Effect of Shaking Table Errors on Specimen Response and Its Correction Measures
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摘要: 为了改善振动台控制精度问题,获取准确的结构响应,首先就振动台特性对结构响应的影响进行了分析,而后基于振动台输入输出响应,提出了实测结构响应修正方法. 试验结果证明:修正方法较好地补偿了振动台误差的影响,修正后的试件响应能代表试件在期望地震动作用下的试件真实响应,因此,该修正方法能提高结构响应精度.Abstract: Shaking table test is the main testing method for engineering structures research, which plays an important role in the research for seismic performance of structures. In order to improve control accuracy problems of shaking tables, and to get the authentic structural response, firstly, the influence of the shaking table dynamic on structural response was analyzed. And then a correction measure was developed based on the input and output of the shaking table. Correction method of better compensation of the influence of vibration table error was provided. And specimen corrected response can be better representative specimen’s response at the desired ground motions. Finally, the performance of the developed method was verified experimentally.
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图 2 不同负载下振动台传递函数
Figure 2. Transfer functions of shaking table with different specimens
图 3 考虑振动台误差前后试件加速度响应对比
Figure 3. Comparison of acceleration with and without the consideration of shaking tabledynamics
图 5 振动台误差对原型试验反应谱的影响(Ms=0.5Mt)
Figure 5. Effect of shaking table errors on the response spectrum of prototype specimen when Ms=0.5Mt
图 6 振动台误差对原型试验反应谱的影响(Ms=1.5Mt)
Figure 6. Effect of shaking table errors on the response spectrum of prototype specimen when Ms=1.5Mt
图 7 振动台误差对缩尺试验反应谱的影响(Ms=1.5Mt)
Figure 7. Effect of shaking table errors on the response spectrum of scaled specimen when Ms=1.5Mt
图 8 振动台误差对非线性结构试验反应谱的影响(Ms=1.5Mt)
Figure 8. Effect of shaking table errors on the response spectrum of nonlinear specimen when Ms=1.5Mt
图 10 振动台再现误差修正流程
Figure 10. Overall process of correction measures for shaking table errors
图 13 修正前后试件响应对比(Ms/Mt=0.7)
Figure 13. Comparison of the specimen response before and after correction when Ms/Mt=0.7
图 14 修正前后试件响应对比(Ms/Mt=1.6)
Figure 14. Comparison of the specimen response before and after correction when Ms/Mt=1.6
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