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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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

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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