# Chapter09:Sheet Pile Walls¶

## Ex9.1:Pg-419¶

In :
#example 9.1

import math
from scipy.optimize import fsolve
sall=30;#sigma allowed
pi=math.pi;
zbar=12.1; # in ft
L1=10.0; # in ft
L2=20.0; #in ft
Gamma=0.12; # in lb/ft^3
Gamma1=0.1294-0.0624; # in lb/ft^3
phi=40*pi/180; # angle given
Ka=(math.tan(pi/4-phi/2))**2;
Kp=(math.tan(pi/4+phi/2))**2;
s1=Gamma*L1*Ka;#sigma1 in Kip/ft
s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in Kip/ft
L3=s2/(Gamma1*(Kp-Ka)); # in ft
print round(L3,2),"is length in ft"
P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;# in Kip/ft
print round(P,2)," is force in kip/ft"
s5=Gamma*L1*Kp+Gamma1*L2*Kp+Gamma*L3*(Kp-Ka);#sigma5 in Kip/ft
print round(s5,2)," is pressure in kip/ft"
A1=s5/(Gamma1*(Kp-Ka)); # in ft^2
A2=8.0*P/(Gamma1*(Kp-Ka)) # in ft^2
A3=6.0*P*(2.0*zbar*(Gamma1*(Kp-Ka))+s5)/(Gamma1*(Kp-Ka))**2.0 # in ft^2
A4=P*(6.0*zbar*s5+4.0*P)/(Gamma1*(Kp-Ka))**2.0 # in ft^2
print "A1,A2,A3,A4 respectively is  ",round(A1,1),round(A2,2),round(A3,2),round(A4,2)
print "slight error due to rounding off error"
#part(b)
def f(x):
return x**4+41.7*x**3-270.5*x**2-13363*x-106863
[x]=fsolve(f,20);
D=1.88+x;
print round(D,2)," is value of D, in ft"
TL=L1+L2+1.3*D;
print round(TL,2)," is total length in ft"
#partc
z=math.sqrt(2*P/(Gamma1*(Kp-Ka)));#zdash
Mmax=P*(z+zbar)-1/2.0*(Gamma1*(Kp-Ka))*z**2*z/3.0;
S=Mmax*12/sall;
print round(S,2),"is section modulus in in^3/ft"

# The answers in the book are different due to approximation while here calculations are precise

1.88 is length in ft
9.96  is force in kip/ft
12.67  is pressure in kip/ft
A1,A2,A3,A4 respectively is   43.2 271.33 13708.16 110880.89
slight error due to rounding off error
21.68  is value of D, in ft
58.19  is total length in ft
70.06 is section modulus in in^3/ft


## Ex9.2:Pg-426¶

In :
#example 9.2
import math
from scipy.optimize import fsolve
sall=172.5*1000;#sigma allowed in KN/m^2
pi=math.pi;
c=47.0;  # in KN/m^2
zbar=1.78; # in m
L1=2.0; #in m
L2=3.0; # in m
Gamma=15.9;  # in KN/m^3
Gamma1=19.33-9.81;  # in KN/m^3
phi=32*pi/180;
Ka=(math.tan(pi/4-phi/2))**2;
Kp=(math.tan(pi/4+phi/2))**2;
s1=Gamma*L1*Ka;#sigma1 in KN/m^2
s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2
P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2; # in  kN/ft
print round(P,2)," is force in kN/m"
def f(x):
return 127.4*x**2-104.4*x-357.15
[x]=fsolve(f,2);
D=x;
print round(D,2)," is value of D in m"
print round(D*1.5,2),"actual D in m"
L4=D*(4*c-(Gamma*L1+Gamma1*L2)-P/D)/4/c;
print round(L4,2)," is length in m"
s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in KN/m^2
s7=4*c+(Gamma*L1+Gamma1*L2);#sigma7 in KN/m^2
z=P/s6;#zdash
Mmax=P*(z+zbar)-1/2.0*s6*z**2; # in KN-m/m
S=Mmax*12.0/sall; #  in m^3/m
print round(S,4)," is section modulus in m**3/m"
print "is slight error due to rounding off error"

# The answers in the book are different due to approximation while here calculations are precise

52.25  is force in kN/m
2.13  is value of D in m
3.2 actual D in m
1.17  is length in m
0.0072  is section modulus in m**3/m
is slight error due to rounding off error


## Ex9.3:Pg-433¶

In :
#example 9.3
import math
from scipy.optimize import fsolve

pi=math.pi;
zbar=2.23;  # in m
L1=2.0; # in m
L2=3.0; # in m
Gamma=15.9; # in KN/m^3
Gamma1=19.33-9.81; # in KN/m^3
phi=32*pi/180;
Ka=(math.tan(pi/4-phi/2))**2;
Kp=(math.tan(pi/4+phi/2))**2;
s1=Gamma*L1*Ka;#sigma1 in KN/m^2
s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2
L3=s2/(Gamma1*(Kp-Ka)); # in m
print round(L3,2),"length in m"
P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;
print round(P,2),"force in kN/m"
def f(x):
return x**3+6.99*x**2-14.55
[x]=fsolve(f,1.4);
D=L3+x;
print round(D,2),"value of D in m"
AL=1.4*D;
print round(AL,2),"actual length in m"
#partb
L4=1.4;
F=P-1/2.0*(Gamma1*(Kp-Ka)*L4**2);
print round(F,2),"anchor force in kN/m"
#partc
def f(x):
return x**2+6.682*x-14.44
[x]=fsolve(f,1.7);
z=x+2;
Mmax=-1/2.0*s1*L1*(x+2/3.0)+F*(x+1)-s1*x*x/2-1/2.0*Ka*Gamma1*x**3/3.0;
print round(Mmax,2)," is maximum moment in kN-m/m"

# The answers in the book are different due to approximation while here calculations are precise

0.66 length in m
58.38 force in kN/m
1.98 value of D in m
2.78 actual length in m
30.88 anchor force in kN/m
43.74  is maximum moment in kN-m/m


## Ex9.4:Pg-439¶

In :
#example 9.4

Gamma=15.9; # in KN/m^3
Gamma1=19.33-9.81; # in KN/m^3
GD=0.23;  # from fig. 9.16
CDL1=1.172; # from fig. 9.19
L1=2; # in m
L2=3; # in m
Dth=(L1+L2)*GD*CDL1;
print round(Dth,2),"theoritical depth in m"
Dac=1.4*Dth;
print round(Dac,2),"actual depth in m"
print "approximate it as 2 m"
#part(b)
CFL1=1.073;
GF=0.07;
Gammaa=(Gamma*L1**2+Gamma1*L2**2+2*Gamma*L1*L2)/(L1+L2)**2; # in KN/m^3
F=Gammaa*(L1+L2)**2*GF*CFL1; # in KN/m
print round(F,2),"force in kN/m"
#part(c)
GM=0.021; # from fig. 9.18
CML1=1.036; # from fig. 9.21
Mmax=Gammaa*(L1+L2)**3*GM*CML1; # in kN-m/m
print round(Mmax,2),"maximum moment in kN-m/m"

1.35 theoritical depth in m
1.89 actual depth in m
approximate it as 2 m
25.54 force in kN/m
36.99 maximum moment in kN-m/m


## Ex9.5:Pg-442¶

In :
#example 9.5
import math
import matplotlib.pyplot as plt
import numpy
Mmax=43.72; # in kN-m/m
sp=["PSA-31","PSA-23"];#sheet pile
H=[7.9,7.9] # in m
I=[4.41e-6,5.63e-6]; # in m^4/m
p=[0.00466,0.00365];
S=[10.8e-5,12.8e-5]; # in m^3/m
Md=[18.63,22.08]; # kn-m/m
Logp=numpy.zeros(2)
k=numpy.zeros(2)
print "SheetPile    I(m**4/m)    H(m)    p\t     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax \n"
for i in range(0,2):
Logp[i]=math.log10(p[i]);
k[i]=Md[i]/Mmax;
print sp[i],"\t   ",I[i],"    ",H[i]," ",p[i],"   ",round(Logp[i],2),"   ",S[i],"   ",Md[i],"\t ",round(k[i],3)

plt.plot(Logp,k);
plt.title("Ex9.5")
plt.xlabel("LogP")
plt.ylabel("Md/Mmax")
plt.show()

SheetPile    I(m**4/m)    H(m)    p	     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax

PSA-31 	    4.41e-06      7.9   0.00466     -2.33     0.000108     18.63 	  0.426
PSA-23 	    5.63e-06      7.9   0.00365     -2.44     0.000128     22.08 	  0.505


## Ex9.6:Pg-445¶

In :
#example 9.6

import math
from scipy.optimize import fsolve

pi=math.pi;
R=0.6;
L1=10.0; # in ft
L2=20.0; #in ft
Gammasat=122.4; # in lb/ft^3
l1=5; # in ft
Gamma=110.0; # in lb/ft^3
C=0.68;
L=L1+L2; # in ft
Gammaw=62.4; # in lb/ft^3
Gammaav=(Gamma*L1+Gamma1*L2)/(L1+L2); # in lb/ft^3
phi=35.0*pi/180;
Ka=(math.tan(pi/4-phi/2))**2;
sa=C*Ka*Gammaav*L;#sigmaa in lb/ft^2
sp=R*sa;#sigmap # in lb/ft^2
def f(x):
return x**2+50*x-1000
[x]=fsolve(f,15);
D=x;#in ft
print round(D,2)," is depth in ft"
R=L/D*(L-2*l1)/(2*L+D-2*l1);
print "value of R=",round(R,2)," is OK\n"
#partb
F=sa*(L-R*D); # in lb/ft
print round(F,2)," is Force in lb/ft"
#partc
Mmax=0.5*sa*L**2*((1-R*D/L)**2-(2*l1/L)*(1-R*D/L)); # in lb-ft/ft
print round(Mmax,2),"maximum moment lb-ft/ft"

# The answers in the book are different due to approximation while here calculations are precise

15.31  is depth in ft
value of R= 0.6  is OK

8821.24  is Force in lb/ft
47693.02 maximum moment lb-ft/ft


## Ex9.7:Pg-451¶

In :
#example 9.7
import math
from scipy.optimize import fsolve

pi=math.pi;
zbar=3.2; # in m
c=41; # in KN/m^2
L1=3; # in m
L2=6; # in m
Gamma=17;# in KN/m^3
Gamma1=20-9.81; # in KN/m^3
phi=35*pi/180;
Ka=(math.tan(pi/4-phi/2))**2;
Kp=(math.tan(pi/4+phi/2))**2;
s1=Gamma*L1*Ka;#sigma1 in kN/m**2
s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in kN/m**2
P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2;
print round(P,2),"Force in kN/m"
s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in kN/m**2
print round(s6,2),"pressure in kN/m**2"
def f(x):
return x**2+15*x-25.43
[x]=fsolve(f,1.6);
D=x; # in m
print round(D,1),"depth in m"
F=P-s6*D; # in kN/m
print round(F,2),"force in kN/m"
print "slight error due to rounding off"

153.36 Force in kN/m
51.86 pressure in kN/m**2
1.5 depth in m
73.61 force in kN/m
slight error due to rounding off


## Ex9.8:pg-458¶

In :
#example 9.8
import math
pi=math.pi;
Gamma=105.0; # in lb/ft^3
Cov=14.0;
B=15/12.0; # in inch
Ka=0.26;
phi=35.0*pi/180; # given angle in degree
H=37.5/12; # in inch
h=15/12.0; # in inch
t=6/12.0; # in inch
Gc=150.0;#gamma concrete in lb/ft^3
W=H*t*Gc; # in lb/ft
k=4.5;#kp*cos(delta)
Pu=1/2.0*Gamma*H**2*(k-Ka*math.cos(phi)); # in lb/ft
print round(Pu,2),"force in lb/ft"
Pus=((Cov+1)/(Cov+H/h))*Pu; # in lb/ft
print round(Pus,2),"force in lb/ft"
Be=0.227*(H+h)+B;
Pu=Pus*Be; # in lb/ft
print round(Pu,2)," is resistance of anchor plate in lb/ft"

# The answers in the book are different due to approximation while here calculations are precise

2197.94 force in lb/ft
1998.12 force in lb/ft
4482.04  is resistance of anchor plate in lb/ft