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