QTZ63塔吊桩基验算书.doc

第21页，共31页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 = 9