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基于拟力法的框架结构静力推覆分析

郝润霞,王谋庭,贾硕,李钢

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郝润霞, 王谋庭, 贾硕, 李钢. 基于拟力法的框架结构静力推覆分析[J]. 江南娱乐网页版入口官网下载安装学报, 2020, 55(5): 1028-1035. doi: 10.3969/j.issn.0258-2724.20180100
引用本文: 郝润霞, 王谋庭, 贾硕, 李钢. 基于拟力法的框架结构静力推覆分析[J]. 江南娱乐网页版入口官网下载安装学报, 2020, 55(5): 1028-1035.doi:10.3969/j.issn.0258-2724.20180100
HAO Runxia, WANG Mouting, JIA Shuo, LI Gang. Static Pushover Analysis of Frame Structure Based on Force Analogy Method[J]. Journal of Southwest Jiaotong University, 2020, 55(5): 1028-1035. doi: 10.3969/j.issn.0258-2724.20180100
Citation: HAO Runxia, WANG Mouting, JIA Shuo, LI Gang. Static Pushover Analysis of Frame Structure Based on Force Analogy Method[J].Journal of Southwest Jiaotong University, 2020, 55(5): 1028-1035.doi:10.3969/j.issn.0258-2724.20180100

基于拟力法的框架结构静力推覆分析

doi:10.3969/j.issn.0258-2724.20180100
基金项目:国家自然科学基金(51422802)
详细信息
    作者简介:

    郝润霞(1972—),女,副教授,研究方向为建筑结构抗震减震,电话:13604721753,E-mail:13604721753@163.com

  • 中图分类号:TU311.4

Static Pushover Analysis of Frame Structure Based on Force Analogy Method

    • 摘要:传统静力推覆分析方法求解结构非线性变形需对结构整体刚度矩阵进行实时地合成与分解,该过程将占用大量计算资源.基于拟力法的纤维梁有限元分析方法进行静力推覆分析,在迭代求解结构非线性变形时,首先对弹性刚度矩阵进行分解,计算出侧向荷载作用下的弹性位移;然后通过反复调用弹性刚度矩阵的分解结果与弹性位移,减少回代计算量;最后采用算法时间复杂度理论定量对比了该方法与传统方法的计算效率,通过一榀八层钢筋混凝土框架结构数值算例,分析比较了两种方法的计算结果与算法时间复杂度. 结果表明:两种方法顶点位移-基底剪力曲线基本吻合,层间位移角与楼层之间的关系曲线也基本一致,两者的最大误差出现在第3层,为3.72%,与传统方法相比,基于拟力法的静力推覆分析方法算法时间复杂度降低了80%,计算效率至少是传统方法的5倍.

    • 图 1基于拟力法的静力推覆分析流程

      Figure 1.Flow chart for static pushover analysis based on force analogy method

      图 2钢筋混凝土一榀框架几何尺寸

      Figure 2.Size of reinforced concrete frame

      图 3顶点位移-基底剪力

      Figure 3.Relationship between roof displacement and base shear

      图 4楼层数与层间位移角关系

      Figure 4.Relationship between story drift ratio and floors

      图 5塑性自由度数与顶点位移关系

      Figure 5.Relationship between number of plastic degrees and roof displacement

      表 1时间复杂度

      Table 1.Time complexity

      计算步骤 计算项目 时间复杂度
      1 $\left( {\Delta { {{X} }_1} } \right)_1^{\left( j \right)} = \Delta \lambda _1^{\left( j \right)}{ {{U} }_{\rm{F} } }$ $n$
      2 $\Delta { {{u} }_0} = \left( { {{K} }_{ {\rm{pr} } }^{\rm{T} } } \right)_1^{\left( j \right)}{\left( {\Delta { {{X} }_1} } \right)^{\left( j \right)} }$ $2\alpha d - d$
      3 $\left( {{{{K}}_{{\rm{pl}}}}} \right)_1^{\left( j \right)} = \left( {{{{K}}_{{\rm{rr}}}}} \right)_1^{\left( j \right)} - \left( {{{K}}_{{\rm{pr}}}^{\rm{T}}} \right)_1^{\left( j \right)}{{K}}_{\rm{e}}^{ - 1}\left( {{{{K}}_{{\rm{pr}}}}} \right)_1^{\left( j \right)}$ $\beta d/2$
      4 $\Delta {{{u}}_1} = \left( {{{K}}_{{\rm{pl}}}^{ - 1}} \right)_1^{\left( j \right)}\Delta {{{u}}_0}$ $m_{\rm{d}}^2d + 8{m_{\rm{d}}}d + d$
      5 $\Delta {{{u}}_2} = \left( {{{{K}}_{{\rm{pr}}}}} \right)_1^{\left( j \right)}\Delta {{{u}}_1}$ $2\alpha d - d$
      6 $\left( {\Delta { {{X} }_2} } \right)_1^{\left( j \right)} = {{K} }_{\rm{e} }^{ - 1}\Delta { {{u} }_2}$ $2mn - n - {m^2}/2 + 1/2$
      7 $\left( {\Delta {{X} } } \right)_1^{\left( j \right)} = \left( {\Delta { {{X} }_1} } \right)_1^{\left( j \right)} + \left( {\Delta { {{X} }_2} } \right)_1^{\left( j \right)}$ $n$
      下载: 导出CSV

      表 2截面配筋

      Table 2.Parameters of sections

      楼层 中柱 边柱
      截面的宽,高/mm 配筋/mm2 截面的宽,高/mm 配筋/mm2 截面的宽,高/mm 配筋/mm2
      1 $300,\;600$ 6874 $700 ,\;700$ 8758 $700 ,\;700$ 11235
      2 $300 ,\;600$ 6874 $700 ,\;700$ 5024 $700 ,\;700$ 5024
      3 $300 ,\;600$ 6874 $700 ,\;700$ 5024 $700 ,\;700$ 5024
      4 $300 ,\;600$ 6874 $700 ,\;700$ 5024 $700 ,\;700$ 5024
      5 $300 ,\;600$ 4688 $600 ,\;600$ 3786 $600 ,\;600$ 3786
      6 $300 ,\;600$ 4688 $600 ,\;600$ 3786 $600 ,\;600$ 3786
      7 $300 ,\;600$ 4688 $600 ,\;600$ 3786 $600 ,\;600$ 3786
      8 $300 ,\;600$ 4688 $600 ,\;600$ 3786 $600 ,\;600$ 3786
       注:配筋指截面中钢筋截面面积.
      下载: 导出CSV
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    • 收稿日期:2018-02-14
    • 修回日期:2020-05-25
    • 网络出版日期:2020-06-19
    • 刊出日期:2020-10-01

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