﻿ 基于构件合力的曲线梁桥最不利激励方向确定
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 同济大学学报(自然科学版)  2019, Vol. 47 Issue (3): 301-308.  DOI: 10.11908/j.issn.0253-374x.2019.03.001 0

### 引用本文

FENG Ruiwei, LAO Tianpeng, DENG Tongfa, YUAN Wancheng. Resultant Response-based Method for Assessing the Critical Excitation Direction of Horizontally Curved Bridges[J]. Journal of Tongji University (Natural Science), 2019, 47(3): 301-308. DOI: 10.11908/j.issn.0253-374x.2019.03.001

### 文章历史

1. 同济大学 土木工程防灾国家重点实验室，上海 200092;
2. 广州大学 广州大学-淡江大学工程结构灾害与控制联合研究中心，广东 广州 510006;
3. 江西理工大学 建筑与测绘工程学院，江西 赣州 341000

Resultant Response-based Method for Assessing the Critical Excitation Direction of Horizontally Curved Bridges
FENG Ruiwei 1, LAO Tianpeng 1, DENG Tongfa 2,3, YUAN Wancheng 1
1. State Key Laboratory of Disaster Reduction in Civil Engineering, Tongji University, Shanghai 200092, China;
2. Guangzhou University-Tamkang University Joint Research Center for Engineering Structure Disaster Prevention and Control, Guangzhou University, Guangzhou 510006, China;
3. School of Architectural and Surveying and Mapping Engineering, Jiangxi University of Science and Technology, Ganzhou 341000, China
Abstract: Some methods for determining the critical excitation direction of structures regard the response along principal axes of components as the judging criterion, which have some limitations in the practical engineering application. To fill in this gap, a resultant response-based method was proposed. Based on the response spectrum method, formulae for this method were derived. A typical horizontally curved bridge was selected as the case project, and the finite element model for this bridge was built. The critical excitation direction of the bridge was assessed by applying the resultant response-based method and the results derived from this method were compared with those obtained from the linear time history analyses at multiple ground motion orientations. Results indicate that the resultant response-based method can comprehensively reflect the variation of mechanical performance of the bridge component with the seismic excitation direction and has advantages of minor computational efforts and higher reliability. When regarding the resultant response as the judging criterion, adopting the excitation directions recommended by the current code for dynamic analyses of curved bridges can satisfy the seismic design requirement of practical engineering.
Key words: horizontally curved bridges    critical excitation direction    resultant force    component    response spectrum method

1 构件合力法的理论推导

 图 1 单向地震动激励下构件坐标轴方向规定 Fig.1 Regulation for coordinate axes of a component subjected to a unidirectional seismic excitation

 $\mathit{\boldsymbol{\delta }}\left( t \right) = \sum\limits_{i = 1}^n {{\mathit{\boldsymbol{\phi }} _i}{\eta _i}\left( t \right)} = \mathit{\boldsymbol{\phi }} \mathit{\boldsymbol{\eta }}\left( t \right)$ (1)

 ${\mathit{\boldsymbol{R}}^2}\left( t \right) = \mathit{\boldsymbol{q}}_2^{\rm{T}}\mathit{\boldsymbol{\delta }}\left( t \right)$ (2)

 ${\mathit{\boldsymbol{R}}^2}\left( t \right) = \mathit{\boldsymbol{q}}_2^{\rm{T}}\sum\limits_{i = 1}^n {{\mathit{\boldsymbol{\phi }} _i}{\eta _i}\left( t \right)} = \sum\limits_{i = 1}^n {{\mathit{\boldsymbol{a}}_i}{\eta _i}\left( t \right)}$ (3)

 ${\mathit{\boldsymbol{a}}_i} = \mathit{\boldsymbol{q}}_2^{\rm{T}}{\mathit{\boldsymbol{\phi }} _i}$ (4)

 $\mathit{\boldsymbol{R}}_i^2 = {\mathit{\boldsymbol{a}}_i}{\eta _{i,\max }}$ (5)

 ${\mathit{\boldsymbol{R}}^3}\left( t \right) = \sum\limits_{i = 1}^n {{\mathit{\boldsymbol{b}}_i}{\eta _i}\left( t \right)}$ (6)
 $\mathit{\boldsymbol{R}}_i^3 = {\mathit{\boldsymbol{b}}_i}{\eta _{i,\max }}$ (7)

 $\begin{array}{*{20}{c}} {\mathit{\boldsymbol{R}}\left( t \right) = {\mathit{\boldsymbol{R}}^2}\left( t \right)\cos \alpha + {\mathit{\boldsymbol{R}}^3}\left( t \right)\sin \alpha = }\\ {\sum\limits_{i = 1}^n {\left( {{\mathit{\boldsymbol{a}}_i}\cos \alpha + {\mathit{\boldsymbol{b}}_i}\sin \alpha } \right){\eta _i}\left( t \right)} } \end{array}$ (8)

(1) 步骤1.将Sa沿X轴方向作用，由式(5)和(7)可知，构件2轴及3轴方向第i阶模态的峰值响应RiX2RiX3

 $\mathit{\boldsymbol{R}}_{iX}^2 = {\mathit{\boldsymbol{a}}_i}{\eta _{iX,\max }}$ (9)
 $\mathit{\boldsymbol{R}}_{iX}^3 = {\mathit{\boldsymbol{b}}_i}{\eta _{iX,\max }}$ (10)

 ${\mathit{\boldsymbol{R}}_X}\left( t \right) = \sum\limits_{i = 1}^n {\left( {{\mathit{\boldsymbol{a}}_i}\cos \alpha + {\mathit{\boldsymbol{b}}_i}\sin \alpha } \right){\eta _{iX}}\left( t \right)}$ (11)

 ${\mathit{\boldsymbol{R}}_{iX}}\left( t \right) = \left( {{\mathit{\boldsymbol{a}}_i}\cos \alpha + {\mathit{\boldsymbol{b}}_i}\sin \alpha } \right){\eta _{iX,\max }}$ (12)

(2) 步骤2.将Sa沿Y轴方向作用，与等式(11)、(12)推导过程相同，构件合响应RY(t)及第i阶模态的合响应峰值RiY可以表达为

 ${\mathit{\boldsymbol{R}}_Y}\left( t \right) = \sum\limits_{i = 1}^n {\left( {{\mathit{\boldsymbol{a}}_i}\cos \alpha + {\mathit{\boldsymbol{b}}_i}\sin \alpha } \right){\eta _{iY}}\left( t \right)}$ (13)
 ${\mathit{\boldsymbol{R}}_{iY}} = \left( {{\mathit{\boldsymbol{a}}_i}\cos \alpha + {\mathit{\boldsymbol{b}}_i}\sin \alpha } \right){\eta _{iY,\max }}$ (14)

(3) 步骤3.将Sa沿任意方向θ作用时，构件第i阶模态的合响应R(t)可以通过式(15)得到:

 $\begin{array}{l} {\mathit{\boldsymbol{R}}_{i\theta }}\left( t \right) = {\mathit{\boldsymbol{R}}_{iX}}\left( t \right)\cos \theta + {\mathit{\boldsymbol{R}}_{iY}}\left( t \right)\sin \theta = \\ \;\;\;\;\;\;\;\;\;\;\;\;\left( {{\mathit{\boldsymbol{a}}_i}\cos \alpha + {\mathit{\boldsymbol{b}}_i}\sin \alpha } \right){\eta _{iX}}\left( t \right)\cos \theta + \\ \;\;\;\;\;\;\;\;\;\;\;\;\left( {{\mathit{\boldsymbol{a}}_i}\cos \alpha + {\mathit{\boldsymbol{b}}_i}\sin \alpha } \right){\eta _{iY}}\left( t \right)\sin \theta \end{array}$ (15)

 ${\eta _{iX}}\left( t \right) = {\gamma _{iX}}{\eta _i}\left( t \right)$ (16)
 ${\eta _{iY}}\left( t \right) = {\gamma _{iY}}{\eta _i}\left( t \right)$ (17)

 $\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{R}}_{i\theta }}\left( t \right) = \left( {{\mathit{\boldsymbol{a}}_i}\cos \alpha + {\mathit{\boldsymbol{b}}_i}\sin \alpha } \right){\eta _{iX,\max }}\cos \theta + }\\ {\left( {{\mathit{\boldsymbol{a}}_i}\cos \alpha + {\mathit{\boldsymbol{b}}_i}\sin \alpha } \right){\eta _{iY,\max }}\sin \theta } \end{array}$ (18)

 $\begin{array}{l} {\mathit{\boldsymbol{R}}_{\theta ,\max }} = \sqrt {\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^n {{\rho _{ij}}{\mathit{\boldsymbol{R}}_{i\theta }}{\mathit{\boldsymbol{R}}_{j\theta }}} } } = \\ \sqrt {\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^n {{\rho _{ij}}\left( {{{\cos }^2}\theta {D_1} + {{\sin }^2}\theta {D_2} + 2\sin \theta \cos \theta {D_3}} \right)} } } \end{array}$ (19)

 ${D_1} = \mathit{\boldsymbol{R}}_{iX}^2\mathit{\boldsymbol{R}}_{jX}^2{\cos ^2}\alpha + 2\mathit{\boldsymbol{R}}_{iX}^2\mathit{\boldsymbol{R}}_{jX}^3{\cos ^2}\alpha \sin \alpha + \mathit{\boldsymbol{R}}_{iX}^3\mathit{\boldsymbol{R}}_{jX}^3{\sin ^2}\alpha$ (20)
 ${D_2} = \mathit{\boldsymbol{R}}_{iY}^2\mathit{\boldsymbol{R}}_{jY}^2{\cos ^2}\alpha + 2\mathit{\boldsymbol{R}}_{iY}^2\mathit{\boldsymbol{R}}_{jY}^3{\cos ^2}\alpha \sin \alpha + \mathit{\boldsymbol{R}}_{iY}^3\mathit{\boldsymbol{R}}_{jY}^3{\sin ^2}\alpha$ (21)
 $\begin{array}{l} {D_3} = \mathit{\boldsymbol{R}}_{iX}^2\mathit{\boldsymbol{R}}_{jY}^2{\cos ^2}\alpha + 2\mathit{\boldsymbol{R}}_{iX}^2\mathit{\boldsymbol{R}}_{jY}^3{\sin ^2}\alpha + \\ \;\;\;\;\;\;\;\mathit{\boldsymbol{R}}_{iX}^2\mathit{\boldsymbol{R}}_{jY}^3\cos \alpha \sin \alpha + \mathit{\boldsymbol{R}}_{iY}^2\mathit{\boldsymbol{R}}_{jX}^2\cos \alpha \sin \alpha \end{array}$ (22)

(1) 地震动最不利激励方向.首先给定θ范围，在每一个给定的θi下，依次循环α，求出地震动激励方向为θi时构件的合响应峰值Rθi, max，再分别比较各个激励方向下的Rθi, max，从而确定构件合响应最大值Rθ, max所对应的激励角度，即为最不利激励方向.

(2) 最大合响应方向.首先给定α的范围，在每一个给定的αi下，依次循环θ，求出各地震动激励方向下构件α方向的合响应峰值Rαi, max，再分别比较构件不同合响应方向下的Rαi, max，从而确定构件合响应最大值Rα, max所对应的响应方向，即为最大合响应方向.

2 数值模型建立及地震动选择 2.1 工程概况及有限元模型

 图 2 曲线梁桥整体布置及截面尺寸(单位：m) Fig.2 Configuration and gross-sectional dimension of the curved bridge(unit:m)

2.2 地震动选择及坐标轴方向规定

 图 3 所选地震动反应谱 Fig.3 Response spectrum for the three ground motions
 图 4 构件坐标轴方向规定 Fig.4 Regulation for principle axes of components
3 构件合力法的应用及结果分析

 图 5 Northridge(Rinaldi)地震沿不同方向激励时构件合响应峰值 Fig.5 Peak resultant responses of components when the Northridge (Rinaldi) earthquake excites along different directions
 图 6 Northridge(Rinaldi)地震作用时构件截面不同方向合响应峰值 Fig.6 Peak resultant responses at various directions of the cross-section of components under the Northridge (Rinaldi) earthquake

 图 7 Northridge(Rinaldi)地震作用时不同激励方向下构件响应分量峰值 Fig.7 Peak seismic responses of components along local axes under the Northridge (Rinaldi) earthquake
4 构件合力法与全角度输入法计算结果比较

 图 8 不同激励方向下2种方法所得结构响应峰值比较 Fig.8 Comparison of structural peak responses between the two methods
 图 9 现行规范规定的曲线桥地震输入方向 Fig.9 Excitation directions for curved bridges in the current codes
5 结论

(1) 对于任何结构，构件合力法只需在其纵、横向进行2次反应谱分析，再通过编制计算程序进行极值分析，便可求得结构的最不利激励方向，相比于全角度输入法(本文算例共进行13次时程分析)计算量更小，对于确定复杂结构的最不利激励方向更加省时.

(2) 构件合力法具有较高的计算精度.针对算例桥梁，构件合力法与全角度输入法计算得到的最不利激励方向相同且地震峰值响应相差最多不超过4%.应用构件合力法能够确定圆形截面桥梁构件的最不利受力部位，可以为实际工程的抗震设计及加固提供参考.

(3) 对于算例桥梁，3条地震波作用下构件合力法与全角度时程分析方法得到的最不利激励方向均为θ=45°；采用规范方法得到结构的最不利激励方向为与构件合力法及时程方法相差26°，但在最不利激励方向下构件响应峰值与时程方法最大相差3.5%，表明按照现行规范规定的地震动激励方向对曲线梁桥进行抗震性能验算能够满足实际工程抗震设计的需要.

(4) 曲线梁桥不同构件的地震响应对地震动激励方向的敏感性有所不同；对于算例桥梁，Northridge(Rinaldi)地震沿不同方向激励时墩底弯矩峰值最大仅相差5.3%，而不同激励方向下支座位移峰值最大相差15.2%.

(5) 地震动频谱特性对该曲线梁桥最不利激励方向的影响较小，在3条不同频谱特性的地震动输入下，该桥的最不利激励方向基本不变.

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