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@ -1396,7 +1396,7 @@ It is informative to examine this response behavior in more detail by reference
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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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## 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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@ -1509,18 +1509,18 @@ G=\frac{p_{o}}{k}\left[\frac{1}{\left(1-\beta^{2}\right)+i\left(2\xi\beta\right)
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$$
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Substituting this complex value of $G$ into the first of Eqs. (3-26) and plotting the resulting two vectors in the complex plane, one obtains the representation shown in Fig. 3-5. Note that these two vectors and their resultant along with phase angle $\theta$ are identical to the corresponding quantities in Fig. 3-2, except that now the set of vectors has been rotated counterclockwise through 90 degrees. This difference in the figures corresponds to the phase angle difference between the harmonic excitations $-i\left(p_{o}/m\right)\,\exp(i\overline{{\omega}}t)$ and $(p_{\!_o}/m)\,\exp(i\overline{{\omega}}t)$ producing the results of Figs. 3-2 and 3-5, respectively. Note that $(p_{o}/m)$ $\sin{\overline{{\omega}}}t$ is the real part of $-i\left(p_{o}/m\right)\,\exp(i\overline{{{\omega}}}t)$ .
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将$G$的这个复数值代入式(3-26)中的第一个，并在复平面上绘制出由此产生的两个向量，即可得到图3-5所示的表示。注意到，这两个向量及其合向量以及相角$\theta$与图3-2中的对应量相同，只是现在这组向量已逆时针旋转了90度。图中这种差异对应于谐波激励$-i\left(p_{o}/m\right)\,\exp(i\overline{{\omega}}t)$和$(p_{\!_o}/m)\,\exp(i\overline{{\omega}}t)$之间的相角差，它们分别产生了图3-2和图3-5的结果。注意到，$(p_{o}/m)$ $\sin{\overline{{\omega}}}t$是$-i\left(p_{o}/m\right)\,\exp(i\overline{{{\omega}}}t)$的实部。
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FIGURE 3-5 Steady-state response using viscous damping.
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It is of interest to consider the balance of forces acting on the mass under the above steady-state harmonic condition whereby the total response, as shown in Fig. 3-5, is
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值得注意的是，在上述稳态谐波条件下，作用在质量上的力的平衡，此时总响应（如图3-5所示）为
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$$
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v_{p}(t)=\rho\,\,\exp[i\left(\overline{{\omega}}t-\theta\right)]
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$$
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having an amplitude $\rho$ as given by Eq. (3-22). Force equilibrium requires that the sum of the inertial, damping, and spring forces equal the applied loading
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具有由式 (3-22) 给出的振幅 $\rho$。力平衡要求惯性力、阻尼力和弹簧力之和等于施加的载荷
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$$
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p(t)=p_{o}\,\exp(i\overline{{\omega}}t)
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$$
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@ -1532,9 +1532,10 @@ $$
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$$
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which along with the applied loading are shown as vectors in the complex plane of Fig. 3-6. Also shown is the closed polygon of forces required for equilibrium in accordance with Eq. (2-1). Note that although the inertial, damping, and spring forces as given in Eqs. (3-30) are in phase with the acceleration, velocity, and displacement motions, respectively, they actually oppose their corresponding motions in accordance with the sign convention given in Fig. $2{-}1b$ which was adopted in Eq. (2-1).
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它们与施加的载荷一起，在图3-6的复平面中以向量形式显示。还显示了根据式(2-1)达到平衡所需的力的闭合多边形。注意，尽管式(3-30)中给出的惯性力、阻尼力和弹簧力分别与加速度、速度和位移运动同相，但根据图2-1b中给出的、并在式(2-1)中采用的符号约定，它们实际上与其对应的运动方向相反。
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FIGURE 3-6 Steady-state harmonic forces using viscous damping: (a) complex plane representation; (b) closed force polygon representation.
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图 3-6 采用粘性阻尼的稳态谐波力：(a) 复平面表示；(b) 闭合力多边形表示。
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Example E3-1. A portable harmonic-loading machine provides an effective means for evaluating the dynamic properties of structures in the field. By operating the machine at two different frequencies and measuring the resulting structural-response amplitude and phase relationship in each case, it is possible to determine the mass, damping, and stiffness of a SDOF structure. In a test of this type on a single-story building, the shaking machine was operated at frequencies of $\overline{{\omega}}_{1}=16\;r a d/s e c$ and $\overline{{\omega}}_{2}=25~r a d/s e c$ , with a force amplitude of $500\,l b$ [ $226.8\,k g]$ in each case. The response amplitudes and phase relationships measured in the two cases were
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@ -1596,16 +1597,16 @@ $$
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\xi=\frac{c}{2\,k/\omega}=\frac{1,125\,(27.9)}{200\times10^{3}}=15.7\%
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$$
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# 3-3 RESONANT RESPONSE
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## 3-3 RESONANT RESPONSE
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From Eq. (3-12), it is apparent that the steady-state response amplitude of an undamped system tends toward infinity as the frequency ratio $\beta$ approaches unity; this tendency can be seen in Fig. 3-3 for the case of $\xi=0$ . For low values of damping, it is seen in this same figure that the maximum steady-state response amplitude occurs at a frequency ratio slightly less than unity. Even so, the condition resulting when the frequency ratio equals unity, i.e., when the frequency of the applied loading equals the undamped natural vibration frequency, is called resonance. From Eq. (3-24) it is seen that the dynamic magnification factor under this condition $\left.\beta=1\right.$ ) is
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从式(3-12)可知，当频率比 $\beta$ 趋近于1时，无阻尼系统的稳态响应幅值趋于无穷大；对于 $\xi=0$ 的情况，这种趋势可以在图3-3中看到。对于低阻尼值，在同一图中可以看到，最大稳态响应幅值出现在略小于1的频率比处。即使如此，当频率比等于1时所产生的条件，即当施加荷载的频率等于无阻尼固有振动频率时，称为共振。从式(3-24)可知，在此条件 ($\beta=1$) 下的动放大系数为
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$$
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D_{\beta=1}=\frac{1}{2\,\xi}
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$$
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To find the maximum or peak value of dynamic magnification factor, one must differentiate Eq. (3-24) with respect to $\beta$ and solve the resulting expression for $\beta$ obtaining
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为了找到动放大系数的最大值或峰值，必须对式 (3-24) 关于 $\beta$ 求导，并求解得到的表达式以获得 $\beta$。
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$$
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\beta_{\mathrm{peak}}=\sqrt{1-2\,\xi^{2}}
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$$
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