@《建筑结构》第五版偏压构件习题

6-13 已知某对称配筋的矩形截面偏心受压短柱，截面尺寸b ×h=400m m ×600mm,承受轴向压力设计值N=1500kN,弯矩设计值M=360k N ·m ，该柱采用的混凝土强度等级为C25,

纵向受力钢筋为HRB400级，试求纵向受力钢筋面积As=A s ′=?选择配筋直径，根数，画配筋断面图（箍筋按构造规定选取）。

解：517.0,/360,/9.112

2===b y c mm N f mm N f ξ ,56040600,400'mm h mm a a s s =-===取

mm N M e 24010150010360360=⨯⨯==，mm mm h e a 2020,30max

=⎥⎦⎤⎢⎣⎡=， =+=a i e e e 0260mm ， η=1，mm a h e e s i 520403002602

=-+=-+= kN N kN bh f b c 150011.13782.1378115517.05604009.11N 01b =<==⨯⨯⨯==ξα

为小偏心受压柱

551.0517.0266560016

.1476418764807800000004.121884517.05604009.110.1)

40560)(517.08.0(5604009.110.143.05201500000517.05604009.110.1101500))((43.023*********=++-=+⨯⨯⨯+--⨯⨯⨯⨯-⨯⨯⨯⨯⨯-⨯=++'----=b c s

b c b c bh f a h bh f Ne bh f N ξαξβαξαξ 2

20201987)40560(360)55.05.01(55.05604009.1115201500000)

()5.01(mm a h f bh f Ne s A As s y c =-⨯⨯-⨯⨯⨯⨯⨯-⨯='-'--='=ξξα

4根直径18mm （As=A s ′=1018mm 2）或3根直径22mm （As=A s ′=1140mm 2

）

0.002bh=0.002×400×600=480mm 2<1018mm 2<1140mm 2

0.006 bh=0.006×400×600=1440mm 2<2×1018mm 2<2×1140mm 2

0.05bh=0.05×400×600=12000mm 2>2×1140mm 2>2×1018mm 2

6-14 已知某矩形截面偏心受压柱截面尺寸b ×h=350m m ×550mm,计算长度l 0=4.80m ，承受轴向压力设计值N=1200kN,弯矩设计，M 1=220k N ·m ，M 2=250k N ·m ，采用C30混凝土和HRB400级纵向钢筋,HPB235级箍筋，，试求按对称配筋钢筋截面面积As=A s ′=? 绘配筋图（取a s =a s ′=40mm ）

解：fc=14.3N/mm 2,fy=fy ’=360N/mm 2,ξb=0.517, h 0=550-40=510mm, C m =0.7+0.3（M 1/M 2）=0.7+0.3*220/250=0.964

ζ1=0.5 fcA/N=0.5×14.3×350×550/1200000=1.14 取ζ1=1

e a =max [550/30,20mm]=20mm, l 0/h=4.80/0.55=8.72

13.11172.8)2010

120010250(13005101)()(130********=⨯⨯⨯+⨯⨯⨯+=++=c a ns h l e N M h ζη M= C m ηns M 2=0.964×1.13×250=265.22kN ·m

e 0=M/N=265.22×106/1200×103=221.29mm.

e i =e 0+e a =221.29+20=241.29mm.

e=e i +h/2-a s ’=241.29+550/2-40=476.29mm

Nb= f c b h 0ξb =14.3×350×510×0.517=1319.668kN >N=1200kN 大偏心

47.0510

3503.14120000001=⨯⨯==bh f N c αξ>2a s ’/h 0=80/510=0.16 2

2230201385550350002.0002.062.611)

40510(360)47.05.01(47.05103503.14129.476101200)

()5.01(mm bh mm a h f bh f Ne s A As s y c =⨯⨯=>=-⨯⨯-⨯⨯⨯⨯⨯-⨯⨯='-'--='=ξξα

6-15矩形截面偏心受压柱尺寸b ×h=500m m ×700mm, 计算长度l 0=4.80m, 承受轴向压力设计值N=2800kN,弯矩设计值M 1=70k N ·m ，M 2=75k N ·m ，采用C25混凝土,HRB500级纵筋和HPB235级箍筋，试求按对称配筋钢筋截面面积As=A s ′=? 绘配筋图

解：fc=11.9N/mm 2,fy=435N/mm 2，fy ’=410N/mm 2,ξb=0.482, h 0=700-45=655mm,

C m =0.7+0.3（M 1/M 2）=0.7+0.3*70/75=0.98 ，l 0/h=4.80/0.7=6.86<15

ζ1=0.5 fcA/N=0.5×11.9×500×655/1200000=1.14 取ζ1=1

e a =max [700/30,20mm]=23.3mm, ζc =0.5 fcA/N=0.5×11.9×500×700/2800000=0.74 35.174.086.6)3.2310

28001075(13006551)()(130********=⨯⨯+⨯⨯⨯+=++=c a ns h l e N M h ζη M= C m ηns M 2=0.98×1.35×75=99.225kN ·m

e 0=M/N=99.225×106/2800×103=35.44mm.

e i =e 0+e a =35.44+23.3=58.74mm. ，e=e i +h/2-a s ’=58.74+700/2-45=363.74mm Nb=

f c b h 0ξb =11.9×500×655×0.482=1878.47kN

e i =58.74mm.<0.3h 0=0.3×655=196mm 按小偏心计算