# Chapter11:Pile Foundations¶

## Ex11.1:Pg-532¶

In [3]:
#example 11.1
import math

#parta
phi=30.0; # angle in degree
pa=2000.0;
q=100*50/1000.0;
Nq=55.0;
Ap=16*16/16/12; # area in ft^2
Qp=Ap*q*Nq; # in kip
qp=0.4*pa*Nq*math.tan(phi*math.pi/180)*Ap; # in lb
print "there is change in answer because of calculation mistake in the book"

#partb
Nsigma=36;
Ap=16*16.0/12.0/12;
q=110*50.0/1000;
Qp=Ap*q*Nsigma*((1+2.0*(1-math.sin(phi*math.pi/180)))/3); # in kip
#partc
Nq=18.4;
Qp=Ap*q*Nq; # in kip

# ANSWER IN THE BOOK IS WRONG

275.0 ultimate load in lb
there is change in answer because of calculation mistake in the book


## Ex11.2:Pg-533¶

In [5]:
#example 11.2

import math
#parta
K=1.3;
f0=0;
Delta=0.8*30; # in ft
D=16.0/12; # in ft
L1=50.0;
p=4*16/12.0; # in ft
Gamma=110/1000.0; # in lb/ft^3
L=15*D; # in ft
sigma=Gamma*L; # in kip/ft^2
f20=K*sigma*math.tan(Delta*math.pi/180); # kip/ft^2
Qs=(f0+f20)/2*(p*L)+f20*p*(L1-L);
#partb
FS=4; # factor of safety
Qp=56.45/3+234.7/3+179.9/3; # in kip
Qu=Qs+Qp; # in kip
Qall=Qu/FS; # in kip
print round(Qall,2),"is allowed load in kip"

271.65 ultimate load in kip
107.17 is allowed load in kip


## Ex11.3:Pg-534¶

In [7]:
#example 11.3
import math
K=0.25;
Ap=16*16.0/12/12; # area in ft^2
phi=30*math.pi/180;
Nq=25;
q=110*50.0/1000; # in kip
sigmao=q/2; # in kip/ft^2
p=4*16.0/12; # in ft
L=50; # in ft
FS=4; # factor of safety
Qu=q*Nq*Ap+K*sigmao*math.tan(0.8*phi)*p*L; # in kip
Qall=Qu/FS; # in kip

81.5 allowed load in kip


## Ex11.4:Pg-535¶

In [9]:
#example 11.4

import math
import numpy
FS=4; # factor of safety
Ap=0.1295; # area in m^2
Nc=9;
cu2=100;
Qp=Ap*Nc*cu2; # in kN
D=[5, 10, 30]; # depth in m
avgD=[2.5, 7.5,20.0]; # average depth in m
sigma=[45.0, 110.5, 228.5]; # in KN/m^2
cu=[30, 30, 100]; # in kN/m^2
alpha=[0.6, 0.9, 0.725];
L=[5, 5, 20]; # in m
p=math.pi*0.406;
Qs=0; # in kN
cusig=numpy.zeros(3)
print round(Qp,2),"bearing capacity in kN"
print  "depth (m)\t avg Depth(m)\t avgVerticalStress(kN/m**2)\t cu(kN/m**2)\t cu/sigma\t alpha\n"
for i in range(0,3):
cusig[i]=cu[i]/sigma[i];
Qs=Qs+alpha[i]*cu[i]*L[i]*p;
print round(D[i],2),"\t \t \t",round(avgD[i],2),"\t \t",round(sigma[i],2),"\t\t\t   ",round(cu[i],2),"\t  ",round(cusig[i],2),"\t\t ",round(alpha[i],2),"\n"
print round(Qs,2),"bearing capacity in kN"
#part2
Lambda=0.136;
L=30;
fav=Lambda*(178.48+2*76.7);
Qs2=p*L*fav; # in kN
#part3
fav1=13;
fav2=31.9;
fav3=93.43;
Qs3=p*(fav1*5+fav2*5+fav3*20); # in kN
print round(Qs3,1),"bearing capacity in kN"
Qsavg=Qs/3+Qs2/3+Qs3/3; # in kN
Qu=Qp+Qsavg # in kN
Qall=Qu/FS; # in kN
print round(Qall,1),"allowed bearing capacity in kN"

116.55 bearing capacity in kN
depth (m)	 avg Depth(m)	 avgVerticalStress(kN/m**2)	 cu(kN/m**2)	 cu/sigma	 alpha

5.0 	 	 	2.5 	 	45.0 			    30.0 	   0.67 		  0.6

10.0 	 	 	7.5 	 	110.5 			    30.0 	   0.27 		  0.9

30.0 	 	 	20.0 	 	228.5 			    100.0 	   0.44 		  0.72

2136.44 bearing capacity in kN
2669.7 bearing capacity in kN
573.6 allowed bearing capacity in kN


## Ex11.5:Pg-538¶

In [17]:
#example 11.5

import numpy
D=[6, 12, 20]; # depth in m
fc=[34.34, 54.94, 70.63]; # in kN/m**2
alpha=[0.84, 0.71, 0.63];
dL=[6, 6, 8]; # in m
p=4*0.305;
Qs=0;
Q=numpy.zeros(3)
print "  depth(m)\t fc(kN/m**2)\t alpha \t \t deltaL(m)\t Q(kN)\n"
for i in range (0,3):
Q[i]=alpha[i]*fc[i]*p*dL[i];
Qs=Q[i]+Qs;
print D[i],"\t\t   ",fc[i],"\t  ",alpha[i],"\t   ",dL[i],"\t\t ",round(Q[i],2)

print round(Qs),"bearing force in kN"

  depth(m)	 fc(kN/m**2)	 alpha 	 	 deltaL(m)	 Q(kN)

6 		    34.34 	   0.84 	    6 		  211.15
12 		    54.94 	   0.71 	    6 		  285.53
20 		    70.63 	   0.63 	    8 		  434.29
931.0 bearing force in kN


## Ex5.6:Pg-545¶

In [21]:
#example 11.6

import math
L=21; # in m
Qwp=502-350; # in kN
Qws=350; # in kN
Ap=0.1045; # area in m^2
Ep=21e6; # in kN/m^2
epsilon=0.62;
Se1=(Qwp+epsilon*Qws)*L/Ap/Ep; # in m
#part2
Iwp=0.85;
qwp=152/Ap;
Es=25e3; # in kN/m^2
D=0.356; # in m
mus=0.35;
Se2=qwp*D/Es*Iwp*(1-mus**2); # in m
#part3
p=1.168;
Iws=2+0.35*math.sqrt(L/D);
Se3=Qws/p/L*D/Es*Iws*(1-mus**2); # in m
Se=Se1+Se2+Se3; # in m
print round(Se*1000,1),"settlement in mm"

19.8 settlement in mm


## Ex11.7:Pg-560¶

In [25]:
#example 11.7

Ep=207e6; # in kN/m^2
Ip=123e-6; # in m^4
nh=12000; # in  kN/m^3
#from table 11.13
xz=0.008;
Ax=2.435;
T=(Ep*Ip/nh)**0.2;
Qg1=xz*Ep*Ip/Ax/T**3;
#part2
Fy=248000;
d1=0.254;
Am=0.772;
Mzmax=Fy*Ip*2/d1; # in Kn-m
Qg2=Mzmax/Am/T; # in kN
if Qg2>Qg1 :
Qg=Qg1;
# there is slight variation in answer in textbook due to approximation

53.27 lateral load in kN


## Ex11.8:Pg-561¶

In [28]:
#example 11.8

import math
#part1
Ep=207e6; # in kN/m^2
Ip=123e-6; # in m^4
nh=12000; # in kN/m^3
#from table 11.1a
xo=0.008; # in m
L=25;
Fy=248000; # yield stress in kN/m^2
D=0.254;
Am=0.772;
Gamma=18.0; # in kN/m^3
phi=35; # in angle
Kp=(math.tan(math.pi/4+phi*math.pi/360))**2;
My=Fy*Ip*2/D; # in kN-m
Qug=140*Kp*D**3*Gamma; # in kN

#part2
Qg1=xo*(Ep*Ip)**0.6*nh**0.4/0.15/L; # in kN

if Qug>Qg1:
Qg=Qg1;

40.2 lateral load in kN


## Ex11.9:Pg-567¶

In [35]:
#example 11.9

import math
Wrh=30*12; # in kip-ft
E=0.8;
Wr=7.5; # in kip
S=1/8.0;
C=0.1;
FS=6; # in factor of safety
n=0.4; # Coefficient of restitution
Wp=12/12.0*12/12.0*80*150+550; # in lb
Wp=Wp/1000.0;
Qu=E*Wrh/(S+C)*(Wr+n**2.0*Wp)/(Wr+Wp); # in kip
Qall=Qu/FS; # in kip
print round(Qall),"allowed bearing capacity in kip"
#part2
He=30*12.0;
L=80*12.0;
Ap=12*12.0; # area in in^2
Ep=3e6/1000.0; # in kip/in^2
FS=4; # factor of safety
Qu=E*He/(S+math.sqrt(E*He*L/2.0/Ap/Ep)); # in kip
Qall2=Qu/FS; # in kip
print round(Qall2),"allowed bearing capacity in kip"

#partc
a=27;
b=1;
He=30;
FS=3; # factor of safety
Qu=a*math.sqrt(E*He)*(b-math.log10(S)); # in kip
Qall3=Qu/FS; # in kip
print round(Qall3),"allowed bearing capacity in kip"

101.0 allowed bearing capacity in kip
104.0 allowed bearing capacity in kip
84.0 allowed bearing capacity in kip


## Ex11.10:Pg-570¶

In [38]:
#example 11.10

Hp=350; # in HP
vp=0.0016; #  in m/s
Sl=0.762e-3; # in m/cycle
f=115; # in Hz
Qu=(0.746*Hp+98*vp)/(vp+Sl*f); # in kN
print round(Qu),"pile load capacity in kN"

2928.0 pile load capacity in kN


## Ex11.11:Pg-578¶

In [42]:
#example 11.11

Lg=9.92; # in ft
Bg=7.0; # in ft
n1=3.0;
Nc=8.75;
n2=4.0/1000;
Ap=14.0**2.0/12.0**2;
cup=1775.0;
a1=0.4;#alpha1
p=4*14.0/12.0;
cu1=1050.0; # in lb/ft^2
L1=15.0;
a2=0.54;#alpha2
cu2=1775.0; # in lb/ft^2
L2=45.0;
FS=4; # factor of safety
Qu=n1*n2*(9*Ap*cup+a1*p*cu1*L1+a2*p*cu2*L2); # in kip
Qu2=Lg*Bg*cup*Nc+2*(Lg+Bg)*(cu1*L1+cu2*L2); # in kip
Qall=Qu/FS; # in kip

4314.0 load in kip


## Ex11.12:Pg-583¶

In [45]:
#example 11.12

import math
z1=21/2.0; # in ft
Lg=9.0; # in ft
Bg=6.0;# in ft
Qg=500*1000.0; # in kip
Cc1=0.3;
Cc2=0.2;
Cc3=0.25;
H2=12;
H3=6;
H1=21;
e1=0.82;
e2=0.7;
e3=0.75;
s1=Qg/(Lg+z1)/(Bg+z1); #sigma1 in lb/ft^3
s2=500*1000/(9+27)/(6+27);#sigma2 in lb/ft^3
s3=500*1000/(9+36)/(6+36);#sigma3 in lb/ft^3

9.6 total settlement in inch