工业与民用建筑_受压构件截面承载力计算

1、1200/30,/25lblh3456P109图5-2 不可采用有内折角的箍筋70/8lb 0/8lb 89 . 0sycAfAfNsA91033 222c y2 4 0 0 1 0 N1 7 2 .0 4 3 1 0 m m0 .9 ( )0 .9 1 ( 1 1 .9 N / m m 3 6 0 N / m m 0 .0 1 )NAff 33222cy2400 10 N172.043 10 mm0.9 ()0.9 1 (11.9N/mm360N/mm 0.01)NAff 414.78hbAmm 0/6.2/0.4513.78lb 0.92311322c2y0 .9 2 4 0 0 1 0

2、N - 0 .9 0 .9 2 3 1 1 .9 N / m m 4 5 0 m m 4 5 0 m m1 3 3 2 m m0 .90 .9 0 .9 2 3 3 6 0 N / m msNfAAf 33222cy2 4 0 01 0N1 7 2 .0 4 31 0m m0 .9() 0 .91(1 1 .9N /m m 3 6 0N /m m 0 .0 1 )NAf f 33 222c y2 4 0 0 1 0 N1 7 2 .0 4 3 1 0 m m0 .9 ( )0 .9 1 ( 1 1 .9 N / m m 3 6 0 N / m m 0 .0 1 )NAff 2min1608mm

3、0.794%0.4%450mm 450mm12131415sAsA16bcusAcuuM0h0hhhsyAf)( a)( b)( c受弯构件受弯构件偏压构件偏压构件1718bb19aiaeeemmheNMe002030/0eaeie2021200)(140011hlhei21bb2211002cysyscySsNf bxf Af AxN ef bxhf Aha0Y 0M 对Asbhx02sax 2311002cysyscySsNf bxf Af AxNef bxhf Aha0Y 0M 对As,ssyyssaaffAA0010112sSycccahAfxhbxfNebhfbxfN0Y 0M 24

4、10bcbNf b h01bhfNcbbbNNbNN2500sysssyahfNeAahAfNe2sax 2sax 261101002cyssscyssscyssNf bxf AAf bhf AAxNef bx hf Aha270010112sSycccahAfxhbxfNebhfbxfN0Y 0M 28293031第 I I 阶 段第 I 阶 段BCANucrOt(a )- 开 裂 荷 载Ncr( I )( I I )( I I I )NNcrNcrNNuNNu(b )NN32fyAsNAsfyAssyAfN 333435363722kkM lM lfSSEIBSBl38limf39skqkBMMMB1kMqME205 . 3160.2 15. 1fssshAEB40limflimf4142tkf43teeqsskcrdcEw08.09.1max