-
1 ei e 由
1
(l0 / 2h)
1 21
1
(7.5) 1 1.03
2
1400ei / h0 1400 520/360 0.3 360
属于大偏心受压
1.03 520 535.6 0.3h0
ei h0 /2 as 535.6 180 40 675.6mm x N / a1 fcb 300 103 /9.6 300 104.17mm
x b h0 2 s 0.55
360 180mm
As
选用
As
Ne a1 fcbx(h0 x/ 2) 300 103 675.6 9.6 300 104.17 (360 52.085)
f y (h0 as) 300 320
2
As.min 240mm
2
2
As As 1058.9mm
f
A
4 20
s s
A1256 mm
4 20
4 20
7-7 『解』
已知条件同题 7－3，设计对称配筋的钢筋数量。
查 表 得 ：
y =360N/mm2
取 s
s =40mm
f
c =14.3N/ mm2
f
y =300N/ mm2
h
0 h as =560mm
-
-
ea e0 M / N 30mm ea 20mm ei e0 50mm l0 / 15 h
10 5
要考虑
1 0.5f3
c A/ N
0.5 14.3 300 600/260 10 4.95 1
取 11 取
2 1.0
(l0 / 1 1 1 h)2
2 1
1
(10)2 1 1.8
1400ei / h0 1400 50/560
ei 1.8 50 90 0.3h0 0.3 560 168mm
属于小偏心受
压 由
N a1 f cbh0 b
b Ne 0.43a1 fcbh0 2 ( b )( ha1 f cbh0
1 0 as )
2600
103 0.518 14.3 300 560
0.518 1.183
2600 103 330 0.43 14.3 300 5602
14.3 400 560
(0.8 0.518) 520 Ne a21 fc bh (1 0.5 ) / fy (h0 as ) 1097.6mm20 As As 0.02bh
选用 4
20
As 1256mm2
抗压承载力验算：
A / bh 1256 / 300 560 0.75 0 3 0 s 0
0 0 0.9 ( fc A fy Nu As ) 0.9 0.98 (14.3 400 600 360 1256) 2.67 10所以平面外承载力满足要求。
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3 kN 2600kN