内容发布更新时间 : 2024/12/23 11:04:59星期一 下面是文章的全部内容请认真阅读。
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取 2
1.0
-
-
1
1
(l0 / 2h)
1 2
1
1
(7.5) 1 1.03
2
1400ei / h0 1400 520/360
ei 1.03 520 535.6 0.3h0 0.3 360 属于大偏心受压 e
ei h0 /2 as
535.6 180 40
675.6mm
取 x
b h0
Ne a1 fcbx(h0 x/ 2) f y A (h0 as ) 由 得 s 0.5 b) / (h0 As Ne a1 fcbh0 2 as ) b(1 fy As min
0.02bh 0.02 300 400
560.9mm2
240mm2
As Asmin
选用 4
14
As 615mm2
a
由 Ne a1 fcbh (1 0.5 ) fy As (h0
h0
20
s )得
s Ne f y As (h0 as )/ a1 fcbh0 2 0.385
1 1 2 s 1 0.48 0.52 0.52 360 187.2 2as
80mm
由
N a1 fc bh0 fy As f y As 得
/
f y As fy
9.6 300 360 0.5 300 615 2343mm2
300
As a1 fcbh0
选用 5
A s 252454mm2 截面配筋图如下:
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3 25
4 14
5 25
4 14
2 14
7-3 已知条件同题 7-1 相同,但受压钢筋已配有 4
Φ 16 的 HRB335级钢筋的纵向钢筋。设计 As 数量。已知矩行截面柱 b=600mm,h=400mm。计算长度混泥土强度等级为 C30,纵筋采用 HRB400级钢筋。,柱上
作用 轴 向力 设 计 值 N=2600KN, 弯 矩设 计 值 M=78KN.m,混泥土强度等级为 C30,钢筋采用 HRB400
级钢筋。设计纵向钢筋 As 及 As/ 的数量,并验算垂直弯矩作用平面的抗压承
载力。 『解』
查 表 得 : f2y =360N/mm
取 s
s =40mm
f
c =14.3N/ mm2
f
y =300N/ mm2
h0 h s =560mm
ea
20mm
a
ei ea e0 50mm
e0 M / N 30mm l0 / 15 h
10 5
要考虑
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1
0.5fc A/ N 0.5 14.3 300 600/260 10 2
1.0
3
4.95 1
取 1
1
取
(l0 / h)2
1 2 1
1
1 1
(10)2 1 1.8
1400ei / h0 1400 50/560
168mm
压
ei 1.8 50 90 0.3h0 0.3 560
属于小偏心受
对 AS 合力中心取矩 e h / 2 as (e0 ea)
300 40
10 250mm
h / 2) / f As Ne a1 f cbh(h0 y (h0 As 0
s
A
103 14.3 300 600 (560 300) / 360
as ) 2600 520 0.02 300 600
360mm2
取
0.02bh
M 1 Ne M
131.184 106 N.mm
选用 2
由
A 16s 402mm2
as f y As(1 as / h0 ) / a1 h0
e
fcbh0 ( b
as
1)
h0
f y As (1 as / h0 ) / a1 f cbh0 ( b 1 )
2
Ne
f 2
a1 fcbh0 2
0.75 2 1 b 1.6 0.518 1.082 ei h0 / 2 as 1.8 50 280 40 330mm
Ne a1 fcb2 (1 / 2) f y A (h0 as )
由 得 s (1 0.5 )/ f y (h0 2As Ne a1 fcbh0 2 as ) 1214.6mm
选用 4
2
As 1256mm 20
0.02bh
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抗压承载力验算: l 0 / h
\\ 7-6
量。 『解』
查表得:
f
10 查表 3-1 得 0.98
0
As / bh0 1256 / 300 560 0.75 0 3 0 0
0.9 ( fc A fy
Nu As ) 0.9 0.98 (14.3 300 600 360 1256) 2.67 103 kN 2600kN
所以平面外承载力满足要求。 截面配筋图如下:
2 16
2Φ 10
4 20
已知条件同题 7-1,设计对称配筋的钢筋数
c =9.6N/ mm
2f
y =300N/mm
2
取 s
s =40mm h0 h s =360mm
a
e0 M / N 0.5m l 0 / h 7.5 5
ea 20mm
ei ea e0 520mm
要考虑
1
0.5fc A/ N 0.5 9.6 300 400/300 10 1.92 取 2
1.0
3
1
取 1
1
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