发布时间 : 星期日 文章混凝土结构设计原理习题集答案更新完毕开始阅读
已知e0=100mm
h400??13.3mm?20mm 3030取ea?20mm
ei?e0?ea?100?20?120mm
取?1?1.0
l0h?3600?9?15，?2?1.040012 1?l0?2??1????1??9?1.0?1.0?1.176?1.0?12ei?120h??1400?1400365h0??1.176
e??ei?h2?as?1.176?120?（2）判别大小偏压 求界限偏心率
''Mb0.5?1fcb?bh0(h??bh0)?0.5(fyAs?fyAs)(h?2as)eob??Nb?1fcb?bh0?fy'As'?fyAs400?35?306.12mm 20.5?1.0?11.9?200?0.550?365?(400?0.550?365)?0.5?(300?226?300?226)?(400?2?35)1.0?11.9?200?0.550?365?146.5mm?又因为
?ei?1.176?120?141.1mm?146.5mm，故为小偏压。
（3）求截面承载力设计值N N??1fcbx?fyAs?''???1fyAs
?b??1x?0.8365 ?1.0?11.9?200?x?300?226??300?226?3123x?149160 （A）0.550?0.8又由N?e??1fcbx?h0???x???f'yA's?h0?a's? 2?得：N?306.12?1.0?11.9?200x(365?0.5x)?300?226?(365?35) 整理得：N?2839x?3.889x?73117
联立（A）(B)两式，解得x?205mm,代入（A）式中得： N?49106N0
2（B）
根据求得的N值，重新求出?1、?值：
?1?0.5fcA0.5?11.9?200?400??0.969 N491060相应?值为1.717，与原来的?1、?值相差不大，故无需重求N值。 8．答案： ⑴求ei、η、e
h0?h?as?700?35?665mm
M246?106e0???307.5mm 3N800?10h700??23.3mm?20mm 3030ea?23.3mm
ei?e0?ea?307.5?23.3?330.8mm
?1?0.5fcA0.5?14.3?130000??1.162?1.0 N800?103?1?1.0 l0h?6800?9.71?15，?2?1.070012 1?l0?2??1????1?(9.71)?1.0?1.0?1.135?1.0?12ei?330.8h??14001400665h0??1.135
e??ei?h2?as?1.135?330.8?（2）判别大小偏压
700?35?690.5mm 2T?fc(b'f?b)h'f?14.3?(350?100)?120?429000N N?T800?103?429?103????0.390??b?0.550
?1fcbh01.0?14.3?100?665属大偏压
x??h0?0.390?665?259.4mm?120mm，中性轴位于腹板内。
（3）计算As和As
'As?'Ne??1fcbx?h0?0.5x???1fcbf?bhfh0?0.5hf''???'?120)2f'y?h0?a's?800?103?690.5?1.0?14.3?100?259.4?(665?0.5?259.4)?1.0?14.3?250?120?(665??
360?(665?35)?415.6mm2'2，As?As?452mm
选用
9．答案： 1组内力 Ⅰ、求解第○
M378.3?106e0???687.8mm 3N550?10h800??26.67mm?20mm 3030ea?26.67mm
ei?e0?ea?687.8?26.67?714.47mm
?1?0.5fcA0.5?14.3?170000??2.21?1.0 3N550?10取?1?1.0
l0h?6800?8.5?15，?2?1.080012 1?l0?2??1????1??(8.5)?1.0?1.0?1.05?12ei?714.5h??1400?1400760h0取??1.05
800e??ei?h?as?1.05?714.5??40?1110.23mm
22（3）判别大小偏压
Nb??1fcbh0?b+?1fc(b'f?b)h'f
?1.0?14.3?100?760?0.518?1.0?14.3?(400?100)?150?1206.5KN
N?550KN?1206.5KN，故属于大偏压。
800e'??ei?h?as'?1.05?714.5??40?390.23mm
22??N??1fc(b'f?b)hf?fcbh0550?103?1.0?14.3?300?150??0
1.0?14.3?100?760'取 x?2as?2?40?80mm
由公式As?A's?N(?ei?h?a's)550?103?390.232??828mm2f'y(ho?a's)360?(760?40)
'假设As?0，由N?fyAs
N550?103则As???1527.8mm2
360fy'取As?min{As As}?828mm2??minA?0.002?(100?800?2?300?150)?340mm2
2组内力 Ⅱ、求解第○（1）求解ei、η、e
M280?106e0???397.3mm 3N704.8?10h800??26.67mm?20mm 3030ea?26.67mm
ei?e0?ea?397.3?26.67?423.97mm
?1?0.5fcA0.5?14.3?170000??1.725?1.0 3N704.8?10?1?1.0 l0h?6800?8.5?15，?2?1.080012 1?l0?2??1????1??8.5?1.0?1.0?1.09?12ei?423.97h??1400?1400760h0??1.09
e??ei?h2?as?1.09?423.97?（2）判别大小偏压
800?40?823.2mm 2N?704.8KN?1206.5KN，属大偏压。
800e'??ei?h?as'?1.09?423.97??40?102.13mm
22??N??1fc(b'f?b)hf?fcbh0704?103?1.0?14.3?300?150??0.056
1.0?14.3?100?760x??h0?0.056?760?42.9mm?2as'?80mm