混凝土结构设计原理第四版-沈蒲生版课后习题答案 下载本文

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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

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

(l0 / h)2

1 2 1

1

1 1

(10)2 1 1.8

1400ei / h0 1400 50/560

168mm

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

0.02bh

M 1 Ne M

131.184 106 N.mm

选用 2

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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