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书籍/力学书籍/力学/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library)/auto/Dynamics of Structures (Ray Clough, Joseph Penzien) (Z-Library).md

@@ -872,9 +872,7 @@ where $G$ is an arbitrary complex constant and $\exp(s t)\equiv e^{s t}$ denotes
 in expressing dynamic loadings and responses; therefore it is useful now to briefly review the complex number concept.  
 
 Considering first the complex constant $G$ , this may be represented as a vector plotted in the complex plane as shown in Fig. 2-4. This sketch demonstrates that the vector may be expressed in terms of its real and imaginary Cartesian components:  
-其中 $G$ 是一个任意复常数，$\exp(s t)\equiv e^{s t}$ 表示指数函数。在随后的讨论中，使用复数
+其中 $G$ 是一个任意复常数，$\exp(s t)\equiv e^{s t}$ 表示指数函数。在随后的讨论中，使用复数来表示动态载荷和响应通常会很方便；因此，现在简要回顾复数概念是很有用的。
-
-来表示动态载荷和响应通常会很方便；因此，现在简要回顾复数概念是很有用的。
 
 首先考虑复常数 $G$，它可以表示为在复平面中绘制的向量，如图 2-4 所示。该示意图表明，该向量可以用其实部和虚部笛卡尔分量来表示：
 $$
@@ -882,25 +880,25 @@ G=G_{R}+i\;G_{I}
 $$  
 
 or alternatively that it may be expressed in polar coordinates using its absolute value $\overline{G}$ (the length of the vector) and its angle $\theta$ , measured counterclockwise from the real axis:  
-
+或者，它可以表示为极坐标形式，使用其绝对值 $\overline{G}$ (向量的长度) 和其角度 $\theta$ (从实轴逆时针测量)：
 $$
 G=\overline{{G}}\,\exp(i\,\theta)
 $$  
 
 In addition, from the trigonometric relations shown in the sketch, it is clear that Eq. (2-22a) also may be written  
-
+此外，从草图中所示的三角关系可以清楚地看出，式 (2-22a) 也可以写成
 $$
 G={\overline{{G}}}\,\cos\theta+i\;{\overline{{G}}}\,\sin\theta
 $$  
 
-Using this expression and noting that $\cos\theta\,=\,\sin\left(\theta\,+\,\frac{\pi}{2}\right)$ and $\sin\theta\,=\,-\cos\,\left(\theta\,+\,$ $\scriptstyle{\frac{\pi}{2}})$ it is easy to show that multiplying a vector by $i$ has the effect of rotating it counterclockwise in the complex plane through an angle of $\scriptstyle{\frac{\pi}{2}}$ radians or 90 degrees. Similarly it may be seen that multiplying by $-i$ rotates the vector $90^{\circ}$ clockwise. Now equating Eq. (2-22c) to Eq. (2-22b), and also noting that a negative imaginary component would be associated with a negative vector angle, leads to Euler’s pair of equations that serve to transform from trigonometric to exponential functions:  
+Using this expression and noting that $\cos\theta\,=\,\sin\left(\theta\,+\,\frac{\pi}{2}\right)$ and $\sin\theta\,=\,-\cos\left(\theta\,+\,\frac{\pi}{2}\right)$  it is easy to show that multiplying a vector by $i$ has the effect of rotating it counterclockwise in the complex plane through an angle of $\scriptstyle{\frac{\pi}{2}}$ radians or 90 degrees. Similarly it may be seen that multiplying by $-i$ rotates the vector $90^{\circ}$ clockwise. Now equating Eq. (2-22c) to Eq. (2-22b), and also noting that a negative imaginary component would be associated with a negative vector angle, leads to Euler’s pair of equations that serve to transform from trigonometric to exponential functions:  
-
+利用此表达式并注意到 $\cos\theta\,=\,\sin\left(\theta\,+\,\frac{\pi}{2}\right)$ 和 $\sin\theta\,=\,-\cos\left(\theta\,+\,\frac{\pi}{2}\right)$，很容易证明将向量乘以 $i$ 的作用是在复平面中将其逆时针旋转 $\scriptstyle{\frac{\pi}{2}}$ 弧度或 90 度。类似地，可以看出乘以 $-i$ 会使向量顺时针旋转 90 度。现在，将方程 (2-22c) 等同于方程 (2-22b)，并注意到负虚分量将与负向量角相关联，从而得到欧拉方程组，用于从三角函数转换为指数函数：
 $$
 \left.\begin{array}{l}{\exp(i\theta)=\cos\theta+i\ \sin\theta}\\ {\exp(-i\theta)=\cos\theta-i\sin\theta}\end{array}\right\}
 $$  
 
 Furthermore, Eqs. (2-23a) may be solved simultaneously to obtain the inverse form of Euler’s equations:  
-
+此外，方程 (2-23a) 可以联立求解，以获得欧拉方程的逆形式：
 $$
 \left.{\begin{array}{l}{\cos\theta={\frac{1}{2}}\ \left[\exp(i\theta)+\exp(-i\theta)\right]}\\ {\sin\theta=-{\frac{i}{2}}\ \left[\exp(i\theta)-\exp(-i\theta)\right]}\end{array}}\right\}
 $$  
@@ -909,13 +907,13 @@ $$
 FIGURE 2-4 Complex constant representation in complex plane.  
 
 To derive a free-vibration response expression, Eq. (2-21) is substituted into Eq. (2-20), leading to  
-
+为了推导自由振动响应表达式，将式(2-21)代入式(2-20)，得到
 $$
 (m\;s^{2}+c\;s+k)\;G\;\exp(s t)=0
 $$  
 
 and after dividing by $m G\exp(s t)$ and introducing the notation  
-
+并除以 $m G\exp(s t)$ 并引入记号
 $$
 \omega^{2}\equiv{\frac{k}{m}}
 $$  
@@ -929,13 +927,15 @@ $$
 The two values of $s$ that satisfy this quadratic expression depend on the value of $c$ relative to the values of $k$ and $m$ ; thus the type of motion given by Eq. (2-21) depends on the amount of damping in the system.  
 
 Considering now the undamped system for which $c=0$ , it is evident that the two values of $s$ given by solving Eq.(2-25) are  
+满足此二次表达式的两个 $s$ 值取决于 $c$ 相对于 $k$ 和 $m$ 的值；因此，方程 (2-21) 给出的运动类型取决于系统中的阻尼量。
 
+现在考虑 $c=0$ 的无阻尼系统，显然，通过求解方程 (2-25) 得到的两个 $s$ 值是
 $$
 s_{1,2}=\pm\;i\,\omega
 $$  
 
 Thus the total response includes two terms of the form of Eq. (2-21), as follows:  
-
+因此，总响应包括形式为式 (2-21) 的两项，如下所示：
 $$
 v(t)=G_{1}~\exp(i\omega t)+G_{2}~\exp(-i\omega t)
 $$  
@@ -943,13 +943,15 @@ $$
 in which the two exponential terms result from the two values of $s$ , and the complex constants $G_{1}$ and $G_{2}$ represent the (as yet) arbitrary amplitudes of the corresponding vibration terms.  
 
 We now establish the relation between these constants by expressing each of them in terms of its real and imaginary components:  
+其中，两个指数项源于 $s$ 的两个值，并且复常数 $G_{1}$ 和 $G_{2}$ 表示对应振动项的（目前）任意幅值。
 
+我们现在通过将这些常数中的每一个表示为其实部和虚部来建立它们之间的关系：
 $$
-{\cal G}_{1}=G_{1R}+i\;G_{1I}\quad\quad;\quad\quad{\cal G}_{2}=G_{2R}+i\;G_{2I}
+{G}_{1}=G_{1R}+i\;G_{1I}\quad\quad;\quad\quad{ G}_{2}=G_{2R}+i\;G_{2I}
 $$  
 
 and by transforming the exponential terms to trigonometric form using Eqs. (2-23a), so that Eq. (2-27) becomes  
-
+并通过使用公式（2-23a）将指数项转换为三角形式，使得公式（2-27）变为
 $$
 v(t)=\left(G_{1R}+i\,G_{1I}\right)\,\left(\,\cos\omega t+i\,\sin\omega t\right)+\left(G_{2R}+i\;G_{2I}\right)\,\left(\,\cos\omega t-i\,\sin\omega t\right)
 $$  
@@ -961,28 +963,31 @@ $$
 $$  
 
 However, this free-vibration response must be real, so the imaginary term (shown in square brackets) must be zero for all values of $t$ , and this condition requires that  
-
+然而，这种自由振动响应必须是实数，因此虚数项（显示在方括号中）对于所有 $t$ 值都必须为零，并且此条件要求
 $$
 G_{1I}=-G_{2I}\equiv G_{I}\qquad\qquad G_{1R}=G_{2R}\equiv G_{R}
 $$  
 
 From this it is seen that $G_{1}$ and $G_{2}$ are a complex conjugate pair:  
-
+由此可见，$G_{1}$ 和 $G_{2}$ 是一对复共轭：
 $$
-{\cal G}_{1}=G_{R}+i\;G_{I}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;G_{2}=G_{R}-i\;G_{I}
+{ G}_{1}=G_{R}+i\;G_{I}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;G_{2}=G_{R}-i\;G_{I}
 $$  
 
 and with these Eq. (2-27) becomes finally  
-
+鉴于此，方程 (2-27) 最终变为
 $$
 v(t)=(G_{R}+i\,G_{I})\;\exp(i\omega t)+(G_{R}-i\,G_{I})\;\exp(-i\omega t)
 $$  
 
 The response given by the first term of Eq. (2-29) is depicted in Fig. 2-5 as a vector representing the complex constant $G_{1}$ rotating in the counterclockwise direction with the angular velocity $\omega$ ; also shown are its real and imaginary constants. It will be noted that the resultant response vector $\left(G_{R}+i\,G_{I}\right)\exp(i\omega t)$ leads vector $G_{R}\exp(i\omega t)$ by the phase angle $\theta$ ; moreover it is evident that the response also can be expressed in terms of the absolute value, $\overline{G}$ , and the combined angle $(\omega t+\theta)$ . Examination of the second term of Eq. (2-29) shows that the response associated with it is entirely equivalent to that shown in Fig. 2-5 except that the resultant vector $\overline{{G}}\exp[-i(\omega t\!+\!\theta)]$ is rotating in the clockwise direction and the phase angle by which it leads the component $G_{R}\exp(-i\omega t)$ also is in the clockwise direction.  
+由方程 (2-29) 的第一项给出的响应在图 2-5 中描绘为一个向量，表示复常数 $G_{1}$ 以角速度 $\omega$ 沿逆时针方向旋转；图中还显示了它的实部和虚部常数。值得注意的是，合响应向量 $\left(G_{R}+i\,G_{I}\right)\exp(i\omega t)$ 以相位角 $\theta$ 超前向量 $G_{R}\exp(i\omega t)$ ；此外，显然响应也可以用绝对值 $\overline{G}$ 和组合角 $(\omega t+\theta)$ 来表示。检查方程 (2-29) 的第二项表明，与之相关的响应与图 2-5 中所示的响应完全等效，不同之处在于合向量 $\overline{{G}}\exp[-i(\omega t\!+\!\theta)]$ 沿顺时针方向旋转，并且它超前分量 $G_{R}\exp(-i\omega t)$ 的相位角也沿顺时针方向。
 
 The two counter-rotating vectors $\overline{{G}}\exp[i(\omega t+\theta)]$ and $\overline{{G}}\exp[-i(\omega t+\theta)]$ that represent the total free-vibration response given by Eq. (2-29) are shown in Fig. 2-6;  
-
+代表由式 (2-29) 给出的总自由振动响应的两个反向旋转向量 $\overline{{G}}\exp[i(\omega t+\theta)]$ 和 $\overline{{G}}\exp[-i(\omega t+\theta)]$ 如 图 2-6 所示；
 ![](971b6f64e38f1091af1b6e11ff5c9363f0593ab047d236d237211c3a358ccd87.jpg)  
+
+
 FIGURE 2-5 Portrayal of first term of Eq. (2-29).  
 
 it is evident here that the imaginary components of the two vectors cancel each other leaving only the real vibratory motion