混凝土结构 第4章计算题及答案
s 0.5 (1 -2 s) 0.5 (1 2 0.243) 0.858
AM/f125 106
s y sh0 300 0.858 465
1044mm2
选用钢筋4 18,As 1017mm2
As 1044 mibh.2% 200 500 200mm2n 0 （2）采用双排配筋 h0 h 60 440mm
2
125 106
s M/ 1fcbh0=1.0 11.9 200 440
2
0.271 1 2 s=1 2 0.271 0.323
s 0.5 (1 -2s) 0.5 (1 2 0.271
） 0.838 A125 106
s M/fy sh0
210 0.838 440
1614mm2 选用钢筋8 16
A2s=1608mm
As 1614 mibhn 0.27% 200 500 27mm02 (3)假定受拉钢筋放两排 a 60 h0 500 60 440mm
M/ 2
225 106
s 1fcbh0=
1.0 11.9 200 4402
0.488
1 2 s=1- 2 0.488 0.845 0.55
故采用双筋矩形截面
取 M2
b 1 1fcbh0
b(1 0.5 b)
1.0 11.9 200 4402
0.55 (1 0.5 0.55) =183.7KN m
A'
'
'
225 106 183.7 106
s M/fy(h0 a) (440 35)
339.9mm2300
As ''
b 1fcbh0/fy Asfy/fy 0.55 1.0 11.9 200 440/300 339.9