Transient Primary Resonance Phase-Frequency Characteristics of Stay Cables With Different Tensions
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摘要: 考虑拉索垂度及几何非线性，研究了不同索力拉索的瞬时相频特性。利用斜拉索面内分布激励下的运动控制方程，采用多尺度法对微分方程进行摄动求解，分别得到面内、外主共振响应的近似解析式，再采用Hilbert变换得到响应与激励的瞬时相位差及其幅值。研究了不同索力下，响应与激励的瞬时相位差的变化规律及其原因。研究表明：面外主共振响应与激励间保持恒定的相位差，而面内响应与激励的瞬时相位差与索弹性参数和垂度等有关，微小的索力变化可能导致瞬时相频特性的明显改变。主要原因是面内响应的近似解中存在两倍频项和漂移项，前者使响应瞬时相位在单个周期内出现两次正负交替，后者决定面内响应与激励瞬时相位差的最大值及其变化规律。Abstract: The transient phase-frequency characteristics of stay cables with different cable forces were studied in view of the cable sag and geometric nonlinearity. The method of multiple scales was used to solve the ordinary differential equations of motion for cables subjected to in-plane distributed excitations, and the approximate analytical expressions of in-plane and out-of-plane primary resonance responses were obtained respectively. Then, the transient phase difference and its amplitude between the response and the excitation were obtained through the Hilbert transform. The rule and reason for the transient phase difference between the response and the excitation under different cable forces were studied. The results show that, the phase difference between the out-of-plane response and the excitation is constant, while for the in-plane one it is related to the elastic parameters and the sag of the cable. A small change in cable tension may result in a significant change in the transient phase-frequency characteristics. The main reason is that there are a twice-frequency term and a drift term in the approximate solution of the in-plane response, the former makes the transient phase of response appear twice positive-negative alternations in a single cycle, and the latter determines the maximum value and the variation law of the transient phase difference between the in-plane response and excitation.
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图 1 分布外激励下斜拉索振动简化模型
Figure 1. Simplified model of vibration of cable under distributed external excitation
表 1 Normandie桥索参数
Table 1. Cable parameters of the Normandie bridge
l/m m/(kg·m−1) H/kN A/m2 E/Pa θ/(°) C 440 136 8 000 1.53×10−2 1.9×1011 17.5 0.001 
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