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    等跨等截面連續梁的解析計算公式

    周毅 岳清瑞 潘旦光 蘭成明

    周毅, 岳清瑞, 潘旦光, 蘭成明. 等跨等截面連續梁的解析計算公式[J]. 工程科學學報. doi: 10.13374/j.issn2095-9389.2023.08.31.002
    引用本文: 周毅, 岳清瑞, 潘旦光, 蘭成明. 等跨等截面連續梁的解析計算公式[J]. 工程科學學報. doi: 10.13374/j.issn2095-9389.2023.08.31.002
    ZHOU Yi, YUE Qingrui, PAN Danguang, LAN Chengming. Analytical formulas for prismatic continuous beams of equal spans[J]. Chinese Journal of Engineering. doi: 10.13374/j.issn2095-9389.2023.08.31.002
    Citation: ZHOU Yi, YUE Qingrui, PAN Danguang, LAN Chengming. Analytical formulas for prismatic continuous beams of equal spans[J]. Chinese Journal of Engineering. doi: 10.13374/j.issn2095-9389.2023.08.31.002

    等跨等截面連續梁的解析計算公式

    doi: 10.13374/j.issn2095-9389.2023.08.31.002
    基金項目: 國家自然科學基金資助項目(52192662); 佛山市科技創新專項資金資助項目(BK20BE008,BK21BE014)
    詳細信息
      通訊作者:

      E-mail: zhouyi@ustb.edu.cn

    • 中圖分類號: TU378.2

    Analytical formulas for prismatic continuous beams of equal spans

    More Information
    • 摘要: 求解等跨等截面連續梁的變形和內力是土木工程領域的典型問題. 基于Euler–Bernoulli梁理論,利用位移法和輔助數列推導出任意跨數的等跨等截面連續梁梁端轉動剛度的解析表達式,進而得到連續梁支點轉角、彎矩在跨中集中荷載、滿跨均布荷載、豎向溫差作用下的通用計算公式. 研究表明:當跨數趨于無窮大時,等跨等截面連續梁的梁端轉動剛度上限值為單跨梁抗彎線剛度的2$ \sqrt{\text{3}} $倍. 不同跨數的等跨等截面連續梁可采用形式統一的解析公式計算支點轉角和彎矩,不同靜力荷載作用結果的區別僅由單跨梁的固端彎矩決定. 所得公式形式簡潔、通用性強、應用方便,能揭示跨數對連續梁力學特性的影響規律,亦可用于分析頂推施工導梁參數優化等實際工程問題.

       

    • 圖  1  n跨連續梁受到梁端彎矩作用的分析模型. (a) M0作用在最左端; (b) M0作用在最右端

      Figure  1.  Analytical model of an n-span continuous beam subjected to the beam-end moment: (a) M0 applied at the leftmost end; (b) M0 applied at the rightmost end

      圖  2  n跨連續梁在第u跨受到荷載作用的分析模型

      Figure  2.  Analytical model of an n-span continuous beam with the uth span subjected to arbitrary loads

      表  1  等跨等截面梁的計算公式

      Table  1.   Formulas for a prismatic continuous beam of equal spans

      Case No. Analytical model Rotation, $ {z_k} $ Moment, $ {M_k} $ Parameter, $ M_1^{{\text{fix}}} $
      1Eq. 23Eq. 24$ M_1^{{\text{fix}}} = {{ - F{l_0}} \mathord{\left/ {\vphantom {{ - F{l_0}} 8}} \right. } 8} $
      2$ M_1^{{\text{fix}}} = {{ - ql_0^2} \mathord{\left/ {\vphantom {{ - ql_0^2} {12}}} \right. } {12}} $
      3$ M_1^{{\text{fix}}} = {{EI\alpha \cdot \Delta T} \mathord{\left/ {\vphantom {{EI\alpha \cdot \Delta T} h}} \right. } h} $
      4Eq. 25Eq. 26$ M_1^{{\text{fix}}} = {{ - F{l_0}} \mathord{\left/ {\vphantom {{ - F{l_0}} 8}} \right. } 8} $
      5$ M_1^{{\text{fix}}} = {{ - ql_0^2} \mathord{\left/ {\vphantom {{ - ql_0^2} {12}}} \right. } {12}} $
      6$ M_1^{{\text{fix}}} = {{EI\alpha \cdot \Delta T} \mathord{\left/ {\vphantom {{EI\alpha \cdot \Delta T} h}} \right. } h} $
      Note: α and h are the linear expansion coefficient and the cross-sectional depth of the beam, respectively. The temperature changes in the top and bottom surfaces of the beam are denoted as t1 and t2, respectively.
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    • 收稿日期:  2022-08-31
    • 網絡出版日期:  2023-08-02

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