?e0?1.39?45?62.55mm?0.3h0?165mm,故可先按照小偏心受压来进行配筋计算。 es??e0?h/2?as?62.55?300?50?312.55mm es'??e0?h/2?as'?62.55?300?50??187.45mm (2)计算所需的纵向钢筋面积
As??min'bh?0.002bh?0.002?300?600?360mm2
现选As为412,As=452mm2 又Ax3?Bx2?Cx?D?0
A??0.5fcdb??0.5?13.8?300??2070 B?fcdbas'?13.8?300?50?207000
C??cuEsA(-Nes'sas'?h0)?0.0033?2?105?452(50?550)?2909.5?103?(?187.45) ?396225775D???cuEsA(sh0?as')h0?0.8?0.0033?2?105?452(550?50)?550 ?6.56304?1010求得x1?551mm ??x/h0?551/550?1.00???h/h0?1.091??,
???0.53b??按小偏压计算,??1.00 ?s??cuE(s?-1)?0.8-1) 1.0?-132MPa(压应力)5?0.0033?2?10(将As?452mm2,?s?-132MPa(压应力)x?551mm Nu?fcdbx?fsd'As'-?sAsAs'?Nu-fcdbx-?sAsfsd'2909.5?103?13.8?300?551?132?452?330'?2084mm2??minbh?0.002?300?600?360mm2
取428的钢筋,As'?2463mm2 取as?as'?50mm
bmin?(250-31.6)?3?32?4?31.6?290.8?b?300mm 2截面复核
(1)垂直弯矩作用平面 垂直弯矩作用平面的长细比
l0yb?6000?20,查附表得:??0.75 300则得:
Nu?0.9?[fcdbh?fsd'(As?As')]?0.9?0.75?[13.8?300?600?330?(452?2463)]?2326.0kN?2654kN
不满足承载力要求。
(2)弯矩作用平面内的复核 大小偏心受压的初步判断
as?as'?50mmAs?452mm2As'?2463mm2h0?h-as?550mm
?1?0.2?2.7e045?0.2?2.7??0.42?1,?1?0.42 h0550?2?1.15?0.01l06000?1.15?0.01??1.05?1,取?2?1.0 h6001e13000l1(0)2?1?2?1??102?0.42?1?1.39 h1300?45/550所以偏心距增大系数??1?h0?e0?1.39?45?62.55mm
es??e0?h/2?as?62.55?300?50?312.55mm es'??e0?h/2?as'?62.55?300?50??187.45mm 假定为大偏压,取?s?fsd
2x?(h0-es)?(h0-es)?2?fsdAses-fsd'As'es'fcdb13.8?300330?452?312.55-330?2463?(-187.45) 2由?(550-312.55)?(550-312.55)?2??237.45?56382.5?96124.4?627.97mm?0.53?550?291.5mm故截面为小偏压。
截面受压区高度x
Ax3?Bx2?Cx?D?0
A?0.5fcdb?0.5?13.8?300?2070
B?fcdb(es?h0)?13.8?300?(312.55?550)??983043C??cuEsAses?fsd'As'es'?0.0033?2?105?452?312.55?330?2463?(?187.45) ?-59117569.5D?-??cuEsAsesh0?-0.8?0.0033?2?105?452?312.55?550 ??4.10?1010
求得x1?582mm??bh0?0.53?550?291.5mm ??x/h0?582/550?1.058???h/h0?600/550?1.091??,
???0.53?b?截面为部分受压的小偏心受压构件,
?0.85?s??cuE(-1)?0.0033?2?10(-1)??161MPa(压应力) s?1.058Nu1?fcdbx?fsd'As'??sAs?13.8?300?582?330?2463?161?452?3295.04kN 因?e0?1.39?45?62.55mm 满足?e0?h?as' 2h600-as'?-50?250mm 22故偏心轴力作用于钢筋As合力点和As’合理点之间
h0'?h?as'?600?500?550mm
e'?h/2?e0?as'?300?45?50?205mm
Nu2?fcdbh(h0'?h/2)?fsd'As(h0'?as)?0e'13.8?300?600?(550?300)?330?452?(550?50)??3393.07kN1.0?205
?Nu2?Nu1?Nu?3393.07kN?N?2645kN,满足承载力要求。
7-10 与非对称布筋的矩形截面偏心受压构件相比,对称布筋设计时的大、小偏心受压的判别方法有何不同之处?P182
答:
对称布筋时:
由于As?As',fsd?fsd',由此可得N?fcdbx
??x/h0?N可直接求出。
fcdbh0然会根据???b,判断为大偏心受压;???b,判断为小偏心受压; 非对称布筋时:
??x/h0无法直接求出。
判断依据为?e0?0.3h0,可先按小偏心受压构件计算;?e0?0.3h0,可先按大偏心受压构件计算。
7-11矩形截面偏心受压构件的截面尺寸为bxh=250mmx300mm,弯矩作用平面内和垂直于弯矩作用平面的计算长度l0=2.2m:C30混凝土和HRR400级钢筋;I类环境条件,设计使用年限50年,安全等级为二级;轴向力设计值Nd=122kN.相应弯矩设计值Md=58.5kN.m,试按对称布筋进行截面设计和截面复核。
解:查表得:
fcd?13.8MPa,fsd?fsd'?330MPa,?0?1.0,cmin?20mmN?Nd??0?122?1.0?122kN,M?Md??0?58.5?1.0?58.5kN?m ?b?0.53,
h30?300?10mm?20mm 30M58.5?106??480mm, 偏心距e0?N122?103在弯矩作用平面内,长细比
l02200??7.33?5 h300设as?as'?45mm,h0?h-as?300?45?255mm
?1?0.2?2.7e0480?0.2?2.7??5.28?1,?1?1 h0255l02200?1.15?0.01??1.08?1,取?2?1.0 h300?2?1.15?0.01所以偏心距增大系数
??1?11300e0(0)2?1?2?1?lh122002?()?1?1?1.02
1300?(480/255)300h0?e0?1.02?480?489.6mm
N122?103????0.139??b?0.53
fcdbh013.8?250?255可按大偏心受压构件设计
由??0.139,h0=255mm,x=?h0=0.139?255=35.4mm<2as’=80mm
es'??e0?h/2?as'?489.6?300?45?384.6mm 2122?103?384.6As???693.6mm2
fsd(h0-as')330?(250?45)现选As为318,As=763mm2>Asmin=0.002bh=0.002?250?300=150mm2
Nes'bmin=2?34.75+2?50+3?20.5?231mm?b?250mm
所以,as和as’取为45mm,取箍筋直径为8mm
c=45-20.5?8?26.75mm?20mm 2
截面复核
1. 在垂直于弯矩作用平面内的复核 长细比
l02200??8.8mm,??0.992 b250(13.8?300?250?330?763)?1148.8kN2. Nu?0.9?(fcdA?fsd'As')?0.9?0.9922.在弯矩作用平面内
as?as'?45mm,As?As'?763mm2
h0?455mm,??1.02,?e0?489.6mmes??e0??as?489.6?150?40?599.6mm
2es'?384.6mm假定为大偏压,?s?fsd
h