QTZ63塔吊桩基验算书.doc

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，7.3、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=0.81.21.591.951.350.20.351.6=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）=1021.94+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 = 301.76 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： 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 = 1.00 = 18.00 kN/m3m = 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)/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 = 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 = 76.8359.1059.10 = 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 = 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= 3359.73 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 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 = (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)= 220.000.0018.00(4.503.00)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）。8.5.3.2轴心荷载作用下地基承载力验算按建筑地基基础设计规范（GB 500072002）下列公式验算：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)/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) = 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 = pjAl(8.2.73)am = (atab)/2(8.2.72)pjmax,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 mFlx 0.7hpftamxH0 = 0.70.971570.002.6501.150= 3237.56 kN，满足要求。Y方向：Aly = 1/4(B2H0B1B2)(A1A2A2H0)= (1/4)(1.5021.154.50)(4.501.5021.15)= 1.45 m2Fly = pjAly = 101.801.45 = 147.87 kNab = minB2H0, B1B2 = min1.5021.15, 4.50 = 3.80 mamy = (atab)/2 = (Bab)/2 = (1.503.80)/2 = 2.65 mFly 0.7hpftamyH0 = 0.70.971570.002.6501.150= 3237.56 kN，满足要求。8.5.3.5基础受压验算计算公式：混凝土结构设计规范（GB 500102002）Fl 1.35clfcAln(7.8.1-1)局部荷载设计值：Fl = 997.31 kN混凝土局部受压面积：Aln = Al = BA = 1.501.50 = 2.25 m2混凝土受压时计算底面积：Ab = minB2A, B1B2min3A, A1A2 = 20.25 m2混凝土受压时强度提高系数：l = sq.(Ab/Al) = sq.(20.25/2.25) = 3.001.35clfcAln= 1.351.003.0016700.002.25= 152178.75 kN Fl = 997.31 kN，满足要求。8.5.3.6基础受弯计算按简明高层钢筋混凝土结构设计手册（第二版）中下列公式验算：M=/48(La)2(2Bb)(pjmaxpjnx)(11.47)M=/48(Bb)2(2La)(pjmaxpjny)(11.48)(1)标准节根部受弯计算：G = 1.35Gk = 1.35486.00 = 656.10kN-截面处弯矩设计值：pjnx = pjmin,x(pjmax,xpjmin,x)(B1B2B)/2/(B1B2)= 44.48(54.0244.48)(4.501.50)/2/4.50= 50.84 kPaM = /48(B1B2B)22(A1A2)A(pjmax,xpjnx)= 1.0000/48(4.501.50)2(24.501.50)(54.0250.84)= 206.43 kNm-截面处弯矩设计值：pjny = pjmin,y(pjmax,ypjmin,y)(A1A2A)/2/(A1A2)= 1.46(97.041.46)(4.501.50)/2/4.50= 65.18 kPaM = /48(A1A2A)22(B1B2)B(pjmax,ypjny)= 1.0849/48(4.501.50)2(24.501.50)(97.0465.18)= 346.46 kNm-截面受弯计算：相对受压区高度： = 0.002079 配筋率： = 0.000096 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： = 0.003492 配筋率： = 0.000162 min = 0.001500 = min = 0.001500计算面积：1800.00 mm2/m8.5.3.7计算结果1X方向弯矩验算结果：计算面积：1800.00 mm2/m采用方案：C20100实配面积：3141.59 mm2/m2Y方向弯矩验算结果：计算面积：1800.00 mm2/m采用方案：C20100实配面积：3141.59 mm2/m