vault backup: 2025-10-07 17:43:40
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.obsidian/plugins/copilot/data.json
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.obsidian/plugins/copilot/data.json
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copilot-conversations/20251003_092718__解释从3-25到3-30的部分.md
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copilot-conversations/20251003_092718__解释从3-25到3-30的部分.md
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---
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epoch: 1759454838874
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modelKey: gemini-2.5-pro|google
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tags:
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- copilot-conversation
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---
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**user**: 解释从3-25到3-30的部分
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[Timestamp: 2025/10/03 09:27:18]
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**user**: [[Chap 3 RESPONSE TO HARMONIC LOADING谐波载荷响应]]解释从公式3-25到公式3-30的部分
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[Timestamp: 2025/10/03 09:28:05]
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**ai**: 好的，我们来逐步解释[[Chap 3 RESPONSE TO HARMONIC LOADING谐波载荷响应]]中从公式3-25到公式3-30的部分。
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这部分内容介绍了一种求解稳态谐波响应的替代方法，即使用复指数形式。这种方法在处理更复杂的谐波载荷（例如周期性载荷的傅里叶级数）时尤其方便和强大。
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---
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### **公式 (3-25): 复指数形式的运动方程**
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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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* **目的**: 这是标准的单自由度（SDOF）有阻尼系统的运动方程，但右侧的谐波载荷 $p(t)$ 被表示为复指数形式。
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* **解释**:
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* 根据欧拉公式，$e^{i\theta} = \cos\theta + i\sin\theta$。因此，一个复指数函数 $\exp[i(\overline{\omega}t)]$ 同时包含了 $\cos(\overline{\omega}t)$ 和 $\sin(\overline{\omega}t)$ 两个分量。
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* 通过使用复数形式的载荷和响应，我们可以将包含相位差的正弦和余弦运算转化为简单的代数运算，从而简化求解过程。
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* $\phi$ 是载荷的初始相位角。在后续的推导中，为了简化，作者假设 $\phi=0$，即载荷为 $p(t) = p_o \exp(i\overline{\omega}t)$。这并不失一般性，因为任何相位差最终都会体现在响应的复振幅中。
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---
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### **公式 (3-26): 假设的特解形式**
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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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* **目的**: 假设系统的稳态响应（特解）也具有与载荷相同的谐波形式。
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* **解释**:
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* 由于系统是线性的，当输入（载荷）是频率为 $\overline{\omega}$ 的谐波时，其稳态输出（响应）也必然是同频率的谐波。
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* $v_p(t) = G \exp(i\overline{\omega}t)$ 是假设的响应形式。这里的 $G$ 是一个**复常数**（Complex Constant）。$G$ 的模 $|G|$ 代表了响应的振幅，而 $G$ 的辐角 $\arg(G)$ 代表了响应相对于载荷的相位差。
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* 下面两行是该响应对时间的一阶和二阶导数，即速度和加速度。求导过程很简单，每次求导都会从指数项中多出一个因子 $i\overline{\omega}$。
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---
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### **公式 (3-27): 求解复常数 G**
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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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* **目的**: 确定复常数 $G$ 的具体表达式。
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* **推导过程**:
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1. 将公式 (3-26) 中的 $v_p, \dot{v}_p, \ddot{v}_p$ 代入到运动方程 (3-25)（其中 $\phi=0$）。
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2. 得到：$(-\overline{\omega}^2 G + 2\xi\omega(i\overline{\omega}G) + \omega^2 G) \exp(i\overline{\omega}t) = \frac{p_o}{m} \exp(i\overline{\omega}t)$。
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3. 消去两边共有的项 $\exp(i\overline{\omega}t)$，得到：$G(-\overline{\omega}^2 + i2\xi\omega\overline{\omega} + \omega^2) = \frac{p_o}{m}$。
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4. 将方程两边同除以 $\omega^2$，并利用关系式 $\beta = \overline{\omega}/\omega$ 和 $k = m\omega^2$：
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$G(-\beta^2 + i2\xi\beta + 1) = \frac{p_o}{m\omega^2} = \frac{p_o}{k}$。
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5. 求解 $G$，得到：$G = \frac{p_o}{k} \frac{1}{(1-\beta^2) + i(2\xi\beta)}$。这就是公式 (3-27) 的第一部分。
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6. 为了将 $G$ 分解为实部和虚部，将分母实数化（乘以其共轭复数 $(1-\beta^2) - i(2\xi\beta)$），就得到了公式的第二部分。
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* **解释**: 这个复数 $G$ 包含了响应的全部信息：
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* **振幅**: $|G| = \frac{p_o}{k} \frac{1}{\sqrt{(1-\beta^2)^2 + (2\xi\beta)^2}}$。
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* **相位**: $\arg(G) = \tan^{-1}\left(\frac{-(2\xi\beta)}{1-\beta^2}\right) = -\theta$。这表示响应滞后于载荷一个角度 $\theta$。
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---
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### **公式 (3-28): 极坐标形式的稳态响应**
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$$
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v_{p}(t)=\rho\,\,\exp[i\left(\overline{{\omega}}t-\theta\right)]\tag {3-28}
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$$
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* **目的**: 将稳态响应 $v_p(t)$ 写成更直观的极坐标形式。
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* **解释**:
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* 这是 $v_p(t) = G \exp(i\overline{\omega}t)$ 的另一种写法。
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* 我们将复常数 $G$ 表示为其极坐标形式 $G = \rho e^{-i\theta}$。
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* $\rho = |G|$ 是响应的振幅，其表达式与之前用三角函数法得到的公式 (3-22) 完全相同。
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* $\theta$ 是响应滞后于载荷的相位角，其表达式与公式 (3-23) 完全相同。
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* 这个公式清晰地表明，稳态响应是一个以角频率 $\overline{\omega}$ 旋转的向量，其振幅为 $\rho$，并且在相位上比载荷 $p(t) = p_o \exp(i\overline{\omega}t)$ 滞后了角度 $\theta$。
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---
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### **公式 (3-29) 和 (3-30): 系统中的各个力**
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$$
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p(t)=p_{o}\,\exp(i\overline{{\omega}}t)\tag {3-29}
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$$
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$$
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\begin{array}{l}{{f_{I_{p}}(t)=m~\ddot{v}_{p}(t)=-m\,\overline{{{\omega}}}^{2}\,\rho~\exp[i\,(\overline{{{\omega}}}t-\theta)]}}\\ {{f_{D_{p}}(t)=c\,\dot{v}_{p}(t)=i\,c\,\overline{{{\omega}}}\,\rho~\exp[i\,(\overline{{{\omega}}}t-\theta)]}}\\ {{f_{S_{p}}(t)=k\,v_{p}(t)=k\,\rho~\exp[i\,(\overline{{{\omega}}}t-\theta)]}}\end{array}\tag {3-30}
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$$
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* **目的**: 在稳态响应下，分析系统中的各个力（惯性力、阻尼力、弹簧力）与外载荷之间的关系。
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* **解释**:
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* **公式 (3-29)** 只是重申了我们所施加的外载荷的复指数形式。
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* **公式 (3-30)** 是将稳态响应 (3-28) 代入到各个力的定义中得到的：
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* **弹簧力 $f_S$**: 与位移 $v_p(t)$ 成正比，因此与位移向量同向。
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* **阻尼力 $f_D$**: 与速度 $\dot{v}_p(t)$ 成正比。由于速度向量 $\dot{v}_p$ 比位移向量 $v_p$ 在复平面上超前 $90^\circ$（因为乘以了因子 $i$），所以阻尼力向量也比位移向量超前 $90^\circ$。
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* **惯性力 $f_I$**: 与加速度 $\ddot{v}_p(t)$ 成正比。由于加速度向量 $\ddot{v}_p$ 比位移向量 $v_p$ 超前 $180^\circ$（因为乘以了因子 $i^2 = -1$），所以惯性力向量与位移向量反向。
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* **物理意义**: 这三个力是系统内部抵抗外载荷的力。在任何时刻，这三个力的矢量和必须与外载荷 $p(t)$ 相等，以满足动态平衡（$f_I + f_D + f_S = p(t)$）。**图3-6** 正是这一平衡关系的图形化表示，它将这四个力向量在复平面上画出，形成一个封闭的力多边形，直观地展示了动态平衡。
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### **总结**
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从公式 (3-25) 到 (3-30) 的推导，展示了如何利用复数代数来求解谐波激励下的单自由度系统响应。其核心思想是：
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1. 将谐波载荷和响应都用复指数表示。
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2. 将微分方程转化为代数方程，求解出响应的复振幅 $G$。
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3. 复振幅 $G$ 的模和辐角分别对应真实响应的振幅和相位。
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4. 最后，通过分析各个力的复数表达式，可以在复平面上清晰地展示系统在稳态振动下的力平衡关系。
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这种方法不仅与之前使用三角函数的求解结果完全一致，而且在数学上更为简洁，尤其是在处理更复杂的振动问题时优势更为明显。
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[Timestamp: 2025/10/03 09:28:52]
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**user**: [[Chap 3 RESPONSE TO HARMONIC LOADING谐波载荷响应]]展示3-3节 RESONANT RESPONSE共振响应 的推导过程
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[Timestamp: 2025/10/03 15:57:51]
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**ai**: 好的，我们来详细解析[[Chap 3 RESPONSE TO HARMONIC LOADING谐波载荷响应]]中第3-3节**共振响应 (RESONANT RESPONSE)** 的推导过程。
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这一节的核心是分析当外部激励频率与系统固有频率相同时（即共振时）系统的响应行为，包括稳态响应和包含瞬态效应的总响应。
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---
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### **1. 稳态响应下的共振 (Steady-State Resonance)**
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共振 (Resonance) 被定义为激励频率与系统的无阻尼固有频率相等的情况，即频率比 $\beta = \overline{\omega}/\omega = 1$。
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#### **推导公式 (3-31): 共振时的动力放大系数**
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* **出发点**: 稳态响应的动力放大系数 $D$ 的通用公式 (3-24):
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$$
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D = \left[(1-\beta^{2})^{2}+(2\xi\beta)^{2}\right]^{-1/2} \tag{3-24}
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$$
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* **推导过程**:
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1. 将共振条件 $\beta = 1$ 代入公式 (3-24)。
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2. 分母中的第一项 $(1-\beta^2)^2$ 变为 $(1-1^2)^2 = 0$。
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3. 分母中的第二项 $(2\xi\beta)^2$ 变为 $(2\xi \cdot 1)^2 = (2\xi)^2$。
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4. 因此，分母变为 $\sqrt{0 + (2\xi)^2} = \sqrt{(2\xi)^2} = 2\xi$。
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5. 所以，在 $\beta=1$ 时的动力放大系数为：
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$$
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D_{\beta=1}=\frac{1}{2\,\xi} \tag{3-31}
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$$
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* **物理意义**: 这个简单的公式表明，在共振时，稳态响应的振幅与阻尼比 $\xi$ 成反比。阻尼越小，振幅越大。当阻尼趋于零时（$\xi \to 0$），理论上稳态振幅趋于无穷大。
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---
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### **2. 峰值响应 (Peak Response)**
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从图3-3可以看出，对于有阻尼的系统，响应的峰值（最大值）并不严格出现在 $\beta=1$ 处，而是略小于1。
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#### **推导公式 (3-32) 和 (3-33): 峰值响应对应的频率比和动力放大系数**
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* **出发点**: 动力放大系数 $D$ 的公式 (3-24)。为了求其最大值，我们需要找到使其分母最小的 $\beta$ 值。令分母的平方为函数 $f(\beta)$:
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$$
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f(\beta) = (1-\beta^2)^2 + (2\xi\beta)^2 = 1 - 2\beta^2 + \beta^4 + 4\xi^2\beta^2
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$$
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* **推导过程**:
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1. 对 $f(\beta)$ 关于 $\beta$ 求导，并令其等于零，以找到极值点：
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$$
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\frac{df}{d\beta} = -4\beta + 4\beta^3 + 8\xi^2\beta = 0
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$$
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2. 由于我们关心的是 $\beta > 0$ 的情况，可以两边同除以 $4\beta$:
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$$
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-1 + \beta^2 + 2\xi^2 = 0
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$$
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3. 解出使响应达到峰值的 $\beta^2$:
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$$
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\beta^2 = 1 - 2\xi^2
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$$
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4. 因此，峰值频率比为：
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$$
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\beta_{\mathrm{peak}}=\sqrt{1-2\,\xi^{2}} \tag{3-32}
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$$
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这个结果只有在 $1-2\xi^2 > 0$ (即 $\xi < 1/\sqrt{2} \approx 0.707$) 时才有意义。
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5. 将 $\beta_{\mathrm{peak}}^2 = 1 - 2\xi^2$ 代回到 $D$ 的分母平方 $f(\beta)$ 中：
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$$
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f(\beta_{\mathrm{peak}}) = (1 - (1-2\xi^2))^2 + 4\xi^2(1-2\xi^2) = (2\xi^2)^2 + 4\xi^2 - 8\xi^4 = 4\xi^4 + 4\xi^2 - 8\xi^4 = 4\xi^2(1-\xi^2)
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$$
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6. 分母为 $\sqrt{f(\beta_{\mathrm{peak}})} = \sqrt{4\xi^2(1-\xi^2)} = 2\xi\sqrt{1-\xi^2}$。
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7. 因此，最大的动力放大系数为：
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$$
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D_{\mathrm{max}}={\frac{1}{2\,\xi\,{\sqrt{1-\xi^{2}}}}} \tag{3-33}
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$$
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* **说明**: 对于结构工程中常见的小阻尼（如 $\xi < 0.2$），$\sqrt{1-2\xi^2}$ 非常接近1，因此峰值响应频率 $\beta_{\mathrm{peak}}$ 与共振频率 $\beta=1$ 非常接近，且 $D_{\mathrm{max}}$ 与 $D_{\beta=1}$ 的值也相差无几。
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---
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### **3. 包含瞬态响应的共振分析**
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为了更完整地理解共振，需要考虑从静止开始的整个响应过程，这包括了瞬态响应和稳态响应。
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#### **推导公式 (3-34): 共振时的总响应方程**
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* **出发点**: 包含瞬态和稳态项的总响应通用公式 (3-19):
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$$
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v(t)=\left[A\,\cos\omega_{D}t+B\,\sin\omega_{D}t\right]\,\,\exp(-\xi\omega t) + \frac{p_{o}}{k}\frac{(1-\beta^{2})\,\sin\overline{{\omega}}t-2\xi\beta\,\cos\overline{{\omega}}t}{(1-\beta^{2})^{2}+(2\xi\beta)^{2}}
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$$
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* **推导过程**:
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1. 将共振条件 $\beta=1$ (即 $\overline{\omega}=\omega$) 代入上式。
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2. 稳态项的分母变为 $(1-1^2)^2 + (2\xi \cdot 1)^2 = (2\xi)^2$。
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3. 稳态项的分子中，$(1-\beta^2)\sin\omega t$ 变为 0。
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4. 稳态项的分子中，$-2\xi\beta\cos\omega t$ 变为 $-2\xi\cos\omega t$。
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5. 因此，稳态项简化为：
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$$
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\frac{p_o}{k} \frac{-2\xi\cos\omega t}{(2\xi)^2} = -\frac{p_o}{k} \frac{\cos\omega t}{2\xi}
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$$
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6. 将简化后的稳态项与瞬态项组合，得到共振时的总响应方程：
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$$
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v(t)=(A\ \cos\omega_{D}t+B\ \sin\omega_{D}t)\ \exp(-\xi\omega t)-{\frac{p_{o}}{k}}\ {\frac{\cos\omega t}{2\,\xi}} \tag{3-34}
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$$
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#### **推导公式 (3-35) 和 (3-36): 从静止开始的共振响应**
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* **条件**: 系统从静止开始，即初始条件为 $v(0)=0$ 和 $\dot{v}(0)=0$。
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* **推导过程**:
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1. **应用 $v(0)=0$**:
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将 $t=0$ 代入 (3-34):
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$v(0) = (A\cos(0) + B\sin(0))e^0 - \frac{p_o}{k}\frac{\cos(0)}{2\xi} = A - \frac{p_o}{k}\frac{1}{2\xi} = 0$
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解得: $A = \frac{p_o}{k}\frac{1}{2\xi}$。
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2. **应用 $\dot{v}(0)=0$**:
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首先对 (3-34) 求导得到速度 $\dot{v}(t)$：
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$\dot{v}(t) = [(-A\omega_D\sin\omega_Dt + B\omega_D\cos\omega_Dt)e^{-\xi\omega t} + (A\cos\omega_Dt + B\sin\omega_Dt)(-\xi\omega)e^{-\xi\omega t}] + \frac{p_o}{k}\frac{\omega\sin\omega t}{2\xi}$
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将 $t=0$ 代入:
|
||||
$\dot{v}(0) = [(0 + B\omega_D)e^0 + (A+0)(-\xi\omega)e^0] + 0 = B\omega_D - A\xi\omega = 0$
|
||||
解得: $B = A\frac{\xi\omega}{\omega_D} = (\frac{p_o}{k}\frac{1}{2\xi})\frac{\xi\omega}{\omega_D} = \frac{p_o}{k}\frac{\omega}{2\omega_D}$。
|
||||
因为 $\omega_D = \omega\sqrt{1-\xi^2}$，所以 $B = \frac{p_o}{k}\frac{1}{2\sqrt{1-\xi^2}}$。
|
||||
这样就得到了公式 (3-35)。
|
||||
3. **代入 A 和 B**:
|
||||
将求得的 A 和 B 代回 (3-34)，并将公共因子 $\frac{p_o}{k}\frac{1}{2\xi}$ 提出，即可得到公式 (3-36)。
|
||||
|
||||
#### **推导公式 (3-37) 和 (3-38): 小阻尼和无阻尼的近似**
|
||||
|
||||
* **推导 (3-37) - 小阻尼近似**:
|
||||
1. 对于结构工程中的小阻尼，$\xi \ll 1$，因此 $\sqrt{1-\xi^2} \approx 1$ 且 $\omega_D \approx \omega$。
|
||||
2. 将此近似代入 (3-36)，并将响应写成响应比 $R(t) = v(t)/(p_o/k)$ 的形式：
|
||||
$R(t) \approx \frac{1}{2\xi} [(\frac{\xi}{1}\sin\omega t + \cos\omega t)e^{-\xi\omega t} - \cos\omega t]$
|
||||
3. 整理上式，将 $\cos\omega t$ 合并：
|
||||
$R(t) \approx \frac{1}{2\xi} [ (e^{-\xi\omega t}-1)\cos\omega t + \xi e^{-\xi\omega t}\sin\omega t ]$
|
||||
这就是公式 (3-37)。
|
||||
|
||||
* **推导 (3-38) - 无阻尼极限**:
|
||||
1. 当 $\xi \to 0$ 时，公式 (3-37) 变为 $0/0$ 的不定式。
|
||||
2. 此时需要使用洛必达法则，对分子和分母同时关于 $\xi$ 求导。
|
||||
3. 令分子为 $N(\xi) = (e^{-\xi\omega t}-1)\cos\omega t + \xi e^{-\xi\omega t}\sin\omega t$。
|
||||
4. 令分母为 $D(\xi) = 2\xi$。
|
||||
5. 求导：
|
||||
$N'(\xi) = (-\omega t e^{-\xi\omega t})\cos\omega t + (e^{-\xi\omega t} - \xi\omega t e^{-\xi\omega t})\sin\omega t$
|
||||
$D'(\xi) = 2$
|
||||
6. 计算极限 $\xi \to 0$:
|
||||
$\lim_{\xi\to 0} \frac{N'(\xi)}{D'(\xi)} = \frac{(-\omega t e^0)\cos\omega t + (e^0 - 0)\sin\omega t}{2} = \frac{-\omega t\cos\omega t + \sin\omega t}{2}$
|
||||
7. 整理后得到无阻尼共振响应比：
|
||||
$$
|
||||
R(t) = \frac{1}{2}(\sin\omega t - \omega t \cos\omega t) \tag{3-38}
|
||||
$$
|
||||
* **物理意义**: 这个公式中的 $\omega t \cos\omega t$ 项表明，无阻尼系统在共振激励下，其振幅会随着时间 $t$ 线性无限增大，这正是共振导致结构破坏的理论原因。
|
||||
[Timestamp: 2025/10/03 15:58:49]
|
||||
@ -0,0 +1,210 @@
|
||||
---
|
||||
epoch: 1759485906614
|
||||
modelKey: gemini-2.5-pro|google
|
||||
tags:
|
||||
- copilot-conversation
|
||||
---
|
||||
|
||||
**user**: [[Chap 8]] 详细描述8-3节GENERALIZED PROPERTIES: DISTRIBUTED FLEXIBILITY 广义特性：分布式柔性 的推导过程
|
||||
[Timestamp: 2025/10/03 18:05:06]
|
||||
|
||||
**ai**: 好的，我们来详细描述一下 [[Chap 8]] 中 8-3 节 "广义特性：分布式柔性" 的推导过程。
|
||||
|
||||
这一节的核心思想是：如何将一个具有无限自由度（例如，一根可以弯曲的梁或塔）的复杂结构，通过一个合理的假设，简化成一个**等效的单自由度（SDOF）系统**进行分析。这个过程的关键在于引入“广义坐标”和“形函数”的概念。
|
||||
|
||||
以下是详细的推导步骤：
|
||||
|
||||
### 第一步：核心假设 - 位移形函数
|
||||
|
||||
对于一个具有分布式柔性的结构（以图 8-2a 中的悬臂塔为例），其在任意位置 `x` 和任意时间 `t` 的位移 `v(x,t)` 是一个复杂的函数。为了将其简化为 SDOF 系统，我们做出一个关键假设：**结构的变形形状是固定不变的，只有变形的幅度随时间变化。**
|
||||
|
||||
这个假设用数学公式表达就是式 **(8-2)**：
|
||||
$$
|
||||
v(x,t)=\psi(x)Z(t)
|
||||
$$
|
||||
这里：
|
||||
- $v(x,t)$ 是在高度 $x$ 处，时间为 $t$ 时的水平位移。
|
||||
- $\psi(x)$ 是**形函数 (shape function)**。它是一个只与位置 $x$ 有关的无量纲函数，描述了结构变形的“形状”或“模式”。一旦选定，它在整个分析过程中保持不变。
|
||||
- $Z(t)$ 是**广义坐标 (generalized coordinate)**。它是一个只与时间 $t$ 有关的函数，代表了整个结构变形的“幅度”。通常，我们可以选择结构上某个特定点（例如塔顶）的位移作为 $Z(t)$。
|
||||
|
||||
通过这个假设，原来需要无限个坐标才能描述的系统，现在只需要一个坐标 $Z(t)$ 就可以完全描述，从而成功地将其简化为了一个广义上的 SDOF 系统。
|
||||
|
||||
### 第二步：推导方法 - 虚功原理
|
||||
|
||||
为了建立这个广义 SDOF 系统的运动方程，教材选用了**虚功原理 (principle of virtual work)**。其核心思想是：对于一个处于平衡状态的系统，所有外力在任意虚位移上所做的功（外虚功 $\delta W_{E}$）等于所有内力在相应虚变形上所做的功（内虚功 $\delta W_{I}$）。
|
||||
|
||||
即式 **(8-4)**：
|
||||
$$
|
||||
\delta W_{E}=\delta W_{I}
|
||||
$$
|
||||
|
||||
### 第三步：计算外虚功 $\delta W_{E}$
|
||||
|
||||
外力包括惯性力、地面运动引起的等效荷载以及顶部的轴向力 $N$。
|
||||
根据式 **(8-6)**，外虚功由三部分组成：
|
||||
$$
|
||||
\delta W_{E}=-\int_{0}^{L}f_{I}(x)\,\delta v(x)\,d x+\int_{0}^{L}p_{\mathrm{eff}}(x,t)\,\delta v(x)\,d x+N\delta e
|
||||
$$
|
||||
1. **惯性力做的功**: 惯性力 $f_I(x,t) = m(x)\ddot{v}(x,t)$ 是抵抗运动的力，所以它做的功是负的。
|
||||
2. **等效荷载做的功**: 对于地震地面运动 $\ddot{v}_g(t)$，其等效于一个作用在结构上的分布荷载 $p_{\mathrm{eff}}(x,t) = -m(x)\ddot{v}_g(t)$。
|
||||
3. **轴向力 $N$ 做的功**: 当结构发生侧向位移 $v(x,t)$ 时，由于弯曲，其轴向会产生一个微小的缩短量 $e(t)$。轴向力 $N$ 在这个缩短量的虚变化 $\delta e$ 上做功。根据式 **(8-11)**，这个虚变化为 $\delta e=\int_{0}^{L}v^{\prime}(x,t)\;\delta v^{\prime}(x)\;d x$。
|
||||
|
||||
### 第四步：计算内虚功 $\delta W_{I}$
|
||||
|
||||
内虚功主要由结构内部的弯矩和阻尼力产生。
|
||||
根据式 **(8-7)**，内虚功是弯矩 $M(x,t)$ 在虚曲率变化 $\delta v^{\prime\prime}(x)$ 上做的功的积分：
|
||||
$$
|
||||
\delta W_{I}(t)=\int_{0}^{L}M(x,t)\ \delta v^{\prime\prime}(x)\ d x
|
||||
$$
|
||||
教材进一步假设了包含阻尼的本构关系（式 8-8），得到了弯矩的表达式（式 8-9）：
|
||||
$$
|
||||
M(x,t)=E I(x)\;[v^{\prime\prime}(x,t)+a_{1}\:\dot{v}^{\prime\prime}(x,t)]
|
||||
$$
|
||||
这表示弯矩由两部分组成：一部分与曲率成正比（弹性恢复力），另一部分与曲率变化率成正比（阻尼力）。
|
||||
|
||||
### 第五步：代入形函数并简化
|
||||
|
||||
这是推导中最关键的数学步骤。我们将第一步的核心假设 $v(x,t)=\psi(x)Z(t)$ 代入到所有相关的物理量中。根据式 **(8-10)**，我们可以得到：
|
||||
- 位移: $v(x,t) = \psi(x)Z(t)$
|
||||
- 速度: $\dot{v}(x,t) = \psi(x)\dot{Z}(t)$
|
||||
- 加速度: $\ddot{v}(x,t) = \psi(x)\ddot{Z}(t)$
|
||||
- 转角: $v'(x,t) = \psi'(x)Z(t)$
|
||||
- 曲率: $v''(x,t) = \psi''(x)Z(t)$
|
||||
- 虚位移: $\delta v(x,t) = \psi(x)\delta Z$
|
||||
- 虚转角: $\delta v'(x,t) = \psi'(x)\delta Z$
|
||||
- 虚曲率: $\delta v''(x,t) = \psi''(x)\delta Z$
|
||||
|
||||
将这些表达式代入第三步和第四步的虚功方程中，经过积分和整理，可以得到式 **(8-12)**：
|
||||
$$
|
||||
\begin{array}{l}{\displaystyle\delta W_{E}=\left[-\ddot{Z}(t)\int_{0}^{L}m(x)\,\psi(x)^{2}\,d x\right.}\\ {\displaystyle\qquad-\left.\ddot{v}_{g}(t)\int_{0}^{L}m(x)\,\psi(x)\,d x+N Z(t)\int_{0}^{L}\psi^{\prime}(x)^{2}\,d x\right]\,\delta Z}\\ {\displaystyle\delta W_{I}=\left[Z(t)\int_{0}^{L}E I(x)\,\psi^{\prime\prime}(x)^{2}\,d x+a_{1}\,\dot{Z}(t)\int_{0}^{L}E I(x)\,\psi^{\prime\prime}(x)^{2}\,d x\right]\,\delta Z}\end{array}
|
||||
$$
|
||||
这里的每一个积分项都有明确的物理意义，它们将结构的分布式物理属性（质量、刚度等）“整合”成了对应于广义坐标 $Z(t)$ 的单个值。
|
||||
|
||||
### 第六步：建立最终的运动方程
|
||||
|
||||
根据虚功原理 $\delta W_{E} = \delta W_{I}$，我们将式 (8-12) 的两个表达式相等，并移项合并：
|
||||
$$
|
||||
\left[ \left(\int m\psi^2 dx\right)\ddot{Z} + \left(a_1\int EI(\psi'')^2 dx\right)\dot{Z} + \left(\int EI(\psi'')^2 dx\right)Z - \left(N\int(\psi')^2 dx\right)Z + \left(\int m\psi dx\right)\ddot{v}_g \right] \delta Z = 0
|
||||
$$
|
||||
由于虚位移 $\delta Z$ 是任意的，不为零，因此括号内的表达式必须为零。这就得到了广义 SDOF 系统的运动方程。为了让它看起来更像标准的 SDOF 方程 $m\ddot{z}+c\dot{z}+kz=p(t)$，我们定义：
|
||||
|
||||
- **广义质量 $m^*$**: $m^{*}=\int_{0}^{L}m(x)~\psi(x)^{2}~d x$
|
||||
- **广义阻尼 $c^*$**: $c^{*}=a_{1}\int_{0}^{L}E I(x)~\psi^{\prime\prime}(x)^{2}~d x$
|
||||
- **广义抗弯刚度 $k^*$**: $k^{*}=\int_{0}^{L}E I(x)~\psi^{\prime\prime}(x)^{2}~d x$
|
||||
- **广义几何刚度 $k_G^*$**: $k_{G}^{*}=N\int_{0}^{L}\psi^{\prime}(x)^{2}~d x$ (注意：压力N会减小总刚度)
|
||||
- **广义等效荷载 $p_{\mathrm{eff}}^{*}(t)$**: $p_{\mathrm{eff}}^{*}(t)=-\ddot{v}_{g}(t)\int_{0}^{L}m(x)~\psi(x)~d x$
|
||||
|
||||
将这些广义属性代入，就得到了最终的运动方程，即式 **(8-13)** 和 **(8-15)**：
|
||||
$$
|
||||
m^{*}\,\ddot{Z}(t)+c^{*}\,\dot{Z}(t)+\overline{{k^{*}}}\;Z(t)=p_{\mathrm{eff}}^{*}(t)
|
||||
$$
|
||||
其中，组合广义刚度 $\overline{{k^{*}}} = k^{*} - k_{G}^{*}$。
|
||||
|
||||
### 总结
|
||||
|
||||
整个推导过程的核心逻辑是：
|
||||
1. **假设**: 假设复杂结构的变形可以用一个固定的**形状函数** $\psi(x)$ 和一个时变的**广义坐标** $Z(t)$ 的乘积来描述。
|
||||
2. **原理**: 应用**虚功原理**，即外力做的虚功等于内力做的虚功。
|
||||
3. **代换**: 将包含 $\psi(x)$ 和 $Z(t)$ 的位移、速度、加速度等表达式代入虚功方程。
|
||||
4. **整合**: 通过对结构全长进行积分，将分布式的物理属性（质量、刚度、阻尼）整合为对应于广义坐标 $Z(t)$ 的**广义质量、广义刚度、广义阻尼**。
|
||||
5. **求解**: 最终得到一个形式上与标准 SDOF 系统完全相同的二阶常微分方程，之后便可以使用前面章节中介绍的所有 SDOF 分析方法进行求解。
|
||||
|
||||
这个方法的巧妙之处在于，通过一个合理的形状假设，将一个偏微分方程问题（描述连续体振动）转化为了一个常微分方程问题，极大地简化了分析。当然，分析结果的精度完全取决于所假设的形函数 $\psi(x)$ 是否接近真实的变形模式。
|
||||
[Timestamp: 2025/10/03 18:05:59]
|
||||
|
||||
**user**: 对于8-3节中的悬臂梁结构，广义坐标选在悬臂梁的固定位置，还是远端位置
|
||||
[Timestamp: 2025/10/07 15:04:18]
|
||||
|
||||
**ai**: 根据您提供的 [[Chap 8]] 笔记内容，对于8-3节中讨论的悬臂梁（或悬臂塔）结构，广义坐标 $Z(t)$ **通常选择在悬臂梁的远端（即自由端、顶端）位置**。
|
||||
|
||||
笔记中有非常明确的说明。在引入公式 (8-2) $v(x,t)=\psi(x)Z(t)$ 之后，紧接着的一段文字解释了广义坐标的选取：
|
||||
|
||||
> Typically the generalized coordinate is selected as the displacement of some convenient reference point in the system, **such as the tip displacement in this tower.** Then the shape function is the dimensionless ratio of the local displacement to this reference displacement:
|
||||
>
|
||||
> 通常，广义坐标被选为系统中某个方便的参考点的位移，**例如这座塔的塔尖位移**。那么，形函数是局部位移与该参考位移的无量纲比：
|
||||
> $$
|
||||
> \psi(x)=\frac{v(x,t)}{Z(t)}\tag {8-3}
|
||||
> $$
|
||||
|
||||
**为什么选择远端（顶端）而不是固定端？**
|
||||
|
||||
1. **物理意义明确**: 悬臂结构的位移在固定端为零，在自由端（顶端）达到最大。选择顶端位移作为广义坐标 $Z(t)$，意味着 $Z(t)$ 直接代表了整个结构在某一时刻的最大变形幅度，这在物理上非常直观。
|
||||
|
||||
2. **方便归一化**: 如果将 $Z(t)$ 定义为顶端（$x=L$）的位移，那么根据形函数的定义，顶端的形函数值 $\psi(L)$ 就会等于1（因为 $\psi(L) = v(L,t) / Z(t) = Z(t) / Z(t) = 1$）。这使得形函数 $\psi(x)$ 成为一个从0（固定端）到1（自由端）变化的无量纲曲线，非常便于计算和理解。
|
||||
|
||||
3. **固定端不可行**: 固定端的位置位移恒为零。如果选择固定端的位移作为广义坐标 $Z(t)$，那么 $Z(t)$ 将永远是0，这将导致整个结构的位移 $v(x,t)$ 永远为0，这显然是不符合实际情况的。
|
||||
|
||||
因此，对于8-3节中的悬臂梁结构，广义坐标 $Z(t)$ 应该选在位移不为零且具有代表性的**远端（自由端）**位置。
|
||||
[Timestamp: 2025/10/07 15:04:38]
|
||||
|
||||
**user**: 解释公式8-17
|
||||
[Timestamp: 2025/10/07 15:11:54]
|
||||
|
||||
**ai**: 好的，我们来详细解释一下公式 **(8-17)**。
|
||||
|
||||
$$
|
||||
N_{\mathrm{cr}}={\frac{\int_{0}^{L}E I(x)\;\psi^{\prime\prime}(x)^{2}\;d x}{\int_{0}^{L}\psi^{\prime}(x)^{2}\;d x}}\tag {8-17}
|
||||
$$
|
||||
|
||||
### 1. 公式的核心作用
|
||||
|
||||
这个公式用于**估算一个受轴向压力作用的柔性构件（如柱子）的临界屈曲荷载 $N_{\mathrm{cr}}$**。
|
||||
|
||||
临界屈曲荷载是指当轴向压力达到这个值时，结构会从稳定状态突然变为不稳定状态，发生侧向的弯曲变形（即“屈曲”）。这个方法被称为**瑞利法 (Rayleigh's method)**，它是一种基于能量原理的近似分析方法。
|
||||
|
||||
### 2. 公式的推导过程
|
||||
|
||||
这个公式的推导基于一个非常直观的物理概念：**屈曲发生在结构的总刚度降为零的时刻**。
|
||||
|
||||
1. **总刚度**: 在8-3节的推导中，我们得到了考虑轴向力 $N$ 影响的组合广义刚度 $\overline{k^{*}}$ (见式 8-16):
|
||||
$$
|
||||
\overline{{k^{*}}} = k^{*} - k_{G}^{*}
|
||||
$$
|
||||
其中：
|
||||
* $k^{*}$ 是**广义抗弯刚度**，代表结构自身抵抗弯曲变形的能力，它总是正值。
|
||||
* $k_{G}^{*}$ 是**广义几何刚度**，代表由轴向压力 $N$ 引起的刚度折减效应（也称为P-Δ效应），它会削弱结构的总刚度。
|
||||
|
||||
2. **屈曲条件**: 当轴向压力 $N$ 逐渐增大，刚度折减效应 $k_{G}^{*}$ 也随之增大。当 $N$ 达到临界值 $N_{\mathrm{cr}}$ 时，刚度折减效应刚好完全抵消了结构自身的抗弯刚度，导致总刚度 $\overline{k^{*}}$ 变为零。此时，结构对任何微小的侧向扰动都失去了抵抗能力，从而发生屈曲。
|
||||
所以，屈曲的条件是：
|
||||
$$
|
||||
\overline{k^{*}} = k^{*} - k_{G}^{*} = 0 \quad \implies \quad k^{*} = k_{G}^{*}
|
||||
$$
|
||||
|
||||
3. **代入表达式**: 我们将 $k^{*}$ 和 $k_{G}^{*}$ 的积分表达式（见式 8-14）代入上述等式：
|
||||
* $k^{*} = \int_{0}^{L}E I(x)~\psi^{\prime\prime}(x)^{2}~d x$
|
||||
* $k_{G}^{*} = N \int_{0}^{L}\psi^{\prime}(x)^{2}~d x$
|
||||
|
||||
代入后得到：
|
||||
$$
|
||||
\int_{0}^{L}E I(x)\;\psi^{\prime\prime}(x)^{2}\;d x = N_{\mathrm{cr}} \int_{0}^{L}\psi^{\prime}(x)^{2}\;d x
|
||||
$$
|
||||
注意这里的 $N$ 已经变成了临界值 $N_{\mathrm{cr}}$。
|
||||
|
||||
4. **求解 $N_{\mathrm{cr}}$**: 将上式进行移项，就可以得到公式 (8-17)：
|
||||
$$
|
||||
N_{\mathrm{cr}}={\frac{\int_{0}^{L}E I(x)\;\psi^{\prime\prime}(x)^{2}\;d x}{\int_{0}^{L}\psi^{\prime}(x)^{2}\;d x}}
|
||||
$$
|
||||
|
||||
### 3. 公式各部分的物理意义
|
||||
|
||||
我们可以把这个公式理解为一个**平衡关系**：
|
||||
|
||||
* **分子 $\int_{0}^{L}E I(x)\;\psi^{\prime\prime}(x)^{2}\;d x$**:
|
||||
* 这正是**广义抗弯刚度 $k^{*}$**。
|
||||
* 它代表了结构**抵抗弯曲的内在能力**。$E I(x)$ 是截面抗弯刚度，$\psi^{\prime\prime}(x)$ 是假定变形形状的曲率。这个积分项综合了整个构件的材料和截面特性，量化了其在发生 $\psi(x)$ 形状的弯曲时所能储存的应变能（即恢复力）。
|
||||
|
||||
* **分母 $\int_{0}^{L}\psi^{\prime}(x)^{2}\;d x$**:
|
||||
* 这个积分项来自于**广义几何刚度 $k_{G}^{*}$**。
|
||||
* 它代表了**几何非线性效应**，即轴向力如何与侧向变形相互作用。$\psi^{\prime}(x)$ 是假定变形形状的转角（斜率）。这个积分项量化了当结构发生 $\psi(x)$ 形状的变形时，由于杆件弯曲导致的轴向缩短效应。这个效应正是轴向压力 $N$ 产生附加弯矩、削弱结构刚度的根源。
|
||||
|
||||
因此，整个公式可以通俗地理解为：
|
||||
$$
|
||||
\text{临界屈曲荷载} = \frac{\text{结构抵抗弯曲的内在能力 (广义抗弯刚度)}}{\text{由单位轴向力引起的几何失稳效应}}
|
||||
$$
|
||||
|
||||
### 4. 关键点与局限性
|
||||
|
||||
* **近似方法**: 瑞利法得到的结果是一个**近似值**。其精度完全取决于你所**假设的形函数 $\psi(x)$** 是否接近真实的屈曲形状。
|
||||
* **上界定理**: 一个非常重要的结论是，任何不正确的、但满足边界条件的形函数 $\psi(x)$ 计算出的 $N_{\mathrm{cr}}$ **总是会大于或等于**真实的临界屈曲荷载。这是因为任何“错误”的形状都需要额外的约束才能维持，这使得系统看起来比实际更“硬”。
|
||||
* **形函数的选择**: 为了得到好的结果，选择的 $\psi(x)$ 必须满足结构的**几何边界条件**（例如，悬臂柱在固定端位移和转角都为零）。如笔记中例子所示，选择一个同时满足力边界条件（如悬臂柱顶端弯矩为零，即曲率为零）的形函数会得到更精确的结果。
|
||||
[Timestamp: 2025/10/07 15:12:28]
|
||||
@ -0,0 +1,149 @@
|
||||
---
|
||||
epoch: 1759829524181
|
||||
modelKey: gemini-2.5-pro|google
|
||||
tags:
|
||||
- copilot-conversation
|
||||
---
|
||||
|
||||
**user**: [[Chap 8]]8-5 到8-7 主要讲了什么内容
|
||||
[Timestamp: 2025/10/07 17:32:04]
|
||||
|
||||
**ai**: 根据您提供的笔记 [[Chap 8]]，8-5 到 8-7 节主要讲解了使用瑞利法（Rayleigh's Method）来近似分析结构的振动频率。
|
||||
|
||||
以下是各节内容的具体概括：
|
||||
|
||||
### 8-5 瑞利法振动分析 (VIBRATION ANALYSIS BY RAYLEIGH’S METHOD)
|
||||
|
||||
这一节介绍了瑞利法的基本概念和原理。
|
||||
* **核心思想**: 将一个复杂的、具有无限自由度的系统（如梁、塔）通过**假设一个振动形状**（shape function, $\psi(x)$）来简化为等效的单自由度（SDOF）系统，从而近似计算其基频。
|
||||
* **两种推导方式**:
|
||||
1. **广义坐标法**: 直接使用前面章节定义的广义质量 $m^*$ 和广义刚度 $k^*$，这两个量都是基于假定的形状函数 $\psi(x)$ 计算得出的。振动频率由公式 $\omega = \sqrt{k^*/m^*}$ 给出。
|
||||
2. **能量守恒原理**: 对于无阻尼自由振动系统，其总能量是守恒的。因此，系统在振动过程中的**最大动能 ($T_{max}$)** 必须等于其**最大势能（应变能, $V_{max}$）**。通过建立这两种能量的表达式并令其相等，就可以求解出振动频率 $\omega$。
|
||||
* **瑞利商 (Rayleigh's Quotient)**: 对于梁结构，通过能量法最终推导出的频率计算公式为：
|
||||
$ \omega^{2}={\frac{\int_{0}^{L}E I(x)\,[\psi^{\prime\prime}(x)]^{2}\,d x}{\int_{0}^{L}m(x)\,[\psi(x)]^{2}\,d x}} $
|
||||
这个公式的分子就是广义刚度 $k^*$，分母是广义质量 $m^*$。
|
||||
|
||||
### 8-6 瑞利法振动形状的选择 (SELECTION OF THE RAYLEIGH VIBRATION SHAPE)
|
||||
|
||||
这一节讨论了如何选择一个合适的振动形状 $\psi(x)$ 以获得更准确的结果。
|
||||
* **关键原则**: 任何不真实的假设形状都会对系统产生额外的约束，使其“变硬”，因此计算出的频率**总是会高于或等于**真实的基频。这意味着，在不同假设形状的计算结果中，**频率最低的那个是最好的近似值**。
|
||||
* **如何选择好的形状**: 一个非常有效的方法是使用结构在**静力荷载作用下的变形曲线**作为假设的振动形状。这个静力荷载应该能模拟振动时的惯性力分布。
|
||||
* **实用方法**: 通常选择结构在**自重（或类似重力的横向荷载）**作用下的静变形曲线作为 $\psi(x)$。这种方法不仅物理意义合理，而且可以简化势能的计算，通常能得到非常精确的结果。
|
||||
|
||||
### 8-7 改进的瑞利法 (IMPROVED RAYLEIGH METHOD)
|
||||
|
||||
这一节介绍了一种系统性的、迭代的程序来进一步提高瑞利法的计算精度。
|
||||
* **基本流程**:
|
||||
1. 从一个初始的、任意选择的形状 $\psi^{(0)}(x)$ 开始（**Method R₀₀**，即标准瑞利法）。
|
||||
2. 根据这个初始形状 $\psi^{(0)}$ 计算出系统振动时的惯性力分布。
|
||||
3. 将这个惯性力作为静力荷载施加到结构上，计算出**新的变形形状** $\psi^{(1)}(x)$。
|
||||
4. 使用新的形状来重新计算频率。
|
||||
* **两种改进方法**:
|
||||
* **Method R₀₁**: 使用初始形状 $\psi^{(0)}$ 计算动能，但使用新形状 $\psi^{(1)}$ 和初始惯性力来计算势能。这种方法避免了对初始形状函数求二阶导数，通常能提高精度。
|
||||
* **Method R₁₁**: 使用新计算出的形状 $\psi^{(1)}$ 来计算**动能和势能**。这是更进一步的改进，结果通常非常接近精确解。
|
||||
* **收敛性**: 这个迭代过程如果重复进行（用 $\psi^{(1)}$ 计算 $\psi^{(2)}$，以此类推），最终会收敛到结构真实的振动形状和精确的振动频率。但在实际应用中，进行一次迭代（即 Method R₁₁）通常就足够了。
|
||||
[Timestamp: 2025/10/07 17:32:37]
|
||||
|
||||
**user**: 详细展开8-7节的推导内容
|
||||
[Timestamp: 2025/10/07 17:37:09]
|
||||
|
||||
**ai**: 好的，我们来详细展开第 8-7 节 “改进的瑞利法” (IMPROVED RAYLEIGH METHOD) 的推导内容。
|
||||
|
||||
这一节的核心思想是：标准的瑞利法依赖于一个**一次性**的形状假设，其精度完全取决于这个假设的好坏。而改进的瑞利法是一个**迭代**过程，它利用上一步计算的结果来获得一个更好的振动形状，从而得到更精确的频率。
|
||||
|
||||
整个推导过程基于能量守恒原理：**最大势能 ($V_{max}$) = 最大动能 ($T_{max}$)**。不同方法的区别在于它们如何计算这两项能量。
|
||||
|
||||
---
|
||||
|
||||
### 1. 基础：标准瑞利法 (Method R₀₀)
|
||||
|
||||
这是我们进行改进的出发点。
|
||||
|
||||
* **第一步：初始形状假设**
|
||||
我们首先假设一个初始的振动形状，记为 $v^{(0)}(x,t)$。上标 `(0)` 代表这是第0次迭代，即我们的初始猜测。
|
||||
$$ v^{(0)}(x,t) = \psi^{(0)}(x) \, Z_{0}^{(0)} \, \sin(\omega t) \quad (8-33) $$
|
||||
其中：
|
||||
* $\psi^{(0)}(x)$ 是初始的、无量纲的**形状函数**。
|
||||
* $Z_{0}^{(0)}$ 是该形状的**振幅**。
|
||||
* $\omega$ 是我们要求解的未知振动频率。
|
||||
|
||||
* **第二步：计算能量**
|
||||
基于这个初始形状，我们计算最大势能和最大动能。
|
||||
* **最大势能 ($V_{max}$)** 来自于结构弯曲产生的应变能，它在位移最大时达到峰值：
|
||||
$$ V_{\mathrm{max}}^{(0)} = \frac{1}{2} \int_{0}^{L} EI(x) \left( \frac{\partial^2 v^{(0)}}{\partial x^2} \right)^2_{\mathrm{max}} dx = \frac{(Z_{0}^{(0)})^2}{2} \int_{0}^{L} EI(x) \, (\psi''^{(0)})^2 \, dx \quad (8-34) $$
|
||||
* **最大动能 ($T_{max}$)** 来自于质量块的运动，它在速度最大时达到峰值。速度为 $\dot{v}^{(0)} = \psi^{(0)}(x) \, Z_{0}^{(0)} \, \omega \cos(\omega t)$，最大速度为 $\omega Z_{0}^{(0)} \psi^{(0)}(x)$。
|
||||
$$ T_{\mathrm{max}}^{(0)} = \frac{1}{2} \int_{0}^{L} m(x) (\dot{v}^{(0)})^2_{\mathrm{max}} dx = \frac{(Z_{0}^{(0)})^2 \omega^2}{2} \int_{0}^{L} m(x) \, (\psi^{(0)})^2 \, dx \quad (8-35) $$
|
||||
|
||||
* **第三步：求解频率**
|
||||
令 $V_{\mathrm{max}}^{(0)} = T_{\mathrm{max}}^{(0)}$，消去公共项 $\frac{(Z_{0}^{(0)})^2}{2}$，解出 $\omega^2$：
|
||||
$$ \omega^2 = \frac{\int_{0}^{L} EI(x) \, (\psi''^{(0)})^2 \, dx}{\int_{0}^{L} m(x) \, (\psi^{(0)})^2 \, dx} \quad (8-36) $$
|
||||
这就是**标准瑞利法 (Method R₀₀)** 的公式。它的主要缺点是需要计算 $\psi''^{(0)}$，如果初始形状 $\psi^{(0)}$ 只是一个粗略的近似，其二阶导数可能会有很大误差。
|
||||
|
||||
---
|
||||
|
||||
### 2. 第一次改进 (Method R₀₁)
|
||||
|
||||
这次改进的核心是**用更物理的方式计算势能**。
|
||||
|
||||
* **第一步：计算等效惯性力**
|
||||
振动是由惯性力引起的。基于我们初始假设的形状 $v^{(0)}$，在位移最大时，加速度也最大，其值为 $\ddot{v}^{(0)} = -\omega^2 v^{(0)}$。因此，产生的惯性力分布为：
|
||||
$$ p^{(0)}(x) = m(x) \cdot (\omega^2 v^{(0)}) = \omega^2 \, m(x) \, \psi^{(0)}(x) \, Z_{0}^{(0)} \quad (8-37) $$
|
||||
这个力 $p^{(0)}(x)$ 是一个与 $\omega^2$ 成正比的**静力荷载**，它代表了振动达到最大位移时惯性力的分布。
|
||||
|
||||
* **第二步：计算新的变形形状**
|
||||
我们将这个惯性力荷载 $p^{(0)}(x)$ 施加到结构上，计算出它引起的**静力变形**，记为 $v^{(1)}(x)$。这个新的变形形状 $v^{(1)}$ 显然比我们最初随便猜的 $v^{(0)}$ 更接近真实的振动形状。
|
||||
由于 $p^{(0)}$ 正比于 $\omega^2$，那么它引起的变形 $v^{(1)}$ 也必然正比于 $\omega^2$。为了方便表示，我们将其写为：
|
||||
$$ v^{(1)} = \omega^2 \, \psi^{(1)} \, \overline{Z}_{0}^{(1)} \quad (8-38) $$
|
||||
这里的 $\psi^{(1)}$ 和 $\overline{Z}_{0}^{(1)}$ 是由荷载 $m(x)\psi^{(0)}Z_0^{(0)}$ 产生的变形形状和幅值。$\omega^2$ 只是一个我们还不知道的比例系数。
|
||||
|
||||
* **第三步：用新方法计算势能**
|
||||
一个物体的应变能（势能）等于使它产生该变形的外力所做的功。因此，最大势能 $V_{max}$ 等于惯性力 $p^{(0)}$ 在变形 $v^{(1)}$ 上所做的功的一半：
|
||||
$$ V_{\mathrm{max}} = \frac{1}{2} \int_{0}^{L} p^{(0)}(x) \, v^{(1)}(x) \, dx $$
|
||||
代入 $p^{(0)}$ 和 $v^{(1)}$ 的表达式：
|
||||
$$ V_{\mathrm{max}} = \frac{1}{2} \int_{0}^{L} \left( \omega^2 m(x) \psi^{(0)} Z_{0}^{(0)} \right) \left( \omega^2 \psi^{(1)} \overline{Z}_{0}^{(1)} \right) dx $$
|
||||
$$ V_{\mathrm{max}} = \frac{Z_{0}^{(0)} \, \overline{Z}_{0}^{(1)}}{2} \, \omega^4 \int_{0}^{L} m(x) \, \psi^{(0)} \, \psi^{(1)} \, dx \quad (8-39) $$
|
||||
这是我们对最大势能的**改进表达式**。
|
||||
|
||||
* **第四步：求解频率 (Method R₀₁)**
|
||||
现在，我们将这个**改进的势能** (Eq. 8-39) 与**原始的动能** (Eq. 8-35) 相等：
|
||||
$$ \frac{Z_{0}^{(0)} \, \overline{Z}_{0}^{(1)}}{2} \, \omega^4 \int_{0}^{L} m(x) \, \psi^{(0)} \, \psi^{(1)} \, dx = \frac{(Z_{0}^{(0)})^2 \omega^2}{2} \int_{0}^{L} m(x) \, (\psi^{(0)})^2 \, dx $$
|
||||
两边消去 $\omega^2$ 和其他公共项，整理后得到：
|
||||
$$ \omega^2 = \frac{Z_{0}^{(0)}}{\overline{Z}_{0}^{(1)}} \, \frac{\int_{0}^{L} m(x) \, (\psi^{(0)})^2 \, dx}{\int_{0}^{L} m(x) \, \psi^{(0)} \, \psi^{(1)} \, dx} \quad (8-40) $$
|
||||
这就是 **Method R₀₁** 的公式。它避免了对 $\psi^{(0)}$ 求二阶导数，通常比 R₀₀ 更精确。
|
||||
|
||||
---
|
||||
|
||||
### 3. 第二次改进 (Method R₁₁)
|
||||
|
||||
这个方法更进一步：既然我们已经得到了一个更好的形状 $v^{(1)}$，为什么不用它来计算**动能**呢？
|
||||
|
||||
* **第一步：使用改进的势能**
|
||||
我们继续使用 Method R₀₁ 中得到的改进势能表达式 (Eq. 8-39)。
|
||||
|
||||
* **第二步：使用改进的形状计算动能**
|
||||
我们基于新的变形 $v^{(1)}$ 来计算最大动能。首先求速度：
|
||||
$$ v^{(1)}(x,t) = (\omega^2 \psi^{(1)} \overline{Z}_{0}^{(1)}) \sin(\omega t) $$
|
||||
$$ \dot{v}^{(1)}(x,t) = (\omega^2 \psi^{(1)} \overline{Z}_{0}^{(1)}) \omega \cos(\omega t) = \omega^3 \psi^{(1)} \overline{Z}_{0}^{(1)} \cos(\omega t) $$
|
||||
最大速度为 $\omega^3 \psi^{(1)} \overline{Z}_{0}^{(1)}$。
|
||||
因此，最大动能为：
|
||||
$$ T_{\mathrm{max}} = \frac{1}{2} \int_{0}^{L} m(x) (\dot{v}^{(1)})^2_{\mathrm{max}} dx = \frac{1}{2} \int_{0}^{L} m(x) \left( \omega^3 \psi^{(1)} \overline{Z}_{0}^{(1)} \right)^2 dx $$
|
||||
$$ T_{\mathrm{max}} = \frac{(\overline{Z}_{0}^{(1)})^2}{2} \, \omega^6 \int_{0}^{L} m(x) \, (\psi^{(1)})^2 \, dx \quad (8-41) $$
|
||||
这是我们对最大动能的**改进表达式**。
|
||||
|
||||
* **第三步：求解频率 (Method R₁₁)**
|
||||
现在，我们将**改进的势能** (Eq. 8-39) 与**改进的动能** (Eq. 8-41) 相等：
|
||||
$$ \frac{Z_{0}^{(0)} \, \overline{Z}_{0}^{(1)}}{2} \, \omega^4 \int_{0}^{L} m(x) \, \psi^{(0)} \, \psi^{(1)} \, dx = \frac{(\overline{Z}_{0}^{(1)})^2}{2} \, \omega^6 \int_{0}^{L} m(x) \, (\psi^{(1)})^2 \, dx $$
|
||||
两边消去 $\omega^4$ 和其他公共项，整理后得到：
|
||||
$$ \omega^2 = \frac{Z_{0}^{(0)}}{\overline{Z}_{0}^{(1)}} \, \frac{\int_{0}^{L} m(x) \, \psi^{(0)} \, \psi^{(1)} \, dx}{\int_{0}^{L} m(x) \, (\psi^{(1)})^2 \, dx} \quad (8-42) $$
|
||||
这就是 **Method R₁₁** 的公式。这个结果通常非常精确，因为它使用了更真实的形状 $\psi^{(1)}$ 来计算系统的两个核心属性（动能和势能）。
|
||||
|
||||
### 总结
|
||||
|
||||
| 方法 | 最大势能 ($V_{max}$) 基于 | 最大动能 ($T_{max}$) 基于 | 优点/特点 |
|
||||
| :--- | :--- | :--- | :--- |
|
||||
| **R₀₀** | 初始形状 $\psi^{(0)}$ 的曲率 | 初始形状 $\psi^{(0)}$ | 简单直接，但精度依赖于初始猜测 |
|
||||
| **R₀₁** | 惯性力 $p^{(0)}$ 在新变形 $v^{(1)}$ 上做的功 | 初始形状 $\psi^{(0)}$ | 避免了求二阶导数，精度更高 |
|
||||
| **R₁₁** | 惯性力 $p^{(0)}$ 在新变形 $v^{(1)}$ 上做的功 | 新形状 $\psi^{(1)}$ | 最精确，因为它在动能和势能计算中都使用了改进后的形状 |
|
||||
|
||||
这个迭代过程（从 $\psi^{(0)}$ 计算 $\psi^{(1)}$，再从 $\psi^{(1)}$ 计算 $\psi^{(2)}$...）如果持续下去，最终会收敛到结构的真实基频振型和精确频率。但在工程应用中，通常一次迭代（即 R₁₁）就足以获得非常高的精度。
|
||||
[Timestamp: 2025/10/07 17:37:56]
|
||||
@ -292,60 +292,62 @@ $$
|
||||
\xi=\frac{c}{2\,k/\omega}=\frac{1,125\,(27.9)}{200\times10^{3}}=15.7\%
|
||||
$$
|
||||
|
||||
## 3-3 RESONANT RESPONSE
|
||||
## 3-3 RESONANT RESPONSE共振响应
|
||||
|
||||
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
|
||||
$$
|
||||
D_{\beta=1}=\frac{1}{2\,\xi}
|
||||
D_{\beta=1}=\frac{1}{2\,\xi}\tag {3-31}
|
||||
$$
|
||||
|
||||
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 $\beta$。
|
||||
$$
|
||||
\beta_{\mathrm{peak}}=\sqrt{1-2\,\xi^{2}}
|
||||
\beta_{\mathrm{peak}}=\sqrt{1-2\,\xi^{2}}\tag {3-32}
|
||||
$$
|
||||
|
||||
(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
|
||||
|
||||
(对于阻尼比$\xi<1/\sqrt{2}$，这会产生正实值)，然后将该频率比值代回式 (3-24)，得到
|
||||
$$
|
||||
D_{\mathrm{max}}={\frac{1}{2\,\xi\,{\sqrt{1-\xi^{2}}}}}={\frac{1}{2\,\xi}}\ {\frac{\omega}{\omega_{D}}}
|
||||
D_{\mathrm{max}}={\frac{1}{2\,\xi\,{\sqrt{1-\xi^{2}}}}}={\frac{1}{2\,\xi}}\ {\frac{\omega}{\omega_{D}}}\tag {3-33}
|
||||
$$
|
||||
|
||||
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
|
||||
对于典型的结构阻尼值，例如 $\xi<0.10$ ，方程 (3-33) 与更简单的方程 (3-31) 之间的差异很小，当 $\xi=0.10$ 时差异为百分之零点五，当 $\xi=0.20$ 时差异为百分之二。
|
||||
|
||||
为了更全面地理解结构对简谐载荷的共振响应特性，有必要考虑通用响应方程 (3-19)，该方程包含瞬态项和稳态项。在共振激励频率 $\begin{array}{r}{\mathcal{\beta}=1\;\ }\end{array}$ ) 下，该方程变为
|
||||
$$
|
||||
v(t)=(A\ \cos\omega_{D}t+B\ \sin\omega_{D}t)\ \exp(-\xi\omega t)-{\frac{p_{o}}{k}}\ {\frac{\cos\omega t}{2\,\xi}}
|
||||
v(t)=(A\ \cos\omega_{D}t+B\ \sin\omega_{D}t)\ \exp(-\xi\omega t)-{\frac{p_{o}}{k}}\ {\frac{\cos\omega t}{2\,\xi}}\tag {3-34}
|
||||
$$
|
||||
|
||||
Assuming that the system starts from rest $[v(0)=\dot{v}(0)=0]$ , the constants are
|
||||
|
||||
假设系统从静止状态开始 $[v(0)=\dot{v}(0)=0]$，常数是
|
||||
$$
|
||||
A={\frac{p_{o}}{k}}\ {\frac{1}{2\,\xi}}\qquad\qquad B={\frac{p_{o}}{k}}\ {\frac{\omega}{2\,\omega_{D}}}={\frac{p_{o}}{k}}\ {\frac{1}{2\,\sqrt{1-\xi^{2}}}}
|
||||
A={\frac{p_{o}}{k}}\ {\frac{1}{2\,\xi}}\qquad\qquad B={\frac{p_{o}}{k}}\ {\frac{\omega}{2\,\omega_{D}}}={\frac{p_{o}}{k}}\ {\frac{1}{2\,\sqrt{1-\xi^{2}}}}\tag {3-35}
|
||||
$$
|
||||
|
||||
Thus Eq. (3-34) becomes
|
||||
|
||||
$$
|
||||
v(t)={\frac{1}{2\,\xi}}\;{\frac{p_{o}}{k}}\;\biggl[\biggl({\frac{\xi}{\sqrt{1-\xi^{2}}}}\,\sin\omega_{D}t+\cos\omega_{D}t\biggr)\;\exp(-\xi\omega t)-\cos\omega t\biggr]
|
||||
v(t)={\frac{1}{2\,\xi}}\;{\frac{p_{o}}{k}}\;\biggl[\biggl({\frac{\xi}{\sqrt{1-\xi^{2}}}}\,\sin\omega_{D}t+\cos\omega_{D}t\biggr)\;\exp(-\xi\omega t)-\cos\omega t\biggr]\tag {3-36}
|
||||
$$
|
||||
|
||||
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
|
||||
|
||||
对于结构系统中预期的阻尼量，项 $\sqrt{1-\xi^{2}}$ 几乎等于1；在这种情况下，该方程可以写成近似形式
|
||||
$$
|
||||
R(t)=\frac{v(t)}{p_{o}/k}\doteq\frac{1}{2\,\xi}\,\left\{\left[\exp(-\xi\omega t)-1\right]\;\cos\omega t+\xi\left[\exp(-\xi\omega t)\right]\,\sin\omega t\right\}
|
||||
R(t)=\frac{v(t)}{p_{o}/k}\doteq\frac{1}{2\,\xi}\,\left\{\left[\exp(-\xi\omega t)-1\right]\;\cos\omega t+\xi\left[\exp(-\xi\omega t)\right]\,\sin\omega t\right\}\tag {3-37}
|
||||
$$
|
||||
|
||||
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
|
||||
|
||||
零阻尼时，这个近似方程是不定式的；但当应用洛必达法则时，无阻尼系统的响应比可求得为
|
||||
$$
|
||||
R(t)\doteq\frac{1}{2}\,\left(\sin\omega t-\omega t\,\,\cos\omega t\right)
|
||||
R(t)\doteq\frac{1}{2}\,\left(\sin\omega t-\omega t\,\,\cos\omega t\right)\tag {3-38}
|
||||
$$
|
||||
|
||||
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.
|
||||
|
||||
这些方程的曲线如图3-7所示。注意到，由于包含 $\sin\omega t$ 的项对响应的贡献很小，图中无阻尼情况下的峰值线性增长，每个周期变化量为 $\pi$；然而，对于阻尼情况，它们按照 $(1/2\xi)[\exp(-\xi\omega t)-1]$ 增长。后者这个包络函数在图3-8中针对离散的阻尼值绘制了其随频率变化的曲线。可以看出，达到稳态水平 $1/2\xi$ 的增长速率随阻尼的增加而增加，并且对于实际感兴趣的阻尼值范围，在相对较少的周期内即可达到接近稳态的水平；例如，对于具有5%临界阻尼的情况，14个周期即可使响应非常接近稳态水平。
|
||||
|
||||
FIGURE 3-7 Response to resonant loading $\upbeta=1$ for at-rest initial conditions.
|
||||
|
||||
@ -353,23 +355,26 @@ FIGURE 3-7 Response to resonant loading $\upbeta=1$ for at-rest initial conditio
|
||||
|
||||
FIGURE 3-8 Rate of buildup of resonant response from rest.
|
||||
|
||||
# 3-4 ACCELEROMETERS AND DISPLACEMENT METERS
|
||||
## 3-4 ACCELEROMETERS AND DISPLACEMENT METERS 加速度计和位移计
|
||||
|
||||
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
|
||||
此时，宜于讨论一类重要的动态测量装置的工作所依据的基本原理。这些是地震仪，它们本质上由一个粘性阻尼振荡器组成，如图3-9所示。该系统安装在一个外壳中，该外壳可以连接到待测量运动的表面上。响应以质量相对于外壳的运动 $v(t)$ 来衡量。
|
||||
|
||||
该系统的运动方程已在方程(2-17)中给出，为
|
||||
$$
|
||||
m\,\ddot{v}(t)+c\,\dot{v}(t)+k\,v(t)=-m\,\ddot{v}_{g}(t)\equiv p_{\mathrm{eff}}(t)
|
||||
$$
|
||||
|
||||
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.,
|
||||
|
||||
其中 ${\ddot{v}}_{g}(t)$ 是壳体支撑的垂直加速度。考虑形式为 $\ddot{v}_{g}(t)=\ddot{v}_{g0}\,\sin\overline{{\omega}}t$ 的简谐支撑加速度，使得 $p_{\mathrm{eff}}(t)=-m\,\ddot{v}_{g0}\,\sin\overline{{\omega}}t,$ 运动 $v(t)$ 的动态稳态响应幅值由式 (3-22) 给出，即
|
||||
$$
|
||||
\rho=\frac{m\,\ddot{v}_{g0}}{k}\;D
|
||||
$$
|
||||
|
||||
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.
|
||||
其中，$D$由式(3-24)给出，并在图3-3中以图形方式呈现。检查此图可知，对于阻尼比$\xi\:=\:0.7$，在频率范围$0~<~\beta~<~0.6$内，$D$的值几乎恒定。因此，从式(3-39)可以清楚地看出，对于高达仪器固有频率约十分之六的施加频率($\omega\;=\;2\pi f\;=\;\sqrt{k/m}$)，该仪器指示的响应几乎与支撑加速度幅值成正比。因此，这种仪器在适当阻尼时，将有效地用作相对低频的加速度计；其适用范围将通过提高其相对于激励频率的固有频率而扩大，即，通过增加弹簧刚度或/和减小质量。加速度计的校准很容易通过以下方式进行：首先将仪器灵敏度轴垂直放置，然后将仪器倒置，并记录由此产生的响应变化，该变化对应于两倍重力加速度。
|
||||
|
||||
|
||||
FIGURE 3-9 Schematic diagram of a typical seismometer.
|
||||
|
||||
@ -0,0 +1,826 @@
|
||||
# Chap 8 GENERALIZEDSINGLE-DEGREE-OF-FREEDOMSYSTEMS
|
||||
|
||||
## 8-1 GENERAL COMMENTS ON SDOF SYSTEMS
|
||||
|
||||
In formulating the SDOF equations of motion and response analysis procedures in the preceding chapters, it has been tacitly assumed that the structure under consideration has a single lumped mass that is constrained so that it can move only in a single fixed direction. In this case it is obvious that the system has only a single degree of freedom and that the response may be expressed in terms of this single displacement quantity.
|
||||
|
||||
However, the analysis of most real systems requires the use of more complicated idealizations, even when they can be included in the generalized single-degree-offreedom category. In this chapter we will discuss these generalized SDOF systems, and in formulating their equations of motion it is convenient to divide them into two categories: (1) assemblages of rigid bodies in which elastic deformations are limited to localized weightless spring elements and (2) systems having distributed flexibility in which the deformations can be continuous throughout the structure, or within some of its components. In both categories, the structure is forced to behave like a SDOF system by the fact that displacements of only a single form or shape are permitted, and the assumed single degree of freedom expresses the amplitude of this permissible displacement configuration.
|
||||
|
||||
For structures in the category of rigid-body assemblages, discussed in Section 8-2, the limitation to a single displacement shape is a consequence of the assemblage configuration; i.e., the rigid bodies are constrained by supports and hinges arranged so that only one form of displacement is possible. The essential step in the analysis of such assemblages is the evaluation of the generalized elastic, damping, and inertial forces in terms of this single form of motion.
|
||||
|
||||
In the case of structures having distributed elasticity, considered in Section 8-3, the SDOF shape restriction is merely an assumption because the distributed elasticity actually permits an infinite variety of displacement patterns to occur. However, when the system motion is limited to a single form of deformation, it has only a single degree of freedom in a mathematical sense. Therefore, when the generalized mass, damping, and stiffness properties associated with this degree of freedom have been evaluated, the structure may be analyzed in exactly the same way as a true SDOF system.
|
||||
|
||||
From these comments it should be evident that the material on analysis of SDOF systems, presented in the preceding chapters, is equally applicable to generalized SDOF systems even though it was presented with reference to simple systems having only a single lumped mass.
|
||||
在前面几章中建立单自由度（SDOF）运动方程和响应分析程序时，默认假设所考虑的结构具有一个单一的集中质量，该质量被约束为只能沿单一固定方向运动。在这种情况下，显然该系统只有一个自由度，并且响应可以用这个单一的位移量来表示。
|
||||
|
||||
然而，大多数实际系统的分析需要使用更复杂的理想化模型，即使它们可以归入广义单自由度类别。在本章中，我们将讨论这些广义SDOF系统，在建立它们的运动方程时，方便将其分为两类：(1) 刚体组合体，其中弹性变形仅限于局部无质量弹簧单元；(2) 具有分布式柔性的系统，其中变形可以在整个结构或其某些部件中连续发生。在这两种类别中，结构被强制表现为SDOF系统，因为只允许单一形式或形状的位移，并且假定的单自由度表示这种允许位移构型的幅值。
|
||||
|
||||
对于第8-2节中讨论的刚体组合体类别的结构，限制为单一位移形状是组合体配置的结果；即，刚体通过支座和铰链进行约束，使其只可能发生一种形式的位移。分析此类组合体的关键步骤是根据这种单一运动形式评估广义弹性力、阻尼力和惯性力。
|
||||
|
||||
对于第8-3节中考虑的具有分布式弹性的结构，SDOF形状限制仅仅是一个假设，因为分布式弹性实际上允许发生无限多种位移模式。然而，当系统运动被限制为单一变形形式时，它在数学意义上只有一个自由度。因此，当与该自由度相关的广义质量、阻尼和刚度特性被评估后，结构可以与真正的SDOF系统完全相同的方式进行分析。
|
||||
|
||||
从这些评论中应该清楚地看出，前面几章中介绍的SDOF系统分析材料同样适用于广义SDOF系统，即使它是参照只有单一集中质量的简单系统进行介绍的。
|
||||
## 8-2 GENERALIZED PROPERTIES: ASSEMBLAGES OF RIGIDBODIES
|
||||
|
||||
In formulating the equations of motion of a rigid-body assemblage, the elastic forces developed during the SDOF displacements can be expressed easily in terms of the displacement amplitude because each elastic element is a discrete spring subjected to a specified deformation. Similarly the damping forces can be expressed in terms of the specified velocities of the attachment points of the discrete dampers. On the other hand, the mass of the rigid bodies need not be localized, and distributed inertial forces generally will result from the assumed accelerations. However, for the purposes of dynamic analysis, it usually is most effective to treat the rigid-body inertial forces as though the mass and the mass moment of inertia were concentrated at the center of mass. The inertial-force resultants which are obtained thereby are entirely equivalent to the distributed inertial forces insofar as the assemblage behavior is concerned. Similarly it is desirable to represent any distributed external loads acting on the rigid bodies by their force resultants. The total mass $m$ and the centroidal mass moment of inertia $j$ of a uniform rod and of uniform plates of various shapes are summarized in Fig. 8-1 for convenient reference.
|
||||
|
||||
在建立刚体组合体的运动方程时，单自由度位移过程中产生的弹性力可以很容易地用位移幅值表示，因为每个弹性元件都是一个承受指定变形的离散弹簧。类似地，阻尼力可以用离散阻尼器附着点的指定速度表示。另一方面，刚体的质量无需集中，并且假定加速度通常会产生分布式惯性力。然而，为了进行动力学分析，通常最有效的方法是将刚体惯性力视为质量和质量惯性矩集中在质心。由此获得的惯性力合力，就组合体行为而言，完全等效于分布式惯性力。类似地，宜于用其力合力来表示作用在刚体上的任何分布式外部载荷。均质杆和各种形状均质板的总质量 $m$ 和质心质量惯性矩 $j$ 总结在图8-1中，以供方便查阅。
|
||||
|
||||
FIGURE 8-1 Rigid-body mass and centroidal mass moment of inertia for uniform rod and uniform plates of unit thickness.
|
||||
|
||||
Example E8-1. A representative example of a rigid-body assemblage, shown in Fig. E8-1, consists of two rigid bars connected by a hinge at $E$ and supported by a pivot at $A$ and a roller at $H$ . Dynamic excitation is provided by a transverse load $p({\boldsymbol{x}},t)$ varying linearly along the length of bar $A B$ . In addition, a constant axial force $N$ acts through the system, and the motion is constrained by discrete springs and dampers located as shown along the lengths of the bars. The mass is distributed uniformly through bar $A B$ , and the weightless bar $B C$ supports a lumped mass $m_{2}$ having a centroidal mass moment of inertia $j_{2}$ .
|
||||
示例 E8-1。如图 E8-1 所示，一个刚体组合的典型示例由两根刚性杆组成，它们在 $E$ 处通过铰链连接，并在 $A$ 处由枢轴支撑，在 $H$ 处由滚子支撑。动态激励由沿杆 $A B$ 长度线性变化的横向载荷 $p({\boldsymbol{x}},t)$ 提供。此外，一个恒定的轴向力 $N$ 作用于该系统，并且运动受到沿杆长度分布的离散弹簧和阻尼器的约束，如图所示。质量均匀分布在杆 $A B$ 上，并且无质量杆 $B C$ 支撑一个具有质心质量惯性矩 $j_{2}$ 的集中质量 $m_{2}$。
|
||||
|
||||
FIGURE E8-1 Example of a rigid-body-assemblage SDOF system.
|
||||
|
||||
Because the two bars are assumed rigid, this system has only a single degree of freedom, and its dynamic response can be expressed with a single equation of motion. This equation could be formulated by direct equilibration (the reader may find this a worthwhile exercise), but because of the complexity of the system, it is more convenient to use a work or energy formulation. A virtualwork analysis will be employed here; although using Hamilton’s principle, as described in Chapter 16, would be equally effective.
|
||||
|
||||
For the form of displacement which may take place in this SDOF structure (Fig. E8-2), the hinge motion $Z(t)$ may be taken as the basic quantity and all other displacements expressed in terms of it; for example, $B B^{\prime}(t)=Z(t)/4$ , ${D D^{\prime}(t)=3\,Z(t)/4}$ , $F F^{\prime}(t)=2\,Z(t)/3$ , etc. The force components acting on the system (exclusive of the axial applied force $N_{\ast}$ which will be discussed later) are also shown in this figure. Each resisting force component can be expressed in terms of $Z(t)$ or its time derivatives, as follows:
|
||||
|
||||
$$
|
||||
\begin{array}{r l}&{f_{\eta}(t)=m\,{\frac{1}{2}}\,{\tilde{Z}}(t)=\pi\,{\frac{\Gamma}{4}}\,{\frac{\dot{\Gamma}}{2}}(t)=2\pi\,{\tilde{Z}}(t)}\\ &{u_{h}(t)=j_{\mathrm{~h~}}{\frac{1}{4}}\,{\tilde{Z}}(t)={\frac{\pi\,H}{4}}\,{\frac{\dot{\Gamma}}{12}}\,{\tilde{Z}}(t)={\frac{4}{3}}\,a^{2}\,m\,{\tilde{Z}}(t)}\\ &{f_{\eta}(t)=m_{2}^{2}\,{\frac{2}{3}}\,(t)}\\ &{u_{h}(t)=j_{\mathrm{~h~}}{\frac{1}{2}}\,\tilde{Z}(t)}\\ &{f_{D_{t}}(t)=\alpha_{\mathrm{~L~}}\Big[{\frac{d}{d t}}\,D^{D}(t)\Big]=\alpha_{\mathrm{~L~}}{\frac{1}{4}}\,\tilde{Z}(t)}\\ &{f_{D_{t}}(t)=\alpha_{\mathrm{~L~}}{\frac{\dot{\Gamma}}{2}}\,(t)}\\ &{f_{S_{t}}(t)=k_{\mathrm{~L~}}\Big[D D^{D}(t)\Big]=k_{\mathrm{~L~}}{\frac{3}{4}}\,Z(t)}\\ &{f_{S_{t}}(t)=k_{\mathrm{~L~}}\Big[G C(t)\Big]=k_{\mathrm{~L~}}{\frac{1}{2}}\,\tilde{Z}(t)}\end{array}
|
||||
$$
|
||||
|
||||
The externally applied lateral load resultant is
|
||||
|
||||
$$
|
||||
p_{1}(t)=8\;\overline{{p}}\;a\;f(t)
|
||||
$$
|
||||
|
||||
In these expressions, $\overline{{m}}$ and $\overline{{p}}$ denote reference values of mass and force, respectively, per unit length and $f(t)$ is a dimensionless time-dependent function which represents the dynamic load variation.
|
||||
|
||||
FIGURE E8-2 SDOF displacements and resultant forces.
|
||||
|
||||
The equation of motion of this system may be established by equating to zero all work done by these force components during an arbitrary virtual displacement $\delta Z$ . The virtual displacements through which the force components move are proportional to $Z(t)$ , as indicated in Fig. E8-2. Thus the total virtual work may be written
|
||||
|
||||
$$
|
||||
\begin{array}{c}{{\delta W(t)=-2a\,\overline{{{\cal{D}}}}\,\ddot{Z}(t)\,\displaystyle\frac{\delta Z}{2}-\frac{4}{3}\,a^{2}\,\overline{{{m}}}\,\ddot{Z}(t)\,\displaystyle\frac{\delta Z}{4a}-m_{2}\,\displaystyle\frac{2\ddot{Z}(t)}{3}\,\frac{2}{3}\,\delta Z}}\\ {{-j_{2}\,\displaystyle\frac{\ddot{Z}(t)}{3a}\,\delta Z-c_{1}\,\displaystyle\frac{\dot{Z}(t)}{4}\,\displaystyle\frac{\delta Z}{4}-c_{2}\,\dot{Z}(t)\,\delta Z-k_{1}\,\displaystyle\frac{3}{4}\,Z(t)\,\displaystyle\frac{3}{4}\,\delta Z}}\\ {{-\,k_{2}\,\displaystyle\frac{Z(t)}{3}\,\displaystyle\frac{\delta Z}{3}+8\overline{{{p}}}\,a\,f(t)\,\displaystyle\frac{2}{3}\,\delta Z=0\quad}}\end{array}
|
||||
$$
|
||||
|
||||
which when simplified becomes
|
||||
|
||||
$$
|
||||
\begin{array}{l}{{\displaystyle\left[\left(a\,\overline{{{m}}}+\frac{a\,\overline{{{m}}}}{3}+\frac{4}{9}\,m_{2}+\frac{j_{2}}{9a^{2}}\right)\ddot{Z}(t)+\left(\frac{c_{1}}{16}+c_{2}\right)\dot{Z}(t)\right.\hfill}}\\ {{\displaystyle\left.\qquad\qquad\qquad+\left(\frac{9}{16}\,k_{1}+\frac{k_{2}}{9}\right)Z(t)-\frac{16}{3}\,\overline{{{p}}}\,a\,f(t)\right]\,\delta Z=0}}\end{array}
|
||||
$$
|
||||
|
||||
Because the virtual displacement $\delta Z$ is arbitrary, the term in square brackets must vanish; thus the final equation of motion becomes
|
||||
|
||||
$$
|
||||
\begin{array}{l}{{\displaystyle\left(\frac{4}{3}\,\overline{{{m}}}\,a+\frac{4}{9}\,m_{2}+\frac{j_{2}}{9a^{2}}\right)\ddot{Z}(t)+\left(\frac{c_{1}}{16}+c_{2}\right)\dot{Z}(t)}}\\ {{\displaystyle\qquad\qquad\qquad\qquad+\,\left(\frac{9}{16}\,k_{1}+\frac{k_{2}}{9}\right)Z(t)=\frac{16}{3}\,\overline{{{p}}}\,a\,f(t)}}\end{array}
|
||||
$$
|
||||
|
||||
This may be written in the simplified form
|
||||
|
||||
$$
|
||||
m^{*}\,\ddot{Z}(t)+c^{*}\,\dot{Z}(t)+k^{*}\,Z(t)=p^{*}(t)
|
||||
$$
|
||||
|
||||
if the new symbols are defined as follows:
|
||||
|
||||
$$
|
||||
\begin{array}{l l}{{m^{*}={\displaystyle{\frac{4}{3}}\,{\overline{{{m}}}}\,a+{\frac{4}{9}}\,m_{2}+{\frac{j_{2}}{9a^{2}}}}}}&{{\qquad\qquad c^{*}={\displaystyle{\frac{1}{16}}}\,c_{1}+c_{2}}}\\ {{\qquad}}&{{\qquad}}&{{\qquad}}\\ {{k^{*}={\displaystyle{\frac{9}{16}}\,k_{1}+{\frac{1}{9}}\,k_{2}}}}&{{\qquad\qquad p^{*}(t)={\displaystyle{\frac{16}{3}}}\,{\overline{{{p}}}}\,a\,f(t)}}\end{array}
|
||||
$$
|
||||
|
||||
These quantities are termed, respectively, the generalized mass, generalized damping, generalized stiffness, and generalized load for this system; they have been evaluated with reference to the generalized coordinate $Z(t)$ , which has been used here to define the displacements of the system.
|
||||
|
||||
Consider now the externally applied axial force $N$ of Fig. E8-1. As may be seen in Fig. E8-3, the virtual work done by this force during the virtual displacement $\delta Z$ is $N\!\delta e$ . The displacement $\delta e$ is made up of two parts, $\delta e_{1}$ and $\delta e_{2}$ , associated with the rotations of the two bars. Considering the influence of bar $A E$ only, it is clear from similar triangles (assuming small deflections) that $\delta e_{1}=\left(Z(t)/4a\right)\delta Z$ . Similarly $\delta e_{2}=\left(Z(t)/3a\right)\delta Z$ , thus the total displacement is
|
||||
|
||||
$$
|
||||
\delta e=\delta e_{1}+\delta e_{2}=\frac{7}{12}\,\frac{Z(t)}{a}\,\delta Z
|
||||
$$
|
||||
|
||||
and the virtual work done by the axial force $N$ is
|
||||
|
||||
$$
|
||||
\delta W_{P}={\frac{7}{12}}\,{\frac{N Z(t)}{a}}\,\delta Z
|
||||
$$
|
||||
|
||||
Adding Eq. (d) and Eq. (a) and carrying out simplifying operations similar to those which led to Eq. (c) shows that only one term in the equation of motion
|
||||
|
||||
|
||||
FIGURE E8-3 Displacement components in the direction of axial force.
|
||||
|
||||
is influenced by the axial force, the generalized stiffness. When the effect of the axial force in this system is included, the combined generalized stiffness k∗is
|
||||
|
||||
$$
|
||||
\overline{{k^{*}}}=k^{*}-\frac{7}{12}\frac{P}{a}=\frac{9}{16}k_{1}+\frac{1}{9}k_{2}-\frac{7}{12}\frac{N}{a}
|
||||
$$
|
||||
|
||||
With this modified generalized-stiffness term, the equation of motion of the complete system of Fig. E8-1, including axial force, is given by an equation similar to Eq. (8-1). The last term in Eq. (e), which is directly proportional to the axial force $N_{\ast}$ often is given the name “geometric stiffness.”
|
||||
|
||||
It is of interest to note that the condition of zero generalized stiffness represents a neutral stability or critical buckling condition in the system. The value of axial force $N_{\mathrm{cr}}$ which would cause buckling of this structure can be found by equating $\overline{{k^{*}}}$ of Eq. (e) to zero:
|
||||
|
||||
$$
|
||||
0=\frac{9}{16}k_{1}+\frac{1}{9}k_{2}-\frac{7}{12}\frac{N_{\mathrm{cr}}}{a}
|
||||
$$
|
||||
|
||||
Thus
|
||||
|
||||
$$
|
||||
N_{\mathrm{cr}}=\Bigl(\frac{27}{28}k_{1}+\frac{4}{21}k_{2}\Bigr)a
|
||||
$$
|
||||
|
||||
In general, compressive axial forces tend to reduce the stiffness of a structural system, while tensile axial forces cause a corresponding increase of stiffness. Such loads can have a significant effect on the response of the structure to dynamic loads, and the resulting change of stiffness should always be evaluated to determine its importance in the given problem. It should be noted that axial force in this and in subsequent discussions refers to a force which acts parallel to the initial undistorted axis of the member; such a force is assumed not to change the direction of its line of action or its magnitude with the motion of the structure.
|
||||
|
||||
Example E8-2. As a second example of the formulation of the equations of motion for a rigid-body assemblage, the system shown in Fig. E8-4 will be considered. The small-amplitude motion of this system can be characterized by the downward displacement of the load point $Z(t)$ , and all the system forces resisting this motion can be expressed in terms of it:
|
||||
|
||||
|
||||
FIGURE E8-4 SDOF plate with dynamic forces.
|
||||
|
||||
$$
|
||||
\begin{array}{l l}{{f_{S}(t)=k\,\displaystyle\frac{b}{a}\,Z(t)\qquad\qquad}}&{{f_{I_{1}}(t)=\gamma\,a\,b\,\displaystyle\frac{1}{2}\,\ddot{Z}(t)}}\\ {{\qquad}}&{{\qquad}}\\ {{f_{I_{2}}(t)=\gamma\,a\,b\,\displaystyle\frac{b}{2a}\,\ddot{Z}(t)\qquad\qquad}}&{{M_{I}(t)=\gamma\,a\,b\,\displaystyle\frac{a^{2}+b^{2}}{12}\,\displaystyle\frac{1}{a}\,\ddot{Z}(t)}}\end{array}
|
||||
$$
|
||||
|
||||
The equation of motion for this simple system can be written directly by expressing the equilibrium of moments about the plate hinge:
|
||||
|
||||
$$
|
||||
f_{S}(t)\,b+f_{I_{1}}(t)\,\frac{a}{2}+f_{I_{2}}(t)\,\frac{b}{2}+M_{I}(t)=p(t)\,a
|
||||
$$
|
||||
|
||||
Dividing by the length $a$ and substituting the above expressions for the forces, this equation becomes
|
||||
|
||||
$$
|
||||
\gamma\,a\,b\left[\frac{1}{12}\left(\frac{b^{2}}{a^{2}}+1\right)+\frac{1}{4}+\frac{b^{2}}{4a^{2}}\right]\ddot{Z}(t)+k\,\frac{b^{2}}{a^{2}}\,Z(t)=p(t)
|
||||
$$
|
||||
|
||||
Finally, it may be written
|
||||
|
||||
$$
|
||||
m^{*}\;\ddot{Z}(t)+k^{*}\;Z(t)=p^{*}(t)
|
||||
$$
|
||||
|
||||
in which
|
||||
|
||||
$$
|
||||
m^{*}=\frac{\gamma\,a\,b}{3}\left(1+\frac{b^{2}}{a^{2}}\right)\qquad k^{*}=k\,\frac{b^{2}}{a^{2}}\qquad p^{*}(t)=p(t)
|
||||
$$
|
||||
|
||||
## 8-3 GENERALIZED PROPERTIES: DISTRIBUTED FLEXIBILITY 广义特性：分布式柔性
|
||||
|
||||
The example of Fig. E8-1 is a true SDOF system in spite of the complex interrelationships of its various components because the two rigid bars are supported so that only one type of displacement pattern is possible. If the bars could deform in flexure, the system would have an infinite number of degrees of freedom. A simple SDOF analysis could still be made, however, if it were assumed that only a single flexural deflection pattern could be developed.
|
||||
|
||||
As an illustration of this method of approximating SDOF behavior in a flexure system actually having infinite degrees of freedom, consider the formulation of an equation of motion for the cantilever tower of Fig. $8{-}2a$ . The essential properties of the tower (excluding damping) are its flexural stiffness $E I(x)$ and its mass per unit of length $m(x)$ . It is assumed to be subjected to horizontal earthquake ground-motion excitation $v_{g}(t)$ , and it supports a constant vertical load $N$ applied at the top.
|
||||
|
||||
To approximate the motion of this system with a single degree of freedom, it is necessary to assume that it will deform only in a single shape. The shape function will be designated $\psi(x)$ , and the amplitude of the motion relative to the moving base will be represented by the generalized coordinate $Z(t)$ ; thus,
|
||||
图 E8-1 的例子是一个真正的单自由度（SDOF）系统，尽管其各个组件之间存在复杂的相互关系，因为两根刚性杆的支撑方式使得只有一种位移模式是可能的。如果杆件能够发生弯曲变形，系统将具有无限个自由度。然而，如果假设只能形成单一的弯曲变形模式，仍然可以进行简单的单自由度（SDOF）分析。
|
||||
|
||||
为了说明这种在实际上具有无限自由度的弯曲系统中近似单自由度（SDOF）行为的方法，考虑图 8-2a 所示悬臂塔的运动方程的建立。该塔的基本特性（不包括阻尼）是其抗弯刚度 $E I(x)$ 和单位长度质量 $m(x)$。假设其受到水平地震地面运动激励 $v_{g}(t)$，并且顶部承受一个恒定的竖向载荷 $N$。
|
||||
|
||||
为了用单自由度近似该系统的运动，必须假设它只以单一形状变形。形状函数将指定为 $\psi(x)$，相对于移动基座的运动幅值将由广义坐标 $Z(t)$ 表示；因此，
|
||||
$$
|
||||
v(x,t)=\psi(x)Z(t)\tag {8-2}
|
||||
$$
|
||||
|
||||
Typically the generalized coordinate is selected as the displacement of some convenient reference point in the system, such as the tip displacement in this tower. Then the shape function is the dimensionless ratio of the local displacement to this reference displacement:
|
||||
通常，广义坐标被选为系统中某个方便的参考点的位移，例如这座塔的塔尖位移。那么，**形函数是局部位移与该参考位移的无量纲比**：
|
||||
$$
|
||||
\psi(x)=\frac{v(x,t)}{Z(t)}\tag {8-3}
|
||||
$$
|
||||
|
||||
The equation of motion of this generalized SDOF system can be formulated conveniently only by work or energy principles, and the principle of virtual work will be used in this case.
|
||||
这个广义单自由度（SDOF）系统的运动方程只能通过功或能量原理方便地建立，在这种情况下，将使用虚功原理。
|
||||
|
||||
(a)
|
||||
FIGURE 8-2 Flexure structure treated as a SDOF system.挠曲结构被视为单自由度系统。
|
||||
|
||||
Since the structure in this example is flexible in flexure, internal virtual work $\delta W_{I}$ is performed by the real internal moments $M(x,t)$ acting through their corresponding virtual changes in curvature $\delta\left[{\frac{\partial^{2}v(x)}{\partial x^{2}}}\right]$ . The virtual-work principle requires that the external virtual work, $\delta W_{E}(t)$ , performed by the external loadings acting through their corresponding virtual displacements be equated to the internal virtual work, i.e.,
|
||||
由于本例中的结构具有弯曲柔性，内虚功 $\delta W_{I}$ 由实际内力矩 $M(x,t)$ 通过其相应的曲率虚变化 $\delta\left[{\frac{\partial^{2}v(x)}{\partial x^{2}}}\right]$ 产生。虚功原理要求，由外部载荷通过其相应的虚位移所做的外虚功 $\delta W_{E}(t)$ 等于内虚功，即
|
||||
$$
|
||||
\delta W_{E}=\delta W_{I}\tag {8-4}
|
||||
$$
|
||||
|
||||
To develop the equation of motion in terms of relative displacement $\boldsymbol{v}(\boldsymbol{x},t)$ , the base of the structure can be treated as fixed while an effective loading $p_{\mathrm{eff}}(x,t)$ is applied as shown in Fig. $8{-}2b$ . The inertial loading is then given by
|
||||
为了推导以相对位移 $\boldsymbol{v}(\boldsymbol{x},t)$ 表示的运动方程，结构基础可以视为固定，同时施加一个有效载荷 $p_{\mathrm{eff}}(x,t)$，如图 $8{-}2b$ 所示。惯性载荷则由下式给出。
|
||||
$$
|
||||
f_{I}(x,t)=m(x)\;\ddot{v}(x,t)\tag {8-5}
|
||||
$$
|
||||
|
||||
Using the full set of external forces, the external virtual work is given by
|
||||
使用全部外力，外部虚功由下式给出
|
||||
$$
|
||||
\delta W_{E}=-\int_{0}^{L}f_{I}(x)\,\delta v(x)\,d x+\int_{0}^{L}p_{\mathrm{eff}}(x,t)\,\delta v(x)\,d x+N\delta e\tag {8-6}
|
||||
$$
|
||||
|
||||
and consistent with the above statement regarding internal virtual work,
|
||||
和与上述关于内部虚功的声明一致，
|
||||
$$
|
||||
\delta W_{I}(t)=\int_{0}^{L}M(x,t)\ \delta v^{\prime\prime}(x)\ d x\tag {8-7}
|
||||
$$
|
||||
|
||||
where $v^{\prime\prime}(x)=\partial^{2}v(x)/\partial x^{2}$ .
|
||||
|
||||
If it is assumed that damping stresses are developed in proportion to the strain velocity, a uniaxial stress-strain relation of the form
|
||||
如果假设阻尼应力与应变速率成正比产生，则单轴应力-应变关系的形式为
|
||||
$$
|
||||
\sigma=E\left[\epsilon+a_{1}\,\dot{\epsilon}\right]\tag {8-8}
|
||||
$$
|
||||
|
||||
may be adopted, where $E$ is Young’s modulus and $a_{1}$ is a damping constant. Then the Euler-Bernouli hypothesis that plane sections remain plane leads to the relation
|
||||
可以采用，其中 $E$ 是杨氏模量，$a_{1}$ 是阻尼常数。接着，平面截面保持平面的欧拉-伯努利假设导致以下关系。
|
||||
$$
|
||||
M(x,t)=E I(x)\;[v^{\prime\prime}(x,t)+a_{1}\:\dot{v}^{\prime\prime}(x,t)]\tag {8-9}
|
||||
$$
|
||||
|
||||
Using this equation, the basic relations may be expressed as follows:
|
||||
利用此方程，基本关系可表示如下：
|
||||
$$
|
||||
\begin{array}{l c}{{v(x,t)=\psi(x)~Z(t)~~~~}}&{{~~~~\dot{v}^{\prime\prime}(x,t)=\psi^{\prime\prime}(x)~\dot{Z}(t)}}\\ {{v^{\prime}(x,t)=\psi^{\prime}(x)~Z(t)~~~~}}&{{~~~~\delta v(x,t)=\psi(x)~\delta Z}}\\ {{v^{\prime\prime}(x,t)=\psi^{\prime\prime}(x)~Z(t)~~~~}}&{{~~~~\delta v^{\prime}(x,t)=\psi^{\prime}(x)~\delta Z}}\\ {{\ddot{v}(x,t)=\psi(x)~\ddot{Z}(t)}}&{{~~~~~\delta v^{\prime\prime}(x,t)=\psi^{\prime\prime}(x)~\delta Z}}\end{array}\tag {8-10}
|
||||
$$
|
||||
|
||||
Also, by analogy with the development of Eq. (d) of Example E8-1, the expressions for axial displacement take the form
|
||||
此外，类似于算例 E8-1 中式 (d) 的推导，轴向位移的表达式具有以下形式
|
||||
$$
|
||||
e(t)={\frac{1}{2}}\int_{0}^{L}[v^{\prime}(x,t)]^{2}\;d x\qquad\delta e=\int_{0}^{L}v^{\prime}(x,t)\;\delta v^{\prime}(x)\;d x\tag {8-11}
|
||||
$$
|
||||
|
||||
Finally, expressions for the external and internal virtual work may be formulated by using Eqs. (8-10) and (8-11):
|
||||
最后，外部虚功和内部虚功的表达式可以利用式 (8-10) 和 (8-11) 建立：
|
||||
$$
|
||||
\begin{array}{l}{\displaystyle\delta W_{E}=\left[-\ddot{Z}(t)\int_{0}^{L}m(x)\,\psi(x)^{2}\,d x\right.}\\ {\displaystyle\qquad-\left.\ddot{v}_{g}(t)\int_{0}^{L}m(x)\,\psi(x)\,d x+N Z(t)\int_{0}^{L}\psi^{\prime}(x)^{2}\,d x\right]\,\delta Z}\\ {\displaystyle\delta W_{I}=\left[Z(t)\int_{0}^{L}E I(x)\,\psi^{\prime\prime}(x)^{2}\,d x+a_{1}\,\dot{Z}(t)\int_{0}^{L}E I(x)\,\psi^{\prime\prime}(x)^{2}\,d x\right]\,\delta Z}\end{array}\tag {8-12}
|
||||
$$
|
||||
|
||||
Equating Eqs. (8-12) in accordance with Eq. (8-4) yields the generalized equation of motion
|
||||
根据方程 (8-4)，将方程 (8-12) 等效，得到广义运动方程。
|
||||
$$
|
||||
m^{*}\,\ddot{Z}(t)+c^{*}\,\dot{Z}(t)+k^{*}\,Z(t)-k_{G}^{*}\,Z(t)=p_{\mathrm{eff}}^{*}(t)\tag {8-13}
|
||||
$$
|
||||
|
||||
where
|
||||
|
||||
$$
|
||||
\begin{array}{l}{{{\displaystyle m^{*}=\int_{0}^{L}m(x)~\psi(x)^{2}~d x=\mathrm{generalized~mass}}}}\\ {{{\displaystyle c^{*}=a_{1}\int_{0}^{L}E I(x)~\psi^{\prime\prime}(x)^{2}~d x=\mathrm{generalized~damping}}}}\\ {{{\displaystyle k^{*}=\int_{0}^{L}E I(x)~\psi^{\prime\prime}(x)^{2}~d x=\mathrm{generalized~flexural~sitfiness}}}}\\ {{{\displaystyle k_{G}^{*}=N\int_{0}^{L}\psi^{\prime}(x)^{2}~d x=\mathrm{generalized~geometric~stiffeess}}}}\\ {{{\displaystyle p_{\mathrm{eff}}^{*}(t)=-\ddot{v}_{g}(t)\int_{0}^{L}m(x)~\psi(x)~d x=\mathrm{generalized~effectiveluad}}}}\end{array}\tag {8-14}
|
||||
$$
|
||||
|
||||
Combining the two stiffness terms, Eq. (8-13) can be written as
|
||||
结合这两个刚度项，式 (8-13) 可写为
|
||||
$$
|
||||
m^{*}\;\ddot{Z}(t)+c^{*}\;\dot{Z}(t)+\overline{{k^{*}}}\;Z(t)=p_{\mathrm{eff}}^{*}(t)\tag {8-15}
|
||||
$$
|
||||
|
||||
in which
|
||||
|
||||
$$
|
||||
\overline{{k^{*}}}=k^{*}-k_{G}^{*}\tag {8-16}
|
||||
$$
|
||||
|
||||
is the combined generalized stiffness.
|
||||
|
||||
The critical buckling load can be calculated for this system by the same method used in Example E8-1, i.e., by equating to zero the combined generalized stiffness and solving for $N_{\mathrm{cr}}$ ; thus, one obtains
|
||||
是组合广义刚度。
|
||||
|
||||
该系统的临界屈曲载荷可以通过与例E8-1中相同的方法计算，即通过将组合广义刚度设为零并求解 $N_{\mathrm{cr}}$；由此，得到
|
||||
$$
|
||||
N_{\mathrm{cr}}={\frac{\int_{0}^{L}E I(x)\;\psi^{\prime\prime}(x)^{2}\;d x}{\int_{0}^{L}\psi^{\prime}(x)^{2}\;d x}}\tag {8-17}
|
||||
$$
|
||||
|
||||
This SDOF approximate analysis of the critical buckling load is called Rayleigh’s method, which is discussed in the context of vibration analysis in Section 8-5. The value determined for the critical load depends, of course, upon the assumed shape function $\psi(x)$ , but a very good approximation will be given by any shape that is consistent with the geometric boundary conditions.
|
||||
这种对临界屈曲载荷的单自由度近似分析称为瑞利法，该方法在第8-5节的振动分析背景下进行了讨论。当然，确定的临界载荷值取决于假设的形状函数 $\psi(x)$，但任何与几何边界条件一致的形状都能给出非常好的近似。
|
||||
|
||||
|
||||
Example E8-3. To provide a numerical example of the formulation of the equation of motion for a SDOF system with distributed flexibility, it will be assumed that the tower of Fig. 8-2 has constant flexural stiffness $E I$ and constant mass distribution $\overline{{m}}$ along its length and damping in accordance with Eq. (8-8). Also, its deflected shape in free vibrations will be assumed as
|
||||
例 E8-3。为了提供一个具有分布式柔度的单自由度（SDOF）系统运动方程建立的数值示例，将假设图 8-2 中的塔架沿其长度方向具有恒定的抗弯刚度 $E I$ 和恒定的质量分布 $\overline{{m}}$，并根据式 (8-8) 具有阻尼。此外，其在自由振动中的变形形状将被假定为
|
||||
$$
|
||||
\psi(x)=1-\cos{\frac{\pi x}{2L}}\tag {a}
|
||||
$$
|
||||
|
||||
which satisfies the geometric boundary conditions $\psi(0)=\psi^{\prime}(0)=0$ . When Eqs. (8-14) are applied, one obtains
|
||||
其满足几何边界条件 $\psi(0)=\psi^{\prime}(0)=0$ 。当应用式(8-14)时，可得到
|
||||
$$
|
||||
\begin{array}{l}{{m^{*}=\displaystyle\frac{\pi}{m}\int_{0}^{L}{\left(1-\cos{\frac{\pi x}{2L}}\right)^{2}d x}=0.228\,\overline{{m}}\,L}}\\ {{\displaystyle c^{*}=a_{1}E I\int_{0}^{L}{\left(\frac{\pi^{2}}{4L^{2}}\,\cos{\frac{\pi x}{2L}}\right)^{2}d x}=\frac{a_{1}\pi^{4}E I}{32L^{3}}}}\\ {{\displaystyle k^{*}=E I\int_{0}^{L}{\left(\frac{\pi^{2}}{4L^{2}}\,\cos{\frac{\pi x}{2L}}\right)^{2}d x}=\frac{\pi^{4}E I}{32L^{3}}}}\\ {{\displaystyle k_{G}^{*}=N\int_{0}^{L}{\left(\frac{\pi}{2L}\,\sin{\frac{\pi x}{2L}}\right)^{2}d x}=\frac{N\pi^{2}}{8L}}}\\ {{\displaystyle p_{\mathrm{eff}}(t)=-\overline{{m}}\,\bar{v}_{g}(t)\int_{0}^{L}{\left(1-\cos{\frac{\pi x}{2L}}\right)d x}=0.364\,\overline{{m}}\,L\,\bar{v}_{g}(t)}}\end{array}\tag {b}
|
||||
$$
|
||||
|
||||
which upon substitution into Eq. (8-13) gives the SDOF equation of motion:
|
||||
将其代入方程 (8-13) 后，得到 SDOF 运动方程：
|
||||
$$
|
||||
\begin{array}{r l r}&{}&{\left(0.228\overline{{m}}L\right)\,\ddot{Z}(t)+\left(\frac{a_{1}\,\pi^{4}\,E I}{32\,L^{3}}\right)\,\dot{Z}(t)+\left(\frac{\pi^{4}\,E I}{32\,L^{3}}-\frac{N\pi^{2}}{8L}\right)\,Z(t)}\\ &{}&{=-0.364\,\overline{{m}}\:L\,\ddot{v}_{g}(t)}\end{array}\tag {c}
|
||||
$$
|
||||
|
||||
In addition, the buckling load for this column subjected to tip load will be evaluated by setting the combined stiffness equal to zero and solving for $N_{\mathrm{cr}}$ , with the following result:
|
||||
此外，对于承受尖端载荷的该柱，将通过将组合刚度设为零并求解 $N_{\mathrm{cr}}$ 来评估其屈曲载荷，得到以下结果：
|
||||
$$
|
||||
N_{\mathrm{cr}}=\frac{\pi^{2}}{4}~\frac{E I}{L^{2}}\tag {d}
|
||||
$$
|
||||
|
||||
This is the true buckling load for an end-loaded uniform cantilever column because the assumed shape function of Eq. (a) is the true buckled shape.
|
||||
|
||||
Of course, one could select a different shape function $\psi(x)$ as long as it satisfies the geometric boundary conditions $\psi(0)=\psi^{\prime}(0)=0$ . For example, if this function were assumed to be of the parabolic form
|
||||
这是端部加载的均匀悬臂柱的真实屈曲载荷，因为式(a)中假定的形状函数是真实的屈曲形状。
|
||||
|
||||
当然，只要它满足几何边界条件 $\psi(0)=\psi^{\prime}(0)=0$，就可以选择一个不同的形状函数 $\psi(x)$。例如，如果假设这个函数是抛物线形式的
|
||||
$$
|
||||
\psi(x)={\frac{x^{2}}{L^{2}}}\tag {e}
|
||||
$$
|
||||
|
||||
the equation of motion obtained by the above procedure would be
|
||||
由上述过程得到的运动方程将是
|
||||
$$
|
||||
\left(0.200\,\overline{{{m}}}\,L\right)\,\ddot{Z}(t)+\left(\frac{4\,a_{1}\,E I}{L^{3}}\right)\,\dot{Z}(t)+\left(\frac{4E I}{L^{3}}-\frac{4N}{3L}\right)\,Z(t)=-\frac{\overline{{{m}}}\,L}{3}\,\ddot{v}_{g}(t)\tag {f}
|
||||
$$
|
||||
|
||||
Setting the combined stiffness equal to zero, the critical load is given as
|
||||
将组合刚度设为零，临界载荷可表示为
|
||||
$$
|
||||
N_{\mathrm{cr}}={\frac{3E I}{L^{2}}}\tag {g}
|
||||
$$
|
||||
|
||||
which is about 22 percent higher than the true value given by Eq. (d).
|
||||
这比由式(d)给出的真值高出约22%。
|
||||
|
||||
When using the Rayleigh method of buckling analysis as given by Eq. (8-17), it should be recognized that assuming any shape other than the true buckled shape will require additional external constraints acting on the system to maintain its equilibrium. These additional external constraints represent a stiffening influence on the system; therefore the critical load computed by a Rayleigh analysis using any shape other than the true one must always be greater than the true critical load. In the above example, it is apparent that the parabolic shape is not a good assumption for this structure, even though it satisfies the geometric boundary conditions, because the constant curvature of this shape implies that the moment is constant along its length. It is obvious here that the moment must vanish at the top of the column, and any assumed shape that satisfies this force boundary condition (i.e., one having zero curvature at the top) will give much better results.
|
||||
在使用式 (8-17) 给出的瑞利屈曲分析法时，应认识到，假设任何非真实屈曲形状的形状，都将需要额外的外部约束作用在系统上以维持其平衡。这些额外的外部约束代表了对系统的刚化效应；因此，通过瑞利分析使用任何非真实形状计算出的临界载荷，必然总是大于真实的临界载荷。在上述示例中，很明显，抛物线形状对于该结构而言并非一个好的假设，尽管它满足几何边界条件，因为这种形状的恒定曲率意味着弯矩沿其长度方向是恒定的。在此，很明显，弯矩在柱顶必须为零，并且任何满足此力边界条件（即顶部曲率为零）的假设形状都将给出更好的结果。
|
||||
## 8-4 EXPRESSIONS FOR GENERALIZED SYSTEMPROPERTIES
|
||||
|
||||
As implied by the preceding examples, the equation of motion for any SDOF system, no matter how complex, can always be reduced to the form
|
||||
正如前面的例子所暗示的，任何单自由度（SDOF）系统的运动方程，无论多么复杂，总是可以简化为以下形式
|
||||
$$
|
||||
m^{*}\;\ddot{Z}(t)+c^{*}\;\dot{Z}(t)+\overline{{k^{*}}}\;Z(t)=p^{*}(t)
|
||||
$$
|
||||
|
||||
in which $Z(t)$ is the single generalized coordinate expressing the motion of the system and the symbols with asterisks represent generalized physical properties corresponding to this coordinate. In general, the values of these properties can be determined by application of either the principle of virtual work, as illustrated by the previous examples, or Hamilton’s principle as illustrated in Chapter 16. However, standardized forms of these expressions can be derived easily which are very useful in practice.
|
||||
其中$Z(t)$是表示系统运动的单一广义坐标，且带星号的符号代表对应于该坐标的广义物理特性。通常，这些特性的值可以通过应用虚功原理（如前面的例子所示）或哈密顿原理（如第16章所示）来确定。然而，这些表达式的标准化形式可以很容易地推导出来，这在实践中非常有用。
|
||||
Consider an arbitrary one-dimensional system, as illustrated by the example in Fig. 8-3, assumed to displace only in a single shape $\psi(x)$ with displacements expressed in terms of the generalized coordinate $Z(t)$ as given by
|
||||
考虑一个任意的一维系统，如图8-3中的示例所示，假设其仅以单一形状 $\psi(x)$ 发生位移，其位移可由广义坐标 $Z(t)$ 表示为
|
||||
|
||||
FIGURE 8-3 Properties of generalized SDOF system: (a) assumed shape; $(b)$ mass properties; (c) damping properties; $(d)$ elastic properties; (e) applied axial loading; $(f)$ applied lateral loading.
|
||||
图 8-3 广义单自由度 (SDOF) 系统的特性：(a) 假定形状；(b) 质量特性；(c) 阻尼特性；(d) 弹性特性；(e) 施加的轴向载荷；(f) 施加的横向载荷。
|
||||
$$
|
||||
v(x,t)=\psi(x)\;Z(t)
|
||||
$$
|
||||
|
||||
Part of the total mass of the system is distributed in accordance with $m(x)$ and the remainder is lumped at discrete locations $i$ $\ '(i=1,2,\ldots)$ as denoted by $m_{i}$ . External damping is provided by distributed dashpots varying in accordance with $c(x)$ and by discrete dashpots as denoted by the $c_{i}$ values, and internal damping is assumed to be present in flexure as controlled by the uniaxial stress-strain relation of Eq. (8-8). The elastic properties of the system result from distributed external springs varying in accordance with $k(x)$ , from discrete springs as denoted by the $k_{i}$ values, and from distributed flexural stiffness given by $E I(x)$ . External loadings are applied to the system in both discrete and distributed forms as indicated by the time-independent axial forces $q(x)$ and $N$ and the time-dependent lateral forces $p({\boldsymbol{x}},t)$ and $p_{i}(t)$ . These loadings produce internal axial force and moment distributions $N(x)$ and $M(x,t)$ , respectively.
|
||||
系统总质量的一部分按照 $m(x)$ 分布，其余部分集中在离散位置 $i$ $(i=1,2,\ldots)$，用 $m_{i}$ 表示。外部阻尼由按照 $c(x)$ 变化的分布式阻尼器以及由 $c_{i}$ 值表示的离散阻尼器提供，并且假定内部阻尼存在于弯曲中，由式 (8-8) 的单轴应力-应变关系控制。系统的弹性特性来源于按照 $k(x)$ 变化的分布式外部弹簧、由 $k_{i}$ 值表示的离散弹簧以及由 $E I(x)$ 给出的分布式弯曲刚度。外部载荷以离散和分布两种形式施加到系统上，如由不随时间变化的轴向力 $q(x)$ 和 $N$ 以及随时间变化的横向力 $p({\boldsymbol{x}},t)$ 和 $p_{i}(t)$ 所示。这些载荷分别产生内部轴向力分布 $N(x)$ 和力矩分布 $M(x,t)$。
|
||||
|
||||
Applying the procedure of virtual work to this general SDOF system in the same manner as it was applied to the previous example solutions, one obtains the following useful expressions for the contributions to the generalized properties:
|
||||
将虚功原理以与应用于先前示例解相同的方式应用于这个通用单自由度系统，即可得到广义属性贡献的以下有用表达式：
|
||||
$$
|
||||
\begin{array}{l}{{m^{\ast}=\displaystyle\int_{0}^{L}{m(x)\ \psi(x)^{2}\ d x}+\sum m_{i}\ \psi_{i}^{2}+\sum j_{i}\ \psi_{i}^{2}}}\\ {{\displaystyle c^{\ast}=\displaystyle\int_{0}^{L}{c(x)\ \psi(x)^{2}\ d x}+a_{1}\int_{0}^{L}{E I(x)\ \psi^{\prime\prime}(x)^{2}\ d x}+\sum\cot\psi_{i}^{2}}}\\ {{\displaystyle\overline{{{k^{\ast}}}}=\displaystyle\int_{0}^{L}{k(x)\ \psi(x)^{2}\ d x}+\int_{0}^{L}{E I(x)\ \psi^{\prime\prime}(x)^{2}\ d x}+\sum\lambda_{i\psi_{i}^{2}}}}\\ {{\displaystyle\qquad-\int_{0}^{L}{N(x)\ \psi^{\prime}(x)^{2}\ d x}}}\\ {{\displaystyle\ p^{\ast}(t)=\displaystyle\int_{0}^{L}{p(x,t)\ \psi(x)\ d x}+\sum p_{i}(t)\ \psi_{i}(x)}}\end{array}
|
||||
$$
|
||||
|
||||
The vectorial nature of the force and displacement quantities in the last of Eqs. (8- 18) must be carefully noted. Only components of the forces in the directions of the corresponding assumed displacements can be included, and the positive sense of each force component must be assigned in accordance with the positive sense of the corresponding displacement.
|
||||
|
||||
The above generalized-coordinate concepts apply equally in the reduction of two-dimensional systems to a single degree of freedom. Consider, for example, the rectangular floor slab shown in Fig. 8-4 subjected to a distributed downward loading $p(x,y,t)$ . If the deflections of this slab are assumed to have the shape $\psi(x,y)$ shown, and if the displacement amplitude at the middle is taken as the generalized coordinate, the displacements may be expressed
|
||||
必须仔细注意式(8-18)最后一个中力和位移量的矢量性。只有沿相应假定位移方向的力分量才能被包含在内，并且每个力分量的正方向必须根据相应位移的正方向来指定。
|
||||
|
||||
上述广义坐标概念在将二维系统简化为单自由度系统时同样适用。例如，考虑图8-4所示的矩形楼板，其承受分布向下载荷$p(x,y,t)$。如果该板的变形被假定具有所示的形状$\psi(x,y)$，并且如果中间的位移幅值被作为广义坐标，则位移可以表示为
|
||||
|
||||
FIGURE 8-4 Simply supported two-dimensional slab treated as a SDOF system.
|
||||
图 8-4 被视为单自由度系统的简支二维板。
|
||||
$$
|
||||
w(x,y,t)=\psi(x,y)\;Z(t)
|
||||
$$
|
||||
|
||||
For a uniform simply supported slab, the shape function might logically be of the form
|
||||
对于均匀简支板，形函数逻辑上可能具有以下形式
|
||||
$$
|
||||
\psi(x,y)=\sin{\frac{\pi x}{a}}\ \sin{\frac{\pi y}{b}}
|
||||
$$
|
||||
|
||||
but any other reasonable shape consistent with the support conditions could be used.
|
||||
|
||||
The generalized properties of this system can be calculated by expressions equivalent to those presented in Eqs. (8-18) for the one-dimensional case; however, the integrations must be carried out here in both the $x$ and $y$ directions. For this specific example, the generalized mass, stiffness, and loading would be given by
|
||||
但是，任何其他与支撑条件一致的合理形状都可以使用。
|
||||
|
||||
该系统的广义特性可以通过与一维情况下公式(8-18)中给出的表达式等效的表达式计算；然而，这里的积分必须在 $x$ 和 $y$ 两个方向上进行。对于这个具体的例子，广义质量、刚度和载荷将由以下公式给出：
|
||||
$$
|
||||
{\begin{array}{l}{m^{*}=\displaystyle\int_{0}^{a}\int_{0}^{b}m(x,y)\;\psi(x,y)^{2}\;d x\,d y}\\ {k^{*}=D\int_{0}^{a}\int_{0}^{b}\left\{\left[{\frac{\partial^{2}\psi(x,y)}{\partial x^{2}}}+{\frac{\partial^{2}\psi(x,y)}{\partial y^{2}}}\right]^{2}\right.}\\ {\left.\qquad\qquad\qquad-2\left(1-\nu\right)\left[{\frac{\partial^{2}\psi(x,y)}{\partial x^{2}}}\,{\frac{\partial^{2}\psi(x,y)}{\partial y^{2}}}-\left({\frac{\partial^{2}\psi(x,y)}{\partial x\,\partial y}}\right)^{2}\right]\right\}d x\,d y}\\ {p^{*}(t)=\displaystyle\int_{0}^{a}\int_{0}^{b}p(x,y)\;\psi(x,y)\;d x\,d y}\end{array}}
|
||||
$$
|
||||
|
||||
where
|
||||
|
||||
$$
|
||||
\begin{array}{l}{{D=E h^{3}/12\left(1-\nu^{2}\right)=\mathrm{{flexural}\;r i g i d i t y\;o f\;t h e\;s l a b}}}\\ {{\nu={\mathrm{Poisson}}^{\ast}{\mathrm{s}}\;{\mathrm{ratio}}}}\\ {{\;h=\mathrm{{plate}}\;\mathrm{{thickness}}}}\end{array}
|
||||
$$
|
||||
|
||||
It also should be evident that the same procedures can easily be extended to three-dimensional systems by assuming an appropriate displacement function in three dimensions. However, the difficulty of selecting a suitable shape increases rapidly with the number of dimensions of the system, and the reliability of the results so obtained is reduced accordingly.
|
||||
同样显而易见的是，通过假设一个合适的三维位移函数，相同的程序可以很容易地扩展到三维系统。然而，选择合适形状的难度随着系统维数的增加而迅速增加，并且由此获得的结果的可靠性也相应降低。
|
||||
## 8-5 VIBRATION ANALYSIS BY RAYLEIGH’S METHOD瑞利法振动分析
|
||||
|
||||
It was pointed out in Section 8-2 that the critical buckling load for a flexural member can be calculated approximately from its generalized elastic and geometric stiffness properties, where these quantities are derived from an assumed buckling shape, and it was also noted that such an assumed shape formulation is generally called Rayleigh’s method. Now Rayleigh’s assumed shape concept will be extended further to develop an approximate method of evaluating the vibration frequency of the member. The essence of the concept is evident immediately from the fact that the SDOF frequency of vibrations is defined as
|
||||
第8-2节指出，受弯构件的临界屈曲载荷可以根据其广义弹性刚度和几何刚度特性近似计算，其中这些量是从假定的屈曲形状推导出来的，并且还指出，这种假定形状的公式化方法通常被称为瑞利法。现在，瑞利假定形状的概念将进一步扩展，以开发一种评估构件振动频率的近似方法。该概念的本质从单自由度（SDOF）振动频率定义为...这一事实中立即显而易见。
|
||||
$$
|
||||
\omega={\sqrt{\frac{k}{m}}}\tag {8-21}
|
||||
$$
|
||||
|
||||
where $k$ and $m$ are the system’s mass and stiffness, respectively. The Rayleigh’s method value of the vibration frequency is given directly by this expression if $k^{*}$ and $m^{*}$ , representing the generalized stiffness and mass associated with a given assumed shape, $\psi(x)$ , are substituted.
|
||||
其中 $k$ 和 $m$ 分别是系统的质量和刚度。如果代入 $k^{*}$ 和 $m^{*}$（它们代表了与给定假设形状 $\psi(x)$ 相关的广义刚度和质量），则瑞利法得到的振动频率值可由该表达式直接给出。
|
||||
|
||||
Although this generalized-coordinate concept may be used to determine approximately the vibration frequency of any structure, it is instructive to examine the frequency analysis problem from another point of view, originated by Lord Rayleigh. The basic concept in the Rayleigh method is the principle of conservation of energy; the energy in a freely vibrating system must remain constant if no damping forces act to absorb it. Consider the free-vibration motion of the undamped spring-mass system shown in Fig. $8{-}5a$ . With an appropriate choice of time origin, the displacement can be expressed (Fig. 8-5b) by
|
||||
尽管这种广义坐标概念可用于近似确定任何结构的振动频率，但从瑞利勋爵提出的另一个角度审视频率分析问题也很有启发性。瑞利法的基本概念是能量守恒原理；如果没有阻尼力作用来吸收能量，则自由振动系统中的能量必须保持不变。考虑图 $8{-}5a$ 所示的无阻尼弹簧-质量系统的自由振动。通过适当选择时间原点，位移可以表示为（如图 8-5b 所示）
|
||||
$$
|
||||
v=v_{0}\,\sin\,\omega t\tag {8-22a}
|
||||
$$
|
||||
|
||||
and the velocity (Fig. 8-5c) by
|
||||
|
||||
$$
|
||||
\dot{v}=v_{0}\ \omega\ \cos\omega t\tag {8-22b}
|
||||
$$
|
||||
|
||||
The potential energy of this system is represented entirely by the strain energy of the spring:
|
||||
该系统的势能完全由弹簧的应变能表示：
|
||||
$$
|
||||
V={\frac{1}{2}}\ k\,v^{2}={\frac{1}{2}}\ k\,v_{0}^{2}\,\sin^{2}\omega t\tag {8-23a}
|
||||
$$
|
||||
|
||||
|
||||
|
||||
while the kinetic energy of the mass is
|
||||
|
||||
$$
|
||||
T={\frac{1}{2}}\ m\,{\dot{v}}^{2}={\frac{1}{2}}\,m\,v_{0}^{2}\,\omega^{2}\,\cos^{2}\omega t\tag {8-23b}
|
||||
$$
|
||||
|
||||
Now considering the time $t\,=\,\pi/2\omega$ , it is clear from Fig. 8-5 [or from Eqs.(8-23)] that the kinetic energy is zero and that the potential energy has reached its maximum value:
|
||||
现考虑时间 $t\,=\,\pi/2\omega$，从图 8-5 [或从公式(8-23)] 可以清楚地看出，动能为零，且势能已达到其最大值：
|
||||
$$
|
||||
V_{\mathrm{max}}={\frac{1}{2}}\;k\,v_{0}^{2}\tag {8-24a}
|
||||
$$
|
||||
|
||||
Similarly, at the time $t=\pi/\omega$ , the potential energy vanishes and the kinetic energy is maximum
|
||||
类似地，在 $t=\pi/\omega$ 时刻，势能为零，动能达到最大值。
|
||||
$$
|
||||
T_{\mathrm{max}}={\frac{1}{2}}\ m\,v_{0}^{2}\,\omega^{2}\tag {8-24b}
|
||||
$$
|
||||
|
||||
Hence, if the total energy in the vibrating system remains constant (as it must in undamped free vibration), it is apparent that the maximum kinetic energy must equal the maximum potential energy, $V_{\mathrm{max}}=T_{\mathrm{max}}$ ; that is,
|
||||
因此，如果振动系统中的总能量保持不变（正如在无阻尼自由振动中必然如此），那么显然，最大动能必须等于最大势能，即 $V_{\mathrm{max}}=T_{\mathrm{max}}$；也就是说，
|
||||
$$
|
||||
{\frac{1}{2}}\;k\,v_{0}^{2}={\frac{1}{2}}\;m\,v_{0}^{2}\,\omega^{2}
|
||||
$$
|
||||
|
||||
from which
|
||||
|
||||
$$
|
||||
\omega^{2}={\frac{k}{m}}
|
||||
$$
|
||||
|
||||
This, of course, is the same frequency expression which was cited earlier; in this case it has been derived by the Rayleigh concept of equating expressions for the maximum strain energy and kinetic energy.
|
||||
这当然是前面引用的相同频率表达式；在这种情况下，它是通过瑞利概念推导出来的，该概念将最大应变能和动能的表达式等同起来。
|
||||
|
||||
There is no advantage to be gained from the application of Rayleigh’s method to vibration analysis of a spring-mass system as described above; its principal use is for the approximate frequency analysis of a system having many degrees of freedom. Consider, for example, the nonuniform simple beam shown in Fig. 8-6. This beam actually has an infinite number of degrees of freedom; that is, it can displace in an infinite variety of displacement patterns. To apply the Rayleigh procedure, it is necessary to assume the shape which the beam will take in its fundamental mode of vibration. As explained above, this assumption may be expressed by Eq. (8-2), or noting the harmonic variation of the generalized coordinate in free vibrations
|
||||
将瑞利方法应用于上述弹簧-质量系统的振动分析并无优势；它的主要用途是用于具有多个自由度系统的近似频率分析。例如，考虑图8-6所示的非均匀简支梁。该梁实际上具有无限个自由度；也就是说，它能够以无限多种位移模式进行位移。为了应用瑞利方法，有必要假设梁在其基频振动模态下的形状。如上所述，该假设可以用式(8-2)表示，或者注意到自由振动中广义坐标的谐波变化
|
||||
$$
|
||||
v(x,t)=\psi(x)\;Z_{0}\;\sin\omega t\tag {8-25}
|
||||
$$
|
||||
|
||||
in which $\psi(x)$ is the shape function, which represents the ratio of the displacement at any point $x$ to the reference displacement or generalized coordinate $Z(t)$ . Equation (8-25) expresses the assumption that the shape of the vibrating beam does not change with time; only the amplitude of motion varies, and it varies harmonically in a freevibration condition.
|
||||
|
||||
The assumption of the shape function $\psi(x)$ effectively reduces the beam to a SDOF system. Thus the frequency of vibration can be found by equating the maximum strain energy developed during the motion to the maximum kinetic energy. The strain energy of this flexural system is given by
|
||||
其中 $\psi(x)$ 是形函数，它表示任意点 $x$ 处的位移与参考位移或广义坐标 $Z(t)$ 之比。方程 (8-25) 表达了振动梁的形状不随时间变化的假设；只有运动幅值发生变化，并且在自由振动条件下呈谐波变化。
|
||||
|
||||
形函数 $\psi(x)$ 的假设有效地将梁简化为单自由度 (SDOF) 系统。因此，可以通过将运动过程中产生的最大应变能与最大动能相等来找到振动频率。该弯曲系统的应变能由下式给出
|
||||
$$
|
||||
V=\frac{1}{2}\,\int_{0}^{L}E I(x)\,\left(\frac{\partial^{2}v}{\partial x^{2}}\right)^{2}\,d x\tag {8-26}
|
||||
$$
|
||||
|
||||
Thus, substituting the assumed shape function of Eq. (8-25) and letting the displacement amplitude take its maximum value leads to
|
||||
因此，将式(8-25)的假设形函数代入，并使位移幅值取其最大值，可得到
|
||||
$$
|
||||
V_{\mathrm{max}}={\frac{1}{2}}\;Z_{0}^{2}\;\int_{0}^{L}E I(x)\;[\psi^{\prime\prime}(x)]^{2}\;d x\tag {8-27}
|
||||
$$
|
||||
|
||||
The kinetic energy of the nonuniformly distributed mass is
|
||||
非均匀分布质量的动能是
|
||||
$$
|
||||
T=\frac{1}{2}\,\int_{0}^{L}m(x)\,(\dot{v})^{2}\,d x\tag {8-28}
|
||||
$$
|
||||
|
||||
|
||||
FIGURE 8-6 Vibration of a nonuniform beam.
|
||||
|
||||
Thus, when Eq.(8-25) is differentiated with respect to time to obtain the velocity and the amplitude is allowed to reach its maximum,
|
||||
因此，当式(8-25)对时间求导以获得速度，并且振幅达到其最大值时，
|
||||
$$
|
||||
T_{\mathrm{max}}={\frac{1}{2}}\;Z_{0}^{2}\,\omega^{2}\;\int_{0}^{L}m(x)\;[\psi(x)]^{2}\,d x\tag {8-29}
|
||||
$$
|
||||
|
||||
Finally, after equating the maximum potential energy to the maximum kinetic energy, the squared frequency is found to be
|
||||
最后，在将最大势能与最大动能相等后，求得频率的平方为
|
||||
$$
|
||||
\omega^{2}={\frac{\int_{0}^{L}E I(x)\,[\psi^{\prime\prime}(x)]^{2}\,d x}{\int_{0}^{L}m(x)\,[\psi(x)]^{2}\,d x}}\tag {8-30}
|
||||
$$
|
||||
|
||||
At this point, it may be noted that the numerator of Eq. (8-30) is merely the generalized stiffness of the beam $k^{*}$ for this assumed displacement shape while the denominator is its generalized mass $m^{*}$ [see Eqs. (8-18)]. Thus Rayleigh’s method leads directly to the generalized form of Eq. (8-21), as is to be expected since it employs the same generalized-coordinate concept to reduce the system to a single degree of freedom.
|
||||
此时，可以注意到，式 (8-30) 的分子仅仅是梁对于这种假定的位移形状的广义刚度 $k^{*}$，而分母是其广义质量 $m^{*}$ [参见式 (8-18)]。因此，瑞利法直接得出式 (8-21) 的广义形式，正如所预期的那样，因为它采用了相同的广义坐标概念将系统简化为单自由度。
|
||||
## 8-6 SELECTION OF THE RAYLEIGH VIBRATION SHAPE
|
||||
|
||||
The accuracy of the vibration frequency obtained by Rayleigh’s method depends entirely on the shape function $\psi(x)$ which is assumed to represent the vibrationmode shape. In principle, any shape may be selected which satisfies the geometric boundary conditions of the beam, that is, which is consistent with the specified support conditions. However, any shape other than the true vibration shape would require the action of additional external constraints to maintain equilibrium; these extra constraints would stiffen the system, adding to its strain energy, and thus would cause an increase in the computed frequency. Consequently, it may be recognized that the true vibration shape will yield the lowest frequency obtainable by Rayleigh’s method, and in choosing between approximate results given by this method, the lowest frequency is always the best approximation.
|
||||
瑞利法获得的振动频率的准确性完全取决于假定代表振动模态形状的形状函数 $\psi(x)$。原则上，可以选择任何满足梁的几何边界条件，即与规定的支承条件一致的形状。然而，任何非真实振动形状都需要额外的外部约束作用来保持平衡；这些额外的约束会使系统刚度增加，增加其应变能，从而导致计算频率的增加。因此，可以认识到，真实振动形状将产生瑞利法可获得的最低频率，并且在该方法给出的近似结果之间进行选择时，最低频率始终是最佳近似值。
|
||||
|
||||
Example E8-4. To illustrate this point, assume that the beam of Fig. 8-6 has uniform mass $\overline{{m}}$ and stiffness $E I$ . As a first approximation for the frequency analysis, assume that the vibration shape is parabolic: $\psi(x)=(x/L)\left(x/L-1\right)$ . Then, $\psi^{\prime\prime}(x)=2/L^{2}$ , and
|
||||
例 E8-4。为了说明这一点，假设图 8-6 中的梁具有均匀质量 $\overline{{m}}$ 和刚度 $E I$。作为频率分析的第一次近似，假设振动形状是抛物线形的：$\psi(x)=(x/L)\left(x/L-1\right)$。那么，$\psi^{\prime\prime}(x)=2/L^{2}$，并且
|
||||
$$
|
||||
V_{\mathrm{max}}={\frac{1}{2}}\,Z_{0}^{2}\,E I\,\int_{0}^{L}\left({\frac{2}{L^{2}}}\right)^{2}d x={\frac{1}{2}}\,Z_{0}^{2}\,{\frac{4E I}{L^{3}}}
|
||||
$$
|
||||
|
||||
while
|
||||
|
||||
$$
|
||||
T_{\mathrm{max}}={\frac{1}{2}}\,Z_{0}^{2}\,\omega^{2}\,{\overline{{m}}}\,\int_{0}^{L}\,\left[{\frac{x}{L}}\left({\frac{x}{L}}-1\right)\right]^{2}\,d x={\frac{1}{2}}\,Z_{0}^{2}\,\omega^{2}\,{\frac{\overline{{m}}L}{30}}
|
||||
$$
|
||||
|
||||
from which
|
||||
|
||||
$$
|
||||
\omega^{2}=\frac{V_{\mathrm{max}}}{\left(1/\omega^{2}\right)T_{\mathrm{max}}}=\frac{120\,E I}{\overline{{m}}L^{4}}
|
||||
$$
|
||||
|
||||
If the shape were assumed to be a sine curve, $\psi(x)\,=\,\sin(\pi x/L)$ , the same type of analysis would lead to the result
|
||||
如果形状假设为正弦曲线，$\psi(x)\,=\,\sin(\pi x/L)$，那么相同类型的分析将得出结果
|
||||
$$
|
||||
\omega^{2}={\frac{E I\,\pi^{4}/2L^{3}}{m\,L/2}}=\pi^{4}\,{\frac{E I}{\overline{{m}}\,L^{4}}}
|
||||
$$
|
||||
|
||||
This second frequency is significantly less than the first (actually almost 20 percent less); thus it is a much better approximation. As a matter of fact, it is the exact answer because the assumed sine-curve shape is the true vibration shape of a uniform simple beam. The first assumption should not be expected to lead to very good results; the assumed parabolic shape implies a uniform bending moment along the span which does not correspond to the simple end-support conditions. It is a valid shape, since it satisfies the geometric requirements of zero end displacements, but is not a realistic assumption.
|
||||
|
||||
The question now arises of how a reasonable deflected shape can be selected in order to ensure good results with Rayleigh’s method (or the equivalent generalizedcoordinate approach described earlier). The concept to be used in selecting the vibration shape is that the displacements in free vibration result from the application of inertial forces and that the inertial forces (which are the product of mass and acceleration) are proportional to the mass distribution and to the displacement amplitude. Thus, the correct vibration shape $\psi_{c}(x)$ is that deflected shape resulting from a loading $p_{c}(x)$ proportional to $m(x)\,\psi_{c}(x)$ . Of course, it is not possible to guess the exact shape $\psi_{c}(x)$ , but the deflection shape computed from the loading $\overline{{p}}(x)=m(x)\,\overline{{\psi}}(x)$ [as shown in Fig. 8-7, where $\overline{{\psi}}(x)$ is any reasonable approximation of the true shape] will provide extremely good accuracy in the solution.
|
||||
|
||||
In general, the evaluation of the generalized coordinate shape on the basis of an assumed shape in this fashion involves more computational effort than is necessary in an approximate analysis. The Rayleigh procedure will give good accuracy with a considerably less refined approach than this. One common assumption is that the inertial loading $\overline{{p}}(\boldsymbol{x})$ (see Fig. 8-7) is merely the weight of the beam, that is, $\overline{{p}}(x)\,=\,m(x)\,g$ , where $m(x)$ is the mass distribution and $g$ is the acceleration of gravity. The frequency then is evaluated on the basis of the deflected shape ${v_{d}}(x)$ resulting from this dead-weight load. The maximum strain energy can be found very simply in this case from the fact that the stored energy must be equal to the work done on the system by the applied loading:
|
||||
第二个频率显著小于第一个频率（实际上几乎小了20%）；因此它是一个更好的近似。事实上，它是精确解，因为假设的正弦曲线形状是均匀简支梁的真实振动形状。第一个假设不应期望能得到很好的结果；假设的抛物线形状意味着沿展向的弯矩是均匀的，这与简单的端部支承条件不符。它是一个有效的形状，因为它满足零端部变形的几何要求，但不是一个现实的假设。
|
||||
|
||||
现在出现了一个问题，即如何选择一个合理的变形形状，以确保瑞利法（或前面描述的等效广义坐标方法）能获得良好的结果。选择振动形状时要使用的概念是，自由振动中的变形是惯性力作用的结果，并且惯性力（质量和加速度的乘积）与质量分布和变形幅值成正比。因此，正确的振动形状 $\psi_{c}(x)$ 是由与 $m(x)\,\psi_{c}(x)$ 成正比的载荷 $p_{c}(x)$ 引起的变形形状。当然，不可能猜出精确的形状 $\psi_{c}(x)$，但由载荷 $\overline{{p}}(x)=m(x)\,\overline{{\psi}}(x)$ [如图8-7所示，其中 $\overline{{\psi}}(x)$ 是真实形状的任何合理近似] 计算出的变形形状将在求解中提供极高的精度。
|
||||
|
||||
通常，以这种方式基于假设形状评估广义坐标形状所涉及的计算工作量超过了近似分析所需的。瑞利法将使用比这粗糙得多的方法提供良好的精度。一个常见的假设是，惯性载荷 $\overline{{p}}(\boldsymbol{x})$ (参见图8-7) 仅仅是梁的重量，也就是说，$\overline{{p}}(x)\,=\,m(x)\,g$，其中 $m(x)$ 是质量分布，$g$ 是重力加速度。然后，频率是根据由此自重载荷产生的变形形状 ${v_{d}}(x)$ 进行评估的。在这种情况下，最大应变能可以非常简单地从以下事实中找到：储存的能量必须等于施加载荷对系统所做的功：
|
||||
|
||||
FIGURE 8-7 Deflected shape resulting from inertial load of assumed shape.
|
||||
|
||||
$$
|
||||
V_{\mathrm{max}}={\frac{1}{2}}\,\int_{0}^{L}{\overline{{p}}}(x)\,v_{d}(x)\,d x={\frac{1}{2}}\,g\,Z_{0}\int_{0}^{L}m(x)\,\psi(x)\,d x
|
||||
$$
|
||||
|
||||
The kinetic energy is given still by Eq. (8-29), in which $\psi(x)=v_{d}(x)/Z_{0}$ is the shape function computed from the dead load. Thus the squared frequency found by equating the strain and kinetic-energy expressions is
|
||||
动能仍由式 (8-29) 给出，其中 $\psi(x)=v_{d}(x)/Z_{0}$ 是由恒载计算得到的形函数。因此，通过使应变和动能表达式相等而得到的频率平方为
|
||||
$$
|
||||
\omega^{2}={\frac{g}{Z_{0}}}\;{\frac{\int_{0}^{L}m(x)\,\psi(x)\,d x}{\int_{0}^{L}m(x)\,[\psi(x)]^{2}\,d x}}=g\;{\frac{\int_{0}^{L}m(x)\,v_{d}(x)\,d x}{\int_{0}^{L}m(x)\,[v_{d}(x)]^{2}\,d x}}
|
||||
$$
|
||||
|
||||
Equation (8-32) is commonly used for the approximate frequency analysis of any type of system. It should be noted that the reference amplitude $Z_{0}$ must be included in the expression if the shape is defined by the dimensionless shape function $\psi(x)$ , but it is not involved if the actual dead-load deflections are used.
|
||||
|
||||
The loading $\overline{{p}}(\boldsymbol{x})$ used to calculate the dead-weight deflection ${v_{d}}(x)$ in Eq. (8-32) is actually a gravitational loading only in cases where the principal vibratory motion is in the vertical direction. For a structure like the vertical cantilever of Fig. $8{-}8a$ , in which the principal motion is horizontal, the loading must be applied laterally, as shown in this figure; in effect it is assumed that gravity acts horizontally for this purpose. An appropriate deflected shape to approximate the symmetrical vibration frequency of the frame of Fig. $8–8b$ could be obtained by applying a vertical gravity load, as shown. However, the fundamental vibrations of this type of structure will generally be in the horizontal direction; to obtain a shape $\psi(x)$ for approximating the lateral vibration frequency, the gravity forces should be applied laterally. Furthermore, in the fundamental mode of vibration of the two-span beam shown in Fig. $8.8c$ , the two spans will deflect in opposite directions. Thus, to obtain a deflected shape for this case, the gravitational forces should be applied in opposite directions in the adjacent spans. A considerably higher vibration frequency would be obtained from the deflected shape resulting from downward loads acting in both spans.
|
||||
方程 (8-32) 常用于任何类型系统的近似频率分析。值得注意的是，如果形状由无量纲形状函数 $\psi(x)$ 定义，则表达式中必须包含参考振幅 $Z_{0}$，但如果使用实际的恒载变形，则不涉及 $Z_{0}$。
|
||||
|
||||
方程 (8-32) 中用于计算自重变形 ${v_{d}}(x)$ 的载荷 $\overline{{p}}(\boldsymbol{x})$ 实际上仅在主振动运动为垂直方向的情况下是重力载荷。对于图 $8{-}8a$ 所示的垂直悬臂梁等结构，其主运动是水平的，载荷必须横向施加，如图所示；实际上，为此目的，假定重力水平作用。通过施加垂直重力载荷，如图所示，可以获得一个合适的变形形状，以近似图 $8–8b$ 所示框架的对称振动频率。然而，这种结构的基本振动通常是水平方向的；为了获得一个近似横向振动频率的形状 $\psi(x)$，重力应横向施加。此外，在图 $8.8c$ 所示的两跨梁的基本振动模态中，两跨将向相反方向变形。因此，为了获得这种情况下的变形形状，重力应在相邻跨中施加相反方向。由两跨均受向下载荷作用产生的变形形状将获得显著更高的振动频率。
|
||||
|
||||
FIGURE 8-8 Assumed shapes resulting from dead loads.
|
||||
|
||||
The reader must be cautioned, however, against spending too much time in computing deflected shapes which will yield extremely accurate results. The principal value of the Rayleigh method is in providing a simple and reliable approximation to the natural frequency. Almost any reasonable shape assumption will give useful results.
|
||||
然而，必须提醒读者，不要花费过多时间去计算那些能产生极其精确结果的变形形状。瑞利方法的主要价值在于为固有频率提供一个简单可靠的近似值。几乎任何合理的形状假设都会给出有用的结果。
|
||||
|
||||
Example E8-5. The use of the Rayleigh method to compute the vibration frequency of a practical system will be illustrated by the analysis of the uniform cantilever beam supporting a weight at midspan, shown in Fig. E8-5. For this study, the vibration shape has been taken to be that produced by a load applied to the end of the cantilever, as shown in the lower sketch. The resulting deflected shape is
|
||||
示例 E8-5。瑞利法计算实际系统振动频率的应用将通过对图 E8-5 所示的均质悬臂梁在中跨支撑重物的分析进行说明。对于本研究，振型被认为是施加在悬臂梁末端的载荷所产生的，如下方草图所示。产生的变形形状为
|
||||
$$
|
||||
v(x)={\frac{p L^{3}}{3E I}}\,\left[{\frac{3x^{2}L-x^{3}}{2L^{3}}}\right]\equiv Z_{0}\,\psi(x)
|
||||
$$
|
||||
|
||||
The maximum potential energy of the beam can be found in this case from
|
||||
梁的最大势能可以在此情况下从
|
||||
$$
|
||||
V_{\mathrm{max}}={\frac{1}{2}}\,p\,Z_{0}={\frac{1}{2}}\,{\frac{3E I}{L^{3}}}\,Z_{0}^{2}
|
||||
$$
|
||||
|
||||
where $Z_{0}$ is the deflection under the load and $p$ has been expressed in terms of this end deflection.
|
||||
其中 $Z_{0}$ 是载荷作用下的变形，并且 $p$ 已经用这个末端变形表示。
|
||||
|
||||
|
||||
The maximum kinetic energy of the beam can be calculated in two parts, considering separately the beam and the supported weight:
|
||||
梁的最大动能可以分两部分计算，分别考虑梁和支撑的重量：
|
||||
$$
|
||||
\begin{array}{l}{{T_{\mathrm{max}}^{B}={\displaystyle{\frac{\omega^{2}}{2}}\,\int_{0}^{L}{\overline{{{m}}}\,v^{2}\,d x}={\displaystyle{\frac{\overline{{{m}}}}{2}}}\omega^{2}\,Z_{0}^{2}\int_{0}^{L}[\psi(x)]^{2}\,d x}}}\\ {{\ \ \ \ \ \ \ ={\displaystyle{\frac{33}{140}}\,{\frac{\overline{{{m}}}L}{2}}\,\omega^{2}\,Z_{0}^{2}}}}\\ {{\ \ \ \ \ \ }}\\ {{T_{\mathrm{max}}^{W}={\displaystyle{\frac{W}{2g}}\omega^{2}\left[v\left(x={\frac{L}{2}}\right)\right]^{2}}={\displaystyle{\frac{W}{2g}}\omega^{2}\left({\frac{5}{16}}\,Z_{0}\right)^{2}}}}\\ {{\ \ \ \ \ \ \ ={\displaystyle{\frac{25}{256}}\,{\frac{W}{2g}}\omega^{2}\,Z_{0}^{2}}}}\end{array}
|
||||
$$
|
||||
|
||||
Hence the total kinetic energy is
|
||||
|
||||
$$
|
||||
T_{\mathrm{max}}=\left({\frac{33}{140}}+{\frac{25}{256}}\,{\frac{W}{\overline{{m}}{L}g}}\right){\frac{\overline{{m}}{L}}{2}}\,\omega^{2}\,Z_{0}^{2}
|
||||
$$
|
||||
|
||||
and equating the maximum kinetic- and potential-energy expressions leads to the frequency equation
|
||||
|
||||
$$
|
||||
\omega^{2}={\frac{3}{\left[{\frac{33}{140}}+{\frac{25}{256}}\,{\frac{W}{\overline{{{m}}}L g}}\right]}}\,{\frac{E I}{\overline{{{m}}}L^{4}}}
|
||||
$$
|
||||
|
||||
## 8-7 IMPROVED RAYLEIGH METHOD
|
||||
|
||||
The idea of using a deflected shape resulting from an inertial loading in a Rayleigh analysis, as described above, can be applied systematically to develop improved versions of the procedure. The standard analysis involves the arbitrary selection of a deflected shape which satisfies the geometric boundary conditions of the structure. For the purposes of this discussion, this initially selected shape will be identified with the superscript zero:
|
||||
如上所述，在瑞利分析中，使用由惯性载荷引起的变形形状的想法可以系统地应用于开发该方法的改进版本。标准分析涉及任意选择一个满足结构几何边界条件的变形形状。为了本次讨论的目的，这个初始选择的形状将用上标零来表示：
|
||||
$$
|
||||
v^{(0)}(x,t)=\psi^{(0)}(x)\,Z_{0}^{(0)}\,\sin\omega t\tag {8-33}
|
||||
$$
|
||||
|
||||
The maximum potential and kinetic energies associated with this shape are then given by
|
||||
与此形状相关的最大势能和动能则由下式给出
|
||||
$$
|
||||
{\displaystyle V_{\mathrm{max}}=\frac{1}{2}\,\int_{0}^{L}E I(x)\left(\frac{\partial^{2}v^{(0)}}{\partial x^{2}}\right)^{2}d x=\frac{(Z_{0}^{(0)})^{2}}{2}\,\int_{0}^{L}E I(x)\,(\psi^{\prime\prime(0)})^{2}\,d x}\tag {8-34}
|
||||
$$
|
||||
$$
|
||||
{\displaystyle T_{\mathrm{max}}=\frac{1}{2}\,\int_{0}^{L}m(x)\,(\dot{v}^{(0)})^{2}\,d x=\frac{(Z_{0}^{(0)})^{2}}{2}\,\omega^{2}\,\int_{0}^{L}m(x)\,(\psi^{(0)})^{2}\,d x}\tag {8-35}
|
||||
$$
|
||||
Method $\underline{{R}}_{00}$ — The standard Rayleigh frequency expression, designated as method $R_{00}$ , is
|
||||
标准瑞利频率表达式，指定为方法 $R_{00}$ ，是
|
||||
$$
|
||||
\omega^{2}=\frac{\int_{0}^{L}E I(x)\,(\psi^{\prime\prime(0)})^{2}\,d x}{\int_{0}^{L}m(x)\,(\psi^{(0)})^{2}\,d x}\tag {8-36}
|
||||
$$
|
||||
|
||||
However, a better approximation of the frequency can be obtained by computing the potential energy from the work done in deflecting the structure by the inertial force associated with the assumed deflection. The distributed inertial force is (at the time of maximum displacement)
|
||||
然而，频率的更好近似可以通过从与假定变形相关的惯性力使结构变形所做的功中计算势能来获得。分布式惯性力是（在最大位移时）
|
||||
$$
|
||||
p^{(0)}(x)=\omega^{2}\,m(x)\,v^{(0)}=Z_{0}^{(0)}\,\omega^{2}\,m(x)\,\psi^{(0)}\tag {8-37}
|
||||
$$
|
||||
|
||||
The deflection produced by this loading may be written
|
||||
由此载荷产生的变形可写为
|
||||
$$
|
||||
v^{(1)}=\omega^{2}\,\frac{v^{(1)}}{\omega^{2}}=\omega^{2}\,\psi^{(1)}\,\frac{Z_{0}^{(1)}}{\omega^{2}}\equiv\omega^{2}\,\psi^{(1)}\,\overline{{{Z}}}_{0}^{(1)}\tag {8-38}
|
||||
$$
|
||||
|
||||
in which $\omega^{2}$ is the unknown squared frequency. It may be looked upon as a proportionality factor in both Eqs. (8-37) and (8-38); it is not combined into the expression because its value is not known. The potential energy of the strain produced by this loading is given by
|
||||
其中 $\omega^{2}$ 是未知的平方频率。它可以被看作是方程 (8-37) 和 (8-38) 中的一个比例因子；它没有被并入表达式中，因为它的值是未知的。由该载荷产生的应变势能由下式给出
|
||||
$$
|
||||
V_{\mathrm{max}}=\frac{1}{2}\,\int_{0}^{L}p^{(0)}\,v^{(1)}\,d x=\frac{Z_{0}^{(0)}\,\overline{{Z}}_{0}^{(1)}}{2}\,\omega^{4}\,\int_{0}^{L}m(x)\,\psi^{(0)}\,\psi^{(1)}\,d x\tag {8-39}
|
||||
$$
|
||||
|
||||
Method $\underline{{R}}_{01}$ — Equating this expression for the potential energy to the kinetic energy given by the originally assumed shape [Eq. (8-35)] leads to the improved Rayleigh frequency expression, here designated as method $R_{01}$ :
|
||||
方法 $\underline{{R}}_{01}$ — 将势能的这个表达式与由最初假设的形状 [式 (8-35)] 给出的动能等同起来，从而得到改进的瑞利频率表达式，此处指定为方法 $R_{01}$ :
|
||||
$$
|
||||
\omega^{2}=\frac{Z_{0}^{(0)}}{\overline{{{Z}}}_{0}^{(1)}}\,\frac{\int_{0}^{L}m(x)\,(\psi^{(0)})^{2}\,d x}{\int_{0}^{L}m(x)\,\psi^{(0)}\,\psi^{(1)}\,d x}\tag {8-40}
|
||||
$$
|
||||
|
||||
This expression often is recommended in preference to Eq. (8-36) because it avoids the differentiation operation required in the standard formula. In general, curvatures $\psi^{\prime\prime}(x)$ associated with an assumed deflected shape will be much less accurate than the shape function $\psi(x)$ , and thus Eq. (8-40), which involves no derivatives, will give improved accuracy.
|
||||
|
||||
However, a still better approximation can be obtained with relatively little additional effort by computing the kinetic energy from the calculated shape ${v}^{(1)}$ rather than from the initial shape ${{v}}^{(0)}$ . In this case the result is
|
||||
这种表达式通常比公式 (8-36) 更受推荐，因为它避免了标准公式中所需的微分运算。通常，与假定变形形状相关的曲率 $\psi^{\prime\prime}(x)$ 将远不如形状函数 $\psi(x)$ 精确，因此，不涉及导数的公式 (8-40) 将提供更高的精度。
|
||||
|
||||
然而，通过从计算出的形状 ${v}^{(1)}$ 而不是从初始形状 ${{v}}^{(0)}$ 计算动能，只需相对较少的额外工作即可获得更好的近似。在这种情况下，结果是
|
||||
$$
|
||||
T_{\mathrm{max}}={\frac{1}{2}}\,\int_{0}^{L}{m(x)\,(\dot{v}^{(1)})^{2}\,d x}={\frac{1}{2}}\,\omega^{6}\,(\overline{{Z}}^{(1)})^{2}\,\int_{0}^{L}{m(x)\,(\psi^{(1)})^{2}\,d x}\tag {8-41}
|
||||
$$
|
||||
|
||||
Method $\underline{{R_{11}}}$ — Equating this to the strain energy of Eq. (8-39) leads to the further improved result (here designated as method $R_{11}$ ):
|
||||
方法 $\underline{{R_{11}}}$ — 将其与式 (8-39) 的应变能相等，得到进一步改进的结果 (此处指定为方法 $R_{11}$ )：
|
||||
$$
|
||||
\omega^{2}={\frac{Z_{0}^{(0)}}{\overline{{Z}}_{0}^{(1)}}}\,{\frac{\int_{0}^{L}m(x)\,\psi^{(0)}\,\psi^{(1)}\,d x}{\int_{0}^{L}m(x)\,(\psi^{(1)})^{2}\,d x}}\tag {8-42}
|
||||
$$
|
||||
|
||||
Further improvement could be made by continuing the process another step, that is, by using the inertial loading associated with $\psi^{(1)}$ to calculate a new shape $\psi^{(2)}$ . In fact, as will be shown later, the process will eventually converge to the exact vibration shape if it is carried through enough cycles and therefore will yield the exact frequency. However, for practical use of the Rayleigh method there is no need to go beyond the improved procedure represented by Eq. (8-42). Also, it should be noted that the generalized-coordinate amplitudes $Z_{0}^{(0)}$ and $\overline{{Z}}_{0}^{(1)}$ in Eqs. (8-40) and (8-42) are arbitrary and can be set to unity if the shape functions $\psi^{(0)}$ and $\psi^{(1)}$ are defined appropriately. However, it is advisable to leave the generalized coordinates in the equations to show that the relative amplitude of ${{v}}^{(0)}$ and ${v}^{(1)}$ is a factor in computing the frequency.
|
||||
进一步的改进可以通过将该过程再推进一步来实现，即，利用与 $\psi^{(1)}$ 相关的惯性载荷来计算新的形状 $\psi^{(2)}$。事实上，正如稍后将展示的，如果该过程经过足够的循环，它最终将收敛到精确的振动形状，从而将得到精确的频率。然而，对于瑞利方法的实际应用，没有必要超出由式 (8-42) 表示的改进程序。此外，应该注意的是，式 (8-40) 和 (8-42) 中的广义坐标幅值 $Z_{0}^{(0)}$ 和 $\overline{{Z}}_{0}^{(1)}$ 是任意的，并且如果形状函数 $\psi^{(0)}$ 和 $\psi^{(1)}$ 定义得当，它们可以设为1。然而，建议将广义坐标保留在方程中，以表明 ${{v}}^{(0)}$ 和 ${v}^{(1)}$ 的相对幅值是计算频率的一个因素。
|
||||
|
||||
Example E8-6. The two improved versions of the Rayleigh method will be demonstrated and compared with the standard method in carrying out the frequency analysis of the three-story frame shown in Fig. E8-6a. The mass of this frame is lumped in the girders, with values as shown, and the columns are assumed to be weightless. Also, the girders are assumed to be rigid, so that the columns in each story act as simple lateral springs with stiffness coefficients as indicated.
|
||||
|
||||
Method $R_{00}\,{-}\,\mathrm{In}$ order to demonstrate the effectiveness of the improvement procedures, a poor choice will be deliberately assumed for the initial vibration shape for the frame. This shape consists of equal displacements for the three stories, as shown in Fig. E8- ${.6b}$ ; thus
|
||||
示例 E8-6。将演示瑞利法的两种改进版本，并将其与标准方法进行比较，以对图 E8-6a 所示的三层框架进行频率分析。该框架的质量集中在梁上，数值如图所示，且假定柱子是无质量的。此外，假定梁是刚性的，使得每层的柱子充当具有所示刚度系数的简单侧向弹簧。
|
||||
|
||||
方法 $R_{00}\,{-}$ 为了证明改进程序的有效性，将故意假设一个不佳的框架初始振动形状。这种形状包括三层相等的位移，如图 E8- ${.6b}$ 所示；因此
|
||||
$$
|
||||
v_{1}^{(0)}=v_{2}^{(0)}=v_{3}^{(0)}=1.0=Z_{0}^{(0)}\,\psi_{i}^{(0)}\qquad\mathrm{where}\qquad\psi_{i}^{(0)}=Z_{0}^{(0)}=1.0
|
||||
$$
|
||||
|
||||
From this shape, the maximum kinetic energy is given by
|
||||
|
||||
$$
|
||||
T_{\mathrm{max}}^{(0)}=\frac{1}{2}\,\sum m_{i}\,(v_{i}^{(0)})^{2}=\frac{1}{2}\,\omega^{2}\,(Z_{0}^{(0)})^{2}\,\sum m_{i}\,(\psi_{i}^{(0)})^{2}=\frac{1}{2}\,\omega^{2}(\psi_{i}^{(0)})^{2},
|
||||
$$
|
||||
|
||||
The maximum potential energy depends on the relative story-to-story deformations $\triangle v_{i}$ and is given by
|
||||
|
||||
$$
|
||||
V_{\mathrm{max}}^{(0)}=\frac{1}{2}\,\sum k_{i}(\triangle v_{i}^{(0)})^{2}=\frac{1}{2}\,(Z_{0}^{(0)})^{2}\,\sum k_{i}(\triangle\psi_{i}^{(0)})^{2}=\frac{1}{2}\,(1,800)
|
||||
$$
|
||||
|
||||
|
||||
FIGURE E8-6 Frame for Rayleigh method frequency analysis: (a) mass and stiffness values; $(b)$ initial assumed shape; (c) deflections resulting from initial inertial forces.
|
||||
|
||||
Hence, when the potential and kinetic energies are equated, the squared frequency is
|
||||
|
||||
$$
|
||||
\omega^{2}={\frac{1,800}{4.5}}=400\ \qquad\omega=20\ r a d/s e c
|
||||
$$
|
||||
|
||||
Method ${\underline{{R}}}_{01}$ — The assumption that the structure behaves as though the columns were rigid above the first story clearly is not reasonable for this frame and can be expected to give a gross overestimate of the frequency. Using the inertial forces associated with these initial deflections to calculate an improved shape, in accordance with the improved method $R_{01}$ , leads to much better results.
|
||||
|
||||
The inertial loads of the initial shape and the deflections they produce are shown in Fig. $\mathrm{E}8–6c$ . The deflections can easily be calculated because the deformation $\triangle v_{i}$ in each story is given by the story shear divided by the story stiffness. The maximum potential energy of this new shape $v_{i}^{(1)}$ may be found as follows:
|
||||
方法 ${\underline{{R}}}_{01}$ — 假设结构在第一层以上柱子是刚性的，这对于这个框架来说显然不合理，并且可以预期会严重高估频率。根据改进的方法 $R_{01}$，使用与这些初始变形相关的惯性力来计算改进的形状，会得到更好的结果。
|
||||
|
||||
初始形状的惯性载荷以及它们产生的变形显示在图 $\mathrm{E}8–6c$ 中。变形可以很容易地计算出来，因为每层的变形 $\triangle v_{i}$ 等于层剪力除以层刚度。这个新形状 $v_{i}^{(1)}$ 的最大势能可以按如下方式找到：
|
||||
$$
|
||||
V_{\mathrm{max}}^{(1)}=\frac{1}{2}\,\sum{p_{i}^{(0)}\,v_{i}^{(1)}}=\frac{\omega^{4}}{2}\,\overline{{Z}}_{0}^{(1)}\,\sum{m_{i}\,\psi_{i}^{(0)}\,\psi_{i}^{(1)}}=\frac{\omega^{4}}{2}\,\overline{{Z}}_{0}^{(1)}(
|
||||
$$
|
||||
|
||||
When this is equated to the kinetic energy found previously, the frequency is
|
||||
|
||||
$$
|
||||
\omega^{2}=\frac{1}{\overline{{Z}}_{0}^{(1)}}\,\frac{4.50}{2.90}=\frac{1}{22.5/3,600}\,\frac{4.5}{2.9}=248\qquad\omega=15.73\;r a d/s e c
|
||||
$$
|
||||
|
||||
It is apparent that this much smaller frequency represents a great improvement over the result obtained by the standard method $R_{00}$ .
|
||||
显然，这个小得多的频率比标准方法$R_{00}$得到的结果有了很大的改进。
|
||||
|
||||
Method $R_{11}$--Still better results can be obtained by using the improved shape ψ1 $\overline{{\psi_{1}^{(1)}}}$ in calculating the kinetic as well as the potential energy. Thus the maximum kinetic energy becomes
|
||||
方法 $R_{11}$--通过使用改进的形状 ψ1 $\overline{{\psi_{1}^{(1)}}}$，在计算动能和势能时可以获得更好的结果。因此，最大动能变为
|
||||
$$
|
||||
T_{\mathrm{max}}^{(1)}=\frac{\omega^{2}}{2}\,(\overline{{Z}}_{0}^{(1)})^{2}\,\sum m_{i}\,(\psi_{i}^{(1)})^{2}=\frac{\omega^{6}}{2}\left(\frac{22.5}{3,600}\right)^{2}
|
||||
$$
|
||||
|
||||
Hence, equating this to the improved potential-energy expression leads to the squared frequency value
|
||||
因此，将此等同于改进的势能表达式，可得到频率的平方值。
|
||||
$$
|
||||
\omega^{2}={\frac{1}{\overline{{{Z}}}_{0}^{(1)}}}\;{\frac{2.90}{2.124}}={\frac{3,600}{22.5}}\;{\frac{2.90}{2.124}}=218\qquad\omega=14.76\;r a d/s e c
|
||||
$$
|
||||
|
||||
This is quite close to the exact first-mode frequency for this structure, $\omega_{1}=$ 14.5 rad/sec, as will be derived in Chapter 11.
|
||||
这与该结构准确的第一个模态频率 $\omega_{1}=$ 14.5 rad/sec 相当接近，该频率将在第11章中推导。
|
||||
|
||||
It is interesting to note that method $R_{11}$ gives the same result here as would be given by Eq. (8-32), where the deflections due to a lateral gravity acceleration are the basis of the analysis. This is because the inertial forces associated with equal story displacements are equivalent to the lateral gravity forces. However, if a more reasonable estimate had been made of the initial shape (rather than equal story deflections), the improved method $R_{11}$ would have given a better result than Eq. (8-32).
|
||||
值得注意的是，方法 $R_{11}$ 在此处给出的结果与式 (8-32) 给出的结果相同，其中由横向重力加速度引起的变形是分析的基础。这是因为与相等层间位移相关的惯性力等效于横向重力。然而，如果对初始形状进行了更合理的估计（而不是相等层间变形），改进方法 $R_{11}$ 将会给出比式 (8-32) 更好的结果。
|
||||
## PROBLEMS
|
||||
|
||||
8-1. For the uniform cantilever tower of Example E8-3, the following expressions for the generalized mass and stiffness were determined:
|
||||
|
||||
$$
|
||||
\begin{array}{l}{{m^{*}=0.228\;\overline{{{m}}}\;L}}\\ {{\displaystyle{k^{*}=\frac{\pi^{4}}{32}\frac{E I}{L^{3}}}}}\end{array}
|
||||
$$
|
||||
|
||||
Based on these expressions, compute the period of vibration for a concrete tower $200~f t$ high, with an outside diameter of $12\ f t$ and wall thickness of $8~i n$ , for which the following properties may be assumed:
|
||||
|
||||
$$
|
||||
\begin{array}{l}{{\overline{{m}}=110\;l b\cdot s e c^{2}/f t^{2}}}\\ {{{\cal E}I=165\times10^{9}\;l b\cdot f t^{2}}}\end{array}
|
||||
$$
|
||||
|
||||
8-2. Assuming that the tower of Prob. 8-1 supports an additional point weight of 400 kips at the top, determine the period of vibration (neglecting the geometric stiffness effect).
|
||||
|
||||
8-3. For the system shown in Fig. P8-1, determine the generalized physical properties $m^{*},\,c^{*},$ $k^{*}$ , and the generalized loading $p^{*}(t)$ , all defined with respect to the displacement coordinate $Z(t)$ . Express the results in terms of the given physical properties and dimensions.
|
||||
|
||||
|
||||
|
||||
FIGURE P8-1
|
||||
|
||||
8-4. Repeat Prob. 8-3 for the structure shown in Fig. P8-2.
|
||||
|
||||
|
||||
|
||||
8-5. Repeat Prob. 8-3 for the structure shown in Fig. P8-3. (Hint: this system has only one dynamic degree of freedom; this is associated with the rotational inertia of the rigid bar of mass $m$ .)
|
||||
|
||||
|
||||
|
||||
FIGURE P8-3
|
||||
|
||||
8-6. The column of Fig. P8-4 is to be treated as a SDOF system by defining its displaced shape as
|
||||
|
||||
$$
|
||||
\psi(x)={\frac{v(x,t)}{Z(t)}}=\left({\frac{x}{L}}\right)^{2}\!\left({\frac{3}{2}}-{\frac{x}{2L}}\right)
|
||||
$$
|
||||
|
||||
Denoting the uniformly distributed mass per unit length by $\overline{{m}}$ , the uniform stiffness by $E I$ , and the uniformly distributed load per unit length by $\overline{{p}}(t)$ , evaluate the generalized physical properties $m^{*}$ and $k^{*}$ and the generalized loading $p^{*}(t)$ .
|
||||
|
||||
|
||||
FIGURE P8-4
|
||||
|
||||
8-7. $(a)$ If a downward load $N$ is applied at the top of the column of Prob. 8- 6, evaluate its combined generalized stiffness $\overline{{k}}^{\ast}$ using the same shape function $\psi(x)$ .
|
||||
|
||||
$(b)$ Repeat part $a$ assuming that the axial force in the column varies linearly along its length as $N(x)=N(1-x/L)$ .
|
||||
|
||||
8-8. Assume that the uniform slab of Fig. 8-4 is square, with side length $a$ , and is simply supported on all four edges.
|
||||
|
||||
$(a)$ If its mass per unit area is $\gamma$ and its flexural rigidity is $D$ , determine its generalized properties $m^{*}$ and $k^{*}$ in terms of the central displacement coordinate $Z(t)$ . Assume the displacement function is
|
||||
|
||||
$$
|
||||
\psi(x,y)=\sin{\frac{\pi x}{a}}\ \sin{\frac{\pi y}{a}}
|
||||
$$
|
||||
|
||||
$(b)$ The uniformly distributed external loading per unit of area is $\overline{{p}}(t)$ . Determine the generalized loading $p^{*}(t)$ based on the displacement function of part $a$ .
|
||||
|
||||
8-9. The outer diameters, height, and material properties of a conical concrete smokestack are shown in Fig. P8-5. Assuming a uniform wall thickness of $8\;i n$ and that the deflected shape is given by
|
||||
|
||||
$$
|
||||
\psi(x)=1-\frac{\cos\pi x}{2L}
|
||||
$$
|
||||
|
||||
compute the generalized mass $m^{*}$ and stiffness $k^{*}$ of the structure. Dividing the height into two equal segments, use Simpson’s rule to evaluate the integrals, including in the summations the integrand values for the bottom, middle, and top sections. For example
|
||||
|
||||
$$
|
||||
m^{*}\doteq\frac{\triangle x}{3}\left(y_{0}+4y_{1}+y_{2}\right)
|
||||
$$
|
||||
|
||||
where $y_{i}=m_{i}\psi_{i}^{2}$ evaluated at level $^{\ast}i$ .”
|
||||
|
||||
|
||||
FIGURE P8-5
|
||||
|
||||
8-10. By Rayleigh’s method, compute the period of vibration of the uniform beam supporting a central mass $m_{1}$ shown in Fig. P8-6. For the assumed shape, use the deflection produced by a central load $p$ ; i.e., $v(x)=p x(3L^{2}-4x^{2})/48E I$ for $0\le x\le L/2$ , symmetric with respect to $x=L/2$ . Consider the cases: $(a)$ $m_{1}=0$ , and $(b)\;m_{1}=3\overline{{m}}L$ .
|
||||
|
||||
|
||||
FIGURE P8-6
|
||||
|
||||
8-11. $(a)$ Determine the period of vibration of the frame shown in Fig. P8-7, assuming the girder to be rigid and the deflected shape of the columns to be that due to a lateral load $p$ acting on the girder $v(x)=p\,(3L^{2}x-x^{3})/12E I$ ; $(b)$ What fraction of the total column weight assumed lumped with the girder weight will give the same period of vibration as was found in part $a?$
|
||||
|
||||
|
||||
FIGURE P8-7
|
||||
|
||||
8-12. The shear building of Fig. P8-8 has its entire mass lumped in the rigid girders. For the given mass and stiffness properties, and assuming a linear initial shape (as shown), evaluate the period of vibration by:
|
||||
|
||||
$(a)$ Rayleigh method $R_{00}$ $(b)$ Rayleigh method $R_{01}$ $(c)$ Rayleigh method $R_{11}$
|
||||
|
||||
|
||||
FIGURE P8-8
|
||||
|
||||
8-13. Repeat Prob. 8-12 if the building properties are $m_{1}~=~1$ , $m_{2}~=~2$ , $m_{3}~=$ $3\;k i p s\cdot s e c^{2}/i n$ and $k_{1}=k_{2}=k_{3}=800~k i p s/i n$ .
|
||||
@ -0,0 +1,175 @@
|
||||
# Chap 9 FORMULATION OF THE MDOF EQUATIONS OF MOTION多自由度运动方程的建立
|
||||
|
||||
## 9-1 SELECTION OF THE DEGREES OF FREEDOM 自由度的选择
|
||||
|
||||
The discussion presented in Chapter 8 has demonstrated how a structure can be represented as a SDOF system the dynamic response of which can be evaluated by the solution of a single differential equation of motion. If the physical properties of the system are such that its motion can be described by a single coordinate and no other motion is possible, then it actually is a SDOF system and the solution of the equation provides the exact dynamic response. On the other hand, if the structure actually has more than one possible mode of displacement and it is reduced mathematically to a SDOF approximation by assuming its deformed shape, the solution of the equation of motion is only an approximation of the true dynamic behavior.
|
||||
|
||||
The quality of the result obtained with a SDOF approximation depends on many factors, principally the spatial distribution and time variation of the loading and the stiffness and mass properties of the structure. If the physical properties of the system constrain it to move most easily with the assumed shape, and if the loading is such as to excite a significant response in this shape, the SDOF solution will probably be a good approximation; otherwise, the true behavior may bear little resemblance to the computed response. One of the greatest disadvantages of the SDOF approximation is that it is difficult to assess the reliability of the results obtained from it.
|
||||
|
||||
In general, the dynamic response of a structure cannot be described adequately by a SDOF model; usually the response includes time variations of the displacement shape as well as its amplitude. Such behavior can be described only in terms of more than one displacement coordinate; that is, the motion must be represented by more than one degree of freedom. As noted in Chapter 1, the degrees of freedom in a discrete-parameter system may be taken as the displacement amplitudes of certain selected points in the structure, or they may be generalized coordinates representing the amplitudes of a specified set of displacement patterns. In the present discussion, the former approach will be adopted; this includes both the finite-element and the lumpedmass type of idealization. The generalized-coordinate procedure will be discussed in Chapter 16.
|
||||
|
||||
In this development of the equations of motion of a general MDOF system, it will be convenient to refer to the general simple beam shown in Fig. 9-1 as a typical example. The discussion applies equally to any type of structure, but the visualization of the physical factors involved in evaluating all the forces acting is simplified for this type of structure.
|
||||
|
||||
The motion of this structure will be assumed to be defined by the displacements of a set of discrete points on the beam: $v_{1}(t),v_{2}(t),...,v_{i}(t),...,v_{N}(t)$ . In principle, these points may be located arbitrarily on the structure; in practice, they should be associated with specific features of the physical properties which may be significant and should be distributed so as to provide a good definition of the deflected shape. The number of degrees of freedom (displacement components) to be considered is left to the discretion of the analyst; greater numbers provide better approximations of the true dynamic behavior, but in many cases excellent results can be obtained with only two or three degrees of freedom. In the beam of Fig. 9-1 only one displacement component has been associated with each nodal point on the beam. It should be noted, however, that several displacement components could be identified with each point; e.g., the rotation $\partial v/\partial x$ and longitudinal motions might be used as additional degrees of freedom at each point.
|
||||
第8章的讨论阐述了结构如何能被表示为一个单自由度（SDOF）系统，其动力响应可以通过求解一个单自由度运动微分方程来评估。如果系统的物理特性使其运动可以用一个坐标来描述，并且没有其他运动可能发生，那么它实际上就是一个SDOF系统，方程的解提供了精确的动力响应。另一方面，如果结构实际上有不止一种可能的位移模态，并且通过假设其变形形状在数学上被简化为SDOF近似，则运动方程的解只是真实动力行为的近似。
|
||||
|
||||
通过SDOF近似获得的结果质量取决于许多因素，主要是荷载的空间分布和时间变化，以及结构的刚度和质量特性。如果系统的物理特性使其最容易以假设的形状运动，并且如果荷载能够激发这种形状的显著响应，则SDOF解可能是一个很好的近似；否则，真实行为可能与计算响应相去甚远。SDOF近似的最大缺点之一是难以评估其结果的可靠性。
|
||||
|
||||
通常，结构的动力响应不能通过SDOF模型充分描述；通常响应包括位移形状的时间变化及其幅值。这种行为只能用一个以上的位移坐标来描述；也就是说，运动必须由一个以上的自由度来表示。如第1章所述，离散参数系统中的自由度可以取为结构中某些选定点的位移幅值，或者它们可以是表示一组特定位移模式幅值的广义坐标。在本次讨论中，将采用前一种方法；这包括有限元和集中质量两种理想化类型。广义坐标方法将在第16章讨论。
|
||||
|
||||
在推导一般多自由度（MDOF）系统的运动方程时，参照图9-1所示的一般简支梁作为典型示例将很方便。该讨论同样适用于任何类型的结构，但对于这种类型的结构，评估所有作用力所涉及的物理因素的可视化得到了简化。
|
||||
|
||||
该结构的运动将假定由梁上离散点集 $v_{1}(t),v_{2}(t),...,v_{i}(t),...,v_{N}(t)$ 的位移来定义。原则上，这些点可以任意放置在结构上；实际上，它们应与可能重要的物理特性的特定特征相关联，并应分布以提供对变形形状的良好定义。要考虑的自由度（位移分量）的数量由分析师自行决定；数量越多，对真实动力行为的近似越好，但在许多情况下，仅用两到三个自由度就能获得出色的结果。在图9-1的梁中，每个节点只关联了一个位移分量。然而，应该注意的是，每个点可以识别出几个位移分量；例如，转动 $\partial v/\partial x$ 和纵向运动可以作为每个点的附加自由度。
|
||||
|
||||
|
||||
## 9-2 DYNAMIC-EQUILIBRIUM CONDITION
|
||||
|
||||
The equation of motion of the system of Fig. 9-1 can be formulated by expressing the equilibrium of the effective forces associated with each of its degrees of freedom. In general four types of forces will be involved at any point $i$ : the externally applied load $p_{i}(t)$ and the forces resulting from the motion, that is, inertia $f_{I i}$ , damping $f_{D i}$ , and elastic $f_{S i}$ . Thus for each of the several degrees of freedom the dynamic equilibrium may be expressed as
|
||||
|
||||
$$
|
||||
\begin{array}{c}{f_{I1}+f_{D1}+f_{S1}=p_{1}(t)}\\ {f_{I2}+f_{D2}+f_{S2}=p_{2}(t)}\\ {f_{I3}+f_{D3}+f_{S3}=p_{3}(t)}\end{array}
|
||||
$$
|
||||
|
||||
or when the force vectors are represented in matrix form,
|
||||
|
||||
$$
|
||||
\mathbf{f}_{I}+\mathbf{f}_{D}+\mathbf{f}_{S}=\mathbf{p}(t)
|
||||
$$
|
||||
|
||||
which is the MDOF equivalent of the SDOF equation (2-1).
|
||||
|
||||
Each of the resisting forces is expressed most conveniently by means of an appropriate set of influence coefficients. Consider, for example, the elastic-force component developed at point 1; this depends in general upon the displacement components developed at all points of the structure:
|
||||
|
||||
$$
|
||||
f_{S1}=k_{11}v_{1}+k_{12}v_{2}+k_{13}v_{3}+\cdot\cdot\cdot+k_{1N}v_{N}
|
||||
$$
|
||||
|
||||
Similarly, the elastic force corresponding to the degree of freedom $v_{2}$ is
|
||||
|
||||
$$
|
||||
f_{S2}=k_{21}v_{1}+k_{22}v_{2}+k_{23}v_{3}+\cdot\cdot\cdot+k_{2N}v_{N}
|
||||
$$
|
||||
|
||||
and, in general,
|
||||
|
||||
$$
|
||||
f_{S i}=k_{i1}v_{1}+k_{i2}v_{2}+k_{i3}v_{3}+\cdot\cdot\cdot+k_{i N}v_{N}
|
||||
$$
|
||||
|
||||
In these expressions it has been tacitly assumed that the structural behavior is linear, so that the principle of superposition applies. The coefficients $k_{i j}$ are called stiffness influence coefficients, defined as follows:
|
||||
|
||||
$$
|
||||
\begin{array}{r}{k_{i j}=\mathrm{~force~corresponding~to~coordinate~}i\mathrm{~d}\mathrm{t}}\\ {\mathrm{~a~unit~}d i s p l a c e m e n t\;\mathrm{of~coordinate~}j\quad\quad}\end{array}
|
||||
$$
|
||||
|
||||
In matrix form, the complete set of elastic-force relationships may be written
|
||||
|
||||
$$
|
||||
\left\{\begin{array}{l}{f_{S1}}\\ {f_{S2}}\\ {\cdot}\\ {\cdot}\\ {f_{S i}}\\ {\cdot}\\ {\cdot}\end{array}\right\}=\left[\begin{array}{l l l l l l l}{k_{11}}&{k_{12}}&{k_{13}}&{\cdots}&{k_{1i}}&{\cdots}&{k_{1N}}\\ {k_{21}}&{k_{22}}&{k_{23}}&{\cdots}&{k_{2i}}&{\cdots}&{k_{2N}}\\ {\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}\\ {k_{i1}}&{k_{i2}}&{k_{i3}}&{\cdots}&{k_{i i}}&{\cdots}&{k_{i N}}\\ {\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}\end{array}\right]\left\{\begin{array}{l}{v_{1}}\\ {v_{2}}\\ {v_{3}}\\ {\cdot}\\ {v_{i}}\\ {\cdot}\\ {\cdot}\end{array}\right\}
|
||||
$$
|
||||
|
||||
or, symbolically,
|
||||
|
||||
$$
|
||||
\mathbf{f}_{S}=\mathbf{k}\ \mathbf{v}
|
||||
$$
|
||||
|
||||
in which the matrix of stiffness coefficients $\mathbf{k}$ is called the stiffness matrix of the structure (for the specified set of displacement coordinates) and $\mathbf{v}$ is the displacement vector representing the displaced shape of the structure.
|
||||
|
||||
If it is assumed that the damping depends on the velocity, that is, the viscous type, the damping forces corresponding to the selected degrees of freedom may be expressed by means of damping influence coefficients in similar fashion. By analogy with Eq. (9-5), the complete set of damping forces is given by
|
||||
|
||||
$$
|
||||
\left\{\begin{array}{c}{f_{D1}}\\ {f_{D2}}\\ {\cdot}\\ {\cdot}\\ {f_{D i}}\\ {\cdot}\end{array}\right\}=\left[\begin{array}{c c c c c c c}{c_{11}}&{c_{12}}&{c_{13}}&{\cdots}&{c_{1i}}&{\cdots}&{c_{1N}}\\ {c_{21}}&{c_{22}}&{c_{23}}&{\cdots}&{c_{2i}}&{\cdots}&{c_{2N}}\\ {\cdots}&{\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}\\ {c_{i1}}&{c_{i2}}&{c_{i3}}&{\cdots}&{c_{i i}}&{\cdots}&{c_{i N}}\\ {\cdots}&{\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}\end{array}\right]\,\left\{\begin{array}{c}{\dot{v}_{1}}\\ {\dot{v}_{2}}\\ {\dot{v}_{3}}\\ {\cdot}\\ {\dot{v}_{i}}\\ {\dot{v}_{i}}\\ {\cdot}\end{array}\right\}\,\,
|
||||
$$
|
||||
|
||||
in which $\dot{v}_{i}$ represents the time rate of change (velocity) of the $i$ displacement coordinate and the coefficients $c_{i j}$ are called damping influence coefficients. The definition of these coefficients is exactly parallel to Eq. (9-4):
|
||||
|
||||
$$
|
||||
\begin{array}{c}{{c_{i j}=\mathrm{force\;corresponding\;to\;coordinate\;}i\mathrm{\;due\;to\;unit}}}\\ {{{\nu e l o c i t y\;\mathrm{of\;coordinate\;}j}}}\end{array}
|
||||
$$
|
||||
|
||||
Symbolically, Eq. (9-7) may be written
|
||||
|
||||
$$
|
||||
\mathbf f_{D}=\mathbf c\,\dot{\mathbf v}
|
||||
$$
|
||||
|
||||
in which the matrix of damping coefficients $\mathbf{c}$ is called the damping matrix of the structure (for the specified degrees of freedom) and $\dot{\mathbf{v}}$ is the velocity vector.
|
||||
|
||||
The inertial forces may be expressed similarly by a set of influence coefficients called the mass coefficients. These represent the relationship between the accelerations of the degrees of freedom and the resulting inertial forces; by analogy with Eq. (9-5), the inertial forces may be expressed as
|
||||
|
||||
$$
|
||||
\left\{\begin{array}{l}{f_{I1}}\\ {f_{I2}}\\ {\cdot}\\ {\cdot}\\ {f_{I i}}\\ {\cdot}\\ {\cdot}\end{array}\right\}=\left[\begin{array}{l l l l l l l}{m_{11}}&{m_{12}}&{m_{13}}&{\cdots}&{m_{1i}}&{\cdots}&{m_{1N}}\\ {m_{21}}&{m_{22}}&{m_{23}}&{\cdots}&{m_{2i}}&{\cdots}&{m_{2N}}\\ {\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}\\ {m_{i1}}&{m_{i2}}&{m_{i3}}&{\cdots}&{m_{i i}}&{\cdots}&{m_{i N}}\\ {\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}\\ {\cdot}&{\cdot}&{\cdot}&{\cdot}&{\cdot}&{\cdot}\\ {m_{i1}}&{m_{i2}}&{m_{i3}}&{\cdots}&{\cdots}&{m_{i i}}&{\cdots}&{\cdots}\\ {\cdots}&{\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}\end{array}\right]=\left\{\begin{array}{l}{\ddot{v_{1}}}\\ {\ddot{v_{2}}}\\ {\ddot{v_{2}}}\\ {\cdot}\\ {\dot{v_{i}}}\\ {\dot{v_{i}}}\\ {\cdot}\end{array}\right\}
|
||||
$$
|
||||
|
||||
where $\ddot{v}_{i}$ is the acceleration of the $i$ displacement coordinate and the coefficients $m_{i j}$ are the mass influence coefficients, defined as follows:
|
||||
|
||||
$$
|
||||
m_{i j}=\mathrm{force\corresponding\to\coordinate\}i\ \mathrm{due}
|
||||
$$
|
||||
|
||||
Symbolically, Eq. (9-10) may be written
|
||||
|
||||
$$
|
||||
\mathbf{f}_{I}=\mathbf{m}\;\ddot{\mathbf{v}}
|
||||
$$
|
||||
|
||||
in which the matrix of mass coefficients $\mathbf{m}$ is called the mass matrix of the structure and $\ddot{\mathbf{v}}$ is its acceleration vector, both defined for the specified set of displacement coordinates.
|
||||
|
||||
Substituting Eqs. (9-6), (9-9), and (9-12) into Eq. (9-2) gives the complete dynamic equilibrium of the structure, considering all degrees of freedom:
|
||||
|
||||
$$
|
||||
\mathbf{m}\,\ddot{\mathbf{v}}(t)+\mathbf{c}\,\dot{\mathbf{v}}(t)+\mathbf{k}\,\mathbf{v}(t)=\mathbf{p}(t)
|
||||
$$
|
||||
|
||||
This equation is the MDOF equivalent of Eq. (2-3); each term of the SDOF equation is represented by a matrix in Eq. (9-13), the order of the matrix corresponding to the number of degrees of freedom used in describing the displacements of the structure. Thus, Eq. (9-13) expresses the $N$ equations of motion which serve to define the response of the MDOF system.
|
||||
|
||||
## 9-3 AXIAL-FORCE EFFECTS
|
||||
|
||||
It was observed in the discussion of SDOF systems that axial forces or any load which may tend to cause buckling of a structure may have a significant effect on the stiffness of the structure. Similar effects may be observed in MDOF systems; the force component acting parallel to the original axis of the members leads to additional load components which act in the direction (and sense) of the nodal displacements and which will be denoted by $\mathbf{f}_{G}$ . When these forces are included, the dynamic-equilibrium expression, Eq. (9-2), becomes
|
||||
|
||||
$$
|
||||
\mathbf{f}_{I}+\mathbf{f}_{D}+\mathbf{f}_{S}-\mathbf{f}_{G}=\mathbf{p}(t)
|
||||
$$
|
||||
|
||||
in which the negative sign results from the fact that the forces $\mathbf{f}_{G}$ are assumed to contribute to the deflection rather than oppose it.
|
||||
|
||||
These forces resulting from axial loads depend on the displacements of the structure and may be expressed by influence coefficients, called the geometric-stiffness
|
||||
|
||||
coefficients, as follows:
|
||||
|
||||
$$
|
||||
\left\{\begin{array}{l}{f_{G1}}\\ {f_{G2}}\\ {\cdot}\\ {\cdot}\\ {f_{G i}}\\ {\cdot}\\ {\cdot}\end{array}\right\}=\left[\begin{array}{c c c c c c c}{k_{G11}}&{k_{G12}}&{k_{G13}}&{\cdot\cdot}&{k_{G1i}}&{\cdot\cdot}&{k_{G1N}}\\ {k_{G21}}&{k_{G22}}&{k_{G23}}&{\cdot\cdot}&{k_{G2i}}&{\cdot\cdot}&{k_{G2N}}\\ {\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}\\ {\vdots}&{k_{G i1}}&{k_{G i2}}&{k_{G i3}}&{\cdot\cdot}&{k_{G i i}}&{\cdot\cdot}&{k_{G i N}}\\ {\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}\end{array}\right]\left\{\begin{array}{l}{v_{1}}\\ {v_{2}}\\ {v_{2}}\\ {\cdot}\\ {v_{i}}\\ {\cdot}\\ {\cdot}\end{array}\right\}
|
||||
$$
|
||||
|
||||
in which the geometric-stiffness influence coefficients $k_{G_{i j}}$ have the following definition:
|
||||
|
||||
$k_{G_{i j}}=\mathrm{force\;corresponding\;to\;coordinate\;}i\mathrm{~due\;to\;unit}$ displacement of coordinate $j$ and resulting from axial-force components in the structure
|
||||
|
||||
Symbolically Eq. (9-15) may be written
|
||||
|
||||
$$
|
||||
\mathbf{f}_{G}=\mathbf{k}_{G}\,\mathbf{v}
|
||||
$$
|
||||
|
||||
where $\mathbf{k}_{G}$ is called the geometric-stiffness matrix of the structure.
|
||||
|
||||
When this expression is introduced, the equation of dynamic equilibrium of the structure [given by Eq. (9-13) without axial-force effects] becomes
|
||||
|
||||
$$
|
||||
\mathbf{m}\,\ddot{\mathbf{v}}(t)+\mathbf{c}\,\dot{\mathbf{v}}(t)+\mathbf{k}\,\mathbf{v}(t)-\mathbf{k}_{G}\,\mathbf{v}(t)=\mathbf{p}(t)
|
||||
$$
|
||||
|
||||
or when it is noted that both the elastic stiffness and the geometric stiffness are multiplied by the displacement vector, the combined stiffness effect can be expressed by a single symbol and Eq. (9-18) written
|
||||
|
||||
$$
|
||||
\mathbf{m}\,\ddot{\mathbf{v}}(t)+\mathbf{c}\,\dot{\mathbf{v}}(t)+\overline{{\mathbf{k}}}\,\mathbf{v}(t)=\mathbf{p}(t)
|
||||
$$
|
||||
|
||||
in which
|
||||
|
||||
$$
|
||||
\overline{{\mathbf{k}}}=\mathbf{k}-\mathbf{k}_{G}
|
||||
$$
|
||||
|
||||
is called the combined stiffness matrix, which includes both elastic and geometric effects. The dynamic properties of the structure are expressed completely by the four influence-coefficient matrices of Eq. (9-18), while the dynamic loading is fully defined by the load vector. The evaluation of these physical-property matrices and the evaluation of the load vector resulting from externally applied forces will be discussed in detail in the following chapter. The effective-load vector resulting from support excitation will be discussed in connection with earthquake-response analysis in Chapter 26.
|
||||
@ -3681,7 +3681,7 @@ FIGURE P7-1
|
||||
|
||||
7-3. Solve Prob. 7-1, assuming the nonlinear elastic force-displacement relation, $f_{S}=12[\textstyle{\frac{2}{3}}v-\textstyle{\frac{1}{3}}(2v/3)^{3}]$ , which is sketched in Fig. $\mathsf{P}7\!-\!1b$ ( $f_{S}$ is in kips, $v$ in $i n)$ .
|
||||
|
||||
GENERALIZEDSINGLE-DEGREE-OF-FREEDOMSYSTEMS
|
||||
# Chap 8 GENERALIZEDSINGLE-DEGREE-OF-FREEDOMSYSTEMS
|
||||
|
||||
# 8-1 GENERAL COMMENTS ON SDOF SYSTEMS
|
||||
|
||||
@ -4466,7 +4466,7 @@ MULTIDEGREEOF
|
||||
FREEDOM
|
||||
SYSTEMS
|
||||
|
||||
FORMULATION OF THE MDOF EQUATIONS OF MOTION
|
||||
# Chap 9 FORMULATION OF THE MDOF EQUATIONS OF MOTION
|
||||
|
||||
# 9-1 SELECTION OF THE DEGREES OF FREEDOM
|
||||
|
||||
|
||||
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