工程地质与地基基础 作业答案

解：

已知V72cm3，m129.1g，ms121.5g则：mwmms7.6g，Vsms/s121.5/2.745cm3，Vw7.6cm3VvVVs27cm3wmw/ms6.3%，mg/V17.9kN/m3satmsVvw121.5271g1020.6kN/m3V72msgwgVssatw10.6kN/m3Vd2-3

msg16.9kN/m3V

解：

d16.2kN/m3，w20%令：V1cm3，msdV/g1.62gmwwms0.324g饱和土VvVw0.324cm3，VsVVv0.676cm3Gssms/Vs2.4wweVv0.48Vs

sat2-5

mswVvg19.4kN/m3V解：

SrVwVwVwVwSrVe/(1e)VveVseV/(1e)Vw1Sr1VVw2e0.950.3710.18m31e10.95e0.95Sr2V0.910.438m31e10.95

Vw0.258m3mw258kg2-6

解：

eVvVveVsVVvVs(1e)VsVsV11e11e2V2V123.18万m3V21e21e12-7

解： eGs(1w)w12.67(10.098)110.656

1.77

Dremaxe0.9430.6560.595，中密

emaxemin0.9430.4613-1

解：

e2Gs(1w)w2.73(10.31)1110.8821.9 25.5 35 sat2e3Gse2.730.882w1019.2kN/m3

1e10.882Gs(1w)w2.74(10.41)1111.111

1.8367.2 sat3e4Gse2.741.111w1018.2kN/m3

1e11.111Gs(1w)w12.72(10.27)110.7711.95132.8 sat43-2

Gse2.720.771w1019.7kN/m3

1e10.771FkGkFk680Gd202125kPa AA42161.9 306.9 解：pkeMk6801.31l20.08mm

FkGk6802042263pkmax4-3

2(FkGk)2(68020422)301kPa

3bk32(20.08)解：1f3tan2(45/2)2ctan(45/2)435.2kPa

4-4

解：

1f3tan2(45/2)2ctan(45/2)210tan(4525/2)236tan(4525/2)630.4kPa12

或者 4-5

3f1tan2(45/2)2ctan(45/2)380tan(4525/2)236tan(4525/2)108.4kPa32，弹性状态，安全

解：

5-3

1f3tan2(45/2)2ctan(45/2)fctanc47.13kPa 30解：paA2cKa1220tan(4518/2)29.06kPa

paBKa11h12cKa1tan2(4518/2)185220tan(4518/2)18.4kPa

x29.06x3.07m 5x18.4Ka2h1tan2(4535/2)18524.4kPa paBpaCKa2(1h12h2)tan(4535/2)(185203)40.6kPa

2A 11Ea18.4(53.07)(24.440.6)3115.3kPa

22153.071ha[18.4(53.07)(3)24.432232

140.631]/115.31.73m2 7-3

B

C 解：fafakdm(d0.5)1851.618.6(1.10.5)202.9kPa

b7-4

Fk1801m

faGd202.9201.1解：fafakdm(d0.5)210118(1.90.5)235.2kPa

A1.2Fk24001.214.6m2

faGd235.2201.9A/n3.2m，lnb4.8m

取nl/b1.5bM850601.440.313mmFG2400204.83.21.96FG6e2400204.83.21.960.313pkmax(1)(1)270.3kPa1.2fa282.2kPa

Al4.83.24.8故取l4.8m，b3.2m e7-5

解：fafakdm(d0.5)1841.115.7(1.350.5)198.7kPa

取正方形基础，bFk7802.2m

1.351.65faGd198.7202下卧层验算：

p0F780Gdmd201.515.71.35170kPa A2.22Es1/Es25，z/b125

p0b21702.22z49.8kPa

(b2ztan)2(2.222tan25)2cz15.71.3518.6258.4kPa

fazfakdm(d0.5)881.158.4(3.350.5)142.7kPa 3.35zcz49.858.4108.2kPafaz，满足

7-7

解：持力层验算：pkFG300200.8166kPafa A2fafakdm(d0.5)1801.017.2(0.80.5)185.2kPa

eM3520.105mmFG3002020.866e60.105pkmaxpk(1)166(1)218.3kPa1.2fa222kPa，满足要求

l2

下卧层验算：Es1/Es23，z/b123

p0pkmd16617.20.8152.2kPa

zp0b152.2282.3kPa

b2ztan222tan23cz17.20.87.2228.2kPa

fazfakdm(d0.5)95128.2(2.80.5)118.2kPa 2.8zcz110.5faz，满足

某一柱基础，采用φ500mm钢筋砼预制管桩，荷载Fk=3700KN，Mk=280KN.m，Hk=58KN（水平荷载作用在地面处），桩长12m，地质条件如下：第一层土为2.00m杂填土，第二层土，粉质粘土，厚7m，桩周摩阻力特征值qsa1=22kPa，第三层土，灰色粘土，厚22.5m，qsa2=32kPa，桩尖平面处土的承载力特征值qpa=1100kPa，试确定 ⑴单桩承载力特征值；⑵确定桩数、桩的布置及承台平面尺寸（s=3d）（承台埋深2m）；⑶ 验算单桩承载力是否满足要求。

解：单桩承载力特征值：

RauqsialiqpaAp0.5(227325)11000.52/4 709.2kNFk37005.2根，取n6根 确定桩数：nRa709.2承台平面尺寸： 桩的中心距s承台边长：lFkMk2mHk3d1.5m

2(sd)2(1.50.5)4m；

bs2d1.50.522.5m

桩的布置及承台底面尺寸如图 桩顶荷载：

y500150050050015001500500xFGk37002042.52Qkk683.3kNRa

n6QkmaxQkkmin(MkHkh)xmaxxi2(280582)1.5683.366 241.5749.3kN1.2Ra851.04kN617.3kN0683.3满足要求。

p1(82118)/2100kPa

某基础基底平均压力p=170kPa，基底处土的自重应力σcd=20kPa。地基土为粘性土，厚度为2m，孔隙比e1=0.8，a=0.25MPa-1，其下为不可压缩的岩层，计算最终沉降量。

p2(6238)/2100150kPae1e20.7290.682sH200054.4mme10.729,e20.6821e110.729