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,/9.1122===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 sb c b c bh f a h bh f Ne bh f N ξαξβαξαξ 220201987)40560(360)55.05.01(55.05604009.1115201500000)()5.01(mm a h f bh f Ne s A As s y c =-⨯⨯-⨯⨯⨯⨯⨯-⨯='-'--='=ξξα4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 20.006 bh=0.006×400×600=1440mm 2<2×1018mm 2<2×1140mm 20.05bh=0.05×400×600=12000mm 2>2×1140mm 2>2×1018mm 26-14 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=0.964ζ1=0.5 fcA/N=0.5×14.3×350×550/1200000=1.14 取ζ1=1e a =max [550/30,20mm]=20mm, l 0/h=4.80/0.55=8.7213.11172.8)2010120010250(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 ·me 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=476.29mmNb= f c b h 0ξb =14.3×350×510×0.517=1319.668kN >N=1200kN 大偏心47.05103503.14120000001=⨯⨯==bh f N c αξ>2a s ’/h 0=80/510=0.16 22230201385550350002.0002.062.611)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 =⨯⨯=>=-⨯⨯-⨯⨯⨯⨯⨯-⨯⨯='-'--='=ξξα6-15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<15ζ1=0.5 fcA/N=0.5×11.9×500×655/1200000=1.14 取ζ1=1e 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)3.231028001075(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 ·me 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 <N=2800kN 小偏心e i =58.74mm.<0.3h 0=0.3×655=196mm 按小偏心计算746.00482.264.0482.0389725057.1955.109766046210184720005.921525482.06555009.110.1)40655)(482.08.0(6555009.110.143.074.3632800000482.06555009.110.1102800))((43.023010120101=+=++-=+⨯⨯⨯+--⨯⨯⨯⨯-⨯⨯⨯⨯⨯-⨯=++'----=b c sb c b c bh f a h bh f Ne bh f N ξαξβαξαξ220201852252150.12333931641018472000)40655(410)75.05.01(75.06555009.11174.3632800000)()5.01(mm a h f bh f Ne s A s y c -=-=-⨯⨯-⨯⨯⨯⨯⨯-⨯='-'--='ξξα 按构造配筋6-16已知承受轴向压力设计值N=800kN,偏心矩e 0=403mm 的矩形截面偏心受压柱，其截面尺寸b ×h=400m m ×600mm ，计算长度l 0=8.90m ，采用C30混凝土，并配有HRB400级纵向钢筋As=A s ′=1964mm 2（每边各4ф25），试校核该柱受压承载力是否满足要求（取a s =a s ′=40mm ）解：fc=14.3N/mm 2,fy=fy ’=360N/mm 2,ξb=0.517, h 0=600-40=560mm, e i =e 0+e a =403+20=423mm.e=e i +h/2-a s ’=423+600/2-40=683mm先按大偏心计算 N= f c b h 0ξ=14.3×400×560ξ=3203200ξ.....45.0238.148.0242.0*448.0*48.048.0042.048.042.0248.217619030002*367660800217619030002187785600*2367660800)5.0(17619030003203200683)()5.01(22220201=±-=+±-==-++-=+-=+-=⨯'-''+-=ξξξξξξξξξξξξξξαs y c a h s A f bh f Ne解得ξ=0.45N u = f c b h 0ξ=14.3×400×560×0.45=1441.44kN >N=800kN6-17 某矩形截面偏心受压短柱，截面尺寸b ×h=300mm ×400m m ，采用C25混凝土，对称配筋，并已知As=A s ′=1520mm 2（每边各4ф22，HRB335级），试求偏心矩使截面恰好为界限破坏时截面所能承受的轴向压力设计值Nb 和相应的弯矩设计值M b.（未变） 解：η=1，f c=11.9N/mm 2，f y=300N/mm 2，ξb=0.55，as=45，h 0=400-45=365mme a =[400/3,20mm]max =20mmN b = f c b h 0ξb =11.9×300×365×0.55=1142275N()mme e a h s Af bh f e N s y b b c b 26.293)45365(152030055.05.0155.03653009.111142275)()5.01(20201=-⨯⨯+⨯-⨯⨯⨯=•'-''+-=•ξξα e=e i +h/2-a s =e 0+e a +h/2-a se 0=e-(e a +h/2-a s )=293.26-20-200+45=118.26mmMb= e 0 N b =118.26×1142275N=135.09kN ·m6-18某矩形截面偏心受压柱截面尺寸b ×h=350m m ×500mm, 计算长度l 0=3.90m ，采用C25混凝土,HRB335级纵筋,且已知采用对称配筋时As=A s ′=509mm 2（每边各2ф18）试求当轴向压力设计值N=1200kN,，e 0=140mm 时，该配筋是否能满足受压承载力的要求？解：fc=11.9N/mm 2,fy=fy ’=300N/mm 2,ξb=0.55, h 0=500-40=460mm,e a =max [500/30,20mm]=20mm,e i =e 0+e a =140+20=160mm.e=e i +h/2-a s ’=160+500/2-40=370mm先按大偏心计算 N= f c b h 0ξ=11.9×350×460 ξ=1915900ξ 146.026.14601915900264134000)5.0(2460191590021915900370)40460(509300)5.0(46019159001915900370)()5.01(2220201+-=⨯⨯+-=⨯⨯⨯-⨯⨯+-⨯=⨯'-''+-=ξξξξξξξξξξξαs y c a h s A f bh f Ne 55.063.02146.044.04.04.00146.04.02=>=⨯+⨯±==--b ξξξξ 故为小偏心受压，须重求ξ1.026.146019159002)3704886464134000()5.0(246019159002191590037064134000)5.0(4601915900)()5.01(488641915900)8.08.01(222020101+-=⨯⨯⨯-+-=⨯⨯⨯+-⨯='-''+-=+=---''+=ξξξξξξξξξξαξξξξαs y c by c a h s A f bh f Ne s A f bh f N574.021.0*44.0*4.04.001.04.02=+±==--ξξξ Nu=1915900ξ+48864=1148.59kN<N=1200kN 轴向承载力不满足要求6-19 某对称I 形截面截面尺寸b ×h=120mm ×600m m,b f =b f ′=500mm, h f =h f ′=120mm 计算长度l 0=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=200k N ·m , 试求采用对称配筋时的纵筋截面面积As=A s ′=?，并会配筋截面图。