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@ -675,12 +675,16 @@ The essential physical properties of any linearly elastic structural or mechanic
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The entire mass $m$ of this system is included in the rigid block which is constrained by rollers so that it can move only in simple translation; thus, the single displacement coordinate $v(t)$ completely defines its position. The elastic resistance to displacement is provided by the weightless spring of stiffness $k$ , while the energy-loss mechanism is represented by the damper $c$ . The external dynamic loading producing the response of this system is the time-varying force $p(t)$ .
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任何承受外部激励源或动态载荷的线性弹性结构或机械系统的基本物理特性是其**质量、弹性特性（柔度或刚度）以及能量损失机制或阻尼**。在SDOF系统最简单的模型中，这些特性中的每一个都被假定集中在一个单一的物理单元中。图2-1a显示了这样一个系统的草图。
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该系统的全部质量$m$包含在刚性块中，该刚性块由滚轮约束，使其只能进行简单的平移；因此，单个位移坐标$v(t)$完全定义了其位置。对位移的弹性阻力由刚度为$k$的无质量弹簧提供，而能量损失机制由阻尼器$c$表示。产生该系统响应的外部动态载荷是时变力$p(t)$。
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FIGURE 2-1 Idealized SDOF system: (a) basic components; $(b)$ forces in equilibrium.
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# 2-2 EQUATION OF MOTION OF THE BASIC DYNAMIC SYSTEM
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The equation of motion for the simple system of Fig. 2-1a is most easily formulated by directly expressing the equilibrium of all forces acting on the mass using d’Alembert’s principle. As shown in Fig. $_{2-1b}$ , the forces acting in the direction of the displacement degree of freedom are the applied load $p(t)$ and the three resisting forces resulting from the motion, i.e., the inertial force $f_{I}(t)$ , the damping force $f_{D}(t)$ , and the spring force $f_{S}(t)$ . The equation of motion is merely an expression of the equilibrium of these forces as given by
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图2-1a所示简单系统的运动方程，最容易通过利用达朗贝尔原理直接表达作用在质量上的所有力的平衡来建立。如图$_{2-1b}$所示，作用在位移自由度方向上的力是外载荷 $p(t)$ 和运动产生的三种阻力，即惯性力 $f_{I}(t)$ 、阻尼力 $f_{D}(t)$ 和弹簧力 $f_{S}(t)$ 。运动方程仅仅是这些力平衡的表达式，如下所示。
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$$
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f_{I}(t)+f_{D}(t)+f_{S}(t)=p(t)
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@ -689,47 +693,49 @@ $$
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Each of the forces represented on the left hand side of this equation is a function of the displacement $v(t)$ or one of its time derivatives. The positive sense of these forces has been deliberately chosen to correspond with the negative-displacement sense so that they oppose a positive applied loading.
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In accordance with d’Alembert’s principle, the inertial force is the product of the mass and acceleration
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方程左侧表示的每个力都是位移 $v(t)$ 或其时间导数之一的函数。这些力的正方向被特意选择为与负位移方向对应，以便它们抵抗正的施加荷载。
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根据达朗贝尔原理，惯性力是质量和加速度的乘积。
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$$
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f_{I}(t)=m\;\ddot{v}(t)
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$$
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Assuming a viscous damping mechanism, the damping force is the product of the damping constant $c$ and the velocity
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假设为粘性阻尼机制，阻尼力是阻尼常数$c$与速度的乘积
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$$
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f_{D}(t)=c\;\dot{v}(t)
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$$
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Finally, the elastic force is the product of the spring stiffness and the displacement
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最后，弹性力是弹簧刚度和位移的乘积
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$$
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f_{S}(t)=k\;v(t)
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$$
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When Eqs. (2-2) are introduced into Eq. (2-1), the equation of motion for this SDOF system is found to be
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将式(2-2)代入式(2-1)中，可得该单自由度系统的运动方程为
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$$
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m\;\ddot{v}(t)+c\;\dot{v}(t)+k\;v(t)=p(t)
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$$
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To introduce an alternative formulation procedure, it is instructive to develop this same equation of motion by a virtual-work approach. If the mass is given a virtual displacement $\delta v$ compatible with the system’s constraints, the total work done by the equilibrium system of forces in Fig. $_{2-1b}$ must equal zero as shown by
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为了引入另一种公式化程序，通过虚功法推导相同的运动方程是有益的。如果质量给定一个与系统约束兼容的虚位移 $\delta v$，则图$_{2-1b}$中平衡力系所做的总功必须等于零，如...所示。
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$$
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-f_{I}(t)~\delta v-f_{D}(t)~\delta v-f_{S}(t)~\delta v+p(t)~\delta v=0
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$$
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in which the negative signs result from the fact that the associated forces act opposite to the sense of the virtual displacement. Substituting Eqs. (2-2) into Eq. (2-4) and factoring out $\delta v$ leads to
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其中负号是由于相关力与虚位移方向相反。将式 (2-2) 代入式 (2-4) 并提出 $\delta v$ 得到
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$$
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\big[-m\;\ddot{v}(t)-c\;\dot{v}(t)-k\;v(t)+p(t)\big]\;\delta v=0
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$$
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Since $\delta v$ is nonzero, the bracket quantity in this equation must equal zero, thus giving the same equation of motion as shown by Eq. (2-3). While a virtual-work formulation has no advantage for this simple system, it will be found very useful for more general types of SDOF systems treated subsequently.
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由于 $\delta v$ 不为零，该方程中的括号量必须等于零，因此得到与式 (2-3) 所示相同的运动方程。虽然虚功公式对于这个简单的系统没有优势，但它将被发现对于随后处理的更一般的单自由度系统非常有用。
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# 2-3 INFLUENCE OF GRAVITATIONAL FORCES
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Consider now the system shown in Fig. $2{-}2a$ , which is the system of Fig. 2-1a rotated through $90^{\circ}$ so that the force of gravity acts in the direction of the displacement. In this case, the system of forces acting in the direction of the displacement degree of freedom is that set shown in Fig. 2-2b. Using Eqs. (2-2), the equilibrium of these forces is given by
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现在考虑图 $2{-}2a$ 所示的系统，它是图 2-1a 的系统旋转了 $90^{\circ}$，以便重力作用在位移方向上。在这种情况下，作用在位移自由度方向上的力系是图 2-2b 所示的集合。使用公式 (2-2)，这些力的平衡由下式给出
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$$
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m\ \ddot{v}(t)+c\ \dot{v}(t)+k\ v(t)=p(t)+W
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$$
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@ -737,7 +743,9 @@ $$
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where $W$ is the weight of the rigid block.
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However, if the total displacement $v(t)$ is expressed as the sum of the static displacement $\triangle_{\mathrm{st}}$ caused by the weight $W$ plus the additional dynamic displacement ${\overline{{v}}}(t)$ as shown in Fig. $2{-}2c$ , i.e.,
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其中 $W$ 是刚性块的重量。
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然而，如果总变形 $v(t)$ 表示为由重量 $W$ 引起的静变形 $\triangle_{\mathrm{st}}$ 加上额外的动变形 ${\overline{{v}}}(t)$ 之和，如图 $2{-}2c$ 所示，即
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$$
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v(t)=\triangle_{\mathrm{st}}+\overline{{v}}(t)
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$$
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@ -764,13 +772,13 @@ m\;\ddot{v}(t)+c\;\dot{v}(t)+k\;\overline{{v}}(t)=p(t)
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$$
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Now by differentiating Eq. (2-7) and noting that $\triangle_{\mathrm{st}}$ does not vary with time, it is evident that $\ddot{v}(t)=\ddot{\overline{{v}}}(t)$ and $\dot{v}(t)=\dot{\overline{{v}}}(t)$ so that Eq. (2-10) can be written
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现在，对式(2-7)进行微分，并注意到$\triangle_{\mathrm{st}}$不随时间变化，因此$\ddot{v}(t)=\ddot{\overline{{v}}}(t)$和$\dot{v}(t)=\dot{\overline{{v}}}(t)$是显而易见的，这样式(2-10)可以写成
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$$
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m\,\ddot{\vec{v}}(t)+c\,\dot{\vec{v}}(t)+k\,\vec{v}(t)=p(t)
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$$
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Comparison of Eqs. (2-11) and (2-3) demonstrates that the equation of motion expressed with reference to the static-equilibrium position of the dynamic system is not affected by gravity forces. For this reason, displacements in all future discussions will be referenced from the static-equilibrium position and will be denoted $v(t)$ (i.e., without the overbar); the displacements which are determined will represent dynamic response. Therefore, total deflections, stresses, etc. are obtained by adding the corresponding static quantities to the results of the dynamic analysis.
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比较式(2-11)和式(2-3)表明，动力系统相对于静平衡位置表示的运动方程不受重力影响。因此，在所有未来的讨论中，位移将以静平衡位置为参考，并表示为$v(t)$（即不带上划线）；所确定的位移将代表动态响应。因此，总变形、应力等通过将相应的静态量加到动态分析结果中获得。
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# 2-4 INFLUENCE OF SUPPORT EXCITATION
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Dynamic stresses and deflections can be induced in a structure not only by a time-varying applied load, as indicated in Figs. 2-1 and 2-2, but also by motions of its support points. Important examples of such excitation are the motions of a building foundation caused by an earthquake or motions of the base support of a piece of equipment due to vibrations of the building in which it is housed. A simplified model of the earthquake-excitation problem is shown in Fig. 2-3, in which the horizontal ground motion caused by the earthquake is indicated by the displacement ${v_{g}}(t)$ of the structure’s base relative to the fixed reference axis.
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@ -778,7 +786,11 @@ Dynamic stresses and deflections can be induced in a structure not only by a tim
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The horizontal girder in this frame is assumed to be rigid and to include all the moving mass of the structure. The vertical columns are assumed to be weightless and inextensible in the vertical (axial) direction, and the resistance to girder displacement provided by each column is represented by its spring constant $k/2$ . The mass thus has a single degree of freedom, $v(t)$ , which is associated with column flexure; the damper $c$ provides a velocity-proportional resistance to the motion in this coordinate.
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As shown in Fig. $2{-}3b$ , the equilibrium of forces for this system can be written as
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结构中产生的动应力（Dynamic stresses）和变形（deflections）不仅可能由图2-1和2-2所示的时变外加载荷引起，也可能由其支撑点的运动引起。此类激励的重要例子包括地震引起的建筑物地基运动，或由于设备所在建筑物的振动而引起的设备基座支撑运动。图2-3显示了地震激励问题的简化模型，其中地震引起的水平地面运动由结构基座相对于固定参考轴的位移 ${v_{g}}(t)$ 表示。
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该框架中的水平大梁被假定为刚性的，并包含结构的所有移动质量。垂直柱被假定为无重且在垂直（轴向）方向上不可伸长的，并且每个柱对大梁位移提供的阻力由其弹簧常数 $k/2$ 表示。因此，该质量具有与柱弯曲相关的单自由度 $v(t)$；阻尼器 $c$ 在此坐标中提供与速度成比例的运动阻力。
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如图 $2{-}3b$ 所示，该系统的力平衡方程可写为
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$$
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f_{I}(t)+f_{D}(t)+f_{S}(t)=0
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$$
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