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# Chap 3 RESPONSE TO HARMONIC LOADING谐波载荷响应
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## 3-1 UNDAMPED SYSTEM无阻尼系统
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### Complementary Solution互补解/齐次解
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Assume the system of Fig. 2-1 is subjected to a harmonically varying load $p(t)$ of sine-wave form having an amplitude $p_{o}$ and circular frequency $\overline{{\omega}}$ as shown by the equation of motion
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假设图2-1所示系统受到一个正弦波形式的简谐载荷 $p(t)$，其幅值为 $p_{o}$，圆频率为 $\overline{{\omega}}$，如运动方程所示。
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$$
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m\;\ddot{v}(t)+c\;\dot{v}(t)+k\;v(t)=p_{_o}\;\sin\overline{{{\omega}}}t \tag {3-1}
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$$
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Before considering this viscously damped case, it is instructive to examine the behavior of an undamped system as controlled by
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在考虑这种粘性阻尼情况之前，研究一个由...控制的无阻尼系统的行为是很有启发性的。
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$$
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m\ \ddot{v}(t)+k\ v(t)=p_{o}\ \sin\overline{{\omega}}t\tag {3-2}
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$$
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which has a complementary solution of the free-vibration form of Eq. (2-31) 其互补解具有式 (2-31) 的自由振动形式
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$$
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v_{c}(t)=A\,\cos\omega t+B\,\sin\omega t\tag {3-3}
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$$
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### Particular Solution 特解
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The general solution must also include the particular solution which depends upon the form of dynamic loading. In this case of harmonic loading, it is reasonable to assume that the corresponding motion is harmonic and in phase with the loading; thus, the particular solution is
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通解还必须包括特解，该特解取决于动载荷的形式。在简谐载荷的情况下，合理假设相应的运动是简谐的并与载荷同相；因此，特解为
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$$
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v_{p}(t)=C\,\sin{\overline{{\omega}}t}\tag {3-4}
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$$
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in which the amplitude $C$ is to be evaluated.
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其中振幅 $C$ 待确定。
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Substituting Eq. (3-4) into Eq. (3-2) gives
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将式 (3-4) 代入式 (3-2) 可得
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$$
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-m\,\overline{{{\omega}}}^{2}\,C\,\sin\overline{{{\omega}}}t+k\,C\,\sin\overline{{{\omega}}}t=p_{o}\,\sin\overline{{{\omega}}}t\tag {3-5}
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$$
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Dividing through by $\sin{\overline{{\omega}}}t$ (which is nonzero in general) and by $k$ and noting that $k/m=\omega^{2}$ , one obtains after some rearrangement
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除以 $\sin{\overline{{\omega}}}t$ (通常不为零) 和 $k$，并注意到 $k/m=\omega^{2}$，经过一些重新整理后，得到
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$$
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C=\frac{p_{o}}{k}\left[\frac{1}{1-\beta^{2}}\right]\tag {3-6}
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$$
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in which $\beta$ is defined as the ratio of the applied loading frequency to the natural free-vibration frequency, i.e.,
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其中 **$\beta$ 定义为施加的载荷频率与固有自由振动频率的比值**，**频率比，代表了外部激励频率与系统固有频率的比值**，即
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$$
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\beta\equiv\overline{{\omega}}\mathrm{~/~}\omega\tag {3-7}
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$$
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### General Solution通解
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The general solution of Eq. (3-2) is now obtained by combining the complementary and particular solutions and making use of Eq. (3-6); thus, one obtains
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方程(3-2)的通解现通过结合互补解和特解，并利用方程(3-6)得到；由此，可得
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$$
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v(t)=v_{c}(t)+v_{p}(t)=A\ \cos{\omega t}+B\ \sin{\omega t}+{\frac{p_{o}}{k}}\left[{\frac{1}{1-\beta^{2}}}\right]\ \sin{\overline{{\omega t}}}\tag {3-8}
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$$
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In this equation, the values of $A$ and $B$ depend on the conditions with which the response was initiated. For the system starting from rest, i.e., $v(0)=\dot{v}(0)=0$ , it is easily shown that
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在这个方程中，$A$ 和 $B$ 的值取决于响应启动的条件。对于从静止状态开始的系统，即 $v(0)=\dot{v}(0)=0$ ，可以很容易地证明
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$$
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A=0~~~~~~~~~~~~B=-{\frac{p_{o}\beta}{k}}\left[{\frac{1}{1-\beta^{2}}}\right]\tag {3-9}
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$$
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in which case the response of Eq. (3-8) becomes
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$$
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v(t)=\frac{p_{o}}{k}\,\left[\frac{1}{1-\beta^{2}}\right]\,\left(\sin\overline{{\omega}}t-\beta\,\sin\omega t\right)\tag {3-10}
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$$
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where $p_{o}/k\,=\,v_{\mathrm{st}}$ is the displacement which would be produced by the load $p_{o}$ applied statically and $1/(1-\beta^{2})$ is the magnification factor (MF) representing the amplification effect of the harmonically applied loading. In this equation, $\sin{\overline{{\omega}}}t$ represents the response component at the frequency of the applied loading; it is called the steady-state response and is directly related to the loading. Also $\beta\sin\omega t$ is the response component at the natural vibration frequency and is the free-vibration effect controlled by the initial conditions. Since in a practical case, damping will cause the last term to vanish eventually, it is termed the transient response. For this hypothetical undamped system, however, this term will not damp out but will continue indefinitely.
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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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$$
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R(t)\equiv{\frac{v(t)}{v_{\mathrm{st}}}}={\frac{v(t)}{p_{o}/k}}\tag {3-11}
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$$
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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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$$
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R(t)=\left[{\frac{1}{1-\beta^{2}}}\right]\left(\sin\overline{{\omega}}t-\beta\sin\omega t\right)\tag {3-12}
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$$
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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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$$
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\ddot{v}(t)+2\,\xi\,\omega\,\dot{v}(t)+\omega^{2}\,v(t)=\frac{p_{\scriptscriptstyle o}}{m}\,\sin\overline{{\omega}}t\tag {3-13}
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$$
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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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$$
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v_{c}(t)=\left[A\,\cos\omega_{D}t+B\,\sin\omega_{D}t\right]\;\exp(-\xi\,\omega\,t)\tag {3-14}
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$$
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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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$$
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v_{p}(t)=G_{1}\,\cos{\overline{{\omega}}t}+G_{2}\,\sin{\overline{{\omega}}t}\tag {3-15}
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$$
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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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$$
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\begin{array}{r l}&{\left[-G_{1}\,\overline{{\omega}}^{2}+G_{2}\,\overline{{\omega}}\left(2\xi\omega\right)+G_{1}\,\omega^{2}\right]\;\cos\overline{{\omega}}t}\\ &{\qquad\qquad\qquad+\left[-G_{2}\,\overline{{\omega}}^{2}-G_{1}\,\overline{{\omega}}\left(2\xi\omega\right)+G_{2}\,\omega^{2}-\frac{p_{o}}{m}\right]\;\sin\overline{{\omega}}t=0}\end{array}\tag {3-16}
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$$
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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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$$
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\begin{array}{l}{{G_{1}\left(1-\beta^{2}\right)+G_{2}\left(2\xi\beta\right)=0}}\\ {{{}}}\\ {{G_{2}\left(1-\beta^{2}\right)-G_{1}\left(2\xi\beta\right)=\frac{p_{o}}{k}}}\end{array}\tag {3-17}
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$$
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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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$$
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\begin{array}{l}{G_{1}=\displaystyle\frac{p_{o}}{k}\left[\frac{-2\xi\beta}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\right]}\\ {G_{2}=\displaystyle\frac{p_{o}}{k}\left[\frac{1-\beta^{2}}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\right]}\end{array}\tag {3-18}
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$$
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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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$$
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\begin{array}{l}{{\displaystyle v(t)=\left[A\,\cos\omega_{D}t+B\,\sin\omega_{D}t\right]\,\,\exp(-\xi\omega t)}}\\ {{\displaystyle\qquad+\,\frac{p_{o}}{k}\biggl[\frac{1}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\biggr]\,\,\Big[(1-\beta^{2})\,\sin\overline{{\omega}}t-2\xi\beta\,\cos\overline{{\omega}}t\Big]}}\end{array}\tag {3-19}
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$$
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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 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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$$
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v_{p}(t)=\frac{p_{o}}{k}\,\left[\frac{1}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\right]\,\left[(1-\beta^{2})\,\sin\overline{{\omega}}t-2\xi\beta\,\cos\overline{{\omega}}t\right]\tag {3-20}
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$$
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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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$$
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v_{p}(t)=\rho\,\sin(\overline{{\omega}}t-\theta)\tag {3-21}
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$$
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having an amplitude
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$$
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\rho=\frac{p_{o}}{k}\,\left[(1-\beta^{2})^{2}+(2\xi\beta)^{2}\right]^{-1/2}\tag {3-22}
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$$
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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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$$
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\theta=\tan^{-1}\left[\frac{2\xi\beta}{1-\beta^{2}}\right]\tag {3-23}
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$$
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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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$$
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D\equiv\frac{\rho}{p_{o}/k}=\left[(1-\beta^{2})^{2}+(2\xi\beta)^{2}\right]^{-1/2}\tag {3-24}
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$$
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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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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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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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$$
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\ddot{v}(t)+2\,\xi\,\omega\,\dot{v}(t)+\omega^{2}\,v(t)=\frac{p_{o}}{m}\;\exp[i\left(\overline{{\omega}}t+\phi\right)]\tag {3-25}
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$$
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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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$$
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\begin{array}{l}{{v_{p}(t)=G\,\exp(i\overline{{{\omega}}}t)}}\\ {{\ }}\\ {{\dot{v}_{p}(t)=i\,\overline{{{\omega}}}\,G\,\exp(i\overline{{{\omega}}}t)}}\\ {{\ }}\\ {{\ddot{v}_{p}(t)=-\overline{{{\omega}}}^{2}\,G\,\exp(i\overline{{{\omega}}}t)}}\end{array}\tag {3-26}
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$$
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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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$$
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G=\frac{p_{o}}{k}\left[\frac{1}{\left(1-\beta^{2}\right)+i\left(2\xi\beta\right)}\right]=\frac{p_{o}}{k}\left[\frac{(1-\beta^{2})-i\left(2\xi\beta\right)}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\right]\tag {3-27}
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$$
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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 $-i\left(p_{o}/m\right)\,\exp(i\overline{{{\omega}}}t)$ .
|
||||
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。
|
||||
|
||||
FIGURE 3-5 Steady-state response using viscous damping.
|
||||
|
||||
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
|
||||
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
|
||||
$$
|
||||
v_{p}(t)=\rho\,\,\exp[i\left(\overline{{\omega}}t-\theta\right)]\tag {3-28}
|
||||
$$
|
||||
|
||||
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
|
||||
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
|
||||
$$
|
||||
p(t)=p_{o}\,\exp(i\overline{{\omega}}t)\tag {3-29}
|
||||
$$
|
||||
|
||||
Using Eq. (3-28), these forces are
|
||||
|
||||
$$
|
||||
\begin{array}{l}{{f_{I_{p}}(t)=m~\ddot{v}_{p}(t)=-m\,\overline{{{\omega}}}^{2}\,\rho~\exp[i\,(\overline{{{\omega}}}t-\theta)]}}\\ {{{}}}\\ {{f_{D_{p}}(t)=c\,\dot{v}_{p}(t)=i\,c\,\overline{{{\omega}}}\,\rho~\exp[i\,(\overline{{{\omega}}}t-\theta)]}}\\ {{{}}}\\ {{f_{S_{p}}(t)=k\,v_{p}(t)=k\,\rho~\exp[i\,(\overline{{{\omega}}}t-\theta)]}}\end{array}\tag {3-30}
|
||||
$$
|
||||
|
||||
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).
|
||||
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。
|
||||
|
||||
FIGURE 3-6 Steady-state harmonic forces using viscous damping: (a) complex plane representation; (b) closed force polygon representation.
|
||||
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。
|
||||
|
||||
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
|
||||
|
||||
$$
|
||||
\begin{array}{l l l}{{\rho_{1}=7.2\times10^{-3}\ i n\ }}&{{[18.3\times10^{-3}\ c m]\quad}}&{{\cos\theta_{1}=0.966}}\\ {{\theta_{1}=15^{\circ}\quad}}&{{\sin\theta_{1}=0.259}}&{{\sin\theta_{1}=0.259}}\\ {{\rho_{2}=14.5\times10^{-3}\ i n\ }}&{{[36.8\times10^{-3}\ c m]\quad}}&{{\cos\theta_{2}=0.574}}\\ {{\theta_{2}=55^{\circ}\quad}}&{{\sin\theta_{2}=0.819}}&{{\sin\theta_{2}=0.813}}\end{array}
|
||||
$$
|
||||
|
||||
To evaluate the dynamic properties from these data, it is convenient to rewrite Eq. (3-22) as
|
||||
|
||||
$$
|
||||
\rho=\frac{p_{o}}{k}\,\frac{1}{1-\beta^{2}}\left\{\frac{1}{1+[2\xi\beta\,/\,(1-\beta^{2})]^{2}}\right\}^{1/2}=\frac{p_{o}}{k}\,\,\frac{\cos\theta}{1-\beta^{2}}
|
||||
$$
|
||||
|
||||
where the trigonometric function has been derived from Eq. (3-23). With further algebraic simplification this becomes
|
||||
|
||||
$$
|
||||
k(1-\beta^{2})=k-\overline{{\omega}}^{2}\,m=\frac{p_{o}\,\cos\theta}{\rho}
|
||||
$$
|
||||
|
||||
Then introducing the two sets of test data leads to the matrix equation
|
||||
|
||||
$$
|
||||
{\left[\begin{array}{l l}{1}&{-16^{2}}\\ {1}&{-25^{2}}\end{array}\right]}\;\;{\left[\begin{array}{l}{k}\\ {m}\end{array}\right]}=500\;l b\;\left[{\begin{array}{l}{{\frac{0.966}{7.2\times10^{-3}}}}\\ {{\frac{0.574}{14.5\times10^{-3}}}}\end{array}}\right]
|
||||
$$
|
||||
|
||||
which can be solved to give
|
||||
|
||||
$$
|
||||
\begin{array}{l}{k=100\times10^{3}\:l b/i n\quad[17.8\times10^{3}\:k g/c m]}\\ {m=128.5\:l b\cdot s e c^{2}/i n\quad[22.95\:k g\cdot s e c^{2}/c m]}\end{array}
|
||||
$$
|
||||
|
||||
Thus,
|
||||
|
||||
$$
|
||||
W=m\,g=49.6\times10^{3}\,l b\quad[22.5\times10^{3}\,k g]
|
||||
$$
|
||||
|
||||
The natural frequency is given by
|
||||
|
||||
$$
|
||||
\omega=\sqrt{\frac{k}{m}}=27.9\;r a d/s e c
|
||||
$$
|
||||
|
||||
To determine the damping coefficient, two expressions for $\cos\theta$ can be derived from Eqs. (a) and (3-23). Equating these and solving for the damping ratio leads to
|
||||
|
||||
$$
|
||||
\xi=\frac{p_{_o}\,\sin\theta}{2\,\beta\,k\,\rho}=\frac{p_{_o}\,\sin\theta}{c_{c}\,\overline{{\omega}}\,\rho}
|
||||
$$
|
||||
|
||||
Thus with the data of the first test
|
||||
|
||||
$$
|
||||
c=\xi\,c_{c}={\frac{500\,(0.259)}{16\,(7.2\times10^{-3})}}=1,125\;l b\cdot s e c/i n\quad[200.9\;k g\cdot s e c/c m]
|
||||
$$
|
||||
|
||||
and the same result (within engineering accuracy) is given by the data of the second test. The damping ratio therefore is
|
||||
|
||||
$$
|
||||
\xi=\frac{c}{2\,k/\omega}=\frac{1,125\,(27.9)}{200\times10^{3}}=15.7\%
|
||||
$$
|
||||
|
||||
## 3-3 RESONANT RESPONSE
|
||||
|
||||
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
|
||||
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
|
||||
$$
|
||||
D_{\beta=1}=\frac{1}{2\,\xi}
|
||||
$$
|
||||
|
||||
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
|
||||
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 $\beta$。
|
||||
$$
|
||||
\beta_{\mathrm{peak}}=\sqrt{1-2\,\xi^{2}}
|
||||
$$
|
||||
|
||||
(which yields positive real values for damping ratios $\xi<1/\sqrt{2})$ , and then substitute this value of frequency ratio back into Eq. (3-24) giving
|
||||
|
||||
$$
|
||||
D_{\mathrm{max}}={\frac{1}{2\,\xi\,{\sqrt{1-\xi^{2}}}}}={\frac{1}{2\,\xi}}\ {\frac{\omega}{\omega_{D}}}
|
||||
$$
|
||||
|
||||
For typical values of structural damping, say $\xi<0.10$ , the difference between Eq. (3- 33) and the simpler Eq. (3-31) is small, the difference being one-half of 1 percent for $\xi=0.10$ and 2 percent for $\xi=0.20$ .
|
||||
|
||||
For a more complete understanding of the nature of the resonant response of a structure to harmonic loading, it is necessary to consider the general response Eq. (3- 19), which includes the transient term as well as the steady-state term. At the resonant exciting frequency $\begin{array}{r}{\mathcal{\beta}=1\;\ }\end{array}$ ), this equation becomes
|
||||
|
||||
$$
|
||||
v(t)=(A\ \cos\omega_{D}t+B\ \sin\omega_{D}t)\ \exp(-\xi\omega t)-{\frac{p_{o}}{k}}\ {\frac{\cos\omega t}{2\,\xi}}
|
||||
$$
|
||||
|
||||
Assuming that the system starts from rest $[v(0)=\dot{v}(0)=0]$ , the constants are
|
||||
|
||||
$$
|
||||
A={\frac{p_{o}}{k}}\ {\frac{1}{2\,\xi}}\qquad\qquad B={\frac{p_{o}}{k}}\ {\frac{\omega}{2\,\omega_{D}}}={\frac{p_{o}}{k}}\ {\frac{1}{2\,\sqrt{1-\xi^{2}}}}
|
||||
$$
|
||||
|
||||
Thus Eq. (3-34) becomes
|
||||
|
||||
$$
|
||||
v(t)={\frac{1}{2\,\xi}}\;{\frac{p_{o}}{k}}\;\biggl[\biggl({\frac{\xi}{\sqrt{1-\xi^{2}}}}\,\sin\omega_{D}t+\cos\omega_{D}t\biggr)\;\exp(-\xi\omega t)-\cos\omega t\biggr]
|
||||
$$
|
||||
|
||||
For the amounts of damping to be expected in structural systems, the term $\sqrt{1-\xi^{2}}$ is nearly equal to unity; in this case, this equation can be written in the approximate form
|
||||
|
||||
$$
|
||||
R(t)=\frac{v(t)}{p_{o}/k}\doteq\frac{1}{2\,\xi}\,\left\{\left[\exp(-\xi\omega t)-1\right]\;\cos\omega t+\xi\left[\exp(-\xi\omega t)\right]\,\sin\omega t\right\}
|
||||
$$
|
||||
|
||||
For zero damping, this approximate equation is indeterminate; but when L’Hospital’s rule is applied, the response ratio for the undamped system is found to be
|
||||
|
||||
$$
|
||||
R(t)\doteq\frac{1}{2}\,\left(\sin\omega t-\omega t\,\,\cos\omega t\right)
|
||||
$$
|
||||
|
||||
Plots of these equations are shown in Fig. 3-7. Note that because the terms containing $\sin\omega t$ contribute little to the response, the peak values in this figure build up linearly for the undamped case, changing by an amount $\pi$ in each cycle; however, they build up in accordance with $(1/2\xi)[\exp(-\xi\omega t)-1]$ for the damped case. This latter envelope function is plotted against frequency in Fig. 3-8 for discrete values of damping. It is seen that the buildup rate toward the steady-state level $1/2\xi$ increases with damping and that buildup to nearly steady-state level occurs in a relatively small number of cycles for values of damping in the practical range of interest; e.g., 14 cycles brings the response very close to the steady-state level for a case having 5 percent of critical damping.
|
||||
|
||||
|
||||
FIGURE 3-7 Response to resonant loading $\upbeta=1$ for at-rest initial conditions.
|
||||
|
||||
|
||||
|
||||
FIGURE 3-8 Rate of buildup of resonant response from rest.
|
||||
|
||||
# 3-4 ACCELEROMETERS AND DISPLACEMENT METERS
|
||||
|
||||
At this point it is convenient to discuss the fundamental principles on which the operation of an important class of dynamic measurement devices is based. These are seismic instruments, which consist essentially of a viscously damped oscillator as shown in Fig. 3-9. The system is mounted in a housing which may be attached to the surface where the motion is to be measured. The response is measured in terms of the motion $v(t)$ of the mass relative to the housing.
|
||||
|
||||
The equation of motion for this system already has been shown in Eq. (2-17) to be
|
||||
|
||||
$$
|
||||
m\,\ddot{v}(t)+c\,\dot{v}(t)+k\,v(t)=-m\,\ddot{v}_{g}(t)\equiv p_{\mathrm{eff}}(t)
|
||||
$$
|
||||
|
||||
where ${\ddot{v}}_{g}(t)$ is the vertical acceleration of the housing support. Considering a harmonic support acceleration of the form $\ddot{v}_{g}(t)=\ddot{v}_{g0}\,\sin\overline{{\omega}}t$ , so that $p_{\mathrm{eff}}(t)=-m\,\ddot{v}_{g0}\,\sin\overline{{\omega}}t,$ the dynamic steady-state response amplitude of motion $v(t)$ is given by Eq. (3-22), i.e.,
|
||||
|
||||
$$
|
||||
\rho=\frac{m\,\ddot{v}_{g0}}{k}\;D
|
||||
$$
|
||||
|
||||
in which $D$ as given by Eq. (3-24) is presented graphically in Fig. 3-3. Examination of this figure shows that for a damping ratio $\xi\:=\:0.7$ , the value of $D$ is nearly constant over the frequency range $0~<~\beta~<~0.6$ . Thus it is clear from Eq. (3- 39) that the response indicated by this instrument is almost directly proportional to the support-acceleration amplitude for applied frequencies up to about six-tenths the natural frequency of the instrument $\left(\omega\;=\;2\pi f\;=\;\sqrt{k/m}\right)$ . Hence, this type of instrument when properly damped will serve effectively as an accelerometer for relatively low frequencies; its range of applicability will be broadened by increasing its natural frequency relative to the exciting frequency, i.e., by increasing the stiffness of the spring and/or decreasing the mass. Calibration of an accelerometer is easily carried out by first placing the instrument with its axis of sensitivity vertically and then turning the instrument upside-down and recording the resulting change of response which corresponds to an acceleration twice that of gravity.
|
||||
|
||||
|
||||
FIGURE 3-9 Schematic diagram of a typical seismometer.
|
||||
|
||||
|
||||
FIGURE 3-10 Response of seismometer to harmonic base displacement.
|
||||
|
||||
Consider now the response of the above described instrument subjected to a harmonic support displacement $v_{g}(t)=v_{g0}\,\sin\overline{{\omega}}t$ . In this case, $\ddot{v}_{g}(t)=-\overline{{\omega}}^{2}\,v_{g0}\,\sin\overline{{\omega}}t$ and the effective loading is $p_{\mathrm{eff}}=m\,\overline{{\omega}}\,v_{g0}\,\sin\overline{{\omega}}t$ . In accordance with Eq. (3-22), the relative-displacement response amplitude is
|
||||
|
||||
$$
|
||||
\rho=\frac{m\,\overline{{{\omega}}}^{2}\,v_{g0}}{k}\;D=v_{g0}\,\beta^{2}\,D
|
||||
$$
|
||||
|
||||
A plot of the response function $\beta^{2}\,D$ is presented in Fig. 3-10. In this case, it is evident that $\beta^{2}\,D$ is essentially constant at frequency ratios $\beta>1$ for a damping ratio $\xi\:=\:0.5$ . Thus, the response of a properly damped instrument is essentially proportional to the base-displacement amplitude for high-frequency support motions; i.e., it will serve as a displacement meter in measuring such motions. Its range of applicability for this purpose will be broadened by reducing the natural frequency, i.e., by reducing the spring stiffness and/or increasing the mass.
|
||||
|
||||
# 3-5 VIBRATION ISOLATION
|
||||
|
||||
Although the subject of vibration isolation is too broad to be discussed thoroughly here, the basic principles involved will be presented as they relate to two types of problems: (1) prevention of harmful vibrations in supporting structures due to oscillatory forces produced by operating equipment and (2) prevention of harmful vibrations in sensitive instruments due to vibrations of their supporting structures.
|
||||
|
||||
|
||||
FIGURE 3-11 SDOF vibration-isolation system (applied loading).
|
||||
|
||||
The first situation is illustrated in Fig. 3-11 where a rotating machine produces an oscillatory vertical force $p_{_o}\sin{\overline{{\omega}}t}$ due to unbalance in its rotating parts. If the machine is mounted on a SDOF spring-damper support system as shown, its steadystate relative-displacement response is given by
|
||||
|
||||
$$
|
||||
v_{p}(t)={\frac{p_{o}}{k}}\,D\,\sin(\overline{{\omega}}t-\theta)
|
||||
$$
|
||||
|
||||
where $D$ is defined by Eq. (3-24). This result assumes, of course, that the support motion induced by total reaction force $f(t)$ is negligible in comparison with the system motion relative to the support.
|
||||
|
||||
Using Eq. (3-41) and its first time derivative, the spring and damping reaction forces become
|
||||
|
||||
$$
|
||||
\begin{array}{l}{f_{S}(t)=k\;v(t)=p_{o}\;D\;\sin(\overline{{\omega}}t-\theta)}\\ {f_{D}(t)=c\;\dot{v}(t)=\frac{c\,p_{o}\,D\overline{{\omega}}}{k}\;\cos(\overline{{\omega}}t-\theta)=2\,\xi\,\beta\,p_{o}\,D\;\cos(\overline{{\omega}}t-\theta)}\end{array}
|
||||
$$
|
||||
|
||||
Since these two forces are $90^{\circ}$ out of phase with each other, it is evident that the amplitude of the total base reaction force is given by
|
||||
|
||||
$$
|
||||
f_{\operatorname*{max}}(t)=[f_{S,\operatorname*{max}}(t)^{2}+f_{D,\operatorname*{max}}(t)^{2}]^{1/2}=p_{o}\,D\,\left[1+(2\xi\beta)^{2}\right]^{1/2}
|
||||
$$
|
||||
|
||||
Thus, the ratio of the maximum base force to the amplitude of the applied force, which is known as the transmissibility (TR) of the support system, becomes
|
||||
|
||||
$$
|
||||
\mathrm{TR}\equiv\frac{f_{\mathrm{max}}(t)}{p_{o}}=D\;\sqrt{1+(2\xi\beta)^{2}}
|
||||
$$
|
||||
|
||||
The second type of situation in which vibration isolation is important is illustrated in Fig. 3-12, where the harmonic support motion $v_{g}(t)$ forces a steady-state relative-displacement response
|
||||
|
||||
$$
|
||||
v_{p}(t)=v_{g0}\,\beta^{2}\,D\,\sin(\overline{{\omega}}t-\theta)
|
||||
$$
|
||||
|
||||
|
||||
|
||||
|
||||
FIGURE 3-13 Vibration-transmissibility ratio (applied loading or support excitation).
|
||||
|
||||
in accordance with Eqs. (3-21) and (3-40). Adding this motion vectorially to the support motion $v_{g}(t)=v_{g0}\,\sin\overline{{\omega}}t$ , the total steady-state response of mass $m$ is given by
|
||||
|
||||
$$
|
||||
v^{t}(t)=v_{g0}\ \sqrt{1+(2\xi\beta)^{2}}\ D\sin(\overline{{\omega}}t-\overline{{\theta}})
|
||||
$$
|
||||
|
||||
in which the phase angle $\overline{{\theta}}$ is of no particular interest in the present discussion. Thus, if the transmissibility in this situation is defined as the ratio of the amplitude of total motion of the mass to the corresponding base-motion amplitude, it is seen that this expression for transmissibility is identical to that given by Eq. (3-44), i.e.,
|
||||
|
||||
$$
|
||||
\mathrm{TR}\equiv\frac{v_{\mathrm{max}}^{t}}{v_{g0}}=D\;\sqrt{1+(2\xi\beta)^{2}}
|
||||
$$
|
||||
|
||||
Note that this transmissibility relation also applies to the acceleration ratio $\left(\ddot{v}_{\mathrm{max}}^{t}/\ddot{v}_{g\mathrm{max}}\right)$ because $\ddot{v}_{\mathrm{max}}^{t}=\overline{{\omega}}^{2}\,v_{\mathrm{max}}^{t}$ and $\ddot{v}_{g\mathrm{max}}=\overline{{\omega}}^{2}\,v_{g0}$ .
|
||||
|
||||
Since the transmissibility relations given by Eqs. (3-44) and (3-47) are identical, the common relation expresses the transmissibility of vibration-isolation systems for both situations described above. This relation is plotted as a function of frequency ratio in Fig. 3-13 for discrete values of damping. Note that all curves pass through the same point at a frequency ratio of $\beta=\sqrt{2}$ . Clearly because of this feature, increasing the damping when $\beta<\sqrt{2}$ increases the effectiveness of the vibration-isolation system, while increasing the damping when $\beta>\sqrt{2}$ decreases the effectiveness. Since the transmissibility values for $\beta>\sqrt{2}$ are generally much lower than those for $\beta<\sqrt{2}$ , one should take advantage of operating in the higher frequency ratio range when it is practical to do so. This is not always possible, however, because in many cases the system must operate below $\beta=\sqrt{2}$ for some intervals of time, and in some cases even operate near the resonant condition $\beta=1$ . The following example illustrates such a condition:
|
||||
|
||||
Example E3-2. Deflections sometimes develop in concrete bridge girders due to creep, and if the bridge consists of a long series of identical spans, these deformations will cause a harmonic excitation in a vehicle traveling over the bridge at constant speed. Of course, the springs and shock absorbers of the car are intended to provide a vibration-isolation system which will limit the vertical motions transmitted from the road to the occupants.
|
||||
|
||||
Figure E3-1 shows a highly idealized model of this type of system, in which the vehicle weight is 4, 000 lb $[1,814\ k g]$ and its spring stiffness is defined by a test which showed that adding $100\,l b$ $\left[45.36\,k g\right]$ caused a deflection of 0.08 in $[0.203~c m]$ . The bridge profile is represented by a sine curve having a wavelength (girder span) of 40 ft $[12.2\,m]$ and a (single) amplitude of $1.2\,i n$ $[3.05\;c m]$ . From these data it is desired to predict the steady-state vertical motions in the car when it is traveling at a speed of 45 mph $[72.4\ k m/h r]$ , assuming that the damping is 40 percent of critical.
|
||||
|
||||
The transmissibility for this case is given by Eq. (3-47); hence the amplitude of vertical motion is
|
||||
|
||||
$$
|
||||
v_{\mathrm{max}}^{t}=v_{g0}\,\left[\frac{1+(2\xi\beta)^{2}}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}\right]^{1/2}
|
||||
$$
|
||||
|
||||
|
||||
FIGURE E3-1 Idealized vehicle traveling over an uneven bridge deck.
|
||||
|
||||
When the car is traveling at $45\ m p h=66\ f t/s e c,$ , the excitation period is
|
||||
|
||||
$$
|
||||
T_{p}={\frac{40\;f t}{66\;f t/s e c}}=0.606\;s e c
|
||||
$$
|
||||
|
||||
while the natural period of the vehicle is
|
||||
|
||||
$$
|
||||
T=\frac{2\pi}{\omega}=2\pi\;\sqrt{\frac{W}{k g}}=0.572\;s e c
|
||||
$$
|
||||
|
||||
Hence $\beta=T/T_{p}=0.572/0.606=0.944$ , and with $\xi=0.4$ the response amplitude is
|
||||
|
||||
$$
|
||||
v_{\mathrm{max}}^{t}=1.2~(1.642)=1.97~i n~~~[5.00~c m]
|
||||
$$
|
||||
|
||||
It also is of interest to note that if there were no damping in the vehicle $(\xi=0)$ , the amplitude would be
|
||||
|
||||
$$
|
||||
v_{\mathrm{max}}^{t}=v_{g0}\left[{\frac{1}{1-\beta^{2}}}\right]={\frac{1.2}{0.11}}=10.9~i n~~~[27.7~c m]
|
||||
$$
|
||||
|
||||
This is beyond the spring range, of course, and thus has little meaning, but it does demonstrate the important function of shock absorbers in limiting the motions resulting from waviness of the road surface.
|
||||
|
||||
When designing a vibration-isolation system which will operate at frequencies above the critical value represented by $\beta=\sqrt{2}$ , it is convenient to express the behavior of the SDOF system in terms of isolation effectiveness (IE) rather than transmissibility. This quantity is defined by
|
||||
|
||||
$$
|
||||
{\mathrm{IE}}\equiv[1-{\mathrm{TR}}]
|
||||
$$
|
||||
|
||||
in which $\mathrm{IE}=1$ represents complete isolation approachable only as $\beta\to\infty$ and IE $\mathit{\Theta}=\left0\right.$ represents no isolation which takes place at $\beta\,=\,{\sqrt{2}}$ . For values of $\beta$ below this critical value, amplification of the motion of the mass takes place; thus, actual vibration isolation can take place only when the system functions at values of $\beta$ greater than $\sqrt{2}$ . In this case the isolation system should have as little damping as possible.
|
||||
|
||||
For small damping, the transmissibility given by Eq. (3-44) or Eq. (3-47), after substitution of Eq. (3-24), can be expressed by the approximate relation
|
||||
|
||||
$$
|
||||
\mathrm{TR}\doteq1/(\beta^{2}-1)
|
||||
$$
|
||||
|
||||
in which case the isolation effectiveness becomes
|
||||
|
||||
$$
|
||||
\mathrm{IE}=\left(\beta^{2}-2\right)\big/\left(\beta^{2}-1\right)
|
||||
$$
|
||||
|
||||
Solving this relation for $\beta^{2}$ , one obtains its inverse form
|
||||
|
||||
$$
|
||||
\beta^{2}=\left(2-\mathrm{IE}\right)/\left(1-\mathrm{IE}\right)
|
||||
$$
|
||||
|
||||
Noting that $\beta^{2}=\overline{{\omega}}^{2}/\omega^{2}=\overline{{\omega}}^{2}\left(m/k\right)=\overline{{\omega}}^{2}\left(W/k g\right)=\overline{{\omega}}^{2}\left(\triangle_{\mathrm{st}}/g\right)\!,$ , where $g$ is the acceleration of gravity and $\triangle_{\mathrm{st}}$ is the static deflection produced by the dead weight $W$ on its spring mounting, Eq. (3-51) can be expressed in the form
|
||||
|
||||
$$
|
||||
\overline{{{f}}}=\frac{\overline{{\omega}}}{2\pi}=\frac{1}{2\pi}\,\sqrt{\frac{g}{\triangle_{\mathrm{st}}}\left[\frac{2-\mathrm{IE}}{1-\mathrm{IE}}\right]}\qquad\qquad0<\mathrm{IE}<1
|
||||
$$
|
||||
|
||||
Frequency $\overline{{f}}$ measured in Hertz $(c y c l e s/s e c)$ , as derived from this expression, is plotted against the static deflection $\triangle_{\mathrm{st}}$ in Fig. 3-14 for discrete values of isolation efficiency IE. Knowing the frequency of impressed excitation $\overline{{f}}$ , one can determine directly from the curves in this figure the support-pad deflection $\triangle_{\mathrm{st}}$ required to achieve any desired level of vibration isolation efficiency (IE), assuming, of course, that the isolation system has little damping. It is apparent that any isolation system must be very flexible to be effective.
|
||||
|
||||
|
||||
FIGURE 3-14 Vibration-isolation design chart.
|
||||
|
||||
Example E3-3. A reciprocating machine weighing 20, $000\,l b$ [9, $072\,k g]$ is known to develop a vertically oriented harmonic force of amplitude
|
||||
|
||||
$500~l b$ $[226.8\ k g]$ at its operating speed of $40\ H z$ . In order to limit the vibrations excited in the building in which this machine is to be installed, it is to be supported by a spring at each corner of its rectangular base. The designer wants to know what support stiffness will be required of each spring to limit the total harmonic force transmitted from the machine to the building to $80\ l b\ [36.3\ k g]$ .
|
||||
|
||||
The transmissibility in this case is $\mathrm{TR}\,=\,80/500\,=\,0.16$ which corresponds to an isolation efficiency of $\mathrm{IE}=1-\mathrm{TR}=0.84\$ . From Fig. 3-14 for ${\overline{{f}}}\,=\,40~H z$ and $\mathrm{{IE}=0.84}$ , one finds that $\triangle_{\mathrm{st}}$ is about 0.045 in $[0.114~c m]$ ; thus, the required stiffness $k$ of each spring is
|
||||
|
||||
$$
|
||||
k={\frac{W}{4\;\triangle_{\mathrm{{st}}}}}={\frac{20}{(4)\left(0.045\right)}}=111\;k i p s/i n\quad[19,823\;k g/c m]
|
||||
$$
|
||||
|
||||
# 3-6 EVALUATION OF VISCOUS-DAMPING RATIO
|
||||
|
||||
In the foregoing discussion of the dynamic response of SDOF systems, it has been assumed that the physical properties consisting of mass, stiffness, and viscous damping are known. While in most cases, the mass and stiffness can be evaluated rather easily using simple physical considerations or generalized expressions as discussed in Chapter 8, it is usually not feasible to determine the damping coefficient by similar means because the basic energy-loss mechanisms in most practical systems are seldom fully understood. In fact, it is probable that the actual energy-loss mechanisms are much more complicated than the simple viscous (velocity proportional) damping force that has been assumed in formulating the SDOF equation of motion. But it generally is possible to determine an appropriate equivalent viscous-damping property by experimental methods. A brief treatment of the methods commonly used for this purpose is presented in the following sections:
|
||||
|
||||
# Free-Vibration Decay Method
|
||||
|
||||
This is the simplest and most frequently used method of finding the viscousdamping ratio $\xi$ through experimental measurements. When the system has been set into free vibration by any means, the damping ratio can be determined from the ratio of two peak displacements measured over $m$ consecutive cycles. As shown in Chapter 2, the damping ratio can be evaluated using
|
||||
|
||||
$$
|
||||
\xi=\frac{\delta_{m}}{2\,\pi\,m\,\left(\omega/\omega_{D}\right)}\doteq\frac{\delta_{m}}{2\,\pi\,m}
|
||||
$$
|
||||
|
||||
where $\delta_{m}\equiv\ln(v_{n}/v_{n+m})$ represents the logarithmic decrement over $m$ cycles and $\omega$ and $\omega_{D}$ are the undamped and damped circular frequencies, respectively. For low values of damping, the approximate relation in Eq. (3-53) can be used which is only 2 percent in error when $\xi\,=\,0.2$ . A major advantage of this free-vibration method is that equipment and instrumentation requirements are minimal; the vibrations can be initiated by any convenient method and only the relative-displacement amplitudes need be measured. If the damping is truly of the linear viscous form as previously assumed, any set of $m$ consecutive cycles will yield the same damping ratio through the use of Eq. (3-53). Unfortunately, however, the damping ratio so obtained often is found to be amplitude dependent, i.e., $m$ consecutive cycles in the earlier portion of high-amplitude free-vibration response will yield a different damping ratio than $m$ consecutive cycles in a later stage of much lower response. Generally it is found in such cases that the damping ratio decreases with decreasing amplitude of free-vibration response. Caution must be exercised in the use of these amplitude-dependent damping ratios for predicting dynamic response.
|
||||
|
||||
# Resonant Amplification Method
|
||||
|
||||
This method of determining the viscous-damping ratio is based on measuring the steady-state amplitudes of relative-displacement response produced by separate harmonic loadings of amplitude $p_{o}$ at discrete values of excitation frequency $\overline{{\omega}}$ over a wide range including the natural frequency. Plotting these measured amplitudes against frequency provides a frequency-response curve of the type shown in Fig. 3-15. Since the peak of the frequency-response curve for a typical low damped structure is quite narrow, it is usually necessary to shorten the intervals of the discrete frequencies in the neighborhood of the peak in order to get good resolution of its shape. As shown by Eqs. (3-32) and (3-33), the actual maximum dynamic magnification factor $D_{\mathrm{max}}\equiv\rho_{\mathrm{max}}/\rho_{0}$ occurs at the excitation frequency $\overline{{\omega}}=\omega\,\sqrt{1-2\xi^{2}}$ and is given by $D_{\mathrm{max}}\,=\,1/2\xi\,\sqrt{1-\xi^{2}}$ ; however, for damping values in the practical range of interest, one can use the approximate relation $D_{\mathrm{max}}\,\doteq\,D\,(\beta\,=\,1)\,=\,1/2\xi$ . The damping ratio can then be determined from the experimental data using
|
||||
|
||||
|
||||
Frequency ratio, $\upbeta$
|
||||
FIGURE 3-15 Frequency-response curve for moderately damped system.
|
||||
|
||||
$$
|
||||
\xi\doteq\rho_{_0}/2\,\rho_{\mathrm{max}}
|
||||
$$
|
||||
|
||||
This method of determining the damping ratio requires only simple instrumentation to measure the dynamic response amplitudes at discrete values of frequency and fairly simple dynamic-loading equipment; however, obtaining the static displacement $\rho_{0}$ may present a problem because the typical harmonic loading system cannot produce a loading at zero frequency. As pointed out above, the damping ratio for practical systems is often amplitude dependent. In this case, the value of $\xi$ obtained by Eq. (3- 54) depends on the amplitude $p_{o}$ of the applied harmonic loading. This dependency should be taken into consideration when specifying an appropriate value of $\xi$ for dynamic analysis purposes.
|
||||
|
||||
# Half-Power (Band-Width) Method
|
||||
|
||||
It is evident from Eq. (3-22), in which $(p_{\!o}/k)\equiv\rho_{\!0}$ , that the frequency-response curve $\rho$ vs. $\beta$ shown in Fig. 3-15 has a shape which is controlled by the amount of damping in the system; therefore, it is possible to derive the damping ratio from many different properties of the curve. One of the most convenient of these is the half-power or band-width method whereby the damping ratio is determined from the frequencies at which the response amplitude $\rho$ is reduced to the level $1/\sqrt{2}$ times its peak value ρmax
|
||||
|
||||
The controlling frequency relation is obtained by setting the response amplitude in Eq. (3-22) equal to $1/\sqrt{2}$ times its peak value given by Eq. (3-33), that is, by setting
|
||||
|
||||
$$
|
||||
\left[(1-\beta^{2})^{2}+(2\xi\beta)^{2}\right]^{-1/2}=(1/\sqrt{2})\,\left[1/2\xi\,\sqrt{1-\xi^{2}}\right]
|
||||
$$
|
||||
|
||||
Squaring both sides of this equation and solving the resulting quadratic equation for $\beta^{2}$ gives
|
||||
|
||||
$$
|
||||
\beta_{1,2}^{2}=1-2\,\xi^{2}\;\mp\;2\,\xi\,\sqrt{1-\xi^{2}}
|
||||
$$
|
||||
|
||||
which, for small values of damping in the practical range of interest, yields the frequency ratios
|
||||
|
||||
$$
|
||||
\beta_{1,2}\doteq1-\xi^{2}\,\mp\,\xi\,\sqrt{1-\xi^{2}}
|
||||
$$
|
||||
|
||||
Subtracting $\beta_{1}$ from $\beta_{2}$ , one obtains
|
||||
|
||||
$$
|
||||
\beta_{2}-\beta_{1}=2\,\xi\,\sqrt{1-\xi^{2}}\doteq2\,\xi
|
||||
$$
|
||||
|
||||
while adding $\beta_{1}$ and $\beta_{2}$ gives
|
||||
|
||||
$$
|
||||
\beta_{2}+\beta_{1}=2\,\left(1-\xi^{2}\right)\doteq2
|
||||
$$
|
||||
|
||||
Combining Eqs. (3-58) and (3-59) yields
|
||||
|
||||
$$
|
||||
\xi=\frac{\beta_{2}-\beta_{1}}{\beta_{2}+\beta_{1}}=\frac{f_{2}-f_{1}}{f_{2}+f_{1}}
|
||||
$$
|
||||
|
||||
where $f_{1}$ and $f_{2}$ are the frequencies at which the amplitudes of response equal $1/\sqrt{2}$ times the maximum amplitude. The use of either Eq. (3-58) or Eq. (3-60) in evaluating the damping ratio is illustrated in Fig. 3-15 where a horizontal line has been drawn across the curve at $1/\sqrt{2}$ times its peak value. It is evident that this method of obtaining the damping ratio avoids the need for obtaining the static displacement $\rho_{0}$ ; however, it does require that the frequency-response curve be obtained accurately at its peak and at the level $\rho_{\mathrm{max}}/\sqrt{2}$ .
|
||||
|
||||
To clarify why the above method is commonly referred to as the half-power method, consider the time-average power input provided by the applied loading, which must equal the corresponding average rate of energy dissipation caused by the damping force $F_{D}(t)\,=\,c\,\,\dot{v}(t)$ . Under the steady-state harmonic condition at frequency $\overline{{\omega}}$ where the displacement response amplitude is $\rho$ , the average rate of energy dissipation is
|
||||
|
||||
$$
|
||||
P_{\mathrm{avg}}={\frac{c{\,\overline{{{\omega}}}}}{2\pi}}\;\int_{0}^{2\pi/\overline{{{\omega}}}}{\dot{v}}(t)^{2}\;d t=c{\,\overline{{{\omega}}}}^{2}\;\left[{\frac{\overline{{{\omega}}}}{2\pi}}\;\int_{0}^{2\pi/\overline{{{\omega}}}}v(t)^{2}\;d t\right]=\xi\,m\,\omega\,{\overline{{{\omega}}}}^{2}\;\rho^{2}
|
||||
$$
|
||||
|
||||
which shows that the corresponding average power input is proportional to $\beta^{2}\rho^{2}$ ; thus, since $\rho_{1}=\rho_{2}=\rho_{\mathrm{peak}}/\sqrt{2}$ , the average power inputs at frequency ratios $\beta_{1}$ and $\beta_{2}$ are
|
||||
|
||||
$$
|
||||
P_{\beta_{1}}=\left(\frac{\beta_{1}}{\beta_{\mathrm{peak}}}\right)^{2}\,\frac{P_{\mathrm{peak}}}{2}\qquad\qquad P_{\beta_{2}}=\left(\frac{\beta_{2}}{\beta_{\mathrm{peak}}}\right)^{2}\,\frac{P_{\mathrm{peak}}}{2}
|
||||
$$
|
||||
|
||||
where $\beta_{\mathrm{peak}}$ is given by Eq. (3-32). While the average power input at $\beta_{1}$ is somewhat less than one-half the peak power input and the average power input at $\beta_{2}$ is somewhat greater, the mean value of these two averaged inputs is very close to one-half the peak average power input.
|
||||
|
||||
Example E3-4. Data from a frequency-response test of a SDOF system have been plotted in Fig. E3-2. The pertinent data for evaluating the damping ratio are shown. The sequence of steps in the analysis after the curve was plotted were as follows:
|
||||
|
||||
|
||||
FIGURE E3-2 Frequency-response experiment to determine damping ratio.
|
||||
|
||||
(2) Construct a horizontal line at $1/\sqrt{2}$ times the peak level.
|
||||
|
||||
(3) Determine the two frequencies at which this horizontal line cuts the response curve; $f_{1}=19.55$ , $f_{2}=20.42\:H z$ .
|
||||
|
||||
(4) The damping ratio is given by
|
||||
|
||||
$$
|
||||
\xi={\frac{f_{2}-f_{1}}{f_{2}+f_{1}}}=0.022
|
||||
$$
|
||||
|
||||
showing 2.2 percent of critical damping in the system.
|
||||
|
||||
# Resonance Energy Loss Per Cycle Method
|
||||
|
||||
If instrumentation is available to measure the phase relationship between the input force and the resulting displacement response, the damping ratio can be evaluated from a steady-state harmonic test conducted only at resonance: $\begin{array}{r}{\beta\,=\,\frac{\overline{{\omega}}}{\omega}\,=\,1}\end{array}$ . This procedure involves establishing resonance by adjusting the input frequency until the displacement response is $90^{\circ}$ out-of-phase with the applied loading. As shown in Fig. 3-6 for $\theta=90^{\circ}$ , the applied loading is exactly balancing the damping force so that if the relationship between the applied loading and the resulting displacement is plotted for one loading cycle as shown in Fig. 3-16, the result can be interpreted as the damping force vs. displacement diagram. If the system truly possesses linear viscous damping, this diagram will be an ellipse as shown by the dashed line in this figure. In this case, the damping ratio can be determined directly from the maximum damping force and the maximum velocity using the relation
|
||||
|
||||
|
||||
FIGURE 3-16 Actual and equivalent damping energy per cycle.
|
||||
|
||||
$$
|
||||
p_{o}=f_{D{\operatorname*{max}}}=c\;{\dot{v}}_{\operatorname*{max}}=2\,\xi\,m\,\omega\,{\dot{v}}_{\operatorname*{max}}=2\,\xi\,m\,\omega^{2}\,\rho
|
||||
$$
|
||||
|
||||
or
|
||||
|
||||
$$
|
||||
\xi=p_{_o}/\,2\,m\,\omega^{2}\,\rho
|
||||
$$
|
||||
|
||||
If damping is not of the linear viscous form previously assumed but is of a nonlinear viscous form, the shape of the applied-force/displacement diagram obtained by the above procedure will not be elliptical; rather, it will be of a different shape as illustrated by the solid line in Fig. 3-16. In this case, the response $v(t)$ will be a distorted harmonic, even though the applied loading remains a pure harmonic. Nevertheless, the energy input per cycle, which equals the damping energy loss per cycle $E_{D}$ , can be obtained as the area under the applied-force/displacement diagram. This permits one to evaluate an equivalent viscous-damping ratio for the corresponding displacement amplitude, which when used in the linear viscous form will dissipate the same amount of energy per cycle as in the real experimental case. This equivalent damping ratio is associated with an elliptical applied-force/displacement diagram having the same area $E_{D}$ as the measured nonelliptical diagram. Making use of Eq. (3-61), this energy equivalence requires that
|
||||
|
||||
$$
|
||||
E_{D}=(2\pi/\omega)\,\,P_{\mathrm{avg}}=(2\pi/\omega)\,\,(\xi_{\mathrm{eq}}\,m\,\omega^{3}\,\rho^{2})
|
||||
$$
|
||||
|
||||
or
|
||||
|
||||
$$
|
||||
\xi_{\mathrm{eq}}=E_{D}/(2\,\pi\,m\,\omega^{2}\,\rho^{2})=E_{D}/(2\,\pi\,k\,\rho^{2})
|
||||
$$
|
||||
|
||||
The latter form of Eq. (3-66) is more convenient here because the stiffness of the structure can be measured by the same instrumentation used to obtain the energy loss per cycle, merely by operating the system very slowly at essentially static conditions. The static-force displacement diagram obtained in this way will be of the form shown in Fig. 3-17, if the structure is linearly elastic. The stiffness is obtained as the slope of the straight line curve.
|
||||
|
||||
|
||||
|
||||
# 3-7 COMPLEX-STIFFNESS DAMPING
|
||||
|
||||
Damping of the linear viscous form discussed above is commonly used because it leads to a convenient form of equation of motion. It has one serious deficiency, however; as seen from Eq. (3-61), the energy loss per cycle
|
||||
|
||||
$$
|
||||
E_{D}=(2\pi/\overline{{\omega}})\,\,P_{\mathrm{avg}}=2\,\pi\,\xi\,m\,\omega\,\overline{{\omega}}\,\rho^{2}
|
||||
$$
|
||||
|
||||
at a fixed amplitude $\rho$ is dependent upon the excitation (or response) frequency $\overline{{\omega}}$ . This dependency is at variance with a great deal of test evidence which indicates that the energy loss per cycle is essentially independent of frequency. It is desirable therefore to model the damping force so as to remove this frequency dependence. This can be accomplished by using the so-called “hysteretic” form of damping in place of viscous damping. Hysteretic damping may be defined as a damping force proportional to the displacement amplitude but in phase with the velocity, and for the case of harmonic motion it may be expressed as
|
||||
|
||||
$$
|
||||
f_{D}(t)=i\;\zeta\;k\;v(t)
|
||||
$$
|
||||
|
||||
where $\zeta$ is the hysteretic damping factor which defines the damping force as a function of the elastic stiffness force, and the imaginary constant $i$ puts the force in phase with the velocity. It is convenient to combine the elastic and damping resistance into the complex stiffness $\hat{k}$ defined as
|
||||
|
||||
$$
|
||||
\hat{k}=k\;(1+i\,\zeta)
|
||||
$$
|
||||
|
||||
leading to the following harmonic forced vibration equation of motion:
|
||||
|
||||
$$
|
||||
m\ \ddot{v}(t)+\hat{k}\ v(t)=p_{_o}\ \mathrm{exp}(i\overline{{\omega}}t)
|
||||
$$
|
||||
|
||||
The particular (or steady-state) solution of Eq. (3-70) is
|
||||
|
||||
$$
|
||||
v_{p}(t)=G\,\exp(i\overline{{\omega}}t)
|
||||
$$
|
||||
|
||||
in which $G$ is a complex constant, and the corresponding acceleration is given by
|
||||
|
||||
$$
|
||||
\ddot{v}_{p}(t)=-\overline{{\omega}}^{2}\;G\;\exp(i\overline{{\omega}}t)
|
||||
$$
|
||||
|
||||
Substituting these expressions into Eq. (3-70) yields
|
||||
|
||||
$$
|
||||
[\;-\;m\,\overline{{{\omega}}}^{2}+\hat{k}]\ G\ \mathrm{exp}(i\overline{{{\omega}}}t)=p_{_o}\ \mathrm{exp}(i\overline{{{\omega}}}t)
|
||||
$$
|
||||
|
||||
from which the value of $G$ is found to be
|
||||
|
||||
$$
|
||||
G=\frac{p_{o}}{k\;\left[\;-\;\frac{m}{k}\\,\overline{{{\omega}}}^{2}+(1+i\,\zeta)\right]}=\frac{p_{o}}{k}\;\frac{1}{\left[(1-\beta^{2})+i\,\zeta\right]}
|
||||
$$
|
||||
|
||||
or in a more convenient complex form
|
||||
|
||||
$$
|
||||
G={\frac{p_{o}}{k}}\left[{\frac{(1-\beta^{2})-i\,\zeta}{(1-\beta^{2})^{2}+\zeta^{2}}}\right]
|
||||
$$
|
||||
|
||||
Substituting this into Eq. (3-71) finally gives the following expression for the steadystate response with hysteretic damping
|
||||
|
||||
$$
|
||||
v_{p}(t)=\frac{p_{o}}{k}\left[\frac{(1-\beta^{2})-i\,\zeta}{(1-\beta^{2})^{2}+\zeta^{2}}\right]\,\exp(i\overline{{\omega}}t)
|
||||
$$
|
||||
|
||||
This response is depicted graphically by its two orthogonal vectors plotted in the complex plane of Fig. 3-18. The resultant of these two vectors gives the response in terms of a single-amplitude vector, namely
|
||||
|
||||
$$
|
||||
v_{p}(t)=\overline{{\rho}}\,\exp\left[\,i\overline{{\omega}}t-\overline{{\theta}}\,\right]
|
||||
$$
|
||||
|
||||
|
||||
FIGURE 3-18 Steady-state displacement response using complex stiffness damping.
|
||||
|
||||
in which
|
||||
|
||||
$$
|
||||
\overline{{\rho}}=\frac{p_{o}}{k}\,\left[(1-\beta^{2})^{2}+\zeta^{2}\right]^{-1/2}
|
||||
$$
|
||||
|
||||
and the response phase angle is
|
||||
|
||||
$$
|
||||
\overline{{\theta}}=\tan^{-1}\left[\frac{\zeta}{(1-\beta^{2})}\right]
|
||||
$$
|
||||
|
||||
Comparing these three equations with Eqs. (3-28), (3-22), and (3-23), respectively, it is evident that the steady-state response provided by hysteretic damping is identical to that with viscous damping if the hysteretic damping factor has the value
|
||||
|
||||
$$
|
||||
\zeta=2\,\xi\,\beta
|
||||
$$
|
||||
|
||||
In this case, the energy loss per cycle at a fixed amplitude $\overline{{\rho}}$ is dependent upon the excitation frequency $\overline{{\omega}}$ exactly as in the case of viscous damping. As will be shown subsequently, this frequency dependence can be removed by making the hysteretic damping factor $\zeta$ frequency independent. In doing so, it is convenient to use Eq. (3- 78) and to adopt the factor given at resonance for which $\beta=1$ ; thus the recommended hysteretic damping factor is $\zeta=2\,\xi$ , and the complex stiffness coefficient given by Eq. (3-69) becomes
|
||||
|
||||
$$
|
||||
\hat{k}=k\;[1+i\,2\xi]
|
||||
$$
|
||||
|
||||
Then as shown by Eqs. (3-76) and (3-77), the response amplitude and phase angle, respectively, are
|
||||
|
||||
$$
|
||||
\begin{array}{l}{\overline{{\rho}}=\frac{p_{0}}{k}\,\left[(1-\beta^{2})^{2}+(2\,\xi)^{2}\right]^{-1/2}}\\ {{\ }}\\ {\overline{{\theta}}=\tan^{-1}\left[\frac{2\,\xi}{(1-\beta^{2})}\right]}\end{array}
|
||||
$$
|
||||
|
||||
This response with hysteretic damping is identical to the viscous-damping response if the system is excited at resonance ( $\begin{array}{r}{\beta=1}\end{array}$ ). However, when $\beta\neq1$ , the two amplitudes differ in accordance with Eqs. (3-22) and (3-80) and the corresponding phase angles differ in accordance with Eqs. (3-23) and (3-81).
|
||||
|
||||
When the complex stiffness is defined in accordance with Eq. (3-69) and when $\zeta=2\xi$ , the damping force component under steady-state harmonic excitation is given by
|
||||
|
||||
$$
|
||||
f_{D}(t)=2\,i\xi\,k\,\overline{{\rho}}\,\left[\,\exp(i\overline{{\omega}}t-\overline{{\theta}})\right]
|
||||
$$
|
||||
|
||||
and the damping energy loss per cycle, $E_{D}$ , can be obtained by integrating the instantaneous power loss
|
||||
|
||||
$$
|
||||
P(t)=f_{D}(t)\;\dot{v}_{p}(t)=2\,\xi\,k\,\overline{{\omega}}\,\overline{{\rho}}^{2}\,\Big[-\exp(i\overline{{\omega}}t-\overline{{\theta}})\Big]^{2}
|
||||
$$
|
||||
|
||||
over one cycle, with the final result
|
||||
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$$
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E_{D}=2\,\pi\,\xi\,m\,\omega^{2}\,\overline{{\rho}}^{2}
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$$
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It is evident that this energy loss per cycle at fixed amplitude $\overline{{\rho}}$ is independent of the excitation frequency, $\overline{{\omega}}$ , thus it is consistent with the desired frequency-independent behavior; for this reason it is recommended that this form of hysteretic damping (complex stiffness damping) be used in most cases for general harmonic response analysis purposes.
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# PROBLEMS
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3-1. Consider the basic structure of Fig. $_{2-1a}$ with zero damping and subjected to harmonic excitation at the frequency ratio $\beta=0.8$ . Including both steady-state and transient effects, plot the response ratio $R(t)$ . Evaluate the response at increments $\overline{{\omega}}\triangle t=80^{\circ}$ and continue the analysis for 10 increments.
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3-2. Consider the basic system of Fig. 2-1a with the following properties: $m=$ $2\ k i p s\cdot s e c^{2}/i n$ and $k=20~k i p s/i n$ . If this system is subjected to resonant harmonic loading $\left(\overline{{\omega}}\,\right=\,\omega\right)$ ) starting from “at rest” conditions, determine the value of the response ratio $R(t)$ after four cycles $\overline{{\omega}}t=8\pi]$ ), assuming: (a) $c=0$ [use Eq. (3-38)] $(b)$ $\begin{array}{r l}&{c=0.5\ k i p s\cdot s e c/i n\ [\mathrm{use}\ \mathrm{Eq.}\ (3{-}37)]}\\ &{c=2.0\ k i p s\cdot s e c/i n\ [\mathrm{use}\ \mathrm{Eq.}\ (3{-}37)]}\end{array}$ (c)
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3-3. Consider the same vehicle and bridge structure of Example E3-2, except with the girder spans reduced to $L=36~f t$ . Determine:
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$(a)$ thevehiclespeedrequiredtoinduceresonanceinthevehiclespringsystem.
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$(b)$ the total amplitude of vertical motion $v_{\mathrm{max}}^{t}$ at resonance.
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$(c)$ the total amplitude of vertical motion $v_{\mathrm{max}}^{t}$ at the speed of 45 mph.
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3-4. A control console containing delicate instrumentation is to be located on the floor of a test laboratory where it has been determined that the floor slab is vibrating vertically with an amplitude of $0.03~i n$ at $20~H z$ . If the weight of the console is $800~l b$ , determine the stiffness of the vibration isolation system required to reduce the vertical-motion amplitude of the console to $0.005\ i n$ .
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3-5. A sieving machine weighs 6, $500\,l b$ , and when operating at full capacity, it exerts a harmonic force on its supports of $700\,l b$ amplitude at $12\,H z$ . After mounting the machine on spring-type vibration isolators, it was found that the harmonic force exerted on the supports had been reduced to a $50\,l b$ amplitude. Determine the spring stiffness $k$ of the isolation system.
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3-6. The structure of Fig. P3-1a can be idealized by the equivalent system of Fig. P3- $1b$ . In order to determine the values of $c$ and $k$ for this mathematical model, the concrete column was subjected to a harmonic load test as shown in Fig. $\mathrm{P}3{\-}1c$ . When operating at a test frequency of $\overline{{\omega}}=10~r a d s/s e c$ , the force-deflection (hysteresis) curve of Fig. P3- $1d$ was obtained. From this data:
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||||
|
||||
$(a)$ determine the stiffness $k$ .
|
||||
$(b)$ assuming a viscous damping mechanism, determine the apparent viscous damping ratio $\xi$ and damping coefficient $c$ .
|
||||
(c) assuming a hysteretic damping mechanism, determine the apparent hys
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|
||||
|
||||
FIGURE P3-1
|
||||
|
||||
tereticdampingfactor $\zeta$ .
|
||||
|
||||
3-7. Suppose that the test of Prob. 3-6 were repeated, using a test frequency $\overline{{\omega}}=$ $20~r a d s/s e c$ , and that the force-deflection curve (Fig. P3-1d) was found to be unchanged. In this case:
|
||||
|
||||
$(a)$ determine the apparent viscous damping values $\xi$ and $c$ . $(b)$ determine the apparent hysteretic damping factor $\zeta$ . $(c)$ Based on these two tests $\overline{{\omega}}=10$ and $\overline{{\omega}}=20\ r a d s/s e c)$ , which type of damping mechanism appears more reasonable — viscous or hysteretic?
|
||||
|
||||
3-8. If the damping of the system of Prob. 3-6 actually were provided by a viscous damper as indicated in Fig. P3- $1b$ , what would be the value of $E_{D}$ obtained in a test performed at $\overline{{\omega}}=20$ rads/sec?
|
||||
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