# Chapter 10 : EARTH AND ROCKFILL DAM¶

## Example 10.1 pg : 502¶

In [1]:
import math
from numpy import array,float64,round

#Given
K = 5.E-4;             				#coefficient of permeability of soil
Bt = 6.;               				#width of top of dam
wb = 146.;             				#width of base of dam
H = 20.;               				#heigth of dam
hw = 2.;               				#heigth of water in reservior
hs1 = 4.;              				#slope on upstream side
hs2 = 3.;              				#slope on downstream side
df = 30.;              				#length of drainage filter

x = wb-df-72+72*0.3;
y = 18.;
s = (x**2+y**2)**0.5-x;

x = array([0, 10, 20, 30, 40, 50, 60, 65.6],dtype=float64);
y = (4.849*x+5.879)**0.5;
y = round(y*1000)/1000;

print "x                       y";
for i  in range(8):
print "%.2f          %.2f"%(x[i],y[i]);
sf = K*s*10000;
sf = round(sf*1000)/1000;
print "Seepage flow per unit length of dam = %.2fD-6 cumecs/metre length of dam."%(sf);

x                       y
0.00          2.42
10.00          7.37
20.00          10.14
30.00          12.30
40.00          14.14
50.00          15.76
60.00          17.23
65.60          18.00
Seepage flow per unit length of dam = 12.12D-6 cumecs/metre length of dam.


## Example 10.2 pg : 502¶

In [3]:

#Given
K = 3.E-3;            				#coefficient of permeability
nd = 25.;              				#number of potential drops
nf = 4.;              				#number of flow channels
lf = 40.;             				#filter length
H = 52.;              				#heigth of dam
fb = 2.;              				#free board

# Calculations
q = K*(H-fb)*nf/(nd*100);

# Results
print "Discharge per meter length of dam = %.5f cumec/metre length."%(q);

Discharge per meter length of dam = 0.00024 cumec/metre length.


## Example 10.3 pg : 503¶

In [5]:
import math

#Given
x = 4.;
#Given scale
An = 14.4;       				#area of N recmath.tangle
At = 6.4;        				#area of T recmath.tangle
Au = 4.9;        				#area of U recmath.tangle
L = 12.6;        				#length of arc;
gamma_m = 19.;    				#unit weigth of soil
gamma_w = 9.81;  				#unit weigth of water
fi = 26.;         				#effective angle(degree)
co = 19.5;       				#cohesion value

# Calculations
#consider 1m length of dam
SumN = An*x**2*gamma_m;
SumT = At*x**2*gamma_m;
SumU = Au*x**2*gamma_w;
Le = x*L;
F = round(F*100)/100;

# Results
print "Factor of safety for slope = %.2f."%(F);

Factor of safety for slope = 1.41.


## Example 10.4 pg : 503¶

In [6]:
import math

#check section for:
#Stability of d/s slope against steady seepage
#Sloughing of u/s slope against sudden drawdown
#Stability of the foundation against shear
#Seepage through body of dam

#Given
#Dimensions
H = 20.;            				#Heigth of dam
Bt = 6.;            				#top width of dam
s1 = 4.;            				#u/s slope
s2 = 3.;            				#d/s slope
fb = 2.;            				#free board
#Properties of materials of dam
gamma_d = 17.27;       				#dry density
wc = 0.15;             				#optimum water content
gamma_s = 21.19;       				#saturated density
gamma_w = 9.81;        				#unit weigth of water
wavg = 19.62;          				#average unit weigth under seepage
theta = 26.;            				#average angle of internal friction(degree)
co = 19.13;            				#average cohesion
K = 5.E-4;              				#coefficient of permeability
#properties of foundation materials
gamma_f = 17.27;      				#average unit weigth
cof = 47.87;          				#average cohesion
fi = 8.;               				#average angle internal friction
t = 6.;                				#thickness of clay
FOSp = 1.5;           				#permissible factor of safety of slope
PS = 8.E-6;            				#permissible seepage
#(a) Stability of d/s slope against steady seepage
An = 302.4;            				#area of N diagram
At = 91.2;             				#area  of T diagram
Au = 98.4;             				#area of U diagram
Le = 60.;               				#length of arc
SumN = An*gamma_s;
SumT = At*gamma_s;
SumU = Au*gamma_w;
F = round(F*100)/100;
print "Parta:"
print "Factor of safety for slope = %.2f."%(F);
print "Safe";

#(b) Sloughing of u/s slope against sudden drawdown
h1 = 15.;
b = 80.;
sav = P/b;
smax = 2*sav;
Ne = (gamma_s-gamma_w)*b*H/2;
fs = R/P;
fs = round(fs*100)/100;
print "Partb:"
print "Factor of safety w.r.t average shear = %.2f."%(fs);
print "Safe";
FS = sr/smax;
FS = round(FS*100)/100;
print "Factor of safety w.r.t maximum shear = %.2f."%(FS);
print "Safe";

#(c) Stability of the foundation against shear
h1 = 26.;
h2 = 6.;
gamma_m = (wavg*(h1-h2)+gamma_f*h2)/h1;
sav = P/b;
smax = 2*sav;
as1 = (s1+s2)/2;
fs = as1/sav;
fs = round(fs*100)/100;
print "Partc:"
print "Factor of safety w.r.t overall shear = %.2f."%(fs);
print "Safe";

gamma_av = (wavg*0.6*H+gamma_f*h2)/(0.6*H)+h2;
fs = s/smax;
fs = round(fs*100)/100;
print "Factor of safety w.r.t overall shear = %.2f."%(fs);
print "Unsafe";

#(d) Seepage through body of dam
s = 2.;     				#measured
q = K*s*100000/100;
print "Partd:"
print " Seepage through body of dam = %.2fD-5 cumecs/m length of dam"%(q);

Parta:
Factor of safety for slope = 1149.17.
Safe
Partb:
Factor of safety w.r.t average shear = 2.16.
Safe
Factor of safety w.r.t maximum shear = 1.24.
Safe
Partc:
Factor of safety w.r.t overall shear = 1.18.
Safe
Factor of safety w.r.t overall shear = 0.69.
Unsafe
Partd:
Seepage through body of dam = 1.00D-5 cumecs/m length of dam


## Example 10.5 pg : 507¶

In [1]:
%matplotlib inline
import math
from matplotlib.pylab import plot,show
from numpy import zeros

#Given
Zb = 1.2;       				#thickness of blanket
Zf = 8;         				#dismath.tance of blanket from foundation
kb = 0.06;      				#coefficient of permeability of blanket material
kf = 72;        				#coefficient of permeability of foundation soil
Hw = 10;         				#heigth of water in reservior
Xd = 40;

a = (kb/(kf*Zb*Zf))**0.5;
Xo = 1.414/a;

#we vary value of x
x = [0, 25, 50, 75, 100, 125, 151.8, 300]
Xr = zeros(8)
ho = zeros(8)
r = zeros(8)

for i in range(8):
e = math.exp(2*a*x[i]);
Xr[i] = (e-1)/(a*(e+1));
ho[i] = Xr[i]*Hw/(Xr[i]+Xd);
r[i] = Xr[i]*100/(Xr[i]+Xd);

print "x                     Xr              ho         reduction qpercent";
for i in range(8):
print "%.2f        %.2f        %.2f        %.2f"%(x[i],Xr[i],ho[i],r[i]);

#graph is plotted between r and x.
#after around 130m length there is only slight increase in head dissipated(ho)
plot(x,r)
show()
L = 130;
print "Thickness of blanket = %.2f m"%(Zb);
print "Length of blanket = %i m."%(L);

The history saving thread hit an unexpected error (OperationalError('disk I/O error',)).History will not be written to the database.
x                     Xr              ho         reduction qpercent
0.00        0.00        0.00        0.00
25.00        24.56        3.80        38.04
50.00        46.67        5.38        53.85
75.00        64.78        6.18        61.83
100.00        78.50        6.62        66.24
125.00        88.28        6.88        68.82
151.80        95.35        7.04        70.45
300.00        106.53        7.27        72.70

Thickness of blanket = 1.20 m
Length of blanket = 130 m.