Chapter12:Drilled-Shaft Foundations

Ex12.1:Pg-609

In [5]:
#example 12.1

import math
Ap=math.pi/4*1.75**2;  # area in m^2
FS=4; # factor of safety
Nq=37.75; 
L=8;
Es=50000.0;
mus=0.265;
pu=100.0;
Db=1.75; # in m
q=6*16.2+2*19.2;
phi=36*math.pi/180.0;
Fqs=1+math.tan(phi);
Fqd=1+2*math.tan(phi)*(1-math.sin(phi))**2*math.atan(L/Db);
Ir=Es/(2*(1+mus)*q*math.tan(phi));
delta=0.005*(1-phi/20*180/math.pi+25/20.0)*q/pu;
Irr=Ir/(1+Ir*delta);
Fqc=math.exp(-3.8*math.tan(phi)+(3.07*math.sin(phi)*math.log10(2*Irr))/(1+math.sin(phi)));
Qp=Ap*(q*(Nq-1)*Fqs*Fqd*Fqc);
Qpall=Qp/FS;
print round(Qpall,2),"allowed load in kN"
print "due to rounding off error there is slight change in answer"
6648.7 allowed load in kN
due to rounding off error there is slight change in answer

Ex12.2:Pg-610

In [8]:
#example 12.2
import math
Ap=math.pi/4*1.75**2; # area in m^2
q=135.6; 
w=0.83;
FS=4; # factor of safety
phi=36; # given angle
Nq=0.21*math.exp(0.17*phi);
Qp=Ap*q*(w*Nq-1); # in kN
Qpall=Qp/FS; # in kN
print round(Qpall),"allowed load in kN"
6383.0 allowed load in kN

Ex12.3:Pg-611

In [10]:
#example 12.3

import math
Ap=math.pi/4*1.5**2; # area in m^2
Db=1.5; # in m
z=3.0;
p=math.pi*1;
Li=6.0;
N60=30.0;
sigmazi=16*z;
Beta=2.0-0.15*z**0.75;
fi=Beta*sigmazi; # in kN/m^2
qp=57.5*N60; # in kN/m^2
qpr=1.27/Db*qp; # in kN/m^2
Qunet=qpr*Ap+fi*p*Li; # in kN
print round(Qunet,2),"allowed load in kN"
#part b
k1=0.315; #from table
k2=12.0/1.5/1000*100.0;
Qunet2=qpr*Ap*k1+fi*p*Li*k2; # in kN
print round(Qunet2,2),"allowed load in kN"

# the answer is slightly different in textbook due to approximation
4081.11 allowed load in kN
2013.14 allowed load in kN

Ex12.4:Pg-617

In [12]:
#example 12.4

Nc=9;
Ap=math.pi/4*1.5**2; # area in m^2
cu=105; # in kN/m^2
Qpnet=Ap*cu*Nc; # in kN
print round(Qpnet),"net ultimate bearing point capacity in kN"
#part2
alpha=0.4;
Ds=1.5; # in m 
p=math.pi*Ds;
Qs=alpha*p*(50*8+105*3); # in kN
print int(Qs),"skin resistance in kN"
#part3
FS=3;  # factor of safety
Qu=Qpnet/FS+Qs/FS; # in kN
print round(Qu,2),"working load in kN"
1670.0 net ultimate bearing point capacity in kN
1347 skin resistance in kN
1005.9 working load in kN

Ex12.5:Pg-618

In [15]:
#example 12.5

import math
cub=3000;
L=20+5; # in ft
Db=4; # in ft
Ap=math.pi/4*Db**2; # area in ft^2
alpha=0.55;
cu1=800; # in lb/ft^2
L1=7; # in ft
L2=5.5; # in ft
cu2=1200; # in lb/ft^2
p=math.pi*2.5;
k=alpha*p*(cu1*L1+cu2*L2);#f*p*deltaLi
j1=6*cub*(1+0.2*L/Db);
j2=9*cub;
qp=min(j1,j2);
Qu=k/1000+qp*Ap/1000; # in kip
print round(Qu),"allowed load in kip"
#part b
k1=0.57; #from table
k2=0.89;
Qunet2=qp*Ap*k1+k*k2; # in kip
print round(Qunet2/1000,2),"allowed load in kip"
392.0 allowed load in kip
240.3 allowed load in kip

Ex12.6:Pg-621

In [17]:
#example 12.6

import math
Qws=1005-250; # in kN
Qwp=250; # in kN
epsilon=0.65; 
L=11; # in m
Ds=1.5; # in m
Es=14000; # in kN/m^2
Ap=math.pi/4*1.5**2; # area in m^2
Ep=21e6; # in kN/m^2
Cp=0.04; # in kN/m^2
Db=1.5;
mus=0.3;
p=math.pi*1.5;
Nc=9;
qp=105*Nc; # in kN/m^2
se1=(Qwp+epsilon*Qws)*L/(Ap*Ep); # in m 
se2=Qwp*Cp/(Db*qp); # in m
Iws=2+0.35*math.sqrt(L/Ds);
se3=Qws/p/L*Ds/Es*(1-mus**2)*Iws; # in m
se=se1+se2+se3; # in m
print round(se*1000,2)," is net settlement in mm"

# the answer is slightly different in textbook due to approximation
11.46  is net settlement in mm

Ex12.7:Pg-628

In [31]:
#example 12.7

import math
import numpy
from scipy.optimize import fsolve
Ds=1.0;
Ep=22e6;
Ri=1.0;
cu=100.0;
Ip=math.pi*Ds**4/64.0;
Qc=7.34*Ds**2*Ep*Ri*(cu/Ep/Ri)**0.6;
print round(Qc,2),"bearing force in kN"
Mc=3.86*Ds**3*Ep*Ri*(cu/Ep/Ri)**0.6;
print round(Mc,2),"bearing moment in kNm"
#from figure
xoQM=0.0046*1;
xoMQ=0.0041*1;
xo=0.5*(xoQM+xoMQ);
print round(xo*1000,2),"net ground line deflection in mm"
#partb
Ip=0.049;
Qg=150.0;
Mg=200.0;
def f(T):
    return 338e-6*T**3+300.6e-6*T**2-0.00435
[x]=fsolve(f,2);
T=x;
k=[0, 0.4, 0.6, 0.8, 1.0, 1.1, 1.25];#z/T
Am=[0, 0.36, 0.52, 0.63, 0.75, 0.765, 0.75];
Bm=[1.0, 0.98, 0.95, 0.9, 0.845, 0.8, 0.73];
print "z/T\t Am\t Bm\t Mz(kN-m)\n"
Mz=numpy.zeros(7)
for i in range(0,7):
    Mz[i]=Am[i]*Qg*T+Bm[i]*Mg;
    print k[i],"\t",round(Am[i],2),"\t",round(Bm[i],2),"\t",round(Mz[i],2)

print round(1*T,2),"depth in m"
#partc
Mmax=400;
sigma=Mmax*Ds/2/Ip;
print round(sigma,2),"tensile stress in kN/m**2"
#partd
#from figure
k=8.5;
L=k*1;
print L,"length in m"

# the answer is slightly different in textbook due to approximation
100615.56 bearing force in kN
52912.27 bearing moment in kNm
4.35 net ground line deflection in mm
z/T	 Am	 Bm	 Mz(kN-m)

0 	0.0 	1.0 	200.0
0.4 	0.36 	0.98 	308.4
0.6 	0.52 	0.95 	352.35
0.8 	0.63 	0.9 	376.69
1.0 	0.75 	0.84 	403.16
1.1 	0.77 	0.8 	398.84
1.25 	0.75 	0.73 	380.16
2.08 depth in m
4081.63 tensile stress in kN/m**2
8.5 length in m

Ex12.8:Pg-634

In [35]:
#example 12.8

qu=3000; # lb/in^2
Ds=3*12; #in inch
L=15*12; # in inch
FS=3; # factor of safety
Ecore=0.36e6; # in lb/in^2
f=min(2.5*qu**0.5,0.15*qu);
Qu=math.pi*Ds*L*f/1000; # in kip
Emass=Ecore*(0.266*80-1.66); # in lb/in^2
Ec=17.9*Emass; # in lb/in^2
Ac=math.pi/4*Ds**2; # area in in^2
If=0.35;
se=Qu*L/Ac/Ec+Qu*If/Ds/Emass;
Qall=Qu/FS; # in kip
print round(Qall),"allowed load in kip"
929.0 allowed load in kip