哈尔滨学院本科毕业论文(设计)
0.5fcA0.5?16.7?5002?1???1.22，取?1?1.22N1765.68?103 lo?15取?2?1.0h判别大小偏心
N1765.68?103????0.44??b?0.55
?1fcbho1.0?16.7?500?465所以属于大偏心受压
??1?l1(o)2?1?2?1?(10.2)2?0.44?1.0?1.20 eh61.421400?1400?i465ho1e=?ei?h?as?1.2?61.42?250?35?288.7mm 21=As?AsNe??1(1?0.5?1)?1fcbho21fy(ho?as)11765.68?103?288.7?0.44(1?0.5?0.44)?1?16.7?500?4652??0300?(465?35)
所以按构造配筋即可
As=As`=0.002×500×500=500mm2 选取4Φ14（As=As`=615mm2） 情况二：Nmax=1863KN M=12.63KN.m
eo?MN?12.63?1031863?6.78mm
?500?ea=max?20,??,20mm
30??ei?eo?ea?6.78+20=26.78mm
loh ?5100?10.2?15；需考虑纵向弯曲影响5000.5fcA0.5?16.7?5002?1???1.22，取?1?1.223N1863?10 lo?15取?2?1.0h判别大小偏心
31
哈尔滨学院本科毕业论文(设计)
N1863?103?2???0.48??b?0.55
?1fcbho1.0?16.7?500?465所以属于大偏心受压
??1?l1(o)2?1?2?1?(10.2)2?0.48?1.22?1.76 eh26.781400?1400?i465ho1e=?ei?h?as?1.76?26.78?250?35?262.13mm 21=As?AsNe??1(1?0.5?1)?1fcbho21fy(ho?as)11863?103?262.13?0.48(1?0.5?0.48)?1?16.7?500?4652??0
300?(465?35)所以按构造配筋即可
As=As`=0.002×500×500=500mm2 选取4Φ14（As=As`=615mm2） 一层中柱B 情况一：
N=1622.51KN Mmax?156.24KN.m 验算轴压比
1622.51?103???0.39?0.9满足要求
16.7?500?500eo?MN?156.24?1031622.51?96.3mm
?500?ea=max?20,??,20mm
30??ei?eo?ea?96.3+20=116.3mm
loh ?5100?10.2?15；需考虑纵向弯曲影响5000.5fcA0.5?16.7?5002?1???1.28，取?1?1.283N1622.51?10 lo?15取?2?1.0h判别大小偏心
32
哈尔滨学院本科毕业论文(设计)
N1622.51?103????0.42??b?0.55
?1fcbho1.0?16.7?500?465所以属于大偏心受压
??1?l1(o)2?1?2?1?(10.2)2?1.28?1.0?1.38 eh116.31400?1400?i465ho1e=?ei?h?as?1.38?116.3?250?35?375.5mm 21=As?AsNe??1(1?0.5?1)?1fcbho21fy(ho?as)11622.51?103?375.5?0.42(1?0.5?0.42)?1?16.7?500?4652??0300?(465?35)
所以按构造配筋即可
As=As`=0.002×500×500=500mm2 选取4Φ14（As=As`=615mm2） 情况二：
M=156.24KN.m Nmax?1622.51KN 验算轴压比
1622.51?103???0.39?0.9满足要求
16.7?500?500eo?MN?156.24?1031622.51?96.3mm
?500?ea=max?20,??,20mm
30??ei?eo?ea?96.3+20=116.3mm
loh ?5100?10.2?15；需考虑纵向弯曲影响5000.5fcA0.5?16.7?5002?1???1.28，取?1?1.283N1622.51?10 lo?15取?2?1.0h判别大小偏心
33
哈尔滨学院本科毕业论文(设计)
N1622.51?103????0.42??b?0.55
?1fcbho1.0?16.7?500?465所以属于大偏心受压
??1?l1(o)2?1?2?1?(10.2)2?1.28?1.0?1.38 eh116.31400?1400?i465ho1e=?ei?h?as?1.38?116.3?250?35?375.5mm 21=As?AsNe??1(1?0.5?1)?1fcbho21fy(ho?as)11622.51?103?375.5?0.42(1?0.5?0.42)?1?16.7?500?4652??0300?(465?35)
所以按构造配筋即可
As=As`=0.002×500×500=500mm2 选取4Φ14（As=As`=615mm2）
第1层框架A轴柱为例，计算箍筋数量 由强剪弱弯思想计算剪力值
Hn=5.1-0.5=4.6m
Mlc?Mbc98.52?113.8Vc??vc?1.1??46.2kN
Hn4.6抗剪计算中， ??Hco4600??5?3 取?=3 2hc2?460Vc46.2?103剪压比：??0.005?0.2
2fcbh02?19.1?500?460满足要求
N?0.3fcbh0?0.3?19.1?500?460?1317.9KN?N?1863kN 取N?1317.9KN
34
哈尔滨学院本科毕业论文(设计)
AsV?1.05/(??1)ftbh0?0.056N?sfyvh046200?1.05/(3?1)?19.1?500?460?0.056?1317.9?103??0
210?460则按构造配筋
查表得，柱端箍筋加密区最小配筋率特征值?v?0.17
?v??vfc19.1?0.17?1.5000?0.7%??min fyv210取四肢Ф10腹式箍，Asv?78.5mm2
s?Asvlsv78.5?8?500??83.73mm 取S=100mm 2l1l2?v500?0.0150箍筋加密区长度为：
?H??5100?max?c,hc??max?,500??850mm
?6??6?加密区箍筋用4Ф10＠100，加密区长度为850mm,按井字形配置。非加密区箍筋用4Ф10＠200。
3.8.2第1层框架B轴柱为例，计算箍筋数量 由强剪弱弯思想计算剪力值
Mlc?Mbc(156.3?121.2)?106Vc??vc?1.1??60.33kN 3Hn4.6?10抗剪计算中，??Hco4600??5?3 取?=3 2hc2?460Vc60.33?103剪压比：??0.007?0.2
2fcbh02?19.1?500?460满足要求
N?0.3fcbh0?0.3?19.1?500?460?1317.9kN?N?1622.51KN 取N?1317.9kN
AsV?1.05/(??1)ftbh0?0.056N??0 sfyvh0
35