Chapter11-Compressibility of Soil¶
Ex1-pg303¶
In [11]:
#evaluvate The elastic settlement at the centre of foundation
Tz=150.
b=1.
l=2.
z=5.*b
Es= (10000*2 + 8000*1 +12000*2)/5
a=4.
H=z
m=l/b
n=2.*H/b
F1=0.641 ##from tables 11.1 and 11.2
F2=0.031
u=0.3
Is= F1 + ((2.-u)/(1.-u))*F2
If=0.71 ##from table 11.3
Sef= Tz *a*b/l *(1-u**2)*Is*If/Es
Ser=0.93*Sef
print'%s %.3f %s'%('The elastic settlement at the centre of foundation =',Ser,'m')
Ex2-pg312¶
In [13]:
#calculate the value of pressure
#find the value of e and plot the graph
## one value of e is done
Gs=2.75
A=30.68
Ms=128.
p=1.
Hs=Ms/(A*Gs*p)
H=2.540
Hv=H-Hs
e=Hv/Hs
print'%s %.3f %s'%('the value of e for give values =',e,'')
import math
%matplotlib inline
import warnings
warnings.filterwarnings('ignore')
from math import log
import numpy
from math import tan
import matplotlib
from matplotlib import pyplot
#given
p=numpy.array([0,0.5,1,2,4,8,16,32])
e=numpy.array([0.671,0.637,0.622,0.599,0.572,0.529,0.464,0.390])
e1=.9
e2=.8
sig1=4.
sig2=2.
#calculations
Cc=(e1-e2)/log(sig1/sig2)
#results
print '%s %.2f %s'%('The value of Cv (cm^2/sec) = ',Cc,'')
pyplot.plot(p,e)
pyplot.xlabel('Pressure (ton/ft^2)')
pyplot.ylabel('void ratio ,e')
pyplot.title('Graph of pressure vs void ratio')
pyplot.show()
Ex3-pg321¶
In [14]:
#find the value compression index
import math
%matplotlib inline
import warnings
warnings.filterwarnings('ignore')
#calculate Compression index
e11=0.9
e21=0.8
T2=4.
T1=2.
Cc= (e11-e21)/math.log10(T2/T1) ## from loading branch
e1=0.67
e2=0.655
Cs=(e1-e2)/math.log10(T2/T1)
k=Cs/Cc
T3=12.
e3=e11-Cc*math.log10(T3/T1)
print'%s %.2f %s'%('Compression index Cc= ',Cc,'')
print'%s %.2f %s'%(' Cs/Cc = ',k,'')
print'%s %.2f %s'%(' e3 = ',e3,'')
#calculate the value of Cv
%matplotlib inline
import warnings
warnings.filterwarnings('ignore')
import math
from math import log
import numpy
from math import tan
import matplotlib
from matplotlib import pyplot
#given
p=numpy.array([.25,.5,1.,2.,4.,8.,16.,8.,4.,2.])
e=numpy.array([1.03,1.02,0.98,0.91,0.79,0.71,0.62,0.635,0.655,0.67])
e1=.9
e2=.8
sig1=4.
sig2=2.
#calculations
Cc=(e1-e2)/log(sig1/sig2)
#results
print 'The value of Cv (cm^2/sec) = ',Cc
pyplot.plot(p,e)
pyplot.xlabel('Pressure (ton/ft^2)')
pyplot.ylabel('void ratio ,e')
pyplot.title('Graph of pressure vs void ratio')
pyplot.show()
Ex4-pg323¶
In [10]:
import math
#calculate Primary Consolidation Sc in three parts
Gd=14.
Gss=18.
Gsc=19.
Gw=9.81
To= 2.*Gd+4.*(Gss-Gw)+2*(Gsc-Gw)
LL=40.
Cc=0.009*(LL-10)
H=4.
T=100.
e=0.8
Sc= Cc*H*math.log10((To+T)/To)/(1.+e)
print'%s %.2f %s'%('a)Primary Consolidation Sc = ',Sc,' m')
Tc=190
Cs=Cc/6
Sc= Cs*H*math.log10((To+T)/To)/(1+e)
print'%s %.2f %s'%(' b)Primary Consolidation Sc =',Sc,'m')
Tc=170
Sc= Cc*H*math.log10((To+T)/Tc)/(1+e)+ Cs*H*math.log10(Tc/To)/(1+e)
print'%s %.3f %s'%(' c)Primary Consolidation Sc =',Sc,' m')
Ex5-pg325¶
In [11]:
#calculate The settlement in the field Sc
Gs=18.
Gw=9.81
H=10.
eo=1.1
To=5.*(Gs-Gw)
T1=48.
T=To+T1
e1=1.045 ## void ratio corresponding to T
e=eo-e1
Sc=H*e/(1.+eo)
print'%s %.2f %s'%('The settlement in the field Sc = ',Sc,' m')
Ex6-pg329¶
In [12]:
import math
#calculate Total consolidation settlement of the clay
T=8.5
eo=0.8
Cc=0.28
To=2650.
T1=970.
C1=0.02
t2=5.
t1=1.5
H=8.5*12
epr=Cc*math.log10((To+T1)/To)
ep=eo-epr
C2=C1/(1.+ep)
Sc=epr*H/(1.+eo)
Ss=C2*H*math.log10(t2/t1)
TS=Sc+Ss
print'%s %.1f %s'%('Total consolidation settlement of the clay =',TS,' in')
Ex7-pg336¶
In [13]:
import math
#calculate t field
##T50 = Cvtlab /H^2 lab = Cvtfield?H^2 fiels
tl=140.
Hf=3.
Hd=0.025/2.
tf=tl*Hf**2/Hd**2
k=tf/(3600.*24.)
print'%s %.1f %s'%('t field = ',k,' days')
Ex8-pg336¶
In [14]:
import math
##Tv is directly proportional to U^2
t1=93.333
U2=30.
U1=50.
t2=t1*U2**2./U1**2.
print'%s %.2f %s'%('t2 =',t2,' days')
Ex9-pg337¶
In [15]:
import math
#evaluvate Cv
#intilization variable
t90=75.*24.*60.*60. ## time in sec
T90=0.848
Hd=1.5*100. ##in cm
Cv=T90*Hd**2/t90
print'%s %.3f %s'%('Cv =',Cv,' cm^2/sec')
Ex10-pg337¶
In [5]:
import math
#calculate K and t60
To=3000. ## lb/ft^2
eo=1.1
e1=0.9
e=eo-e1
ea=(eo+e1)/2.
T1=3000. ## lb/ft^2
T=1. ## in
t = 2. ## min
m=(e/T1)/(1.+ea)
U=50.
Tv=0.197
Gw=62.4 ##lb/ft^3
Cv=Tv*(T/(2.*12.)**2)/t
k=Cv*m*Gw *10**7
print'%s %.3f %s'%('a)k = ',k,' x10^-7 ft/min')
U=60
Tv=0.286
H=6
t60=Tv*H**2/(Cv*60*24)
print'%s %.1f %s'%(' b)t60 =',t60,' days')
Ex11-pg344¶
In [16]:
#calculate the value of Cv
import math
#Cv
t50=19
Hd=2.24/2
Cv=0.197*Hd**2/t50
print'%s %.3f %s'%('Cv = ',Cv,' cm^2/min')
%matplotlib inline
import warnings
warnings.filterwarnings('ignore')
import math
import numpy
from math import tan
import matplotlib
from matplotlib import pyplot
#given
t=numpy.array([.02,.1,.25,.5,1,2.,4.,8.,16.,30.,60.,120.,240.,480.,960.,1440.])
gauge=numpy.array([3975.,4082.,4102.,4128.,4166.,4224.,4298.,4420.,4572.,4737.,4923.,5080.,5207.,5283.,5334.,5364.])
Hdr=2.24
t50=19.
#calculations
Cv=.197*(Hdr/2)**2 /t50/60.
leng=len(t)
logt=numpy.zeros(leng)
for i in range(0,leng):
logt[i]=math.log(t[i])
#results
print 'The value of Cv (cm^2/sec) = ',Cv
pyplot.plot(logt,gauge)
pyplot.xlabel('Time(min) - log scale')
pyplot.ylabel('Dial reading (cm)')
pyplot.title('Graph of dial reading vs time')
pyplot.show()
Ex12-pg346¶
In [8]:
import math
#calculate SC
LL=40.
Cc=0.009*(LL-10)
H=10.*12.
eo=1.0
Gss=120.
Gsc=110.
Gd=100.
To=10.*Gd +10.*(Gss-62.4)+10.*(Gsc-62.4)/2.
Tt=0.408
Tm=0.232
Tb=0.019
Tav= (Tt+4.*Tm+Tb)/6.
Sc=Cc*H*math.log10((To+Tav*1000.)/To)/(1.+eo)
print'%s %.3f %s'%('Sc =',Sc,' in')
Ex13-pg356¶
In [7]:
#intilization variable
#Calculate total primary
import math
H = 6.
Cc = 0.28
eo = 0.9
Cv = 0.36
To=210.
Tp=115.
Sc= Cc*H*math.log10((To+Tp)/To)/(1+eo)
t2=9.
Hd=3.
Tv=Cv*t2/Hd**2
U=0.67
Tf=0.677*Tp
print'%s %.1f %s'%('Tf =',Tf,' kN/m^2')