Ⅰ934220404254×15810934213910KN
因为b20mbc2h04002×106025200mm所以
A1l2hv2h0b42082106×2108m2
A2bch0h0h0bc2b2204106×106106042222148m2
1柱边截面
fPjmaxA115810×1081775KN查表的β
p098C15ft091Nmm207β
pA207×098×091×148000092390KNPjmaxA115810×1081775KN满足要求
2变阶截面算法同上面的一样同样满足要求。4配筋验算
ⅠⅠ截面
MI148PjmaxPj2bbcPjmaxPjblhc2
1481581125762×204158112576×2408228025KNm
AsIMI09fyh02802509×210000×10613989mm2ⅠⅠ截面
MI148PjmaxPj2bb1PjmaxPjbll12
1481581125762×2115158112576×241552190KNm
AsIMI09fyh019009×210000×1069484mm2
比较AsI和AsI应该按AsI来配筋取10φ16As1538ⅡⅡ截面
MⅡ24
1Pjbbc22lhc124×1257620422×40811805KNmAsⅡMⅡ09fyh01180509×210000×1060012596mm2ⅡⅡ截面
MⅡ24
1Pjbb122ll1124×12576211522×41553615KNmAsⅡMⅡ09fyh0361509×210000×0710012272mm2
按构造配筋取25φ10As1963mm2满足要求
fr