1、QTZ63塔吊桩基验算书2、QTZ63塔吊基础承载力验算6#塔吊为QTZ63塔吊,塔吊为独立状态计算,分工况和非工况两种状态分别进行塔吊基础的受力分析。7.1、塔机概况塔吊型号:QTZ63,塔吊安装高度H=27.68m,塔身宽度B=1.60m,自重F1=314.58kN,最大起重荷载F2=58.8kN,基础以上土的厚度D=0.00m,塔吊基础混凝土强度等级:C35基础厚度Hc=1.30m,基础宽度Bc=5.20m,7.2、桩基概况查国家标准图集03SG409可得,PHC400A95-21为C80混凝土,桩身结构竖向承载力设计值R=1650kN。现场桩基间距a=2.50m,桩直径=0.40m,7
2、.3、桩基荷载计算分析7.3.1自重荷载以及起重荷载塔吊自重G0=314.58kN; 起重臂自重G1=47.4kN;小车和吊钩自重G2=5.15kN;平衡臂自重G3=17.5kN;平衡块自重G4=147kN;塔吊最大起重荷载Qmax=58.8kN;塔吊最小起重荷载Qmax=9.8kN;塔基自重标准值:Fk1=531.63kN;基础自重标准值:Gk=800kN;起重荷载标准值:Fqk=58.8kN;7.3.2风荷载计算7.3.2.1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.
3、6=0.45kN/m塔机所受风荷载水平合力标准值:Fvk= qsk H=0.4530=13.53kN基础顶面风荷载产生的力矩标准值:Msk=0.5 Fvk H=0.513.5330=203kN m7.3.2.2非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=0.55kN/m)qsk=0.8z sz oo BH/H=0.81.21.591.951.350.550.351.6=1.24kN/m塔机所受风荷载水平合力标准值:Fvk= qsk H=1.2430=37.2kN基础顶面风荷载产生的力矩标准值:Msk=0.5 Fvk H=0.537.230=558kN m7.3.3塔机的倾覆力矩塔
4、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 =47.421.56=1021.94 kN m2、最大起重荷载产生的力矩标准值:M2=58.810.97=645.04 kN m3、小车产生的力矩标准值:M3=5.1510.97=56.5 kN m4、平衡臂产生的力矩标准值:M4=-17.57.24=-126.7 kN m5、平衡产生的力矩标准值:M5=-14712=-1764 kN m7.3.4综合分析计算7.3.4.1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)=10
5、21.94+56.5-126.7-1764+0.9(645.04+203)=-49.24 kN m2、水平荷载标准值:Fvk=13.53kN3、竖向荷载标准值:塔基自重标准值:Fk1=531.63kN;基础自重标准值:Gk=800kN;起重荷载标准值:Fqk=58.8kN;7.3.4.2非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 + M4 +M5 +Msk=1021.94-126.7-1764+558=-301.76 kN m无起重荷载,小车收拢于塔身边,故没有力矩M2 、M3 。2、水平荷载标准值:Fvk= qsk H=1.2430=37.2kN3、竖向荷载标
6、准值:塔基自重标准值:Fk1=531.63kN;基础自重标准值:Gk=800kN;Fk= Fk1+ Gk =531.63+800=1331.63 kN比较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。7.4 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。7.4.1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:(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。7.4.2桩身轴心受压承载力验算,查
7、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。7.5 塔吊基础承载力验算7.5.1示意图7.5.2相关数据1几何参数:B1 = 2600 mm;A1 = 2600 mm;H1 = 1200 mm;B = 1600 mm;A = 1600 mm;B2 = 2600 mm;A2 = 2600 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 = 1331.63 kN;Mgyk = 301.76 kNm;Vgxk = 37.20 kN (2)作用在基础底部的弯矩标准值Myk = 301
8、.76 kNmVxk = 37.20 kNm绕Y轴弯矩: M0yk = MykVxk(H1H2) = 301.7637.201.20 = 346.40 kNm(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 = 1.40F = rgFgkrqFqk = 1597.96 kNMy = rgMgykrqMqyk = 362.11 kNmVx = rgVgxkrqVqxk = 44.64 kN (4)作用在基础底部的弯矩设计值绕Y轴弯矩: M0y = MyVxH1 = 362.1144.641.20 = 415.68 kNm3材料信息:混凝土: C35;钢筋:
9、 HRB335(20MnSi)4基础几何特性:底面积:S = (A1A2)(B1B2) = 5.205.20 = 27.04 m2绕X轴抵抗矩:Wx = (1/6)(B1+B2)(A1+A2)2 = (1/6)5.205.202 = 23.43 m3绕Y轴抵抗矩:Wy = (1/6)(A1+A2)(B1+B2)2 = (1/6)5.205.202 = 23.43 m37.5.3计算过程7.5.3.1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
10、= 1.00 = 18.00 kN/m3 m = 18.00 kN/m3b = 5.20 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 = 0.5mfa = fakb(b3)dm(d0.5)= 220.000.0018.00(5.203.00)1.0018.00(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200 kPa的要求)。7.5.3.2轴心荷载作用下地基承载力验算计算公式:按建筑地基基础设计规范(GB 500072002)下列公式
11、验算:pk = (FkGk)/A (5.2.21)Fk = FgkFqk = 1331.630.00 = 1331.63 kNGk = 20Sd = 2027.041.20 = 648.96 kNpk = (FkGk)/S = (1331.63648.96)/27.04 = 73.25 kPa fa,满足要求。7.5.3.3偏心荷载作用下地基承载力验算计算公式:按建筑地基基础设计规范(GB 500072002)下列公式验算:当eb/6时,pkmax = (FkGk)/AMk/W (5.2.22)pkmin = (FkGk)/AMk/W (5.2.23)当eb/6时,pkmax = 2(FkGk
12、)/3la (5.2.24)X方向:偏心距exk = M0yk/(FkGk) = 346.40/(1331.63648.96) = 0.17 me = exk = 0.17 m (B1B2)/6 = 5.20/6 = 0.87 mpkmaxX = (FkGk)/SM0yk/Wy= (1331.63648.96)/27.04346.40/23.43 = 88.03 kPa 1.2fa = 1.2232.60 = 279.12 kPa,满足要求。7.5.3.4基础抗冲切验算计算公式:按建筑地基基础设计规范(GB 500072002)下列公式验算:Fl 0.7hpftamh0 (8.2.71)Fl
13、= pjAl (8.2.73)am = (atab)/2 (8.2.72)pjmax,x = F/SM0y/Wy = 1597.96/27.04415.68/23.43 = 76.83 kPapjmin,x = F/SM0y/Wy = 1597.96/27.04415.68/23.43 = 41.36 kPapjmax,y = F/SM0x/Wx = 1597.96/27.040.00/23.43 = 59.10 kPapjmin,y = F/SM0x/Wx = 1597.96/27.040.00/23.43 = 59.10 kPapj = pjmax,xpjmax,yF/S = 76.835
14、9.1059.10 = 76.83 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)(1.6021.155.20)(5.201.6021.15)= 2.96 m2Flx = pjAlx = 76.832.96 = 227.24 kNab = minA2H0, A1A2 = min1.6021.15, 5.20 = 3.90 mamx = (atab)/2 = (Aab)/2 = (1.603.90)/2 = 2.75 mFlx 0.7hpftamxH0 =
15、0.70.971570.002.7501.150= 3359.73 kN,满足要求。Y方向:Aly = 1/4(B2H0B1B2)(A1A2A2H0)= (1/4)(1.6021.155.20)(5.201.6021.15)= 2.96 m2Fly = pjAly = 76.832.96 = 227.24 kNab = minB2H0, B1B2 = min1.6021.15, 5.20 = 3.90 mamy = (atab)/2 = (Bab)/2 = (1.603.90)/2 = 2.75 mFly 0.7hpftamyH0 = 0.70.971570.002.7501.150= 335
16、9.73 kN,满足要求。7.5.3.5基础受压验算计算公式:混凝土结构设计规范(GB 500102010)Fl 1.35clfcAln (7.8.1-1)局部荷载设计值:Fl = 1597.96 kN混凝土局部受压面积:Aln = Al = BA = 1.601.60 = 2.56 m2混凝土受压时计算底面积:Ab = minB2A, B1B2min3A, A1A2 = 23.04 m2混凝土受压时强度提高系数:l = sq.(Ab/Al) = sq.(23.04/2.56) = 3.001.35clfcAln= 1.351.003.0016700.002.56= 173145.60 kN
17、Fl = 1597.96 kN,满足要求。7.5.3.6基础受弯计算计算公式:按建筑地基基础设计规范(GB 500072002)下列公式验算:M=a12(2la)(pmaxp2G/A)(pmaxp)l/12 (8.2.74)M=(la)2(2bb)(pmaxpmin2G/A)/48 (8.2.75)(1)基础根部受弯计算:G = 1.35Gk = 1.35648.96 = 876.10kNX方向受弯截面基底反力设计值:pminx = (FG)/SM0y/Wy = (1597.96876.10)/27.04415.68/23.43 = 73.76 kPapmaxx = (FG)/SM0y/Wy
18、= (1597.96876.10)/27.04415.68/23.43 = 109.23 kPapnx = pminx(pmaxxpminx)(2B1B)/2(B1B2)= 73.76(109.2373.76)6.80/(25.20)= 96.95 kPa-截面处弯矩设计值:M= (B1B2)/2B/222(A1A2)A(pmaxxpnx2G/S)(pmaxxpnx)(A1A2)/12= (5.20/21.60/2)2(25.201.60)(109.2396.952876.10/27.04)(109.2396.95)5.20)/12= 475.34 kN.m-截面处弯矩设计值:M= (A1A2A)22(B1B2)B(pmaxxpminx2G/S)/48= (5.201.60)2(25.201.60)(109.2373.762876.10/27.04)/48= 382.94 kN.m-截面受弯计算:相对受压区高度: = 0.004148 配筋率: = 0.000231 min = 0.001500 , = min = 0.001500;计算面积:1800.00 mm2/m-截面受弯计算:相对受压区高度: = 0.003340 配筋率: = 0.000186 6m,按 b = 6m如果 d 0.5m,按 d = 0.5mfa = fakb(b3)dm(d0.5)= 220.00
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