#example 5.1
#first solution
B1=2.5; # in ft
B2=B1;
z=12.5; # in ft
L1=5; # in ft
L2=L1;
m=B1/z;
n=B2/z;
#from table 5.2 of the values using m,n
q=2000; # in lb/ft^2
I=0.0328;
deltasigma=q*4*I; # in lb/ft**2
print round(deltasigma,2),"change in pressure in lb/ft**2"
#second solution
Ic=0.131;#from table
deltasigma=q*Ic; # in lb/ft**2
print round(deltasigma,2),"change in pressure in lb/ft**2"
#example 5.2
qo=100; # in KN/m^2
H1=3; # in m
H2=5; # in m
#from table
IaH2=0.126;
IaH1=0.175;
deltasigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1)); # in kN/m**2
print round(deltasigma,2),"change in pressure in kN/m**2"
TS=4*deltasigma; # in kN/m**2
print round(TS,2),"total change in pressure in kN/m**2"
#example 5.3
H=7;
Gamma=17.5; # in KN/m^3
q0=Gamma*H # in KN/m^2
print q0," is pressure change in kN/m**2"
#part2
#from figure
Ileftside=0.445;
Irightside=0.445;
deltasigma=q0*(Ileftside+Irightside); # in KN/m^2
print round(deltasigma,2),"is change in stress in kN/m**2"
#partc
#from figure 5.11
I=0.24;#I'
Dsigma1=43.75*I;#deltasigma1 in KN/m^2
I2=0.495;#I'
Dsigma2=I2*q0;#deltasigma2 in KN/m^2
I3=0.335;#I'
Dsigma3=I3*78.75;#deltasigma3 in KN/m^2
Dsigma=Dsigma1+Dsigma2-Dsigma3; # in KN/m^2
print round(Dsigma,2),"is total stress increase in A in kN/m**2"
#example 5.4
zbar=5;
mus=0.3;
F1=0.641;
F2=0.031;
z1=2.0; # in m
z2=1.0; # in m
z3=2.0; # in m
Es1=10000; # in kN/m**2
Es2=8000; # in kN/m**2
Es3=12000;# in kN/m**2
qo=150; # in KN/m^2
#from table 5.4
If=0.709;
Es=(Es1*z1+Es2*z2+Es3*z3)/zbar; # in kN/m**2
print Es," is modulus of elasticity in kN/m**2"
Is=F1+(2-mus)/(1-mus)*F2;
Sc=qo*(1.0/Es-mus**2.0/Es)*Is*If*2;
Scrigid=0.93*Sc; # in m
print round(Scrigid*1000,2),"is settlement in mm"
#example 5.5
import math
B=5; # in ft
L=10; # in ft
Ef=2.3e6; # in lb/in^2
Eo=1400.0; # in lb/in^2
k=25.0; # in lb/in^2/ft
t=1.0;
mus=0.3;
Df=5.0; # in ft
qo=5000.0; # in lb/ft^2
Ig=0.69;
Be=math.sqrt(4*B*L/math.pi);
If=math.pi/4+1/(4.6+10*(Ef/(Eo+2*Be/2*k))*(2*t/Be)**3);
Ie=1-1/(3.5*math.exp(1.22*mus-0.4)*(Be/Df+1.6));
Se=qo*Be*Ig*If*Ie/Eo*(1-mus**2)/144; # in ft
print round(Se*12,2),"settlement in inches"
#example 5.6
import math
import numpy
q=3.06; # in lb/in^2
qbar=25; # in lb/in^2
C1=1-0.5*(q/(qbar-q));
Sum=0;
C2=1+0.2*math.log10(10/0.1);
L=[1, 2, 3, 4, 5];
Dz=[48, 48, 96, 48, 144]; # in inch
Es=[750, 1250, 1250, 1000, 2000]; # in lb/in^2
z=[24, 72, 144, 216, 312]; # in inch
Iz=[0.275, 0.425, 0.417, 0.292, 0.125];
k=numpy.zeros(5)
print "Layer No.\t deltaz (in)\t Es(lb/in**2)\t z to the middle of the layer (in) Iz at the middle of the layer Iz/delta(z) \n"
for i in range(0,5):
k[i]=Iz[i]/Es[i]*Dz[i];
print L[i],"\t \t ",Dz[i],"\t\t ",Es[i],"\t\t ",z[i]," \t\t\t\t\t ",Iz[i],"\t\t ",round(k[i],3)
Sum=Sum+k[i];
Se=C1*C2*(qbar-q)*Sum; # in inch
print round(Se,2),"settlement in inches"
#example 5.7
Df=1.0; # in m
B=1.75; # in m
L=1.75 # in m
qnet=120.0; # in KN/m^2
N60=10.0;# in m
alpha1=0.14 # for normally consolated sand
alpha2=1.71/(N60)**1.4 # for normally consolated sand
alpha3=1.0 # for normally consolated sand
Se=0.3*alpha1*alpha2*alpha3*(qnet/100)*((B/0.3)**0.7)*((1.25*(L/B)/(0.25+(L/B))))**2
print round(Se*1000,2),"settlement in mm"
#example 5.8
Df=1; # in m
B=1.75; # in m
qnet=120; # in KN/m^2
N60=10; # in m
Fd=1+0.33*Df/B;
Se=2*qnet/N60/Fd*(B/(B+0.3))**2; # in mm
print round(Se,2),"settlement in mm"
#example 5.9
Ny=23.76;
Nq=16.51;
q=3*110.0; # in lb/ft^2
Gamma=110.0; # in lb/ft^3
B=4.0; # in ft
Nqe=0.63*Nq;
Nye=0.4*Ny;
que=q*Nqe+1/2.0*Gamma*B*Nye; # in lb/ft^2
print round(que,2)," is bearing capacity in lb/ft**2"
#part 2
V=0.4; # in ft/sec
A=0.32; # given in question
g=9.81; # acceleration constant in m/sec^2
kh=0.26;
k=0.92;#tan(alphae)
Seq=0.174*k*V**2/A/g*kh**-4/A**-4; # in m
print round(Seq,3),"settelement in m"
print round(Seq*39.57,2),"settlement in inches"
#example 5.10
import math
Cc=0.32;
Hc=2.5;
eo=0.8;
sigmao=2.5*16.5+0.5*(17.5-9.81)+1.25*(16-9.81); # in kN/m^2
m1=[2, 2, 2];
z=[2, 3.25, 4.5];
n1=[4, 6.5, 9];
Ic=[0.19, 0.085, 0.045];
Dsigma=[28.5, 12.75, 6.75];#deltasigma
print ("m1\t z(m)\t n1\t Ic\t Dsigma \n");
for i in range(0,3):
print round(m1[i],2),"\t ",round(z[i],2),"\t ",round(n1[i],2),"\t ",round(Ic[i],2),"\t ",round(Dsigma[i],2)
Dsigmaav=1/6.0*(Dsigma[0]+4*Dsigma[1]+Dsigma[2]);
Sc=Cc*Hc/(1+eo)*math.log10((sigmao+Dsigmaav)/sigmao);
print round(Sc*1000,2),"settlement in mm"
#partb
B=1.0; # in m
L=2.0; # in m
z=0.5+1.5; # in m
B=B+z; # in m
L=L+z; # in m
A=0.6; # given in question
#from table
kcr=0.78; # by data
Sep=kcr*Sc;
print round(Sep*1000,2),"settlement in mm"
#example 5.11
import numpy
N60=(3+7+12+12+16)/5.0;
B=[2, 2.25, 2.3]; # in m
Fd=[1.248, 1.22, 1.215];
Qoac=102000*9.81/1000;#actual Qo
Se=25; # in mm
qnet=numpy.zeros(3)
Qo=numpy.zeros(3) # in kN
print "B(m)\t Fd\t qnet(kN/m**2)\t \t Qo \n"
for i in range(0,3):
qnet[i]=10/0.08*(B[i]+0.3)**2/(B[i])**2*Fd[i]*Se/25;
Qo[i]=qnet[i]*B[i]**2;
print B[i],"\t",Fd[i]," \t ",round(qnet[i],2),"\t\t ",Qo[i],"\n"
print int(Qoac),"value of Qo in kN"
print "since Qo is 1000 kN thus B is equal to 2.3 m from the table"