CONCULTANTS ASIA LTD.81.2.1.3.结o 弯矩轴力剪力弯矩轴力剪力M N V M'N'V'kN.mkN kN kN.m kN kN 121402.5720.71402.5720.7正截面受弯ULS4. 1.混混凝=C35混凝f c =16.7N/mm 混凝f t=1.57N/mm2.普纵向=HRB400纵向f y =f y=N/钢筋E s =210000N/mm3.箍箍筋=HRB400箍筋f yv=N/5.截面r =m 保护=mm 纵向r s =m 纵向n=根纵向φz =mm 箍筋φg =mm箍筋间距s =mm采用普通箍筋619280991极限状态结构分析值结构设计值7030280.416根据中国人民共和国国家标准GB50010-2010 (混凝土结构设计规范)桩身为沿周边均匀配筋的圆形截面。

1402.50720.70.5正截面受弯ULS 内力工况受力状态12100C35HRB400HRB400CONCULTANTS ASIA LTD.全部A s'=n pf z2/4=30x=18473mm2构件MAUNSELLCONCULTANTS ASIA LTD.11.3 钻孔灌注桩设计A =p r2=πx =0.79m 2纵向ρs '=As'/A =1847=2.34%<3%OK全部A s =n pf z 2/4=30x=18473mm2A)正弯距M =1928KNm系数α1=1.00解方αα E.0.令aα1fcA(1=αx1x16.=求得α=受拉αt=内力[M ]=2/3a 1fcArsin E.0.=2/3 x 1 x + 360x =Nm m =KN m>M'=1928KNOKB)受弯矩M =1928.00KNm 剪力V =991.00KN 截面b = 1.76r = 1.76=880.00mm 截面h 0= 1.6r= 1.6=800.00mm步骤混凝βc=1.00截面h w=h 0=800.00mm截面[V ]=0.25b c f c bh 0当h w /b =2939.20k N 6.3.=0.2b c f cbh 0当h w /b =2351.36k N6.3.=线性内插当4<h w∵h w /b =800 /880=∴[V ]=0.25bc fcbh0=KN>V'=991.00KN OK步骤0.7f=0.7 6.3.=KN<V'=KN NOT OK0.280.702020.802939.200.91-1189584.372.02E+09773.70991.00CONCULTANTS ASIA LTD.步骤预加V p=KN 同一A sv=2pf2g/4=2xπ=226.19mm2斜截V cs=0.7f t bh0+f yv A6.3.=(0.7 x1.57 x=1425.14KN 截面[V]=Vp+Vcs=0+6.3.=1425.14KN>V'=991.00KN OK 步骤截面h=2r=2x=1000mm6.3.箍筋s max=150当150<h且V>0.表9.200300<h250500<h300h>800200150<h且V≤300300<h350500<h400h>800=mm>s=mm OK箍筋ρsv =A sv/(bs)=226.=0.26%0.24=0.24=0.10%9.2.∵0.7f>V'，箍筋ρsv>0.24配有9. 2.[S]=min(15φ=mm>s=100mm OK 结论=步骤斜截根据V=∑6.3.非预A sb=0.00mm 2预应A pb=0.00mm 2斜截c=V*s/(fyv*A sv)=1216.99mm受拉A sl=A s a t=0.7x=12844.43mm2纵向z=0.9h0=0.9x800=720.00mm 不需进行斜截面受弯承载力计算100 300不需要满足4000.00CONCULTANTS ASIA LTD.斜截面上箍筋的合力至斜截面受压区合力点的距离z sv=c/2=1216.99/2=608.50mm斜截面受弯承载能力[M]=(f y A sl+f py A p)z+∑f y A sb z sb+∑f py A pb zpb+∑f yv A sv zsv6.3.9-1=((360x 12844. 43+0)7 20+0+0 +360x2 26.19x 608x12 16.99/ 100)/1 0^6=3932.30KNm>M'=1928.00KN OKC)最大裂缝宽度计算最大裂缝宽度限值w lim=0.3mm表3.4.5构件受力特征系数a cr= 1.9表7.1.2-1有效受拉混凝土截面面积A te=0.5bh7.1.2=0.5x880x80=352000mm2按有7.1.r te=(As+A p)/A te=(atA s'+A p)/A te=(0.7x18472.56+0)/35200=0.0367受拉钢筋应力值s sq=M q/(0.87Ash o)7.1.4-3 =1402.5x10^6/[0.87x0.7x18472.56x(800-0.416)]=156.97MPa裂缝间纵向受拉钢筋应变不均匀系数f=1.1-0.65f tk/r tes sq7.1.2-2=1. 1-0. 65 x1 .5 7/ 0. 03 67 /1 56 .9 7∴= 1.0000&< 1.0最大裂缝宽度w max=a cr xfxs skx(1.9xc+0.08xd eq/r te)/E s7.1.2-1 =1.9x0.923x156.97x(1.9x70+0.08x28/0.037)/210000=mm<0.3mm满足裂缝宽度要求0.28。