Chapter15-Slope Stability

Ex1-pg518

In [2]:
import math
#a.The factor of safety against sliding along the soil-rock interface.
#b.The height,H, that will give a factor of safety (Fs) of 2 against sliding alongthe soil-rock interface.
Gs=17.8
Gw=9.81
C=10.
c=20.
b=15.
H=6.
G=Gs-Gw
Fs= C/(Gs*H*math.cos(b/57.3)*math.cos(b/57.3)*math.tan(b/57.3))+G*math.tan(c/57.3)/(Gs*math.tan(b/57.3))
print'%s %.2f %s'%('a)The factor of safety = ',Fs,' ')
Fs=2.
H=2.247/(Fs-0.61)
print'%s %.2f %s'%(' b)H= ',H,' m')

a)The factor of safety =  0.98  
 b)H=  1.62  m

Ex3-pg529

In [3]:
import math
#a.Determine the maximum depth up to which the excavation can be carried out.
#b.Find the radius,r, of the critical circle when the factor of safety is equal to 1(Part a).
#c. Find the distance  . BC
Cu=40.
G=17.5
b=60.
a=35.
c=72.5
m=0.195
Hc=Cu/(G*m)
r=Hc/(2.*math.sin(a/57.3)*math.sin((c/2)/57.3))
BC=Hc*((1./math.tan(a/57.3))-(1./math.tan(b/57.3)))
print'%s %.1f %s'%('a)The maximum depth Hc = ',Hc,' m')
print'%s %.2f %s'%(' b)The radius, r = ',r,' m')
print'%s %.3f %s'%(' c)The distance BC.= ',BC,' m')

a)The maximum depth Hc =  11.7  m
 b)The radius, r =  17.28  m
 c)The distance BC.=  9.973  m

Ex4-pg531

In [1]:
import math
#a.Determine the undrained cohesion of the clay (Figure 15.13).
#b.What was the nature of the critical circle?
#c. With reference to the toe of the slope, at what distance did the surface of sliding intersect the bottom of the excavation?
Gs=17.29
d=9.15
d1=6.1
D=d/d1
a=40.
m=0.175
b=40.
H=6.1
Cu=H*Gs*m
print'%s %.1f %s'%('a)The undrained cohesion of the clay Cu = ',Cu,' kN/m^2')
print(' b)The nature of the critical circle is midpointcircle')
d=1.5
b=40.
n=0.9
D1=n*H
print'%s %.1f %s'%(' c)Distance = ',D1,' m')

a)The undrained cohesion of the clay Cu =  18.5  kN/m^2
 b)The nature of the critical circle is midpointcircle
 c)Distance =  5.5  m

Ex5-pg534

In [5]:
import math
#a.Determine the maximum depth up to which the cut could be made.
#b.How deep should the cut be made if a factor of safety of 2 against sliding is required
Fs=1.
b=56.
Kh=0.25
M=3.66
Cu=500.
G=100.
Hc=Cu*M/G
print'%s %.1f %s'%('a)The maximum depth =',Hc,' ft')
Fs=2.
H=Cu*M/(G*Fs)
print'%s %.1f %s'%(' b)H= ',H,' ft')

a)The maximum depth = 18.3  ft
 b)H=  9.2  ft

Ex6-pg541

In [3]:
import math
#a.Find the critical height of the slope.
#b.If the height of the slope is 10 m, determine the factor of safety with respect to strength.
b=45.
c=20.
C=24.
G=18.9
m=0.06
Hc=C/(G*m)
Cd=G*Hc*m
Fc=C/Cd
print'%s %.1f %s'%('a)Critical height of slope = ',Hc,'')
#calculate the factor of safety using spencers solution
import math
%matplotlib inline
import warnings
warnings.filterwarnings('ignore')
import numpy
from math import tan
import matplotlib
from matplotlib import pyplot
phid=numpy.array([20,15,10,5])*math.pi/180.
mx=numpy.array([.06,.083,.105,.136])
cdx=numpy.array([11.34,15.69,19.85,25.7])
m=.06
g=18.9
cd=24.
#calculations
Hcr=cd/g/m
leng=len(phid)
Fcd=numpy.zeros(leng)
Fphi=numpy.zeros(leng)
tanphid=numpy.zeros(leng)
for i in range(0,leng):
	tanphid[i]=math.tan(phid[i])
	Fphi[i]= tan(phid[0])/tan(phid[i])
	Fcd[i]=cd/cdx[i]


#results
print'%s %.2f %s'%('The value of Hcr (in m) = ',Hcr,'')
print 'from graph, Fss=1.4'
pyplot.plot(Fcd,Fphi)
pyplot.xlabel('Fc assumed')
pyplot.ylabel('Fc calculated')
pyplot.title('Graph of Fc assumed vs Fc calculated')
pyplot.show()

a)Critical height of slope =  21.2 
The value of Hcr (in m) =  21.16 
from graph, Fss=1.4

Ex7-pg544

In [7]:
# using Michalowski’s solution.
import math
FSs=1.
c=20.
G=18.9
C=24.
Hcr=C/(G*math.tan(c/57.3)*0.17)
print'%s %.1f %s'%('a)Critical height Hc = ',Hcr,' m')
H=10.
k=C/(G*H*math.tan(c/57.3))
Fs=4.*math.tan(c/57.3)
print'%s %.1f %s'%(' b)Fs = ',Fs,'')

a)Critical height Hc =  20.5  m
 b)Fs =  1.5

Ex8-pg560

In [8]:
import math
# Determine the factor ofsafety,Fs . Use Table 15.3.
W=22.4
C=20.
a=70.
s=math.sin(a/57.3)
c=math.cos(a/57.3)
l=2.924
Wn=W*s
Wn1=W*c
##doing this to all values
F1=30.501
F2=776.75
F3=1638.
Fs=(F1*C+F3*math.tan(C/57.3))/F2
print'%s %.2f %s'%('Fs = ',Fs,'')

Fs =  1.55

Ex9-pg560

In [9]:
import math
#using Michalowski’s solution
C=20.
G=18.5
r=0.25
H=21.62
C=25.
b= math.atan(0.5)
##from table 15.3 
m=1.624
n=1.338
Fs=m-n*r
print'%s %.1f %s'%(' The value of Fs for D= 1 is',Fs,'')

 The value of Fs for D= 1 is 1.3

Ex10-pg561

In [2]:
#calculate the factor of safety using spencers solution
import math
%matplotlib inline
import warnings
warnings.filterwarnings('ignore')
import numpy
from math import tan
import matplotlib
from matplotlib import pyplot
beta=numpy.array([26.57,26.57,26.57,26.57])
Fs=numpy.array([1.1,1.2,1.3,1.4])
phid=25*math.pi/180. #degrees
#calculations
print 'From spencers graphs,'
cd=numpy.array([0.0455,0.0417,0.0385,0.0357])
phia=numpy.array([18.,19.,20.,21.])*math.pi/180.
leng=len(phia)
Fss=numpy.zeros(leng)
for i in range(0,leng):
	Fss[i]= tan(phid)/tan(phia[i])

#results
print 'From graph, a footing of dimensions Fs=1.3'
pyplot.plot(Fs,Fss)
pyplot.xlabel('Fs assumed')
pyplot.ylabel('Fs calculated')
pyplot.title('Graph of Fs assumed vs Fs calculated')
pyplot.show()

From spencers graphs,
From graph, a footing of dimensions Fs=1.3

Ex11-pg561

In [10]:
import math
#using Michalowski’s solution
C=20.
G=18.5
H=21.62
c=25.
r=0.25
Fs=3.1*math.tan(c/57.3)
print'%s %.1f %s'%('Fs = ',Fs,'')

Fs =  1.4