《建筑结构》第五版偏压构件习题

已知某对称配筋的矩形截面偏心受压短柱，截面尺寸b×h=400mm×600mm,压力设计值N=1500kN,弯矩设计值M=360kN·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=As配筋断面图（箍筋按构造规定选取。解：fc

11.9N/mm2,fy

360N/mm2,b

0.517取a a'40mm,hs s

60040560mm,e M

360106

240mm

h,20mm

20mm，ee

260mm，0 N 1500103

a 30

max

i 0 a=，eei

ha2

26030040520mmN fbh 11.94005600.5171378115.21378.11kNN1500kNb 1 c 0 b为小偏心受压柱 N1

fbhc 0

b Ne1

fbh2c 0

bf(1 b

)(h0

as

1 c 0 15001031.011.94005600.517

0.51715000005200.431.011.94005602(0.80.517)(56040)

1.011.9400560 121884.4780000000641876480147.16

0.5170.551AsAs

Ne1

fbh

2)f (ha )y 0 s1500000520111.940056020.550.50.55) 987mm2360(56040)4根直径18mm（As=As′=1018mm2）或3根直径22mm（As=As′=1140mm2）0.002bh=0.002×400×600=480mm2<1018mm2<1140mm20.006bh=0.006×400×600=1440mm2<2×1018mm2<2×1140mm20.05bh=0.05×400×600=12000mm2>2×1140mm2>2×1018mm2已知某矩形截面偏心受压柱截面尺寸b×h=350mm×550mm,计算长度l0=4.80m轴向压力设计值N=1200kN,弯矩设计，M1=220kN·m，M2=250kN·m，采用C30混凝土和HRB400,HPB235As=As′=?绘配筋图（a=a=40mm）s s解：fc=14.3N/mm2,fy=fy’=360N/mm2,ξb=0.517,Cm=0.7+0.3（M1/M2）=0.7+0.3*220/250=0.9641=0.5fcA/N=0.5×14.3×350×550/1200000=1.14取1=1e=[550/30,20mm]=20mm,l0/h=4.80/0.55=8.72amaxl0 1 h ( )2l0

1 8.722111.13510ns M h c510

2501061300( 2eN a

) 1300( 20)1200103M=CmηM2=0.964×1.13×250=265.22kN·mnse0=M/N=265.22×106/1200×103=221.29mm.e=e0+e=221.29+20=241.29mm.i ae=ei+h/2-as’=241.29+550/2-40=476.29mmc 0 Nb=fbhξ=14.3×350×510c 0 N

1200000

0.47>2a’/h

=80/510=0.16fbh1 c

14.3350510 s 0AsAs

Ne1

fbh

2)f (ha )y 0 s1200103476.29114.335051020.470.50.47)360(51040)611.62mm20.002bh0.002350550385mm2b×h=500mm×700mm,l0=4.80m,计值N=2800kN,弯矩设计值M1=70kN·m，M2=75kN·m，采用C25,HRB500级纵筋HPB235级箍筋，试求按对称配筋钢筋截面面积As=As′=?绘配筋图解：fc=11.9N/mm2,fy=435N/mm2，fy’=410N/mm2,ξb=0.482,h0=700-45=655mm,Cm=0.7+0.3（M1/M2）=0.7+0.3*70/75=0.98，l0/h=4.80/0.7=6.86<151=0.5fcA/N=0.5×11.9×500×655/1200000=1.14取1=1e=[700/30,20mm]=23.3mm,=0.5fcA/N=0.5×11.9×500×700/2800000=0.74amax ch l

655 1

0 ( )2

1 6.8620.741.35ns M h c

751061300( 2eN a

) 1300( 2800103M=CmηM2=0.98×1.35×75=99.225kN·mnse0=M/N=99.225×106/2800×103=35.44mm.e

=35.44+23.3=58.74mm.，e=e+h/2-a’=58.74+700/2-45=363.74mmi i Nb=fbhξ=11.9×500×655×0.482=1878.47kN<N=2800kN小偏心c 0 be=58.74mm.<0.3h=0.3×655=196mm 按小偏心计算i 0 N1

fbhc 0

b Ne0.431

fbhc 0

bf(1

)(h0

as

1 c 0 28001031.011.95006550.4822800000363.740.431.011.950065521.011.9500655(0.80.482)(65540)

0.482 921525.510184720001097660462.5195.57

0.4820.264.04820.746As

Ne1

fbh

2)f (h a )y 0 s2800000363.74111.950065520.750.50.75)410(65540)10184720001233393164. 852mm2252150按构造配筋已知承受轴向压力设计值N=800kN,偏心矩e0=403mml0=8.90m，采用C30混凝土，并配有HRB400钢筋As=s=1964m（每边各42，试校核该柱受压承载力是否满足要求（取a=a′s s=40mm）解：fc=14.3N/mm2,fy=fy’=360N/mm2,ξb=0.517,e=403+20=423mm.i ae+h/2-s=423+600/2-40=683mm先按大偏心计算N=fbhξ=14.3×400×560ξ=3203200ξcNe

0fbh2)fAs(h a)1 c 0 y 0 s68332032001761903000(0.52)367660800 2*367660800* 2 2176190300048220.4220.4200.48 0.480.48 0.48*0.484*0.42

0.481.38

0.45.....2 2解得ξ=0.45N=fbhξ=14.3×400×560×0.45=1441.44kN >N=800kNu c 0b×h=300mm×400mm，采用C25混凝土，对称配筋，并已知As=s′=1520m（每边各2HRB335级限破坏时截面所能承受的轴向压力设计值NbMb.（未变）解：η=1，fc=11.9N/mm2，fy=300N/mm2，ξb=0.55，as=45，h0=400-45=365mmea=[400/3,20mm]max=20mmNb=cbh0b=11.9×30×365×0.55=1142275NN efbh

)fAs(h a)b 1 c 0 b

b y 0 s 1142275e11.930036520.5510.50.55e293.26mmi s 0 a i s 0 a

3001520(36545)0 a e=e-(e+h/2-a0 a Mb=e0Nb=118.26×1142275N=135.09kN·m某矩形截面偏心受压柱截面尺寸b×h=350mm×500mm,l0=3.90mC25,HRB335As=As′=509mm2（2ф18）试求当轴向压力设计值N=1200kN,，e0=140mm解：fc=11.9N/mm2,fy=fy’=300N/mm2,ξb=0.55,e=[500/30,20mm]=20mm,amaxe

=140+20=160mm.i ae=ei+h/2-as’=160+500/2-40=370mm先按大偏心计算N=fbhξ=11.9×350×460ξ=1915900ξc 0Nefbh2)fAs(ha)1 c 0 y 0 s370191590019159002)300509(46040) 37019159002 2( 0.5 2)1915900460 19159004601.6220.14620.40.14600.4 0.40.40.4 0.40.440.1462 故为小偏心受压，须重求ξ

0.55N1

fbhc

fAs(1y

0.80.8

)191590048864Ne

bfbh2)fAs(h

a)1915900460(0.52)641340001 c 0 y 0 s 37019159002 (6413400048864370) 2( 0.5 2)1915900460 19159004601.6220.10.4 0.4*0.44*0.4 0.4*0.44*0.1 0.5742Nu=1915900ξ+48864=1148.59kN<N=1200kN 轴向承载力不满足要求I形截面截面尺寸b×h=120mm×600mm,b=b′=500mm,h=h=120mmf f f f度l0=5.0m采用C30混凝土,HRB400级纵筋和HPB235级箍筋，承受轴向压力设计值N=125kN,弯矩设计值M=200kN·m,试求采用对称配筋时的纵筋截面面积As=As′=?，并会配筋截面图。l0已知一承受轴向压力设计值N=1500kN,M=150kN·mI形截面,计算长度=14.20m，截面尺寸b×h=150mm×800mm,b=b=400mm,h=hl0f f f fC25混凝土,HRB335级纵筋和HPB235级箍筋，试求采用对称配筋时的纵筋截面面积As=As′=?，并绘配筋截面图。0柱净高H=4.80m,计算长l0=6.30m,采用C25,HRB335级纵筋和HPB235计值,0取解：已知矩形截面柱b×h=400mm×600mm，采用C30,HRB5００级钢筋对称配筋，计算长度=4.0m,1M=３３kN=1３0k2M３５k，N=２5０0kNAs=As′=?解：f1N/m2,fy=435N/mm2fy’=410N/mm,ξb=0.482,as=as′=4m＝５６０ｍｍ，c ０04000/6006.67hN

fbh

f

A1940565113781N1371kNN153kNb 1c 0b y

ys

＜Ｎ＝2500kN该柱在这两组内力作用下均为小偏心受压，M1≈M2，N1<N2故第二组内力对截面配筋起控制作用：c 0.5fc1 12

A/N0.511.9400600/250