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竖向荷载作用下的内力计算

第6章 竖向荷载作用下内力计算§6.1 框架结构的荷载计算§6.1.1.板传荷载计算计算单元见下图所示:因为楼板为整体现浇,本板选用双向板,可沿四角点沿45°线将区格分为小块,每个板上的荷载传给与之相邻的梁,板传至梁上的三角形或梯形荷载可等效为均布荷载。

一.A ~B, (C ~E)轴间框架梁:屋面板传荷载:恒载:()()[]++⨯-⨯⨯3226.6/25.26.6/25.22125.2KN/m 06.7 ()()[]m KN /44.226.6/5.16.6/5.1215.106.732=+⨯-⨯⨯ 活载:()()[]++⨯-⨯⨯3226.6/25.26.6/25.22125.2KN/m 2 ()()[]m KN /36.66.6/5.16.6/5.1215.1232=+⨯-⨯⨯ 楼面板传荷载:恒载:()()[]++⨯-⨯⨯3226.6/25.26.6/25.22125.2.1KN/m 4 ()()[]m KN /03.136.6/5.16.6/5.1215.11.432=+⨯-⨯⨯活载:()()[]++⨯-⨯⨯3226.6/25.26.6/25.22125.2.5KN/m 2 ()()[]m KN /95.76.6/5.16.6/5.1215.15.232=+⨯-⨯⨯ 梁自重:3.34KN/mA ~B, (C ~E)轴间框架梁均布荷载为:屋 面 梁:恒载=梁自重+板传荷载=3.34 KN/m+22.44 KN/m=25.78 KN/m活载=板传荷载=6.36 KN/m楼面板传荷载:恒载=梁自重+板传荷载=3.34 KN/m+13.03 KN/m=116.37 KN/m活载=板传荷载=7.95 KN/m二. B ~C 轴间框架梁:屋面板传荷载:恒载:()()[]++⨯-⨯⨯3222.7/25.22.7/25.22125.2.06KN/m 7 ()[]m KN .10.142.7/5.12.7/5.1215.1.6KN/m 0322=+⨯-⨯⨯ 活载:()()[]++⨯-⨯⨯322.7/25.22.7/25.22125.22()[]m KN .17.42.7/5.12.7/5.1215.1.3KN/m 0322=+⨯-⨯⨯ 楼面板传荷载:恒载:()()[]++⨯-⨯⨯3222.7/25.22.7/25.22125.2.1KN/m 4 ()[]m KN .38.132.7/5.12.7/5.1215.1.1KN/m 4322=+⨯-⨯⨯ 活载:()()[]++⨯-⨯⨯3222.7/25.22.7/25.22125.2.5KN/m 2 ()[]m KN .16.82.7/5.12.7/5.1215.1.5KN/m 2322=+⨯-⨯⨯ 梁自重:3.34KN/mB ~C 轴间框架梁均布荷载为:屋 面 梁:恒载=梁自重+板传荷载=3.34 KN/m+14.10 KN/m=17.44 KN/m活载=板传荷载=4.17 KN/m楼面板传荷载:恒载=梁自重+板传荷载=3.34 KN/m+13.38KN/m=16.72KN/m活载=板传荷载=8.16 KN/m三.A 轴柱纵向集中荷载计算:顶层柱:顶层柱恒载=女儿墙+梁自重+板传荷载=KN 5.1475.328/525.206.72.334.235.4.6726=⨯⨯⨯⨯+⨯+⨯顶层柱活载=板传荷载=KN 28.18)5.15.125.225.22(28/52=⨯+⨯⨯⨯标准层柱恒载=墙自重+梁自重+板荷载 =KN 01.72)5.125.2(28/51.42.334.3)5.075.3(76.522=+⨯⨯⨯+⨯+-⨯准层柱活载=板传荷载=KN m m KN 8.288.428/54.2/2=⨯⨯⨯⨯○a 基础顶面荷载=底层外纵墙自重+基础自重=KN 85.22)55.075.3(75.3)55.075.3(11.14=-⨯+-⨯四.C 柱纵向集中力计算:顶层柱荷载=梁自重+板传荷载=3.13×(3.75-0.5)+++⨯⨯⨯)225.125.2(28/506.70.6×1.5×5/8×2×1.5=120.91KN顶层柱活载=板传荷载==KN 78.315.128/55.13.0)5.1225.2(28/5222=⨯⨯⨯⨯+⨯+⨯⨯⨯标准柱恒载=墙+梁自重+板传荷载=m KN /32.114)5.1225.2(28/51.4)5.075.4(113.319.2922=⨯+⨯⨯⨯+-⨯+标准层活载=板传荷载=KN 70.45328/55.15.25.428/525.25.2=⨯⨯⨯⨯+⨯⨯⨯⨯基础顶面恒载=底层内纵墙自重+基础自重=KN 95.58)55.075.3(75.395.46=-⨯+(3).框架柱自重:柱自重: 底层:边柱1.2×0.55m ×0.55m ×253/KN m ×5.4m=49.01KN中柱1.2×0.55m ×0.55m ×253/KN m ×5.4m=49.01KN标准层: 边柱1.2×0.5m ×0.5m ×253/KN m ×3.6m=27KN中柱1.2×0.5m ×0.5m ×253/KN m ×3.6m=27KN顶层: 边柱1.2×0.55m ×0.55m ×253/KN m ×3.9m=35.39KN中柱1.2×0.55m ×0.55m ×253/KN m ×3.9m=35.39KN§6.2恒荷载作用下框架的内力§6.2.1.恒荷载作用下框架的弯矩计算恒荷载作用下框架的受荷简图如图6-3所示,由于A E 二轴的纵梁外边线分别与该柱的外边线齐平,故此二轴上的竖向荷载与柱轴线偏心,且偏心距离为75mm。

则均布恒载和集中荷载偏心引起的固端弯矩构成框架节点不平衡弯矩。

一.恒荷载作用下框架可按下面公式求得:均载M 1: 21/12ab M ql =- (61)-21/12ba M ql =集中荷载M e :M e =-Fe (62)-故:m KN M B A .58.936.678.2512/1266-=⨯⨯-= m KN M A B .58.9366=m KN M C B .34.752.744.1712/1266-=⨯⨯-= m KN M B C .34.7566=m KN M B A .42.596.642.5912/1255-=⨯⨯-= m KN M A B .24.5955=m KN M C B .23.722.772.1612/1255-=⨯⨯-= m KN M B C .23.7255-= m KN M e .06.18125.05.1146=⨯=m KN M e .11.9125.085.725=⨯=m KN M e .01.9125.001.724=⨯=所以梁固端弯矩为1M M M e n F +=恒荷载作用下框架的受荷简图如图6-3所示:图6-3竖向受荷总图:注:1.图中各值的单位为KN, 2.图中数值均为标准值3.图中括号数值为活荷载图6-4:恒载作用下的受荷简图(2).根据梁,柱相对线刚度,算出各节点的弯矩分配系数ij μ:/()ij c b i i i μ=∑+∑ (63)-分配系数如图6-5 , 图6-6所示:A 柱:底层 335.0)977.00.1942.0/(977.00=++=下柱i342.0)977.00.1942.0/(0.1=++=上柱i323.0)977.00.1942.0/(942.0=++=左梁i标准层:340.0)0.10.1942.0/(0.1=++=上柱i340.0)0.10.1942.0/(0.1=++=下柱i32.0)0.10.1942.0/(942.0=++=左梁i五层: 410.0)0.1942.035.1/(35.1=++=上柱i304.0)0.1942.035.1/(0.1=++=下柱i286.0)0.1942.035.1/(9420=++=。

i 左梁顶层: 590.0)35.1942.0/(35.1=+=下柱i41.0)942.035.1/(942.0=+=左梁iB 柱:底层: 250.0)0.1977.0864.0942.0/(942.0=+++=右梁i264.0)0.1977.0864.0942.0/(0.1=+++=上柱i 258.0)0.1977.0864.0942.0/(977.0=+++=下柱i 228.0)0.1977.0864.0942.0/(864.0=+++=左梁i标准层:63.0)864.00.10.1942.0/(0.1=+++=下柱i63.0)864.00.10.1942.0/(0.1=+++=上柱i247.0)864.00.10.1942.0/(942.0=+++=右梁i227.0)864.00.10.1942.0/(864.0=+++=左梁i顶层:241.0)942.0864.035.10.1/(0.1=+++=下柱i325.0)942.0864.035.10.1/(35.1=+++=上柱i227.0)942.0864.035.10.1/(942.0=+++=右梁i207.0)942.0864.035.10.1/(864.0=+++=左梁i顶层: 428.0)942.0864.035.1/(35.1=++=下柱i274.0)942.0864.035.10.1/(864.0=+++=左梁i294.0)942.0864.035.10.1/(942.0=+++=右梁i三.恒荷载作用下的弯矩剪力计算,根据简图(6-4)梁:A M 0∑= 21/2.0A B B M M ql Q l ---=/1/2B A B Q M M l ql =--B M 0∑= 21/2.0A B A M M ql Q l -+-=/1/2A A B Q M M l ql =-+(6-4) 柱:C M 0∑= .0C D D M M Q h ---=()/D C D Q M M h =-+D M 0∑= .0C D C M M Q h ---=()/C C D Q M M h =-+(6-5)四.恒荷载作用下的边跨框架的轴力计算,包括连梁传来的荷载及柱自重.N6=114.5+25.78×6.6/2=199.574KNN5= N6+72.81+16.37×7.2/2+35.39=361.795KNN4= N5+72.01+16.37×6.6/2+27=514.826KNN3= N4+72.01+16.37×6.6/2+27=667.857KNN2= N3+72.01+16.37×6.6/2+27=820.888KNN1= N2+72.01+16.37×6.6/2+27=973.92KN恒荷载作用下的中跨框架的轴力计算:N6=120.91+17.44×7.2/2+25.78×6.6/2=268.77KNN5= N6+114.32+16.37×6.6/2+16.72×7.2/2+35.39=532.69KNN4= N5+114.32+16.37×6.6/2+16.72×7.2/2+27=788.24KNN3= N4+114.32+16.37×6.6/2+16.72×7.2/2+27=1043.75KNN2= N3+114.32+16.37×6.6/2+16.72×7.2/2+27=1299.29KNN1= N2+114.32+16.37×6.6/2+16.72×7.2/2+27=1554.82KN图6-7 恒荷载作用下的计算简图五.弯矩分配及传递弯矩二次分配法比分层法作了更进一步的简化。

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