Chapter 1:Geotechnical Properties of Soil

Ex1.1:Pg-10

In [9]:
#example 1.1

V=0.25; # ft^3
W=30.8; #lb
Wd=28.2; # weight dried lb
Gs=2.7;
Gammaw=62.4; #lb/ft^3
Gamma=W/V;
print Gamma," is moist unit weight in lb/ft^3"
w=(W-Wd)/W;
print round(w*100,2),"is moisture content in %"
Gammad=Wd/V;
print Gammad, "is dry unit weight in lb/ft^3"
Vs=Wd/Gs/Gammaw;
Vv=V-Vs;
e=Vv/Vs;
print round(e,3)," is void ratio"
n=e/(1+e);
print round(n,2),"is porosity"
Vw=(W-Wd)/Gammaw;
S=Vw/Vv;
print round(S*100,2),"is saturation in %"

# The answers in the book are different due to approximation while here calculations are precise

123.2  is moist unit weight in lb/ft^3
8.44 is moisture content in %
112.8 is dry unit weight in lb/ft^3
0.494  is void ratio
0.33 is porosity
50.43 is saturation in %

Ex1.2:Pg-11

In [12]:
#example 1.2

e=0.72;
w=12.0/100; #moisture content
Gs=2.72;
Gammaw=9.81;#kN/m^3
Gammad=Gs*Gammaw/(1+e);
print round(Gammad,2),"= dry unit weight in kN/m^3"
Gamma=Gs*Gammaw*(1+w)/(1+e);
print round(Gamma,2)," = moist unit weight in kN/m^3"
Gammasat=(Gs+e)*Gammaw/(1+e);
wa=Gammasat-Gamma;#water added
print round(wa,2)," = water added in kN/m^3"

15.51 = dry unit weight in kN/m^3
17.38  = moist unit weight in kN/m^3
2.24  = water added in kN/m^3

Ex1.3:Pg-12

In [14]:
#example 1.3
from scipy.optimize import fsolve
gmax=17.1; # Gammadmax
Dr=0.7;
w=8.0/100;#moisture content
gmin=14.2; #Gammadmin
def f(x):
    return (x-14.2)/(17.1-14.2)*17.1/x-0.7 
x=fsolve(f,16);#solving for gammad
Gamma=x[0]*(1+w);
print round(Gamma,2),"moist unit weight in kN/m^3"

17.4 moist unit weight in kN/m^3

Ex1.7:Pg-38

In [27]:
#example 1.7
import math
#part (a)
e1=0.92;
e2=0.86;
Hc=2.8; # in m
s2=212.0;#sigma2dash Load in kN/m2
s1=140.0;#sigma1dash Load in kN/m2
Cc=(e1-e2)/math.log10(s2/s1);
Sc=Cc*Hc/(1+e1)*math.log10(s2/s1);
print Sc*1000,"consolidated depth in mm"
#part (b)
# from Figure (1.21):
Sct=40.0;# in mm
T50=0.197;
t=4.5; # in MIN
Cr=T50*12.7**2.0/t;
U=Sct/Sc*100.0/1000;
H=Hc/2;
Tv=math.pi/4*U**2.0/100**2;
t=Tv*H**2.0/Cr*1000.0**2/60.0/24;
print round(t,1)," is time required in days"

87.5 consolidated depth in mm
31.6  is time required in days

Ex1.8:Pg-42

In [28]:
#example 1.8

Cv=7.061; # in mm^2/min
tc=15*24*60.0;
H=2.8/2*1000.0;
Scmax=87.5; # consolidation
Tc=Cv*tc/H**2;
tv=31.6*24*60;
Tv=Cv*tv/H**2;
#from figure 1.28
Sct=Scmax*0.36;
print Sct,"is consolidation in 31.6 days in mm"

31.5 is consolidation in 31.6 days in mm