4 + 18 X1.5 X3 X3 + 18 X0.5 X1.52 X12468.75⏺[3]有一地基如图所示。地面下 2m深处有一层厚0.8m的软土夹层,在地基上用当地材料修筑土堤,高5m,边坡1: 2,土堤与地基均为砂质粘土, = 20° c = 20kPa, = 20kN/m;软土夹层的 =18kN/m, = 0, c = 10kPa,试问该土堤是否会沿软土夹层滑动,安全系数有多大 ?
4 + 18 X1.5 X3 X3 + 18 X0.5 X1.52 X1
2468.75
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[3]有一地基如图所示。地面下 2m深处有一层厚0.8m的软土夹层,在地基上用当地材料修筑土堤,
高5m,边坡1: 2,土堤与地基均为砂质粘土, = 20° c = 20kPa, = 20kN/m;软土夹层的 =
18kN/m, = 0, c = 10kPa,试问该土堤是否会沿软土夹层滑动,安全系数有多大 ?
题目解答
答案
解:(1)求作用于AB面上的Ea
Ka=tg2(45 2)=tg.45 詈)=0.49,
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K p = tg2(45 +-) = tg2(45 + 牙) = 2.04
PpD =2c Kp = 2 X20.2.04 = 57.13kPa
PpC = hKp +2c K p = 20 X2 X2.04 + 2 X20 2.04 = 138.73kPa
Ep = 0.5X( 57.13+ 138.73) X 2= 195.86kN/m
(3 )求BC面上作用的抗滑力 Tf
Lbc = 5X 2= 10m, W = 0.5X( 2 + 7)X 10X 20= 900kN/m
Tf = f Lbc = cLbc + Lpctg = cLbc +Wtg = 10 X 10= 100kN/m
Ep +Tf
二 K = - =( 195.86 + 100) /84.04 = 3.52,该土堤不会沿软土夹层滑动。
Ea
[4]试计算以下几种情况下土条的稳定安全系数。
(1) = 17kN/m3, = 18° C = 10kPa; (2) = 17kN/m 3, sat = 19kN/m3, c = 10kPa, w = 10kN/m3
(3)= 18kN/m3, c = 5kPa, 24 , u = 5kPa
(4 )土条被浸润线和下游水位分成三部分,各部分土条高度分别为⏺
土条重度分别为 =19kN/m , sat = 20kN/m 3,土条宽度 b = 5m,弧长 I = 5.77m , = 30° C =
10kPa, 10 。
解:(1)滑动面弧长 1= 1 + 4 =』5 , cos = 2/ 1 + 4 = 0.894, sin =1/」5 = 0.447 , 119kN/m
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