边坡稳定分析与计算例题
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边坡稳定分析与计算例题
边坡工程计算例题
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?
Solution ⑴ Equation is β
φ
ββtan tan tan cos 2+⋅⋅⋅=
H r C F s ,
Given βφ,,,,H r C
We have 24.1=s F (2) Equation is β
ββ
φ
tan cos )tan tan (2⋅⋅-⋅=
s F r C
H ,
Given βφ,,,,s F r C We have m H 11.1=
Roc
Soi
3/7.15m kN =γ ⎩⎨
⎧︒
==15/6.92
φm kN c β
H
︒
=25β
2. A cut is to be made in a soil that has 316.5/kN m γ=,229/c kN m =, and 15φ=︒. The side of the cut slope will make an angle of 45°with the horizontal. What depth of the cut slope will have a factor of safety,S FS , of 3?
Solution We are given 15φ=︒ and 229/c kN m =.If 3C FS =, then
C FS and FS φshould both be equal to 3. We have
c d
c FS c =
Or
2299.67/3
d C S c c c kN m FS FS =
=== Similarly,
tan tan d
FS φφ
φ=
tan tan tan15
tan 3
d s FS FS φφφφ=
== Or
1tan15tan 5.13d φ-⎡⎤
==︒⎢
⎥⎣⎦
Substituting the preceding values of d c and d φinto equation gives
()()4sin cos 49.67sin 45cos5.17.11cos 16.51cos 45 5.1d d d c H m βφγβφ⎡⎤⎡⎤
⨯==≈⎢⎥⎢⎥----⎣⎦⎣⎦
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P 3。
解:余推力i s i i i i R F T P P -+=--11ϕ,其中s F 为安全系数1.05 则111R T F P s -= =1.05*1200-5500
=7100N
22112R T F P P s -+=ϕ
=7100*0.733+1.05*1700-1900 =4054.3N 33223R T F P P s -+=ϕ =4054.3*1+1.05*2400-2700 =3874.3N
则滑坡推力为3874.3N
下
滑
力 T
(K
N/m ) 抗滑
力R (K N/m )
滑面 倾角 θ 传
递系数ϕ
①
12 000 5
500
45︒ 0.
73
3
② 17 000 19
000 17︒ 1
③
2 400 2
700
17︒
P
3
4. 某岩性边坡为平面破坏形式,已知滑面AB 上的C=20kPa , ,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:
代入得:
5. 岩质边坡坡角35°,重度 ,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 ,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
取
得:
代入上述数值得:L=93m 为极限长度,
则,岩坡极限高度:
⎪
⎭
⎫ ⎝⎛+=2452ϕγ tg C Z O m
tg Z O 77.26025
202=⨯= 3/3.25m kN =γ 30=ϕ()3
1
2
cos 49.0⎥⎦⎤⎢⎣⎡-=ϕααγπtg tg t EI L 3
12
1t I =
()
3
2
2cos 6ϕααγπtg tg Et L -=()m L H 53sin =⋅=α
30=ϕ