土力学课后习题部分答案重庆大学出版社11.docx

1、土力学课后习题部分答案重庆大学出版社111-1 解( 1 )A 曲线：卵石或碎石 ( 20mm)占 100 77=23% 砾粒 ( 20-2mm )占 77 50=27% 砂粒 (2-0.075mm)占 5010=40% 粉粒 ( d1-3 解V =V , e =V /V =1v s v s =G /(1+ e) = 2.68 10 /(1+1) =13.4kN/ m3Vv = Vw sat =(Gs +Vv ) w /(1+ e) = (ds +e) w /(1+e) = (2.68+1) 10/(1+1) =18.4kN/m3w=Ww/Ws =Vw w/VsGs w =1v /Gs =1/

2、 2.68 = 37.3%1-4 解设V = 1m3W = V = 11kNVs =Ws /(ds w)=11/(2.7 10) =0.407m3V =V -V =1- 0.407 = 0.593m3w=W /W W = w W = 0.5 11 = 5.5kNw s w sVw =Ww/ w =5.5/10=0.55m3V =V -V =0.593-0.55= 0.043m31-5 解：饱和土 e=wGs sat =(Gs + e) w /(1+ e)即 e = 0.4G 18 = (G + e)/(1+ e) 10以上两式联立求解 解得 e =1.05 Gs = 2.651-6解w=W /

3、W =(W -W )/W 可求得w = 16%时，W = 21.64N 时土粒的质量W =W /(1+ w) = 21.64 /(1+ 0.16) =18.655N其中水量为W = w W = 0.16 18.655 = 2.985N当w = 25%时，土中水量为W = w W = 0.25 18.655 = 4.664N因此增加的水量 W = 4.664 - 2.985 = 1.679N1-7 解：设饱和土为V =1m3 V = V +V =1饱和土w = Ww /Ws =Vw/VsGs =0.6由上两式联立可得： Vs = 0.382 Vw = 0.618W =V G =0.382 2.7

4、 10=10.314kNW = w W = 0.60 10.31 = 6.186kN当w =15%时， 需要土W =(w+1)W =1.15 10.314=11.86kN其中水量 W = wW =0.15 10.314=1.547kN故增加水的重量为 W = 6.186 -1.547 = 4.64kN1-8 解：e=1 e=V /V V =V V =1m3 V =V = 0.5m3Sr =Vw /Vv Vw =Sr Vv = 0.8 0.5 = 0.4m3当水蒸发掉一半时Vw =Vw /2=0.4/2=0.2m3 Vv =Vw /sr = 0.2 / 0.9V = 0.5 + 0.2 / 0.

5、9 = 0.722 = 72.2%1-9 证： / -1 = (W /V) /(W /V)-1=W/W -1=(W-W )/W =W /W =w1-10 解：e=G (1+ w)/ -1= (2.65 10 (1+0.073)/17.1-1= 0.663 =W /V = 29.85 / 2000 = 0.0149N / cm 3 = 14.9kN / m3 dmin =GS / w /(1+ emax ) 14.9=2.65 10/(1+emax) emax = 0.779 =W /V = 32.5 / 2000 = 0.0163N / cm 3 = 16.3kN / m3 dmin =GS

6、/ w /(1+ emin ) 16.3=2.65 10/(1+emin) emin = 0.626Dr = (emax -e)/(emax - emin ) = (0.770 - 0.663) /(0.779 - 0.626) = 0.758S =G /e = 2.65 0.073 / 0.663 = 29.2%1-11 解：在压实以前 n = e /(1+ e) = 0.72 /(1+ 0.72) = 0.419在1m2的地基面积中，土粒体积为V = (1- n)V = (1- 0.419) 5 = 2.91m3 在压实以后，土粒体积减少0.3m3n =(V -V )/V = (5-0.3

7、-2.91)/(5-0.3) = 0.381e=n/(1-n) = 0.381 /(1- 0.381) = 0.616此时Dr = (emax -e)/(emax - emin ) = (0.75 - 0.616) /(0.75 - 0.52) = 0.5831-12 解：m = 23-15=8g w=m /m = 8 / 15 = 0.533 = 53.3%I = w -w =40-24=16 I =(w-w )/(w -w ) = (53 - 24) /16 = 1.81310 IP 17 粉质粘土 IL 1 流态1-13 解：(1)A土I =30 B土IP =11 故A土粘粒含量多(2)

8、由 = Gs w(1+w)/(1+e)=ds w(1+w)/(1+w Gs /Sr) 因为是饱和土 S =100% =1A 土 = 2.69 10(1 + 0.53) /(1 + 0.53 2.69) = 16.97kN / m3B土 = 2.71 10(1 + 0.26) /(1 + 0.26 2.71) = 20.03kN/m3B 土的天然重度大(3) d = /(1+ w)A 土 =16.97 /(1+ 0.53) = 11.09kN / m3B土 = 20.03 /(1+ 0.26) =15.89kN / m3B 土的干重度大(4)因为是饱和土 e = w GA 土 e=0.53 2.

9、69=1.43B 土 e = 0.26 2.71= 0.70A 土的孔隙比大1-14 解：因为大于2mm 粒径的土粒占总质量的(11+5)=16% 50%。按表1-7 颗粒级配所示属砂土，而2mm 粒径的颗粒只占总质量 16% 25% ，故该土样不是砾砂，而0.5mm 粒径的 颗粒占总质量(16 + 39) = 55% 50% ，故该土样定名为粗砂。第二章 地基的应力部分习题解答3.1 细砂、粗砂分界线处： = 16.4 3 = 49.2kPaen粗砂 n= e e= n= 2/3 1+e1-n =GS w(1+w)=2.65 10 (1+0.2)=19.08kN/m3 1+e 1+2/3水位

10、线处 = 49.2 +19.08 2.4 = 94.8kPa = 94.8+ (19.9 -10) 2.1=115.6kPa粘土w= mw = Vw =0.6e=Vw = 0.6 2.73 = 1.638 ms GS VS VS = (GS +e) w = (2.73+1.638) 10 =16.56kN /m3sat 1+e 1+ 1.638粘土与泥岩分界处 =115.6+(16.56-10) 3 =135.28kPa泥岩顶面 =135.28+10 5.1 =186.28kPa结论：随着地下水位的下降，土有效应力增大，使得变形会增加。3-3 解：（1）填土前总应力 （kpa） 孔隙水压力 u

11、（kpa） 有效应力 （kpa）中砂3m193=5757中砂3m57+203=117103=30117-30=87粉土3m117+183=17130+103=60171-60=111粗砂3m171+203=23160+103=90231-90=1412）刚填土后（假定一次将土骤然填上）总应力 （kpa） 孔隙水压力 u（kpa） 有效应力 （kpa）中砂3m9090+5714790-090147-0147中砂3m90+117207103=30207-30177粉土3m171+9026190+3012090+60150207-12087261-150111粗砂3m231+903211066010

12、990261-60201321-902313）填土后土层完全固结总应力 （kpa） 孔隙水压力 u（kpa） 有效应力 （kpa）中砂3m9014790147中砂3m20730177粉土3m261602011粗砂3m321902313.4 对基地面积为3m 6m， P = 200kPa点号L/bz/b C Z= 4 CP00200.250200121/1.50.228182.4222/1.50.170136323/1.50.12096424/1.50.08668.8525/1.50.06350.46240.04838.4对基地面积为1m 2m， P = 200kPa点号L/bz/b C Z= 4 CP00200.250200121/0.50.120962