 10-7 Node subjected to real angular acceleration and virtual translation.
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If the beam were subjected to a unit angular acceleration of the left end, $\ddot{v}_{3}=$ $\ddot{\theta}_{a}=1$ , accelerations would be developed along its length, as follows:
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如果梁的左端受到单位角加速度，即 $\ddot{v}_{3}=$ $\ddot{\theta}_{a}=1$ ，则沿其长度会产生加速度，如下所示：
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$$
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\ddot{v}(x)=\psi_{3}(x)\,\ddot{v}_{3}
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$$
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which can be obtained by taking the second derivative of Eq. (10-17). By d’Alembert’s principle, the inertial force resisting this acceleration is
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可通过对式 (10-17) 求二阶导数得到。根据达朗贝尔原理，抵抗此加速度的惯性力是
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$$
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f_{I}(x)=m(x)\;\ddot{v}(x)=m(x)\,\psi_{3}(x)\,\ddot{v}_{3}
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$$
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Now the mass influence coefficients associated with this acceleration are defined as the nodal inertial forces which it produces; these can be evaluated from the distributed inertial force of Eq. (10-26) by the principle of virtual displacements. For example, the vertical force at the left end can be evaluated by introducing a vertical virtual displacement and equating the work done by the external nodal force $p_{a}$ to the work done on the distributed inertial forces $f_{I}(x)$ . Thus
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现在，与此加速度相关的质量影响系数被定义为它产生的节点惯性力；这些可以根据方程 (10-26) 的分布惯性力通过虚位移原理进行评估。例如，左端的垂直力可以通过引入一个垂直虚位移，并将外部节点力 $p_{a}$ 所做的功与分布惯性力 $f_{I}(x)$ 所做的功相等来计算。因此
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$$
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p_{a}\;\delta v_{a}=\int_{0}^{L}f_{I}(x)\,\delta v(x)\,d x
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$$
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Expressing the vertical virtual displacement in terms of the interpolation function and substituting Eq. (10-26) lead finally to
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用插值函数表示竖向虚位移，并代入式 (10-26)，最终得到
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$$
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m_{13}=\int_{0}^{L}m(x)\,\psi_{1}(x)\,\psi_{3}(x)\,d x
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$$
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It should be noted in Fig. 10-7 that the mass influence coefficient represents the inertial force opposing the acceleration, but that it is numerically equal to the external force producing the acceleration.
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From Eq. (10-27) it is evident that any mass influence coefficient $m_{i j}$ of an arbitrary beam segment can be evaluated by the equivalent expression
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应注意图10-7中，质量影响系数代表抵抗加速度的惯性力，但其数值上等于产生加速度的外部力。
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从式(10-27)可知，任意梁段的任意质量影响系数 $m_{i j}$ 可以通过等效表达式进行计算。
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$$
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m_{i j}=\int_{0}^{L}m(x)\,\psi_{i}(x)\,\psi_{j}(x)\,d x
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$$
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The symmetric form of this equation shows that the mass matrix (like the stiffness matrix) is symmetric; that is, $m_{i j}=m_{j i}$ ; also it may be noted that this expression is equivalent to the corresponding term in the first of Eqs. (8-18) in the case where $i=j$ . When the mass coefficients are computed in this way, using the same interpolation functions which are used for calculating the stiffness coefficients, the result is called the consistent-mass matrix. In general, the cubic hermitian polynomials of Eqs. (10- 16) are used for evaluating the mass coefficients of any straight beam segment. In the special case of a beam with uniformly distributed mass the results are
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该方程的对称形式表明质量矩阵（如刚度矩阵）是对称的；即，$m_{i j}=m_{j i}$；还可以注意到，当$i=j$时，该表达式等效于式(8-18)中第一个方程的相应项。当以这种方式计算质量系数时，使用与计算刚度系数相同的插值函数，结果称为一致质量矩阵。通常，式(10-16)中的三次Hermite多项式用于评估任何直梁段的质量系数。在质量均匀分布的梁的特殊情况下，结果为
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$$
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\left\{\begin{array}{c}{f_{I1}}\\ {f_{I2}}\\ {f_{I3}}\\ {f_{I4}}\end{array}\right\}=\frac{\overline{{m}}L}{420}\begin{array}{r l r l}{\mathrm{~156~}}&{54}&{22L}&{-13L}\\ {\mathrm{~54~}}&{156}&{13L}&{-22L}\\ {\mathrm{~22~}}&{13L}&{4L^{2}}&{-3L^{2}}\\ {\mathrm{~}}&{\mathrm{~-22~}}&{-3L^{2}}&{4L^{2}}\end{array}\right\}\begin{array}{c}{\langle\ddot{v}_{1}}\\ {\ddot{v}_{2}}\\ {\ddot{v}_{3}}\\ {\ddot{v}_{4}}\end{array}
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$$
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The dynamic analysis of a consistent-mass system generally requires considerably more computational effort than a lumped-mass system does, for two reasons: (1) the lumped-mass matrix is diagonal, while the consistent-mass matrix has many off-diagonal terms (leading to what is called mass coupling); (2) the rotational degrees of freedom can be eliminated from a lumped-mass analysis (by static condensation, explained later), whereas all rotational and translational degrees of freedom must be included in a consistent-mass analysis.
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Example E10-2. The structure of Example E10-1, shown again in Fig. E10-2a, will be used to illustrate the evaluation of the structural mass
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当结构单元的质量系数被评估后，可以通过与从单元刚度建立刚度矩阵所描述的完全相同的叠加过程来建立完整单元组合体的质量矩阵[式 (10-23)]。所得质量矩阵通常将具有与刚度矩阵相同的构型，即非零项的排列。
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一致质量系统的动力分析通常比集中质量系统需要显著更多的计算工作量，原因有二：(1) 集中质量矩阵是对角的，而一致质量矩阵有许多非对角项（导致所谓的质量耦合）；(2) 旋转自由度可以从集中质量分析中消除（通过静力凝聚，稍后解释），而所有旋转和平移自由度都必须包含在一致质量分析中。
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FIGURE E10-2
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Analysis of lumped- and consistent-mass matrices: (a) uniform mass in members; (b) lumping of mass at member ends; (c) forces due to acceleration $\ddot{v}_{1}=1$ (consistent); $(d)$ forces due to acceleration $\ddot{v}_{2}=1$ (consistent).
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matrix. First the lumped-mass procedure is used: half the mass of each member is lumped at the ends of the members, as shown in Fig. E10-2b. The sum of the four contributions at the girder level then acts in the sidesway degree of freedom $m_{11}$ ; no mass coefficients are associated with the other degrees of freedom because these point masses have no rotational inertia.
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FIGURE E10-2 Analysis of lumped- and consistent-mass matrices: (a) uniform mass in members; (b) lumping of mass at member ends; (c) forces due to acceleration $\ddot{v}_{1}=1$ (consistent); $(d)$ forces due to acceleration $\ddot{v}_{2}=1$ (consistent).
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Example E10-2. The structure of Example E10-1, shown again in Fig. E10-2a, will be used to illustrate the evaluation of the structural mass matrix. First the lumped-mass procedure is used: half the mass of each member is lumped at the ends of the members, as shown in Fig. E10-2b. The sum of the four contributions at the girder level then acts in the sidesway degree of freedom $m_{11}$ ; no mass coefficients are associated with the other degrees of freedom because these point masses have no rotational inertia.
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The consistent-mass matrix is obtained by applying unit accelerations to each degree of freedom in succession while constraining the others and determining the resulting inertial forces from the coefficients of Eq. (10-29). Considering first the sidesway acceleration, as shown in Fig. E10-2c, it must be noted that the coefficients of Eq. (10-29) account only for the transverse inertia of the columns. The inertia of the girder due to the acceleration parallel to its axis must be added as a rigid-body mass (3mL), as shown.
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The joint rotational acceleration induces only accelerations transverse to the members, and the resulting girder and column contributions are given by Eq. (10-29), as shown in Fig. E10-2d. The final mass matrices, from the lumpedand consistent-mass formulations, are
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示例 E10-2。图 E10-2a 中再次显示的示例 E10-1 的结构将用于说明结构质量矩阵的评估。首先采用集中质量法：每个构件一半的质量集中在构件的两端，如图 E10-2b 所示。梁层面上四个贡献的总和然后作用于侧移自由度 $m_{11}$；没有质量系数与其它自由度相关联，因为这些质点没有转动惯量。通过依次对每个自由度施加单位加速度，同时约束其他自由度，并根据方程 (10-29) 的系数确定由此产生的惯性力，从而获得一致质量矩阵。首先考虑侧移加速度，如图 E10-2c 所示，必须指出方程 (10-29) 的系数仅考虑柱的横向惯性。由于平行于其轴线的加速度引起的梁的惯性必须作为刚体质量 (3mL) 添加，如图所示。节点转动加速度仅引起构件的横向加速度，由此产生的梁和柱的贡献由方程 (10-29) 给出，如图 E10-2d 所示。根据集中质量和一致质量公式得到的最终质量矩阵为
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$$
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\begin{array}{r}{\mathbf{m}=\frac{\overline{{m}}L}{210}\begin{array}{l}{\left[840\quad0\quad0\right]}\\ {\left[\begin{array}{l l l}{0}&{0}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{0}\end{array}\right]}\end{array}\quad\mathbf{m}=\frac{\overline{{m}}L}{210}\begin{array}{l}{\left[786\quad11L\quad11L\right]}\\ {11L\quad26L^{2}\quad-18L^{2}}\\ {\left[11L\quad-18L^{2}\quad26L^{2}\right]}\end{array}}\end{array}
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$$
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