上海交通大学：《结构概念设计 Conceptual design》教学资源（课程讲义）Chapter 7 Overall Design of Vertical Subsystem（VS）7.5. Deflections

7.4.Rigid Frame Subsystem The approximate analysis will be possible to provide an overall scheme and sizing of members for the computer analysis of structure in detail design stage where precise analysis is necessary

7.4. Rigid Frame Subsystem The approximate analysis will be possible to provide an overall scheme and sizing of members for the computer analysis of structure in detail design stage where precise analysis is necessary

7.4.1 Rigid frame under vertical loads With simplified moment distribution methods

7.4.1 Rigid frame under vertical loads With simplified moment distribution methods

The simplified analysis is based on assumptions: -neglecting side-way due to asymmetry in loading and stiffness, -uniform I values for each member, -elastic behavior of all members and joints, -member affected only by the member connected directly to it

The simplified analysis is based on assumptions: -neglecting side-way due to asymmetry in loading and stiffness, -uniform I values for each member, -elastic behavior of all members and joints, -member affected only by the member connected directly to it

简化弯矩分配法计算竖向荷载下刚架内力 两端固定约束一 获得基本弯矩图 释放部分约束 按节点约束程度 分配节点弯矩 按各杆件线刚度 传递节点弯矩 按远端约束程度 由两端弯矩+荷载反算剪力---得剪力分布 由Footprint.算柱轴力--得轴力分布

简化弯矩分配法计算竖向荷载下刚架内力 两端固定约束-----获得基本弯矩图 释放部分约束-----按节点约束程度 分配节点弯矩-----按各杆件线刚度 传递节点弯矩-----按远端约束程度 由两端弯矩+荷载反算剪力-----得剪力分布 由Footprint算柱轴力---得轴力分布

Fig.7-11 Example of moment distribution in frame 铰 铰 (M=0) (M=0) 09 90 K=2 K=2 其它杆件 10 30 20411111B20 30 10 D 其它杆件 (M=1/3) K=3 20 -100 K=3 -100 20 K=3 (Mc=1/3) +30 +30 -70 -70 K=2 10 K=2 10 n n 固定 固定 (Mc=1/2) (M=1/2) 注：1)K是各杆件的相对刚度 2)ΣK是A点或B点处所有杆件的总刚度（上图中为10）： 节点处约束杆件K(上图中为70%)： 3》转动约束正比于比值：节点处所有杆件 3)由传递系数Mc确定该杆远端的弯矩

Fig.7-11 Example of moment distribution in frame 3）由传递系数 Mc确定该杆远端的弯矩 （上图中为70%）； 节点处所有杆件 K 节点处约束杆件 K 3）转动约束正比于比值： 2） K 是 A 点或 B点处所有杆件的总刚度（上图中为10）； 注：1）K 是各杆件的相对刚度 固定 ( Mc 1/2) ( Mc 0 ) 铰 铰 ( Mc 0 ) ( Mc 1/3) 30 10 其它杆件 K 3 其它杆件 ( Mc 1/3) K 3 10 30 B 70 30 30 70 -100 -100 ) K 2 K 2 0 20 10 K 3 20 K 2 A 10 20 0 20 K 2 ( Mc 1/2 固定 C D

Fig.7-11 Example of moment distribution in frame 70 70 30 30 20 20 20 20 10 10 80 10 10

Fig.7-11 Example of moment distribution in frame 10 10 10 70 30 20 20 20 20 70 30 10 80

Fig.7-10 Rigid frame rough calculation 情况 弯矩传递系数(Mc) 在梁中产生a转角时 由远端固定程度决定 所需的力矩(M) 的刚度系数(FFF) 1.完全固定 Mc=1/2 M=1 FFF=1 无转动 0M>3/4 1>FFF>3/4 (如7/8) (如7/8) 0<a'<a/2 Mc=0 3.铰接 M=3/4 FFF=3/4 a=a/2

Fig. 7-10 Rigid frame rough calculation

7.4.2 Rigid frame under horizontal loads (横向荷载作用下刚架内力简化计算) -Portal Method（算弯矩+剪力) solving the moment and shear distribution at the columns and beams over the frame -Cantilever Method=Footprint(算轴力) solving the axial forces in the columns caused by the overturn moments

7.4.2 Rigid frame under horizontal loads （横向荷载作用下刚架内力简化计算） -Portal Method（算弯矩+剪力） solving the moment and shear distribution at the columns and beams over the frame -Cantilever Method=Footprint（算轴力） solving the axial forces in the columns caused by the overturn moments

Fig.7-18 Cantilever method for column forces 本 COLUMN#'S: 1 NA % d

Fig. 7-18 Cantilever method for column forces

7.4.2 Rigid frame under horizontal loads -Portal Method solving the moment and shear distribution at the columns and beams over the frame

7.4.2 Rigid frame under horizontal loads -Portal Method solving the moment and shear distribution at the columns and beams over the frame