边坡稳定分析与计算例题Word文档格式.docx
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FSc2
Cd
Or
—=多=9.67kN/m2
FScFSs3
Similarly,
lctan$
tan'
FSs
tan15
3
一丁
SubstitutingtheprecedingvaluesoCdanddintoequationgives
H斗士甞
y[1-cos(B-*d)
49.67
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P3o
解:
余推力R二RJi」"
Fs-Ri,其中Fs为安全系数1.05则Ri=Fs「-Ri
=1.05*1200-5500
=7100N
P2二R1FsT2-R2
=7100*0.733+1.05*1700-1900
=4054.3N
P3二P22FsT3-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?
单一滑动面滑动时,后部张裂缝深度的理论公式为:
代入得:
r2CfQW)
Zotg145
O.2
2x20q
Zotg60-2.77m
O25a
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?
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
[h2EI
L=:
0.49t;
cosa(tga-tg®
)
"
A
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,内摩檫角」=26,计算此边坡的稳定
系数。
①作用于BC上的静水压力V=0.56gZW=0.5X1X9.8X102=490kN
②作用于AB上的静水压力
U为U二
H-Z
0.5\gZwww=0.5X1X9.8X10XsinP
40~10=3133kN
sin28
③AB=(H-Z)十sinB=
(45-15)
-sin28°
=63.9m
G=(H+Z)XABXcosBX0.5X=(45+15)Xcos28°
X0.5X25=331kN
边坡稳定性系数为VU-vsifjCjAB
gsin卩+VcosP
=(331100cos28°
-3133000-490000sin28°
)tg26+80000^63.9
=9.8sin28490000cos28
=2.452
7.某一滑坡下卧稳定基岩,断面如图所示。
滑块各块重量分别为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,:
4=10。
滑面上内摩擦角均为15,粘聚力c为5.0kPa。
滑块长度l1=15m,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。
(1)计算各滑块抗滑力、下滑力和传递系数:
下滑力T=(WP)sin:
i;
抗滑力R=(WP)cos:
也;
ch;
传递系数=cosCi4-aj)-tg■■sin(〉iJ-ai);
将已知值分别代入上式,可得:
第一滑块:
T1=(700+900)Xsin40°
=1600X0.64=1024kNR1=(700+900)Xcos40°
tg15°
+5X15=403kN"
=cos(-40°
)-tg15°
sin(-40°
)=0.94-0.09=0.938
第二滑块:
T2=(2400+500)Xsin20°
=2900X0.34=992kN
R2=(2400+500)Xcos20°
Xtg15°
+5X15=805kN
书2=cos(40°
-20°
sin(40°
)=0.94-0.09=0.85
第三滑块:
T3=1500Xsin(-5°
)=-131kN
R3=1500Xcos(-5°
)xtg15°
+5X9=445kN
书3=cos(20°
+5°
)-tg15°
sin(20°
)=0.793
第四滑块:
T4=1800Xsin10°
=312kN
R4=1800Xcos10°
+5X14=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=1024X1.05-403=672kN
F2=992X1.05-805+0.938X672=867kN
F3=-131X1.05-445+0.85X867=154kN
F4=312X1.05-545+0.793X154=-95kN
F4<
0,边坡稳定。
当Fs=1.5时,计算可得:
F1=1024X1.5-403=1133kN
F2=992X1.5-805+0.938X1133=1746kN
F3=-131X1.5-445+0.85X1746=1236kN
F4=312X1.5-545+0.973X1236=1126kN
F4>
0,安全系数不能达到1.5。
8.UseLimitEquilibriumEquationstoanalysethestabilityofaslopesubjecttoaplanarinstability.Thedesignslope(inrock2.7g/cc)is30mhighanddipsduesouth
at75.
Basecase:
*=30
*c=150kPa
*slipplanedips40duesouthanddaylightsabovethetoeoftheslope
1)ProvideaplotofFSversusslipplanedip(keepallotherbasecaseparametersconstant).
2)ProvidetwoplotsofFSversusslipplanefrictionangle(Q=10°
to40©
)onthesamegraph,onewithc=0andtheotherwithc=150kPa(keepotherbasecaseparametersconstant).
3)Assumewaterpressureexistsalongtheslipplane-withatriangularpressure
distribution.ProvideaplotofFSversuspeakhydraulicheadforpressurehead
rangefrom0to10m.
4)Assumeyoucanaddasinglerowofhighcapacitycableboltsatmid-heightofthe
slope.Eachcablehasaworkingloadof2000kNandisspaced2mapart(into
page).Thecablesareinstalledperpendiculartotheslopestrike.Assume
worst-casewaterconditionsinthetensioncrack.ProvideaplotofFSversus
cableplunge;
includebothupholesanddownholes(keepotherparameters
constant).
5)Youhavejustcompletedasimplesensitivitystudy.Commentonthefindings
whatdidyoulearnfromyourplots,whatarethecontrollingparameter(s)?
Put
someintelligentwordsonpaper,neatly!
Avoidstatingtheobvious(e.g.steeper
slipplaneshavelowerfactorsofsafety)asyourmainconclusion.
6)DiscusshowyouwoulddoaMonteCarlosimulationtodeterminetheprobabilityoffailure.WhatwouldbetheadvantagesanddisadvantagesoftheanalysisyouperformedusingyourexcelspreadsheetcomparedtoaanalysisusingaMonteCarlomethod?
9.Ablockofrockliesonaslopeasshown.Calculatethefactorofsafetyagainst
slidingforthisblock.Iftheslopeandrockbecomecompletelysubmergedbywaterinareservoir,recalculatethefactorofsafety.Forbothcases,assumetheshearstrengthatthebaseoftheblockisgovernedbyafrictionangleof32plusacohesion
of100kPa.Thewidthoftheblockintothepageis3mandthedensityoftherockis2400kg/m3.
C=100KPa,
SolutionLength=3m,Height=2m,Width=3m,:
-40,=32
density=2400kg/m
Volume=332=18m
Weight=240018=42300kg
C汽L+W汽cosaxtan©
Fs=
W=<
sina
1003423.36cos40tan32
423.36汇sin40°
-1.85
Whentheblockiscompletelysubmergedbywate:
V=12w(Zwcos40)2;
U=12LwZw;
其中:
Zw=2m,w=9.81KN/m2,L=3m
V^、9.812.347-11.51KN
U39.812=29.43KN
CL亠〔Wcos-U-Vsin「tanFs二
sina+Vcosa
_1003423.36cos40-29.43-11.51sin40tan32
423.36汉sin40°
+11.51汇cos403
-1.71