1、平衡块自重G4=81kN;塔吊最大起重荷载Qmax=39.2kN;塔吊最小起重荷载Qmax=7.84kN;塔基自重标准值:Fk1=331.09kN;基础自重标准值:Gk=500kN;起重荷载标准值:Fqk=39.2kN;8.3.2风荷载计算8.3.2.1工作状态下风荷载标准值塔机所受风均布线荷载标准值:(o=0.2kN/m)qsk=0.8z sz oo BH/H=0.81.21.591.951.350.20.351.5=0.422kN/m塔机所受风荷载水平合力标准值:Fvk=qskH=0.42239=16.46kN基础顶面风荷载产生的力矩标准值:Msk=0.5FvkH=0.516.4639=3
2、21kN m8.3.2.2非工作状态下风荷载标准值(o=0.55kN/mqsk=0.8z sz oo BH/H=0.80.551.5=1.3kN/mFvk=qskH=1.339=50.27kNMsk=0.5Fvk50.2739=980.27kN 8.3.3塔机的倾覆力矩塔机自身的倾覆力矩,向起重臂方向为正,向平衡臂的方向为负。1、大臂自重产生的力矩标准值:M1 =30.320=606kN 2、最大起重荷载产生的力矩标准值:M2=39.210=392kN 3、小车产生的力矩标准值:M3=2.8610=28.6kN 4、平衡臂产生的力矩标准值:M4=-15.056=-90.3kN 5、平衡产生的力
3、矩标准值:M5=-8111=-891kN 8.3.4综合分析计算8.3.4.1工作状态下塔基对基础顶面的作用1、标准组合的倾覆力矩标准值:Mk=M1 +M3+ M4 +M5 +0.9(M2 +Msk)=606+28.6-90.3-891+0.9(392+321)=295kN 2、水平荷载标准值:Fvk=16.46kN3、竖向荷载标准值:8.3.4.2非工作状态下塔基对基础顶面的作用Mk=M1 +M4 +M5 +Msk=606-90.3-891+980.27=604.79kN 无起重荷载,小车收拢于塔身边,故没有力矩M2 、M3 。Fk=Fk1+Gk =331.09+500=831.09kN比较
4、以上工况和非工况的计算,可知本例塔机在非工作状态时对于基础传递的倾覆力矩最大,故应该按照非工作状态的荷载组合进行塔吊基础承载力验算。8.4桩基承载力验算倾覆力矩按照最不利的对角线方向作用,取最不利的非工作状态荷载进行验算。8.4.1桩基竖向荷载验算1、轴心竖向力作用下:(以最不利情况塔吊基础底部只有两根桩进行验算),满足要求。2、偏心竖向力作用下:8.4.2桩身轴心受压承载力验算,查国家标准图集03SG409可得,PHC400A95-21桩身结构竖向承载力设计值R=1650kN。,轴心受压承载力符合设计要求。8.5塔吊基础承载力验算8.5.1示意图8.5.2相关数据1几何参数:B1 = 225
5、0 mm;A1 = 2250 mm;H1 = 1200 mm;B = 1500 mm;A = 1500 mm;B2 = 2250 mm;A2 = 2250 mm基础埋深d = 1.20 m2荷载值:(1)作用在基础顶部的标准值荷载Fgk = 831.09 kN;Mgxk = 604.79 kNm;Vgxk = 50.27 kN(2)作用在基础底部的弯矩标准值Mxk = 604.79 kNVxk = 50.27 kN绕X轴弯矩: M0xk= 604.79 kN绕Y轴弯矩: M0yk= 60.32 kN(3)作用在基础顶部的基本组合荷载不变荷载分项系数rg = 1.20 活荷载分项系数rq = 1
6、.40F = rgFgkrqFqk = 997.31 kNMx = rgMgxkrqMqxk = 725.75 kNVx = rgVgxkrqVqxk = 60.32 kN(4)作用在基础底部的弯矩设计值 M0x= 725.75 kN M0y = VxH1 =60.321.20 = 72.39 kN3材料信息:混凝土:C35 钢筋:HRB400(20MnSiV、20MnSiNb、20MnTi)4基础几何特性:底面积:S = (A1A2)(B1B2) = 4.504.50 = 20.25 m2绕X轴抵抗矩:Wx = (1/6)(B1+B2)(A1+A2)2 = (1/6)4.504.502 =
7、15.19 m3绕Y轴抵抗矩:Wy = (1/6)(A1+A2)(B1+B2)2 = (1/6)8.5.3计算过程8.5.3.1修正地基承载力按建筑地基基础设计规范(GB 500072002)下列公式验算:fa = fakb(b3)dm(d0.5) (式5.2.4)式中:fak = 220.00 kPab = 0.00,d = 1.00 = 18.00 kN/m3 m = 18.00 kN/m3b = 4.50 m, d = 1.20 m如果 b 3m,按 b = 3m, 如果 b 6m,按 b = 6m如果 d 0.5m,按 d = 0.5m(d0.5)= 220.000.0018.00(4
8、.503.00)1.00(1.200.50)= 232.60 kPa修正后的地基承载力特征值 fa = 232.60 kPa(满足塔吊基础说明书不得低于200kPa的要求)。8.5.3.2轴心荷载作用下地基承载力验算pk = (FkGk)/A (5.2.21)Fk = FgkFqk = 831.090.00 = 831.09 kNGk = 20Sd = 2020.251.20 = 486.00 kNpk = (FkGk)/S = (831.09486.00)/20.25 = 65.04 kPa fa,满足要求。8.5.3.3偏心荷载作用下地基承载力验算当eb/6时,pkmax = (FkGk)
9、/AMk/W (5.2.22)pkmin = (FkGk)/AMk/W (5.2.23)当eb/6时,pkmax = 2(FkGk)/3la (5.2.24)X、Y方向同时受弯。偏心距exk = M0yk/(FkGk) = 60.32/(831.09486.00) = 0.05 me = exk = 0.05 m (B1B2)/6 = 4.50/6 = 0.75 mpkmaxX = (FkGk)/SM0yk/Wy= (831.09486.00)/20.2560.32/15.19 = 69.01 kPa偏心距eyk = M0xk/(FkGk) = 604.79/(831.09486.00) =
10、0.46 me = eyk = 0.46 m (A1A2)/6 = 4.50/6 = 0.75 mpkmaxY = (FkGk)/SM0xk/Wx= (831.09486.00)/20.25604.79/15.19 = 104.86 kPapkmax = pkmaxXpkmaxY(FkGk)/S = 69.01104.8665.04 = 108.83 kPa 1.2fa = 1.2232.60 = 279.12 kPa,满足要求。8.5.3.4基础抗冲切验算Fl 0.7hpftamh0 (8.2.71)Fl = pjAl (8.2.73)am = (atab)/2 (8.2.72)pjmax,
11、x = F/SM0y/Wy = 997.31/20.2572.39/15.19 = 54.02 kPapjmin,x = F/SM0y/Wy = 997.31/20.2572.39/15.19 = 44.48 kPapjmax,y = F/SM0x/Wx = 997.31/20.25725.75/15.19 = 97.04 kPapjmin,y = F/SM0x/Wx = 997.31/20.25725.75/15.19 = 1.46 kPapj = pjmax,xpjmax,yF/S = 54.0297.0449.25 = 101.80 kPa(1)标准节对基础的冲切验算:H0 = H1H2as = 1.200.000.05 = 1.15 mX方向:Alx = 1/4(A2H0A1A2)(B1B2B2H0)= (1/4)(1.5021.154.50)(4.501.5021.15)= 1.45 m2Flx = pjAlx = 101.801.45 = 147.87 kNab = minA2H0, A1A2 = min1.5021.15, 4.50 = 3.80 mamx = (atab)/2 = (Aab)/2 = (1.503.80)/2 = 2.65
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