结构设计原理第六章受压构件习题及答案

As?2512????1.57%，因此?min?0.4%???1.57%?3%，

As400?400因此符合配筋率要求。

2、解：设as?as??40mm，则ho?600?40?560mm

设该构件为大偏心构件，则令???b?0.55 求得：

Nb??1fc?bbho?fy?As??fyAs ?1.0?14.3?0.55?600?560?300?1964?300?1256 ?2855.0kN>1500kN故该构件属于大偏心受压构件

x?N?fy?As??fyAs

?1fcb1500?103?300?1964?300?1256 ?则： 1.0?14.3?600 ?150.1mm2as??80mm?x?150.1mm

lo/h?3000/600?5，则??1.0

e?x???1fcbx?ho???fy?As?ho?as?2??N??150.1??1.0?14.3?600?150.1??560???300?1964??560?40?2?? ? 1500?103 ?620.6mm

因：

hh600e??ei??as??ei?e??as?620.6??40?360.6mm222

eo?ei?ea?360.6?20?340.6mm

则：

?3M?Neo?1500?340.6?10=510.9kN?m

3、解：设as?as??40mm，则ho?600?40?560mm

100?106eo?M/N?=28.6mm

3500?103ea?h/30?20mm

ei?eo?ea?28.6?20?48.6mm lo/h?3000/600?5，则??1.0

e??ei?h?as?1.0?48.6?600/2?40?308.6mm 2因为对称配筋，则

Nb??1fc?bbho ?1.0?14.3?0.55?600?560 ?2642.6kN

??N??1fc?bbho??bNe?0.43?1fc?bbho2??1fcbho???1??b?ho?as3500?103?1.0?14.3?0.55?600?560 ??0.55

3500?103?308.6?0.43?1.0?14.3?0.55?600?560?560?1.0?14.3?600?560?0.8?0.55???560?40? ?0.654??

As?As??Ne???1?0.5???1fcbho2fy?ho?as??? ?3500?103?308.6?0.654??1?0.5?0.654??1.0?14.3?600?560?560300??560?40?

?0则：As?As??0.002bh?0.002?600?600?720mm2