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地下连续墙设计计算

6667设计计算已知条件:(1)土压力系数计算主动土压力系数:22 =0.84/2)=tan=0.70 (45°—10°K=tan)(45°—φ/2a1a1122=0.72=0.52 45°—18°/2K=tan)(45°—φ/2)=tan(a22a222=0.71°—19.2°/2K=tan)(45°—φ/2)=tan=0.64 (45a33a322=0.70—18.9/2)=tan (45°—φ/2)=tan=0.52 (45°K a4a4422=0.72=0.41 φ/2)=tan (45°—19.2/2K=tan)(45°—a5a55被动土压力系数:22=1.40 )=tan (45°+19.2°/2)=1.98 K=tan (45°+φ/2p1p15(2)水平荷载和水平抗力的计算水平荷载计算:=20×0.59-2×10×0.84=-5kPa e=qk-2C a1a0c=(20+18×2.5)×0.59-+2-2×10×0.84=21.55kPah)Kqe=(1a10ab1a1上c=(20+18×2.5)×0.36-K-22×19×0.6=0.6kPa(e=q)+h2a20ab1a2下c=+(20+18×2.5+19.9×1.1)×0.36=e(q+-h)hK-22×19×2a2aca2012上0.6=8.48kPac =(20+18×2.5+19.9×1.1)×h)K-20.64-2×44q=(e×+h0.8=+3a321aca30下14.79kPa-c)×)++qh+hh×1.418.81.119.92.520+18=-K2(×+×+(=e3a321a3ad03-2×44×0.8=2.05kPa上0.64c =(20+18×2.5+19.9×1.1+18.8×1.4)×e=(q+h-+h+h)K24a402a41ad3下0.34-2×21×0.59=13.71kPac (20+18×2.5+19.9×1.1h+)K-2+18.8×1.4q=e(h+h+h+=4a42aea44031上+19.9×0.5)×0.34-2×21×0.59=17.09kPac (20+18×2.5+19.9×-K)++++(=eqhhhh21.1+18.8×1.4=5a5a5ae43210下+19.9×0.5)×0.41-2×20×0.64=24.9kPac (20+18×2.5+19.9×h)K-21.1=e(q++h+h18.8+h+h=+5a524a51af5301.4+19.9×0.5+19.9×0.96)×0.41-2×20×0.64=32.73kPa上×水平抗力计算:c=2×20×e=21.57=58.8kPa5p1p上地面超载q=20kpaa素填土0.34m0.6kpa22.95kpa b粘性土h=6.460m c8.48kpa粘性13.71kp2.05kp粉24.90kp17.09kp基坑底粘性32.73kp62.8kp粘性147.98kp46.93kpa?E(3)墙后净土压力?E=×22.95×2.16+×(0.6+8.48)×1.1+×2.05×1.4+×(13.71+17.09)×0.5+×(24.90+32.73)×0.96+×0.8×62.8=91.70kPa?E作用点离地面的距离)4(.121121222221.4?2.05????1.1??+?8.48??22.95?2.161.1??0.63223232?h??222167.?00.0.?5?294.10.?5+??091?3.971?13.7?2322?a91.70112191.70122??0.96?(32.73?24.90)32?=0.6m91.70?hk?2ck?1.74?19.9?2.46?2=e?20?1.57?147.9815p1p上p1?hk?2ck?18.5?1.74?1.5=e1?2?43?1.23?154.392ppp1下2565?hk?2ck?18.5?(h?1.74)?1.51?2?43?1.23?27.935h=e?57.17d26d6pp2上???E=62.8?1.74?(?1.74?hh?1.74)?(147.98?62.8)?1.74h1.2E??p211dpaap220112??(27.94h?57.17?154.39)7154.394)?(h??(1.74)?1.74?h1.?+ddd231112??1.2?1.0?[?22.95?2.16?(?2.16?h?5.7)?0.6?1.1?1.74)h?(?dd3321111(?1.1?4.6?h)??(8.48?0.6)?1.1?(?1.1?4.6?h)??1.4?2.05dd2232111?(?1.4?h?3.2)?13.71?0.5?(?0.5?h?2.7)?(17.09?13.71)?0.5dd322111?(?0.5?2.7?h)?24.9?0.96?(?0.96?1.74?h)??(32.73?24.9)dd32211?0.96?(?0.96?1.74?h)?32.73?1.74?(?1.74?h?1.74)?46.93?dd3212?1.74)]?(hd232h?3h?2.25h?36.11?0=ddd解得h=2.72m 取2.8md??E?E有pa1?(62.8?48.554h?62.8)?h?66.58?32.73h0002解得h=1.15m0.=254.82-53.94=200.88kN·m所以最大弯矩M= h-h pamax8102?M????0.025s22?fbh14.3?600?96501c?=0.9873查表得s=A2mm700??所以S?965?300h?f0.98738M102?0ys2)mm?7633?18(A选用S2?163?18地下连续墙的稳定性分析(1)墙体内部稳定性验算土层的按土层厚度的加权平均值:γ=m=19.13KN/=C k=17.87kPaφ=k=20.39°采用圆弧滑动简单条分法进行验算,经试算确定最危险滑裂面的半径为r=20m,取土条宽度b=0.1r=2m.计算稳定安全系数如下:在excel中?????tanqb?lc?)cos(223.09?1628.99iki0iiii?1.73?1.3?K=???1069.285)sinq(b?ii0i满足整体稳定性要求。

地连墙外部稳定性分析A. 抗滑动稳定分析???)k?2ckss(E?q?rh??hiaiaiaxivi?91.70?2?4?733.6F作用在墙底断面上抗滑动i?tans?qB)F?(w hi?50.61?tan?(1485.54200)?2??4105.48F4105.48i??5.6?1.2?K则满足要求。

抗滑安全系数h E733.6axB. 抗倾覆稳定性验算B10?(1485.54?20?10)?2??16855.4KN?mM?(w?qB)?s?抗倾覆力矩vw226.46H???m??1565.01KN??s(q?rh)k?2cks?91.70?2?4M倾覆力矩??haiviiaii033M16855.4w???10.77?1.3K抗倾覆安全系数则满足要求p M1565.010C. 基坑抗隆起稳定性验算??h?N d0c?1.6进行验算由公式?(h?h)?q d NN=5.14;—地基承载力系数,取公式中cc?—抗剪强度,由十字板剪切试验或三轴不固结不排水试验确定(KPa);03?m/kN);土的重度(—h—支护结构嵌固深度(m);d h—基坑开挖深度(m);q—地面载荷(KPa)。

??hN?5.14?120?19.9?8.3dc0??2.49?1.6则?(h?h)?q19.9?(6.46?8.3)?20d满足要求(1)工程量计算地下连续墙墙宽为600mm,基坑长100m,宽50m,桩长19m。

3m342019?0.6?300?混凝土方量:30025??450(t2?)193.85??钢筋::0.11916??2?3001.58??85.7(t):0.23m 300=288×冠梁:体积:0.81.2×?16:10×300=3000m钢筋:4.74(t)=1.58kg/m×3000m= M.?6:2×0.8×300/0.4 +(0.8+1.2)×2×300/0.3 =5120mM=5120×0.222 kg/m=1.14(t)腰梁长度:300m(2)各分项工程费10费用(见表4-9钢筋工程大于所示)表4-9锚杆、灌注桩钢筋工程费冠梁费用(见表4-10所示)表 4-10 冠梁工程费地下连续墙灌注费用(见表4-11所示)表 4-11 地下连续墙灌注费(3)工程费用计算(表4-12)工程直接费:54349.6+468000.7+953880.7+154173.0+131923.3+721616.3=2853943.8元工程间接费:施工现场管理费=直接成本×3%=24638.6元施工单位管理费=直接成本×5%=135697.3元工程间接费=91618.3+152697.2=216315.4元总成本=直接成本+间接成本=2853943.8 +216315.5=3068259.1元企业利润=总成本×4%=120730.3元税金=总成本×3.91%=114661.5元劳动保险基费=总成本×1%=29834.6元表4-12 工程费用表用(元2853943.8216315.4114661.5120730.330368.63323265工程总费用=3323265.0元。

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