# Chapter05:Shallow Foundations: Allowable Bearing Capacity and Settlement¶

## Ex5.1:Pg-212¶

In [1]:
#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"

262.4 change in pressure in lb/ft**2
262.0 change in pressure in lb/ft**2


## Ex5.2:Pg-215¶

In [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"

5.25 change in pressure in kN/m**2
21.0 total change in pressure in kN/m**2


## Ex5.3:Pg-217¶

In [4]:
#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"

122.5  is pressure change in kN/m**2
109.03 is change in stress in kN/m**2
44.76 is total stress increase in A in kN/m**2


## Ex5.4:Pg-228¶

In [9]:
#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"

10400.0   is modulus of elasticity in kN/m**2
12.4 is settlement in mm


## Ex5.5:Pg-234¶

In [12]:
#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"

1.07 settlement in inches


## Ex5.6:238¶

In [18]:
#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"

Layer No.	 deltaz (in)	 Es(lb/in**2)	 z to the middle of the layer (in)  Iz at the middle of the layer  Iz/delta(z)

1 	 	  48 		    750 		  24  					   0.275 		   0.018
2 	 	  48 		    1250 		  72  					   0.425 		   0.016
3 	 	  96 		    1250 		  144  					   0.417 		   0.032
4 	 	  48 		    1000 		  216  					   0.292 		   0.014
5 	 	  144 		    2000 		  312  					   0.125 		   0.009
2.54 settlement in inches


## Ex5.7:Pg-244¶

In [27]:
#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"

11.79 settlement in mm


## Ex5.8:Pg-245¶

In [29]:
#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"

14.71 settlement in mm


## Ex5.9:Pg-251¶

In [4]:
#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"

5523.31  is bearing capacity in lb/ft**2
0.019 settelement in m
0.74 settlement in inches


## Ex5.10:Pg-256¶

In [12]:
#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"

m1	 z(m)	 n1	 Ic	 Dsigma

2.0 	  2.0 	  4.0 	  0.19 	  28.5
2.0 	  3.25 	  6.5 	  0.09 	  12.75
2.0 	  4.5 	  9.0 	  0.04 	  6.75
46.45 settlement in mm
36.23 settlement in mm


## Ex5.11:Pg-262¶

In [24]:
#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"

B(m)	 Fd	 qnet(kN/m**2)	 	 Qo

2 	1.248  	  206.31 		  825.24

2.25 	1.22  	  195.88 		  991.63125

2.3 	1.215  	  194.08 		  1026.675

1000 value of Qo in kN
since Qo is 1000 kN thus B is equal to 2.3 m from the table