# Chapter 03:Shallow Foundations: Ultimate bearing capacity¶

## Ex3.1:Pg-130¶

In [2]:
#example 3.1
# From Table 3.1
Nc=17.69;
Nq=7.44;
Ny=3.64;
q=3*115;
Gamma=115.0; #lb/ft**3
c=320;
B=5.0;#ft
FS=4;#factor of safety
qu=1.3*c*Nc+q*Nq+0.4*Gamma*B*Ny
qall=qu/FS; #q allowed
Q=qall*B**2;
print Q,"is allowable gross load in lb"

# the answer is slightly different in textbook due to approximation but here answer are precise

67269.0 is allowable gross load in lb


## Ex3.2:Pg-134¶

In [24]:
#example 3.2

from scipy.optimize import fsolve
import math
Gamma=105.0;#lb/ft**3
Gammasat=118.0;#lb/ft**3
FS=3.0;
pa=2014.125;#lb/ft**2
Depth=[5,10,15,20,25]; # in ft
N60=[4,6,6,10,5]; # in  blow/ft
sigmao=[0,0,0,0,0]; # in  lb/ft^2
phi=[0,0,0,0,0] # in degree
Gammaw=62.4;
s=0;
print "depth (ft)\tN60\t  \tstress(lb/ft**2)\t phi(degrees)\n"
for i in  range(0,5):
sigmao[i]=2*Gamma+(Depth[i]-2)*(Gammasat-Gammaw);
phi[i]=math.sqrt(20*N60[i]*math.sqrt(pa/sigmao[i]))+20;
print " ",Depth[i],"\t      ",N60[i],"\t\t   ",sigmao[i],"  \t \t  \t",round(phi[i],1)," \n"
s=phi[i]+s

avgphi=s/(i+1)

print round(avgphi),"average friction angle in degrees"
#using graph get the values of other terms in terms of B and solve for B
def f(x):
return-150000/x**2+5263.9+5527.1/x+228.3*x
x=fsolve(f,4);
print round(x[0],1)," is the width in ft"

depth (ft)	N60	  	stress(lb/ft**2)	 phi(degrees)

5 	       4 		    376.8   	 	  	33.6

10 	       6 		    654.8   	 	  	34.5

15 	       6 		    932.8   	 	  	33.3

20 	       10 		    1210.8   	 	  	36.1

25 	       5 		    1488.8   	 	  	30.8

34.0 average friction angle in degrees
4.5  is the width in ft


## Ex3.3:Pg-144¶

In [29]:
#example 3.3

import math
phi=25.0; #degrees
Es=620.0; #kN/m**2
Gamma=18.0;#kN/m**2
Df=0.6;# in m
B=0.6; # in m
L=1.2; # in m
Fqc=0.347;
Nq=10.66;
Nc=20.72;
Ngamma=10.88;
mu=0.3; # Poisson's ratio
Fyd=1.0;
c=48.0;#kN/m**2
q=Gamma*(Df+B/2);
Ir=Es/(2*(1+mu)*(c+q*math.tan(phi*math.pi/180.0)));
print round(Ir,2)," is value of Ir"
Fcc=Fqc-(1-Fqc)/(Nq*math.tan(phi*math.pi/180.0));
Fcs=1+Nq/Nc*B/L;
Fqs=1+B/L*math.tan(phi*math.pi/180.0);
Fys=1-0.4*B/L;
Fcd=1+0.4*Df/B;
Fqd=1+2.0*math.tan(phi*math.pi/180.0)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;
q1=0.6*18;
Fyc=Fqc;
qu=c*Nc*Fcs*Fcd*Fcc+q1*Nq*Fqs*Fqd*Fqc+1.0/2*Gamma*Ngamma*Fys*Fyd*Fyc;
print round(qu,2),"is ultimate bearing capacity in kN/m**2"

# the answer is slightly different in textbook due to approximation but here answer are precise

4.29  is value of Ir
469.24 is ultimate bearing capacity in kN/m**2


## Ex3.4:Pg-156¶

In [34]:
#example 3.4
import math
q=110*4.0; #lb/ft**2
Nq=33.3;
phi=35.0; # in degree
Df=4.0; # in ft
B=6.0; # in ft
Gamma=110.0;#lb/ft**3
Ngamma=48.03; #lb/ft**3
B1=6-2*0.5; # in ft
Fqi=1;
Fyi=1;
Fyd=1;
Fqs=1;
Fys=1;
Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;
qu=q*Nq*Fqs*Fqd*Fqi+1/2.0*B1*Gamma*Ngamma*Fys*Fyd*Fyi;
Qult=B1*1*qu;
print round(Qult,2)," is ultimate bearing capacity in lb/ft"
print round(Qult/2000.0,2)," is ultimate bearing capacity in ton/ft"

# the answer is slightly different in textbook due to approximation but here answer are precise

151738.23  is ultimate bearing capacity in lb/ft
75.87  is ultimate bearing capacity in ton/ft


## Ex3.5:Pg-158¶

In [38]:
#example 3.5

e=0.5; # in ft
B=6; # in ft
k=e/B;
Gamma=110; # in lb/ft^3
q=440;
print "get the values of Nqe and Nye from the figure from the value of e/B"
Nye=26.8;
Nqe=33.4;
Qult=B*1*(q*Nqe+1/2.0*Gamma*B*Nye);
print round(Qult,2)," is ultimate bearing capacity in lb/ft"
print round(Qult/2000.0,2)," is ultimate bearing capacity in ton/ft"

get the values of Nqe and Nye from the figure from the value of e/B
141240.0  is ultimate bearing capacity in lb/ft
70.62  is ultimate bearing capacity in ton/ft


## Ex3.6:Pg-159¶

In [41]:
#example 3.6

Df=0.7; # in m
#from table
Nq=18.4;
Ny=22.4;
q=12.6;
phi=30; #angle in degree
L=1.5;# in m
Fyd=1;
Gamma=18; # in KN/m^3
L1=0.85*1.5; # in m
L2=0.21*1.5; # in m
B=1.5; # in m
A=1/2.0*(L1+L2)*B;
B1=A/L1; #B'
Fqs=1+B1/L1*math.tan(phi*math.pi/180);
Fys=1-0.4*B1/L1;
Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180))**2*Df/B;
Qult=A*(q*Nq*Fqs*Fqd+1/2.0*Gamma*B1*Ny*Fys*Fyd);
print round(Qult,2)," is ultimate load in kN"

# the answer is slightly different in textbook due to approximation but here answer are precise

605.45  is ultimate load in kN


## Ex3.7:Pg-161¶

In [45]:
#example 3.7

e=0.15; # in m
B=1.5; # in m
Fqs=1.0;
L=1.5;# in m
Gamma=18.0; # in KN/m^3
q=0.7*18;
#from table
Nqe=18.4;
Nye=11.58;
Fys=1+(2*e/B-0.68)*(B/L)+(0.43-3/2.0*e/B)*(B/L)**2;
Qult=B*L*(q*Nqe*Fqs+1/2.0*L*Gamma*Nye*Fys);
print round(Qult,2),"is ultimate load in kN"

803.03 is ultimate load in kN


## Ex3.8:Pg-163¶

In [48]:
#example 3.8

q=16.0;# in kN/m^2
Nqei=14.2;
Gamma=16.0 # in kN/m^3
B=1.5;# in m
Nyet=20.0;
Qult=B*(Nqei*q+1/2.0*Gamma*B*Nyet);
print round(Qult,2)," is ultimate load in kN/m"

700.8  is ultimate load in kN/m