vault backup: 2025-09-28 09:26:49
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@ -1380,78 +1380,80 @@ where $p_{o}/k\,=\,v_{\mathrm{st}}$ is the displacement which would be produced
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其中，$p_{o}/k\,=\,v_{\mathrm{st}}$ 是由静载荷 $p_{o}$ 产生的位移，而 $1/(1-\beta^{2})$ 是表示简谐载荷放大效应的放大系数 (MF)。在该方程中，$\sin{\overline{{\omega}}}t$ 表示施加载荷频率下的响应分量；它被称为稳态响应，并与载荷直接相关。此外，$\beta\sin\omega t$ 是固有振动频率下的响应分量，并且是由初始条件控制的自由振动效应。由于在实际情况中，阻尼最终会使最后一项消失，因此它被称为瞬态响应。然而，对于这个假设的无阻尼系统，这一项不会衰减，而是会无限期地持续下去。
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Response Ratio — A convenient measure of the influence of dynamic loading is provided by the ratio of the dynamic displacement response to the displacement produced by static application of load $p_{o}$ , i.e.,
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响应比 — 衡量动载荷影响的一个便捷方法是，动位移响应与载荷 $p_{o}$ 静态作用下产生的位移之比，即：
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$$
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R(t)\equiv{\frac{v(t)}{v_{\mathrm{st}}}}={\frac{v(t)}{p_{o}/k}}
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$$
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From Eq. (3-10) it is evident that the response ratio resulting from the sine-wave loading of an undamped system starting from rest is
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从式(3-10)可知，无阻尼系统从静止状态开始，在正弦波载荷作用下产生的响应比为
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$$
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R(t)=\left[{\frac{1}{1-\beta^{2}}}\right]\left(\sin\overline{{\omega}}t-\beta\sin\omega t\right)
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$$
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It is informative to examine this response behavior in more detail by reference to Fig. 3-1. Figure $3{-}1a$ represents the steady-state component of response while Fig. 3- $1b$ represents the so-called transient response. In this example, it is assumed that $\beta\,=\,2/3$ , that is, the applied loading frequency is two-thirds of the free-vibration frequency. The total response $R(t)$ , i.e., the sum of both types of response, is shown in Fig. $3.1c$ . Two points are of interest: (1) the tendency for the two components to get in phase and then out of phase again, causing a “beating” effect in the total response; and (2) the zero slope of total response at time $t=0$ , showing that the initial velocity of the transient response is just sufficient to cancel the initial velocity of the steady-state response; thus, it satisfies the specified initial condition $\dot{v}(0)=0$ .
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参照图3-1，详细考察这种响应行为很有意义。图3-1a表示响应的稳态分量，而图3-1b表示所谓的瞬态响应。在本例中，假设$\beta\,=\,2/3$，即施加的载荷频率是自由振动频率的三分之二。总响应$R(t)$，即两种响应之和，显示在图3.1c中。有两点值得关注：(1) 两个分量趋于同相然后再次异相的趋势，导致总响应中出现“拍频”效应；(2) 总响应在$t=0$时刻的零斜率，表明瞬态响应的初始速度恰好足以抵消稳态响应的初始速度；因此，它满足了指定的初始条件$\dot{v}(0)=0$。
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FIGURE 3-1 Response ratio produced by sine wave excitation starting from at-rest initial conditions: (a) steady state; $(b)$ transient; $(c)$ total $R(t)$ .
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图3-1 从静止初始条件开始的正弦波激励产生的响应比：(a) 稳态；(b) 瞬态；(c) 总 $R(t)$。
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# 3-2 SYSTEM WITH VISCOUS DAMPING
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Returning to the equation of motion including viscous damping, Eq. (3-1), dividing by $m$ , and noting that $c/m=2\,\xi\,\omega$ leads to
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回到包含黏性阻尼的运动方程，式 (3-1)，除以 $m$，并注意到 $c/m=2\,\xi\,\omega$，可得
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$$
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\ddot{v}(t)+2\,\xi\,\omega\,\dot{v}(t)+\omega^{2}\,v(t)=\frac{p_{\scriptscriptstyle o}}{m}\,\sin\overline{{\omega}}t
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$$
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The complementary solution of this equation is the damped free-vibration response given by Eq. (2-48), i.e.,
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该方程的齐次解是阻尼自由振动响应，由式 (2-48) 给出，即
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$$
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v_{c}(t)=\left[A\,\cos\omega_{D}t+B\,\sin\omega_{D}t\right]\;\exp(-\xi\,\omega\,t)
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$$
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The particular solution to Eq. (3-13) is of the form
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方程 (3-13) 的特解形式为
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$$
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v_{p}(t)=G_{1}\,\cos{\varpi t}+G_{2}\,\sin{\overline{{\omega}}t}
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v_{p}(t)=G_{1}\,\cos{\overline{{\omega}}t}+G_{2}\,\sin{\overline{{\omega}}t}
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$$
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in which the cosine term is required as well as the sine term because, in general, the response of a damped system is not in phase with the loading.
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在其中，需要余弦项以及正弦项，因为通常情况下，阻尼系统的响应与载荷不同相位。
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Substituting Eq. (3-15) into Eq. (3-13) and separating the multiples of $\cos{\overline{{\omega t}}}$ from the multiples of $\sin{\overline{{\omega}}}t$ leads to
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将式(3-15)代入式(3-13)，并分离$\cos{\overline{{\omega t}}}$的倍数项与$\sin{\overline{{\omega}}}t$的倍数项，可得
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$$
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\begin{array}{r l}&{\left[-G_{1}\,\overline{{\omega}}^{2}+G_{2}\,\overline{{\omega}}\left(2\xi\omega\right)+G_{1}\,\omega^{2}\right]\;\cos\overline{{\omega}}t}\\ &{\qquad\qquad\qquad+\left[-G_{2}\,\overline{{\omega}}^{2}-G_{1}\,\overline{{\omega}}\left(2\xi\omega\right)+G_{2}\,\omega^{2}-\frac{p_{o}}{m}\right]\;\sin\overline{{\omega}}t=0}\end{array}
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$$
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In order to satisfy this equation for all values of $t$ , it is necessary that each of the two square bracket quantities equal zero; thus, one obtains
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为了使该方程对所有 $t$ 值都成立，必须使两个方括号内的量都等于零；因此，得到
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$$
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\begin{array}{l}{{G_{1}\left(1-\beta^{2}\right)+G_{2}\left(2\xi\beta\right)=0}}\\ {{{}}}\\ {{G_{2}\left(1-\beta^{2}\right)-G_{1}\left(2\xi\beta\right)=\frac{p_{o}}{k}}}\end{array}
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$$
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in which $\beta$ is the frequency ratio given by Eq. (3-7). Solving these two equations simultaneously yields
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其中 $\beta$ 是由式 (3-7) 给出的频率比。同时求解这两个方程得到
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$$
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\begin{array}{l}{G_{1}=\displaystyle\frac{p_{o}}{k}\left[\frac{-2\xi\beta}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\right]}\\ {G_{2}=\displaystyle\frac{p_{o}}{k}\left[\frac{1-\beta^{2}}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\right]}\end{array}
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$$
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Introducing these expressions into Eq. (3-15) and combining the result with the complementary solution of Eq. (3-14), the total response is obtained in the form
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将这些表达式代入式(3-15)，并将结果与式(3-14)的互补解结合，得到总响应，其形式为
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$$
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\begin{array}{l}{{\displaystyle v(t)=\left[A\,\cos\omega_{D}t+B\,\sin\omega_{D}t\right]\,\,\exp(-\xi\omega t)}}\\ {{\displaystyle\qquad+\,\frac{p_{o}}{k}\biggl[\frac{1}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\biggr]\,\,\Big[(1-\beta^{2})\,\sin\overline{{\omega}}t-2\xi\beta\,\cos\overline{{\omega}}t\Big]}}\end{array}
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$$
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The first term on the right hand side of this equation represents the transient response, which damps out in accordance with $\exp(-\xi\omega t)$ , while the second term represents the steady-state harmonic response, which will continue indefinitely. The constants $A$ and $B$ can be evaluated for any given initial conditions, $v(0)$ and $\dot{v}(0)$ . However, since the transient response damps out quickly, it is usually of little interest; therefore, the evaluation of constants $A$ and $B$ will not be pursued here.
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该方程右侧的第一项代表瞬态响应，它根据 $\exp(-\xi\omega t)$ 衰减，而第二项代表稳态谐波响应，它将无限期地持续下去。常数 $A$ 和 $B$ 可以根据任何给定的初始条件 $v(0)$ 和 $\dot{v}(0)$ 进行评估。然而，由于瞬态响应衰减很快，它通常很少受到关注；因此，本文将不讨论常数 $A$ 和 $B$ 的评估。
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Steady-State Harmonic Response — Of great interest, however, is the steadystate harmonic response given by the second term of Eq. (3-19)
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Steady-State Harmonic Response — Of great interest, however, is the steady-state harmonic response given by the second term of Eq. (3-19)
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稳态谐波响应 — 然而，非常令人感兴趣的是由式 (3-19) 的第二项给出的稳态谐波响应。
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$$
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v_{p}(t)=\frac{p_{o}}{k}\,\left[\frac{1}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\right]\,\left[(1-\beta^{2})\,\sin\overline{{\omega}}t-2\xi\beta\,\cos\overline{{\omega}}t\right]
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$$
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This steady-state displacement behavior can be interpreted easily by plotting two corresponding rotating vectors in the complex plane as shown in Fig. 3-2, where their components along the real axis are identical to the two terms in Eq. (3-20). The real component of the resultant vector, $-\rho\,i\,\exp[i(\overline{{\omega}}t-\theta)]$ , gives the steady-state response in the form
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这种稳态位移行为可以通过在复平面中绘制两个对应的旋转矢量（如图3-2所示）来轻松解释，其中它们沿实轴的分量与式(3-20)中的两项相同。合成矢量 $-\rho\,i\,\exp[i(\overline{{\omega}}t-\theta)]$ 的实部给出了形式为...的稳态响应
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$$
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v_{p}(t)=\rho\,\sin(\overline{{\omega}}t-\theta)
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$$
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@ -1467,41 +1469,41 @@ $$
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FIGURE 3-2 Steady-state displacement response.
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and a phase angle, $\theta$ , by which the response lags behind the applied loading
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以及一个相位角 $\theta$，响应滞后于施加的载荷
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$$
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\theta=\tan^{-1}\left[\frac{2\xi\beta}{1-\beta^{2}}\right]
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$$
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It should be understood that this phase angle is limited to the range $0<\theta<180^{\circ}$ .
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应当理解，该相位角限于$0<\theta<180^{\circ}$的范围。
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The ratio of the resultant harmonic response amplitude to the static displacement which would be produced by the force $p_{o}$ will be called the dynamic magnification factor $D$ ; thus
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合成谐波响应幅值与由力 $p_{o}$ 产生的静位移的比值将被称为动力放大系数 ；因此
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$$
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D\equiv\frac{\rho}{p_{o}/k}=\left[(1-\beta^{2})^{2}+(2\xi\beta)^{2}\right]^{-1/2}
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$$
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It is seen that both the dynamic magnification factor $D$ and the phase angle $\theta$ vary with the frequency ratio $\beta$ and the damping ratio $\xi$ . Plots of $D$ vs. $\beta$ and $\theta$ vs. $\beta$ are shown in Figs. 3-3 and 3-4, respectively, for discrete values of damping ratio, $\xi$ .
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At this point it is instructive to solve for the steady-state harmonic response once again using an exponential form of solution. Consider the general case of harmonic
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可以看出，动放大系数 $D$ 和相角 $\theta$ 都随频率比 $\beta$ 和阻尼比 $\xi$ 的变化而变化。图3-3和图3-4分别给出了不同离散阻尼比 $\xi$ 值下的 $D-\beta$ 关系图和 $\theta-\beta$ 关系图。
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loading expressed in exponential form:
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At this point it is instructive to solve for the steady-state harmonic response once again using an exponential form of solution. Consider the general case of harmonic loading expressed in exponential form:
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此时，再次使用指数形式的解来求解稳态谐波响应是很有启发性的。考虑以指数形式表示的谐波载荷的一般情形：
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$$
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\ddot{v}(t)+2\,\xi\,\omega\,\dot{v}(t)+\omega^{2}\,v(t)=\frac{p_{o}}{m}\;\exp[i\left(\overline{{\omega}}t+\phi\right)]
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$$
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where $\phi$ is an arbitrary phase angle in the harmonic loading function. In dealing with completely general harmonic loads, especially for the case of periodic loading where the excitation is expressed as a series of harmonic terms, it is essential to define the input phase angle for each harmonic; however, this usually is accomplished most conveniently by expressing the input in complex number form rather than by amplitude and phase angle. In this chapter only a single harmonic loading term will be considered; therefore, its phase angle is arbitrarily taken to be zero for simplicity, so it need not be included in the loading expression.
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其中$\phi$是谐波载荷函数中的一个任意相位角。在处理完全一般的谐波载荷时，特别是对于激励表示为一系列谐波项的周期性载荷情况，定义每个谐波的输入相位角至关重要；然而，这通常通过将输入表示为复数形式而不是通过幅值和相位角来最方便地实现。在本章中，将只考虑一个单独的谐波载荷项；因此，为了简化，其相位角被任意地取为零，所以它不需要包含在载荷表达式中。
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The particular solution of Eq. (3-25) and its first and second time derivatives are
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Eq. (3-25) 的特解及其一阶和二阶时间导数是
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$$
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\begin{array}{l}{{v_{p}(t)=G\,\exp(i\overline{{{\omega}}}t)}}\\ {{\ }}\\ {{\dot{v}_{p}(t)=i\,\overline{{{\omega}}}\,G\,\exp(i\overline{{{\omega}}}t)}}\\ {{\ }}\\ {{\ddot{v}_{p}(t)=-\overline{{{\omega}}}^{2}\,G\,\exp(i\overline{{{\omega}}}t)}}\end{array}
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$$
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where $G$ is a complex constant. To evaluate $G$ , substitute Eqs. (3-26) into Eq. (3-25), cancel out the quantity $\exp(i{\overline{{\omega}}}t)$ common to each term, substitute $k/\omega^{2}$ for $m$ and $\beta$ for $\overline{{\omega}}/\omega$ , and solve for $G$ yielding
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其中 $G$ 是一个复常数。为求解 $G$，将式 (3-26) 代入式 (3-25)，消去各项共有的量 $\exp(i{\overline{{\omega}}}t)$，将 $k/\omega^{2}$ 代替 $m$，将 $\beta$ 代替 $\overline{{\omega}}/\omega$，并求解 $G$，得到
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$$
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G=\frac{p_{o}}{k}\left[\frac{1}{\left(1-\beta^{2}\right)+i\left(2\xi\beta\right)}\right]=\frac{p_{o}}{k}\left[\frac{(1-\beta^{2})-i\left(2\xi\beta\right)}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\right]
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$$
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