1、排架结构单独基础设计 基础工程课程设计一.方案选取要求设计实验大厅排架柱下单独基础。该实验大厅为河流泥沙与水利工程的室内模型实验大厅,全长84m,跨度24m,采用钢筋混凝土排架结构共15跨,排架间距6m,在大厅中部设有一道宽6cm的伸缩缝。在柱基础顶面处排架柱截面尺寸为30cm80cm,柱截面轴线通过该截面形心。实验大厅排架及柱基础布置平面图,见图1;图1 排架及柱基础布置平面图综上,设计方案采用柱下独立基础。二.确定每层土的地基承载力 地基土层分布及土的物理性质指标如下表1所示:表1土层层底标高(地面46.5m)现场鉴别层厚(m)土的物理性质指标(1)43.50杂填土3.00=18kN/m3
2、, qs=11 kN/m2(2)41.50粉质粘土,可塑2.00=19kN/m3,G=2.71,w=26.2%,wL=31%,wp=19%,Es=8.5MPa(3)39.40粉土, 饱和,软塑2.10sat=20kN/m3,G=2.70,w=26%,wL=28%,wp=18%,Es=7.5MPa(4)38.20饱和软粘土1.20sat=18.7kN/m3,G=2.71,w=40%,wL=42.6%,wp=24.1%,1-2=0.95 MPa-1(5)未揭露粘土,饱和, 可塑8.00sat=20.5kN/m3,G=2.68, w=24.5%,wL=38%, wp=20%, Es=9.8MPa,
3、1-2=0.163MPa-1 由以上表格的各参数计算查表可得各土层的地基承载力基本值如下表2所示:表2土层土的 重度(kN/m3)土的饱 和重度(kN/m3)水 的 重 度(kN/m3)水 的 密 度 103 (kg/m3)g (N/kg)土的密度 103 (kg/m3)土粒相对密度 G含水量w孔隙比e液性 指标 IL土的承载力基本值(kpa)1181019.81.842191019.81.942.710.2620.760.312553201019.81.022.70.260.700.28270418.71019.80.892.710.40.970.426170520.51019.81.072
4、.680.2450.600.38342三.柱下独立基础设计柱下独立基础分3种形式,J1:单排架基础;J2:双排架基础;J3:横墙立柱基础。在此只计算J1和J2型基础形式。各基础受力形式及受力大小如下图2所示:图2 排架剖面图在基础顶面的荷载J1:P=1140kN, M=400kNm, H=150 kN J1:P=1320kN, M=560kNm, H=200kN说明: (1)A-A为单排架剖面,B-B为大厅伸缩缝双排架剖面; (2)J1为单排架基础,J2为双排架基础;(3)地面绝对标高为46.5m。. J1基础设计由于第一层土为杂填土,不宜作为持力层,因此,将第二层土设为持力层。杂填土开挖2m
5、,基础埋深为1m。基底标高为43.5m。1.初步确定基础地面尺寸: 由于基础埋深大于0.5m,因此,须进行修正。查表得d=1.6,地基承载力特征值:m=18kN/m3fa=fak+dm(d-0.5) =255+1.618(1-0.5) =269.4kPa考虑荷载偏心,将基底面积初步扩大20%,则,基底面积A=1.2Pk/(fa-Gd)=1.21.351140/(269.4-201.0)=7.4m2取基底长短边之比n=l/b=2,于是,b=(A/n)1/2=(7.4/2)1/2=1.9m,l=nb2=2.01.9=3.8m实际施工中取b=2.0m,l=4.0m.因b=2.03.0,fa无须作宽度
6、修正。2.计算基底净反力:用C25混凝土,HPB235级钢筋,ft=1.27N/mm2,fy=210N/mm2.垫层采用C10混凝土。基底净反力:pj=Pk/bl=1.35 1140/(24)=192.4kPa净偏心距:e0=(Mk+Hkh)/Pk=(4001.35+1501.350.8)/(11401.35)=0.429mml/6=4/6=0.667mm基底最大净反力设计值:pjmax=Pk(1+6e0/l)/(lb)=11401.35(1+60.429/4)/(24)=316.1kPa1.2fa=1.2269.4=323.28kPa基底最小净反力设计值: pjmin=Pk(1-6e0/l)
7、/(lb)=11401.35(1-60.429/4)/(24)=68.58kPa因此,基础底面压力满足要求。基础尺寸图如下图3所示:3.基础高度(1)柱边截面将基础设为2级阶梯型,上面一级的截面尺寸为b1l1=1m2m初步设定基础高为h=800mm,h0=750,每阶高400mm.柱的截面尺寸为bcac=0.3m0.8m,如上图3所示,则: bc+2h0=0.3+20.75=1.8534.209kN即满足冲切破坏条件:pjmax(l/2-ac/2-h0)b-(b/2-bc/2-h0)20.7hpft(bc+h0)h0(2)变阶处截面b1=1.0m,l1=2.0m,h1=400mm,h01=35
8、0mmb1+2h01=1+20.35=1.7m403.82kpa即满足冲切条件:pjmax(l/2-l1/2-h01)b-(b/2-b1/2-h01)20.7hpft(b1+h01)h014.配筋计算 计算基础长边方向的弯矩设计值,取-截面 M =(pjmax+ pj)(2b+bc)+( pjmax- pj)b(l-ac)2/48 =(316.1+192.4)(22+0.3)+( 316.1-192.4)2(4-0.8)2/48 =519.24kNm As= M/(0.9fyh0)=519.24106/(0.9210750)=3663mm2-截面:M =(pjmax+ pj)(2b+b1)+(
9、 pjmax- pj)b(l-l1)2/48 =(316.1+192.4)(22+1.0)+( 316.1-192.4)2(4-2.0)2/48 =232.49kNm As= M/(0.9fyh01)=232.49106/(0.9210350)=3515mm2比较As和As,应按As配筋,现于2.0m宽范围内配筋1916100,实配钢筋面积为As=3820.9mm23663mm2计算基础短边方向的弯矩:取-截面,前已经算得pj=192.4kpa, M=pj(bbc)2(2lac)/24Hkh =192.4(20.3)2(24.00.8)/24+0.81.35150 =365.88kNm As=
10、 M/(0.9 fyh0)=365.88106/0.9210750=2581mm2-截面:M=pj(bb1)2(2ll1)/24Hkh=192.4(21.0)2(24.02.0)/24+0.81.35150 =242.17 kNmAs= M/(0.9 fyh01)=242.17106/0.9210350=3661mm2比较As和As,应按As配筋。现于4.0m宽度范围内配筋1916200,实配钢筋面积为As=3820.9mm23661mm2,配筋图如下图4所示:5.软弱下卧层承载力验算 作用在软弱下卧层顶面处的附加应力与自重应力之和不应超过它的承载力特征值,即: z+czfaz其中,cz181
11、192.0(2010)2.177kpazlb(pk-cd)/(l+2ztan)(b+2ztan)cd=181.0=18.0kpapk=(Pk+Gk)/A=(1.351140+201.02.04.0)/(2.04.0)=213.375kpaz=4.1m.Es1=7.5Mpa,Es2=(1+e2)/a2其中,e2=(G2w -sat)/( sat-w)=(2.7110-18.7)/(18.7-10)=0.966则:Es2=(1+e2)/a2(10.966)/0.95=2.07MpaEs1/ Es2=3.6,z/b=4.1/2.0=2.050.5查表得:24。故 zlb(pk-cd)/(l+2zta
12、n)(b+2ztan) 4.02.0(213.37518.0)/(4.024.1tan24。)(2.0+24.1tan24。) 36.2kpaz+cz=36.2+77=113.2kpa269.4kpa因此,软弱下卧层的顶面承载力满足要求。.J2基础设计与J1基础一样,将第二层土设为持力层。杂填土开挖2m,基础埋深为1m。基底标高为43.5m。1.初步确定基础地面尺寸由于基础埋深大于0.5m,因此,须对地基承载力进行修正。查表得d=1.6,地基承载力特征值:m=18kN/m3fa=fak+dm(d-0.5) =255+1.618(1-0.5) =269.4kpa考虑荷载偏心,将基底面积初步扩大2
13、0%,则: 基底面积A=1.2Pk/(fa-Gd)=1.21.351140/(269.4-201.0)=7.4m2取基底长短边之比n=l/b=2,于是,b=(A/n)1/2=(7.4/2)1/2=1.9m,l=nb=2.01.9=3.8m实际施工中取b=2.0m,l=4.0m.因b=2.03.0,fa无须作宽度修正。2.计算基底净反力:用C25混凝土,HPB235级钢筋,ft=1.27N/mm2,fy=210N/mm2.垫层采用C10混凝土。基底净反力:pj=Pk/(bl)=1.35 1320/(2.24.4)=184.09kPa净偏心距e0=(Mk+Hkh)/Pk=(5601.35+2001
14、.350.8)/(13201.35)=0.545mml/6=4/6=0.667mm基底最大净反力设计值pjmax=Pk(1+6e0/l)/(lb)=13201.35(1+60.545/4.4)/(2.24.4)=320.9kPa1.2fa=1.2269.4=323.28kPa基底最小净反力设计值: pjmin=Pk(1-6e0/l)/(lb)=13201.35(1-60.545/4.4)/(2.24.4)=42.28kPa因此,基础底面压力满足要求。基础尺寸图5如下所示:3.基础高度(1)柱边截面将基础设为2级阶梯型,上面一级的截面尺寸为b1l1=1m2m初步设定基础高为h=800mm,h0=
15、750mm,上一阶高300mm,下一阶高h1=500mm。柱的截面尺寸为bcac=0.3m0.8m,如上图5所示,则: bc+2h0=0.3+20.75=1.8728.443kN即满足冲切破坏条件:pjmax(l/2-ac/2-h0)b-(b/2-bc/2-h0)20.7hpft(bc+h0)h0(2)变阶处截面b1=1.1m,l1=2.2m,h1=500mm,h01=450mmb1+2h01=1.1+20.35=1.8m455.678kpa即满足冲切条件:pjmax(l/2-l1/2-h01)b-(b/2-b1/2-h01)20.7hpft(b1+h01)h014. 配筋计算 计算基础长边方
16、向的弯矩设计值,取-截面:M=(pjmax+pj)(2b+bc)+(pjmax-pj)b(l-ac)2/48 =(320.9+184.09)(22.2+0.3)+(320.9-184.09)2.2(4.4-0.8)2/48 =722.1kNmAs= M/(0.9fyh0)=722.1106/(0.9210750)=5094mm2-截面:M=(pjmax+ pj)(2b+b1)+( pjmax- pj)b(l-l1)2/48 =(320.9+184.09)(22.2+1.1)+(320.9-184.09)2.2(4.4-2.2)2/48=310.41kNmAs= M/(0.9fyh01)=310
17、.41106/(0.9210450)=3650mm2比较As和As,应按As配筋,现于2.0m宽范围内配筋1720,实配钢筋面积As=5338mm25094mm2。计算基础短边方向的弯矩:取-截面,前已经算得pj=184.09kpa,M=pj(bbc)2(2lac)/24Hkh =184.09(2.20.3)2(24.40.8)/24+0.81.35200=481.83kNmAs= M/(0.9 fyh0)=481.83106/0.9210750=3399mm2-截面:M=pj(bb1)2(2ll1)/24Hkh=184.09(2.21.1)2(24.42.2)/24+0.81.35200 =
18、318.1 kNmAs= M/(0.9 fyh01)=318.1106/0.9210450=3740mm2比较As和As,应按As配筋。现于4.0m宽度范围内配筋1922,实配钢筋面积为As=3820.9mm23740mm2配筋图如下图6所示:5.软弱下卧层承载力验算 作用在软弱下卧层顶面处的附加应力与自重应力之和不应超过它的承载力特征值,即: z+czfaz其中,cz181192.0(2010)2.177kpazlb(pk-cd)/(l+2ztan)(b+2ztan)cd=181.0=18.0kpapk=(Pk+Gk)/A=(1.351320+201.02.24.4)/(2.24.4)=20
19、4.1kpa软弱土层顶面到基底的距离:z=4.1m.Es1=7.5Mpa,Es2=(1+e2)/a2其中,e2=(G2w -sat)/( sat-w)=(2.7110-18.7)/(18.7-10)=0.966则:Es2=(1+e2)/a2(10.966)/0.95=2.07MpaEs1/ Es2=3.6,z/b=4.1/2.2=1.860.5查表得:24。故 zlb(pk-cd)/(l+2ztan)(b+2ztan) 4.42.2(204.118.0)/(4.424.1tan24。)(2.2+24.1tan24。) 38.24kpaz+cz=38.24+77=115.24kpafa=269.
20、4kpa因此,软弱下卧层的顶面承载力满足要求。基础的平面布置图如图7所示:四相邻的J1、J2两柱基的沉降验算J1基础: 基底受力形式为梯形分布,pjmax=316.1kPa, pjmin=68.58kPa,则中心点的沉降计算可以将其转化为受均布荷载的形式,即均布荷载p0=68.58+(316.1-68.58)/2=192.34kPa。J2基础: 和J1基础一样,J2基础基底受力形式亦为梯形分布,pjmax=320.9kPa, pjmin=42.28kPa,则中心点的沉降计算可以将其转化为受均布荷载的形式,即均布荷载p0=42.28+(320.9-42.28)/2=181.59kPa。基础沉降采
21、用分层总和法规范修正公式,如下:S=s. J1基础的沉降(包括自身沉降和附近基础对其影响的沉降):J1基础在自身作用下的沉降量:验算基础中心点的沉降量,此时,l=4.0/2=2.0,b=2.0/2=1.0,计算点深度z,及z/b,l/b, 的查表以及计算过程见下表3:表3 点深度 zi(m)层厚 Hi(m)基础长 l(m)基础宽b(m)基底平均附加应力p0l(m)b(m)l/bz/bizii0042192.3421200.25011142192.3421210.2340.23422142192.3421220.19580.391633142192.3421230.16190.485744.11
22、.142192.342124.10.134050.54960555.31.242192.342125.30.112050.59386566.3142192.342126.30.09830.6192977.3142192.342127.30.08940.6526288.3142192.342128.30.08110.6731399.3142192.342129.30.07310.679831010.3142192.3421210.30.06610.680831111.3142192.3421211.30.06090.688171212.3142192.3421212.30.05730.70479
23、1313.3142192.3421213.30.0550.73151414.3142192.3421214.30.05130.73359合计zii-zi-1i-1Esi(Mpa)(zii-zi-1i-1)/Esi4(zii-zi-1i-1)/Esis1(mm)86.980.2348.50.0275294120.1101176470.15768.50.0185411760.0741647060.09417.50.0125466670.0501866670.0639057.50.0085206670.0340826670.044262.070.0213816430.085526570.025425
24、9.80.0025943880.0103775510.033339.80.003401020.0136040820.020519.80.0020928570.0083714290.00679.80.0006836730.0027346940.0019.80.0001020410.0004081630.007349.80.000748980.0029959180.016629.80.0016959180.0067836730.026719.80.002725510.0109020410.002099.80.0002132650.0008530610.411108869由上表知:S1=s =1.1
25、192.340.41 =86.98mm相邻J1基础的J2基础对J1基础沉降的影响由柱基础平面布置图知,与J1基础相邻的J2基础只有一个,因此,对J1基础沉降有影响的基础有一个J1和一个J2基础,但按两个J2基础对J1基础的影响验算更倾于安全,所以,在此,就按照J1基础受两个J2基础沉降影响验算。 如下图8:计算时,基础中心点o可看成是四个相等矩形面1和另四个相等矩形面2的公共角点,其长宽比l/b分别为7.1/2.2=3.23,4.9/2.2=2.23,查表得到2和3,同样按公式S=s计算沉降量,计算过程见下表4:表4l2b2l2/b2z/b2i2l3b3l3/b3z/b3i3zii47.12.
26、23.22727200.254.92.22.22727272700.2507.12.23.2272720.4545450.2474.92.22.2272727270.4545454550.24707.12.23.2272720.9090910.23794.92.22.2272727270.9090909090.23430.00727.12.23.2272721.3636360.22114.92.22.2272727271.3636363640.22010.0037.12.23.2272721.8636360.20664.92.22.2272727271.8636363640.20440.009
27、027.12.23.2272722.4090910.18624.92.22.2272727272.4090909090.18260.019087.12.23.2272722.8636360.17394.92.22.2272727272.8636363640.16980.025837.12.23.2272723.3181820.164.92.22.2272727273.3181818180.15290.051837.12.23.2272723.7727270.14824.92.22.2272727273.7727272730.1430.043167.12.23.2272724.2272730.1
28、3964.92.22.2272727274.2272727270.13440.048367.12.23.2272724.6818180.13194.92.22.2272727274.6818181820.12630.057687.12.23.2272725.1363640.12334.92.22.2272727275.1363636360.11590.083627.12.23.2272725.5909090.11564.92.22.2272727275.5909090910.1090.081187.12.23.2272726.0454550.11014.92.22.2272727276.045
29、4545450.10340.089117.12.23.2272726.50.10394.92.22.2272727276.50.09980.05863zii4-zi-1(i-1)4(zii4-zi-1(i-1)4)/Esi4(zii4-zi-1(i-1)4)/Esis27.9782492420000.00720.0008470590.003388235-0.0042-0.00056-0.002240.006020.0008026670.0032106670.010060.0048599030.0194396140.006750.0006887760.0027551020.0260.002653
30、0610.010612245-0.00867-0.000884694-0.0035387760.00520.0005306120.0021224490.009320.000951020.0038040820.025940.0026469390.010587755-0.00244-0.00024898-0.0009959180.007930.0008091840.003236735-0.03048-0.003110204-0.012440816合计0.039941373由上表知:S2=s =1.1181.590.039941373 =7.978mm因此,总的沉降量为S=S1+S2=86.98+7
31、.978=94.958mm120mm.即沉降满足要求。.J2基础的沉降(包括自身沉降和附近基础对其影响的沉降):J2基础在自身作用下的沉降量:验算基础中心点的沉降量,此时,l=4.4/2=2.2,b=2.2/2=1.1,计算点深度z,及z/b,l/b, 值的查表以及计算过程见下表5:表5点深度 zi (m)层厚 Hi(m)基础 长度 l(m)p0 (kPa)l (m)b (m)l/bz/bizii004.4222.752.21.1200.2501114.4222.752.21.120.910.240.242214.4222.752.21.121.820.200.413314.4222.752.
32、21.122.730.170.5144.11.14.4222.752.21.123.730.140.5955.31.24.4222.752.21.124.820.120.6466.314.4222.752.21.125.730.110.6777.314.4222.752.21.126.640.090.6988.314.4222.752.21.127.550.090.7199.314.4222.752.21.128.450.080.721010.314.4222.752.21.129.360.070.741111.314.4222.752.21.1210.270.070.751212.314.4
33、222.752.21.1211.180.060.751313.314.4222.752.21.1212.090.060.761414.314.4222.752.21.1213.000.050.75合计zii-zi-1i-1Esi(zii-zi-1i-1)/Esi4(zii-zi-1i-1)/Esi沉降量s86.3170.23728.50.0279058820.1116235290.16968.50.0199529410.0798117650.10687.50.014240.056960.073527.50.0098026670.0392106670.0512.070.0246376810.09
34、85507250.030949.80.0031571430.0126285710.022989.80.0023448980.0093795920.013469.80.0013734690.0054938780.018979.80.0019357140.0077428570.018169.80.0018530610.0074122450.003179.80.0003234690.0012938780.003279.80.0003336730.0013346940.013029.80.0013285710.005314286-0.011349.8-0.001157143-0.004628571合计
35、0.432128114由上表知:S1=s =1.1181.590.432128 =86.32mm相邻J2基础的J1基础对J2基础沉降的影响如下图9所示:由柱基础平面布置图知,与J2基础相邻的J1基础有2个,因此,对J2基础沉降有影响的基础有2个J1基础,计算时,基础中心点o可看成是四个相等矩形面1和另四个相等矩形面2的公共角点,其长宽比l/b分别为7.1/2.0=2.367,4.9/2.0=2.45,再根据z/b值查表得到2和3,同样按公式S=s计算沉降量,计算过程见下表6:表6l2b2l2/b2z/b2i2l3b3l3/b3z/b3i3zii47.122.36666700.254.922.4
36、500.2507.122.3666670.50.2474.922.450.50.24680.00027.122.36666710.2354.922.4510.23460.00087.122.3666671.50.2174.922.451.50.21670.00157.122.3666672.050.24.922.452.050.19820.007387.122.3666672.650.1814.922.452.650.17790.013787.122.3666673.150.1634.922.453.150.16020.01897.122.3666673.650.1544.922.453.650.150.027747.122.3666674.150.1414.922.454.150.13650.03403