1. Consider the infinite slope shown in figure.
(1) Determine the factor of safety against sliding along the soil-rock interface given H = 2.4m.
(2) What height, H, will give a factor of safety, F s, of 2 against sliding along the soil-rock interface?
Rock
Solution
Soil
⑴ Equation is
Fs
t a n-
Given C,r,H ,,: We have Fs =1.24 (2) Equati on is
t a n t
r (Fs
)c o s 1 tan：
Given C,r,Fs,
：
We have H = 1.11m
2. A cut is to be made in a soil that has = 16.5kN /m3,c = 29kN /m2, and = 15 . The side of the cut slope will make an an gle of 45
°with the horiz on tal. What
depth of the cut slope will have a factor of safety, FSS, of 3? Solution We are given
=15 and c = 29kN / m2 .If FSc =3 , then
FSC andFS should both be equal to 3. We have
FSc
2
Cd
Or
—=多=9.67kN /m2
FSC
FSS
3
Similarly,
lc tan $
tan ' FSs
tan15 3
Or
二 tan
J
tan15
一丁
Substituting the preceding values oCd and d into equation gives
H斗士甞
y [1 -cos（B -*d ）
4 9.67 16.5
-sin45cos5.^fc7.1m
1 -cos（45 -5.1 ）
3. 某滑坡的滑面为折线，其断面和力学参数如图和表所示，拟设计抗滑结构物, 取安全系数为1.05，计算作用在抗滑结构物上的滑坡推力
P3 o
下滑力 抗滑力 滑面 倾传递 系角 数 T (KN/m) ① 12 000 R (KN/m) 5 500 0 0.73 45 3 ② ③
17 000 2 400 19 000 2 700 17 1 17 解：余推力R二R Ji」\"Fs -Ri ,其中Fs为安全系数1.05 则 Ri = Fs「- Ri
=1.05*1200-5500 =7100N
P2 二 R 1 FsT2 - R2
=7100*0.733+1.05*1700-1900 =4054.3N
P3 二 P2 2 FsT3 - R3
=4054.3*1+1.05*2400-2700 =3874.3N
则滑坡推力为3874.3N
4. 某岩性边坡为平面破坏形式，已知滑面AB上的C=20kPa甲=30「，当滑面
上岩体滑动时，滑动体后部张裂缝 CE的深度为多少米？ 解：单一滑动面滑动时，后部张裂缝深度的理论公式为:
r 2C f Q W ) Z O tg 145 O
. 2
代入得：
ZO
O
2x20 25 a
q
tg60 - 2.77m
5. 岩质边坡坡角35°，重度Y=25.3kN/m‘，岩层为顺坡，倾角与坡角相同， 厚度t=0.63m，弹性模量E=350MPa内摩擦角「=3^，则根据欧拉定理计算 此岩坡的极限高度为多少米？
解：根据欧拉定理，边坡顺向岩层不发生曲折破坏的极限长度计算式为
1
L = -------- : ---------------------
0.49t ；cosa(tga -tg® )
[
H2EI
\"A
2
Et2
6 cos：鋼
代入上述数值得：L=93m为极限长度,
6 •已探明某岩石边坡的滑面为 AB,坡顶裂缝 DC深z =15m，裂缝内水深
Zw =10m ,坡高H =45m ,坡角:=60 ,滑坡倾角」28 ,岩石容重
咐=25kN/m3，滑面粘结力c=80kPa， 系
数
内摩檫角」=26，计算此边坡的稳定
。
解：①作用于BC上的静水压力V =0.56gZW=0.5 X 1X 9.8 X 10=490kN
2
H -Z ②作用于AB上的静水压力 U为U二 0.5\\gZw w w =0.5 X 1X 9.8 X 10X sin P
4010
sin 28
~
=3133kN
③ AB = (H-Z)十 sin B = (45-15) -sin28 ° =63.9m
G=(H+Z)X AB X cos B X 0.5 X = (45+15)X cos28°X 0.5 X 25=331kN
边坡稳定性系数为 VU -vsif j Cj AB g sin 卩 +V cosP
=(331100cos28°- 3133000- 490000sin 28 °) tg 26+80000^ 63.9 = 9.8sin 28 490000cos28
=2.452
7.
稳定基岩，断面如图所示。滑块各块重量分别为
某一滑坡下卧
W.= 700kN ,
W2 =2400kN，W3 =1500kN，W4 =1800kN。外荷载 P? = 500kN，R = 900kN 分
别作用在第一块、第二 块上，其作用线通过相应块的重心。滑面角 宀=40，
-20，
: ^-5，: =10。滑面上内摩擦角均为15，粘聚力c为5.0kPa。
4
l15m
滑块长度1 =， l2 =15m，l3 =9m， l4 =14m。试计算滑坡推力并判断其稳 定性（安全系数Fs取1.05 ）能否达到1.5。
解：(1)计算各滑块抗滑力、下滑力和传递系数：
下滑力 T =(W P)sin : i ； 抗滑力 R =(W P)cos ：也；ch ; 传递系数
= cosC i 4 -aj) -tg ■■ sin(〉iJ -ai);
将已知值分别代入上式，可得：
第一滑块：T1=(700+900)X sin40° =1600X 0.64=1024kN R1=(700+900)X
cos40° tg15° +5X 15=403kN \"=cos(-40° ) - tg15° sin(-40° )=0.94-0.09=0.938
第二滑块：T2=(2400+500) X sin20° =2900 X 0.34=992kN
R2=(2400+500)X COS20°X tg15° +5X 15=805kN
书 2=COS(40° -20° ) - tg15° sin(40° -20° )=0.94-0.09=0.85
第三滑块：T3=1500X sin (-5°) =-131kN
R3=1500X COS (-5°)x tg15° +5X 9=445kN
书 3=COS(20° +5° )-tg15° sin(20° +5° )= 0.793
第四滑块：T4=1800X sin 10° =312kN
R4=1800X COS10°X tg15° +5X 14=545kN
(2)计算滑坡推力
滑坡推力 Fi二TK -R宀i。
当FS=1.05,由上式计算可得：
F1=1024X 1.05-403=672kN
F2=992 X 1.05-805+0.938X 672=867kN F3=-131 X 1.05-445+0.85X 867=154kN F4=312X 1.05-545+0.793X 154= -95kN F4<0,边坡稳定。
当FS=1.5时，计算可得：
F1=1024X 1.5 -403=1133kN
F2=992 X 1.5-805+0.938X 1133=1746kN F3=-131 X 1.5-445+0.85X 1746=1236kN F4=312X 1.5-545+0.973X 1236= 1126kN F4>0,安全系数不能达到1.5。
8. Use Limit Equilibrium Equations to analyse the stability of a slope subject to a pla nar in
stability. The desig n slope (in rock 2.7 g/cc) is 30 m high and dips due south at 75 . Base case: *
= 30 * C = 150 kPa
* slip pla ne dips 40 due south and daylights above the toe of the slope
1) Provide a plot of FS versus slip pla ne dip (keep all other base case parameters con sta nt).
2) Provide two plots of FS versus slip plane friction angle ( Q = 10° to 40©) on the same graph, one with c = 0 and the other with c = 150kPa (keep other base case parameters con sta nt).
3) Assume water pressure exists along the slip pla ne range from 0 to 10 m.
-with a tria ngular pressure
distributi on. Provide a plot of FS versus peak hydraulic head for pressure head
4) Assume you can add a sin gle row of high capacity cable bolts at mid-height of the slope. Each cable has a work ing load of 2000 kN and is spaced 2 m apart (into page). The cables are in stalled perpe ndicular to the slope strike. worst-case water con diti ons in the tension crack. con sta nt).
Assume
Provide a plot of FS versus
cable plun ge; in clude both upholes and dow nholes (keep other parameters
5) You have just completed a simple sen sitivity study. Comme nt on the findings what did you lear n from your plots, what are the con trolli ng parameter(s)? some in tellige nt words on paper, n eatly!
slip pla nes have lower factors of safety) as your main con clusi on.
Put
Avoid stati ng the obvious (e.g. steeper
6) Discuss how you would do a Monte Carlo simulation to determine the probability of failure. What would be the adva ntages and disadva ntages of the an alysis you performed using your excel spreadsheet compared to a an alysis using a Monte Carlo method?
9. A block of rock lies on a slope as show n. Calculate the factor of safety aga inst
slidi ng for this block. If the slope and rock become completely submerged by water in a reservoir, recalculate the factor of safety. For both cases, assume the shear stre ngth at the base of the block is gover ned by a fricti on an gle of 32 kg/m3.
plus a cohesi on
of 100 kPa. The width of the block into the page is 3 m and the den sity of the rock is 2400
Solution
Length=3m , Height=2m , Width=3m，： - 40 ,
=32
C=100KPa ,
den sity=2400kg/m
3
Volume = 3 3 2 = 18m 3
Weight = 2400 18 = 42300kg
Gravity 二 42300 9.8 = 423.36KN
C 汽 L +W 1000 W = 423.36 汇 s in 40° -1.85 Whe n the block is completely submerged by wate: (Zwcos40 )2 ； U = 12 L w Zw ; 其中：Zw =2m, w=9.81KN/m2, L =3m V^、9.81 2.347 -11.51KN U 3 9.81 2 =29.43KN C L 亠〔W cos W = -1.71 V = 1 2 w 因篇幅问题不能全部显示，请点此查看更多更全内容