双自由度系统

无阻尼自由振动

柔度法振动方程

$$\begin{aligned} y_1 = -m_1\ddot{y}_1 \delta_{11} -m_2\ddot{y}_2 \delta_{12} \\ y_2 = -m_1\ddot{y}_1 \delta_{21} -m_2\ddot{y}_2 \delta_{22} \end{aligned}$$

设 $y_1$ , $y_2$ 为如下形式：

$$\begin{aligned} y_1 = \rho_1 \sin(\omega t + \alpha) \\ y_2 = \rho_2 \sin(\omega t + \alpha) \end{aligned}$$

代入并消去共同项 $\sin(\omega t + \alpha)$ 后得到：

$$\begin{aligned} \rho_1 = m_1\delta_{11} \omega^2 \rho_1 + m_2\delta_{12} \omega^2 \rho_2 \\ \rho_2 = m_1\delta_{21} \omega^2 \rho_1 + m_2\delta_{22} \omega^2 \rho_2 \end{aligned}$$

为使方程有解，则要求行列式为 $0$ ：

$$\begin{aligned} \left| \begin{matrix} \displaystyle m_1\delta_{11}-\frac{1}{\omega^2} & m_2\delta_{12} \\ m_1\delta_{21} &\displaystyle m_2\delta_{22}-\frac{1}{\omega^2} \end{matrix} \right| = 0 \end{aligned}$$

设 $\displaystyle \lambda = \frac{1}{\omega^2}$ ，可解得 $\lambda_1$ ， $\lambda_2$ ，并可分别求出一个 $\displaystyle\frac{\rho_1}{\rho_2}$ ，对应不同的振型。

无阻尼受迫振动

$$\begin{aligned} y_1 = [p_1(t)-m_1\ddot{y}_1] \delta_{11} + [p_2(t)-m_2\ddot{y}_2] \delta_{12} \\ y_2 = [p_1(t)-m_1\ddot{y}_1] \delta_{21} + [p_2(t)-m_2\ddot{y}_2] \delta_{22} \end{aligned}$$

这里考虑同频率简谐振动

$$\begin{aligned} p_1 = P_1 \sin(\theta t) \\ p_2 = P_2 \sin(\theta t) \end{aligned}$$

再考虑到

$$\begin{aligned} \Delta_1 = P_1\delta_{11} + P_2\delta_{12} \\ \Delta_2 = P_1\delta_{21} + P_2\delta_{22} \end{aligned}$$

则原方程可化为

$$\begin{aligned} \rho_1 = m_1\delta_{11} \omega^2 \rho_1 + m_2\delta_{12} \omega^2 \rho_2 + \Delta_1 \\ \rho_2 = m_1\delta_{21} \omega^2 \rho_1 + m_2\delta_{22} \omega^2 \rho_2 + \Delta_2 \end{aligned}$$

由线性代数克莱默法则得出方程的解。

同频同相位

外力和惯性力同时取到最大值。由此可计算结构最大的弯矩，并得出动力放大系数 $\text{MF}$ 。