单向分层总和法计算基础中点最终沉降量已知柱下单独方形基础，基础底面尺寸为，埋深2m,作用于基础上(设计地面标高处)的轴向荷载N=1250kN，有关地 基勘察资料与基础剖面详见下图。试用单向分层总和法计算基础中点最终沉降量。解：按单向分层总和法计算(1)计算地基土的自重应力。z自基底标高起算。当 z=0m, o =19.5X2=39(kPa) sDz=1m, o =39+19.5Xl=58.5(kPa)sz1z=2m, o =58.5+20Xl=78.5(kPa)sz1z=3m, o =78.5+20Xl=98.5(kPa)sz1z=4m, o =98.5+(20-10)Xl=108.5(kPa)sz1z=5m, o =108.5+(2010)Xl=118.5(kPa)sz1z=6m, o =118.5+18.5Xl=137(kPa)sz1z=7m, o =137+18.5Xl=155.5(kPa)sz1(2)基底压力计算。基础底面以上，基础与填土的混合容重取Y =20kN/m3。 0=240kPa1250 + 2.5 x2.5x 2x202.5 x2.53)基底附加压力计算。= 240 -19.5 x 2.0 = 20 lkPa(4) 基础中点下地基中竖向附加应力计算。用角点法计算，L/B=1, o =4Kp,查附加应力系数表得K。 zi si 0si(5) 确定沉降计算深度 zn考虑第层土压缩性比第层土大，经计算后确定z =7m，见下表。n例题4-1计算表格1z(m)zB/2Ks0z(kPa)0sz (kPa)0 /ozsz(%)zn(m)000.250 02013910.80.199 9160.758.521.60.112 390.2978.532.40.064 251.6298.843.20.040 132.24108.529.7154.00.027 021.71118.518.3264.80.019 315.5213711.3375.60.014 811.90155.57.6按7m计(6)计算基础中点最终沉降量。利用勘察资料中的e-p曲线，求按单向分层总和法公式计算结果见下表。例题 4-1 计算表格 2z(m)%(kPa)(kPa)H(cm)158.5160.7278.590.29398.551.624108.532.245118.521.71613715.527155.511.90039201100100100100100100100自重应力平均值(kPa)48.7568.5088.50103.5113.5127.5146.25附加应力平均值(kPa)180.85125.5070.9641.9326.9818.6213.71b 闊 + bg(kPa)229.6194159.46145.43140.48146.12159.96(kPa-1)(kPa)0.710.640.6350.630.630.690.680.640.610.620.620.620.680.670.0003870.0002390.0002110.0002380.0003710.0005370.0007294418686177496848439331472304(cm)4.091.830.920.610.610.590.59(cm)5.926.847.458.068.659.24注：本资料素材和资料部分来自网络，仅供参考。请预览后才下载，期待您的好评与关注！）