塔吊根底计划(四桩)计划书工程称号：华力特大年夜厦体例单元：深圳超卓工程1.计划参数〔1〕全然参数采纳1台QZT63A(5510)塔式起重机，塔身尺寸承台面标高-1.50m；采纳预应力管桩根底，地下水位1〕塔吊根底受力状况1.70m,地下室开挖深度为-5.00m；现场空中标高0.00m，-3.00m。根底荷载.荷载工况P(kN)M(kNm)FkFhMMZ任务外形950.0030.0070.001600.00340.00非任务外形850.001800.000MF=根底顶面所受垂直力kF=根底顶面所受程度力hFkM=根底顶面所受颠覆力矩Mz=根底所受扭矩MzFh塔吊根底受力表示图比拟桩根底塔吊的任务外形跟非任务外形的受力状况，塔吊根底按非任务外形计划如图F=850.00kN,F=70.00kNkh.M=1800.00+70.00×1.40=1898.00kNm，，F=850.00×1.35=1147.50kN,F=70.00×1.35=94.50kNkh.M=(1800.00+70.00×1.40)×1.35=2562.30kNmk2〕桩顶以下岩土力学材料序号厚度L极限侧阻力标极限端阻力规范qsikiλqsikii地层称号抗拔系数λi(m)0.200.20准值qkPa)值q(kPa)pk(kN/m)(kN/m)3.20sik(1砂质粘土桩长40.009000.008.008.000.40∑qsik*Li∑λiqsik*Li3.203〕根底计划要紧参数根底桩采纳4根φ500预应力管桩,桩顶标高-2.90m；桩混凝土品级24C25,f=11.90N/mm,E=2.80×10CC2=2.00×10N/mm2225tys=1.27N/mm,桩长0.20m,壁厚125mm；钢筋HRB335，f=300.00N/mm,EN/mm；f承台尺寸长(a)=5.00m,宽(b)=5.00m,高(h)=1.50m；桩核心与承台核心2.00m,承台面标高-1.50m；承台混22=25kN/m3凝土品级C35，f=1.57N/mm,f=16.70N/mm,γtC砼G=abhγ=5.00×5.00×1.50×25=937.50kNk砼塔吊根底尺寸表示图2.桩顶沾染效应计划〔1〕竖向力1〕轴心竖向力沾染下Nk=(Fkk＋G)/n=(850.00+937.50)/4=446.88kN2〕公正竖向力沾染下.0.5=2.83m依照Mx沾染在对角线进展计划，M＋G)/n±M/Σy=782.22kN,N=111.54kN(基桩不接受竖向拉力)(2)程度力H=F/n=70.00/4=17.50kNx=Mk=1898.00kNm，yi=2.00Nk=(Fkkxyii2=(850.00+937.50)/4±(1898.00×2.83)/(2×2.832)=446.88±335.34Nkmaxkminikh3.单桩赞成承载力特征值计划管桩外径d=500mm=0.50m，内径d=0.16×0.40=0.06〔1〕单桩竖向极限承载力规范值计划1=500-2×125=250mm=0.25m，h=0.20bhb/d=0.20/0.50=0.40,λpA=π(d-d)/4=3.14×(0.50-0.25)/4=0.15m,A=πd/4=3.14×0.25/4=0.05m222122222jpl1Q=u∑q=πd∑q=3.14×0.50×8.00=12.56kNsksikisikiQpk=qpk(Aj+λpApl)=9000.00×(0.15+0.06×0.05)=1377.00kN，Q=Q＋Q=12.56+1377.00=1389.56kNukskpkR=1/KQ=1/2×1389.56=694.78kNauk(2)桩基竖向承载力计划1〕轴心竖向力沾染下N=446.88kN＜R=694.78kN,竖向承载力满意央求。ka2〕公正竖向力沾染下N=782.22kN＜R=1.2×694.78=833.74kN,竖向承载力满意央求。kmaxa4.桩基程度承载力验算(1)单桩程度承载力特征值计划44)/64=3.14×(0.50-0.25)/64=0.0029m444I=π(d-d1I=2.80×10×0.0029=81200kN.m27EI=Ec34查表得:m=6.00×10kN/m,X=0.010moab=0.9(1.5d+0.5)=1.13m=1130mmo0.20.2=0.61oα=(mb/ECI)=(6.00×1000×1.13/81200)αL=0.61×0.20=0.12＜4,按αL=0.12,查表得:υ=0.053x33=0.75×(αEI/υ)χ=0.75×(0.61×81200/0.053)×0.01=2608.14kNxoa〔2〕桩基程度承载力计划RHaH=17.50kN＜R=2608.14kN,程度承载力满意央求。ikha5.抗拔桩基承载力验算〔1〕抗拔极限承载力规范值计划T=1/nuΣλqL=1/4×(2.00×2+0.50)×4×3.20=14.40kNgk1isikiTuk=ΣλiqsikuiLi=3.20×3.14×0.50=5.02kN(2)抗拔承载力计划G=G＋G=5.00×5.00×0×18.80/4＋5.00×5.00×0.10×(18.80-10)/4=5.50kNgpgp1gp2G=G＋G=0.15×0.10×25＋0.15×0.10×(25-10)=0.60kNpp1p2Tgk/2+G/2+Ggp=14.40/2+5.50=12.70kNTukp=5.02/2+0.60=3.11kN因为基桩不接受竖向拉力,故基桩呈全体性跟非全体性毁坏的抗拔承载力满意央求。6.抗颠覆验算a=2.00+0.50/2=2.25m,b=2.00×2+0.50/2=4.25m1i颠覆力矩M=M＋Fh=1800+70.00×(5.00-1.50)=2045.00kNh.m抗颠覆力矩M=〔F＋G〕a＋2(T/2+G)bi倾抗kkiukp.=(850.00+937.50)×2.25+2×(5.02/2+0.60)×4.25=4048.31kNmM/M=4048.31/2045.00=1.98抗倾抗颠覆验算1.98＞1.6,满意央求。7.桩身承载力验算(1)正截面受压...
