# geotecha - A software suite for geotechncial engineering
# Copyright (C) 2018 Rohan T. Walker (rtrwalker@gmail.com)
#
# This program is free software: you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program. If not, see http://www.gnu.org/licenses/gpl.html.
"""
Deng et al (2013) and (2014), "Consolidation by vertical drains when
the discharge capacity varies with depth and time".
"""
from __future__ import print_function, division
import numpy as np
from matplotlib import pyplot as plt
import geotecha.consolidation.smear_zones as smear_zones
[docs]def dengetal2013(z, t, rw, re, A1=1, A2=0, A3=0, H=1, rs=None, ks=None,
kw0=1e10, kh=1, mv=0.1, gamw=10, ui=1):
"""Radial consolidation with depth and time dependent well resistance
Implementation of [1]_.
Features:
- Radial flow to drain (no vertical flow in soil)
- Vertical flow in drain.
- Linear depth variation of drain permeability in time.
- Exponential decrease in drain peremability with time.
- Uses approximate (like Hansbo) method to solve pore pressure in drain.
- Radially averaged pore pressure at depth and time in soil.
Drain pereability is kw = kw0 * (A1 - A2 * z / H) * exp(-A3 * t)
Parameters
----------
z : float or 1d array/list of float
Depth.
t : float or 1d array/list of float
Time.
rw : float
Drain radius.
re : float
Drain influence radius.
A1 : float, optional
Parameter controlling depth dependance of well resistance.
Default A1=1.
A2 : float, optional
Parameter controlling depth dependance of well resistance.
Default A2=0.
A3 : float, optional
Parameter controlling time dependance of well resistance.
Default A3=0.
H : float, optional
Drainage path length. Default H=1.
rs : float, optional
Drain influence radius. Default rs=None i.e. no smear zone.
ks : float, optional
Smear zone permeability. Default ks=None, i.e. no smear zone.
kw0 : float, optional
Initial well permeability. Default kw0=1e10 i.e. ideal drain.
kh : float, optional
Horizontal coefficient of permeability. Default kh=1.
mv : float, optional
Volume compressibility. Default mv=0.1.
gamw : float, optional
Unit weight of water. Default gamw=10.
ui : float, optional
Initial uniform pore water pressure. Default ui=1.
Returns
-------
por : 2d array of float
Pore pressure at depth and time. `por` is an array of size
(len(z), len(t)).
References
----------
.. [1] Deng, Yue-Bao, Kang-He Xie, and Meng-Meng Lu. 2013. 'Consolidation
by Vertical Drains When the Discharge Capacity Varies
with Depth and Time'. Computers and Geotechnics 48 (March): 1-8.
doi:10.1016/j.compgeo.2012.09.012.
"""
if A1 < 0:
raise ValueError("A1 must be greater than 0. "
"You have A1={}".format(A1))
if A2 > A1:
raise ValueError("A2 must be less than A1. "
"You have A1={}, A2={}".format(A1, A2))
if A3 < 0:
raise ValueError("A3 must be greater than or equal to 0. "
"You have A3={}".format(A3))
t = np.atleast_1d(t)
z = np.atleast_1d(z)
ch = kh / mv / gamw
n = re / rw
qw0 = kw0 * np.pi * rw**2
Th = ch * t / 4 / re**2
if ks is None or rs is None:
mu0 = smear_zones.mu_ideal(n)
else:
s = rs / rw
kap = kh / ks
mu0 = smear_zones.mu_constant(n, s, kap)
if A3 == 0 and A2 == 0:
# qw is constant
mus = mu0 + np.pi * kh / (qw0 * A1) * (2 * H * z - z ** 2)
por = ui * np.exp(-8 * Th[None, :] / mus[:, None])
return por
if A3 == 0 and A2 > 0:
# qw varies with depth
mu1 = A2 * z / H + (A1 - A2)*np.log(1 - A2 / A1 * z / H)
mu1 *= 2 * np.pi * kh * H**2 / (qw0 * A2**2)
mu1 += mu0
por = ui * np.exp(-8 * Th[None, :] / mu1[:, None])
return por
if A3 > 0 and A2 == 0:
# qw varies with time
a3 = A3 * 4 * re**2 / ch
alp0 = qw0 * A1 * mu0 / (np.pi * kh * (2*H*z - z**2))
por = (ui * ((1+alp0[:, None] * np.exp(-a3 * Th[None, :])) /
(1+alp0[:, None])) ** (8/(a3 * mu0)))
return por
if A3 > 0 and A2 > 0:
# qw varies with depth and time:
a3 = A3 * 4 * re**2 / ch
alp = mu0 * qw0 * A2**2 / (2 * np.pi * H**2 * kh)
alp /= A2* z / H + (A1 - A2) * np.log(1 - A2 / A1 * z / H)
por = (ui * ((1+alp[:, None] * np.exp(-a3 * Th[None, :])) /
(1+alp[:, None])) ** (8/(a3 * mu0)))
return por
[docs]def dengetal2014(z, t, rw, re, A3=0, H=1, rs=None, ks=None,
kw0=1e10, kh=1, mv=0.1, gamw=10, ui=1, nterms=100):
"""Radial consolidation with time dependent well resistance
An implementation of [1]_
Features:
- Single layer, soil properties constant over time.
- Instant load uniform with depth.
- Radial flow to drain (no vertical flow in soil)
- Vertical flow in drain.
- Exponential decrease in drain peremability with time.
- Uses rigorous (infintite sum ) method to solve pore pressure in drain.
- Radially averaged pore pressure at depth and time in soil.
Drain permeability is kw = kw0 * exp(-A3 * t)
Parameters
----------
z : float or 1d array/list of float
Depth.
t : float or 1d array/list of float
Time.
rw : float
Drain radius.
re : float
Drain influence radius.
A3 : float, optional
Parameter controlling time dependance of well resistance.
Default A3=0.
H : float, optional
Drainage path length. Default H=1.
rs : float, optional
Drain influence radius. Default rs=None i.e. no smear zone.
ks : float, optional
Smear zone permeability. Default ks=None, i.e. no smear zone.
kw0 : float, optional
Initial well permeability. Default kw0=1e10 i.e. ideal drain.
kh : float, optional
Horizontal coefficient of permeability. Default kh=1.
mv : float, optional
Volume compressibility. Default mv=0.1.
gamw : float, optional
Unit weight of water. Default gamw=10.
ui : float, optional
Initial uniform pore water pressure. Default ui=1.
nterms : int, optional
number of summation terms, default = 100.
Returns
-------
por : 2d array of float
Pore pressure at depth and time. `por` is an array of size
(len(z), len(t)).
References
----------
.. [1] Deng, Yue-Bao, Gan-Bin Liu, Meng-Meng Lu,
and Kang-he Xie. 'Consolidation Behavior of Soft Deposits
Considering the Variation of Prefabricated Vertical Drain
Discharge Capacity'. Computers and Geotechnics 62
(October 2014): 310-16. doi:10.1016/j.compgeo.2014.08.006.
"""
if A3 < 0:
raise ValueError("A3 must be greater than or equal to 0. "
"You have A3={}".format(A3))
t = np.atleast_1d(t)
z = np.atleast_1d(z)
ch = kh / mv / gamw
n = re / rw
qw0 = kw0 * np.pi * rw**2
Th = ch * t / 4 / re**2
if ks is None or rs is None:
mu0 = smear_zones.mu_ideal(n)
else:
s = rs / rw
kap = kh / ks
mu0 = smear_zones.mu_constant(n, s, kap)
a3 = A3 * 4 * re**2 / ch
G0 = np.pi * kh * H**2 / 4 / qw0
M = ((2 * np.arange(nterms) + 1) / 2 * np.pi)
C0 = M**2*mu0/8/G0 * n**2/ (n**2-1)
alp0 = qw0 * mu0 / (np.pi * kh * (2*H*z - z**2))
Th = Th[np.newaxis, :, np.newaxis]
M = M[np.newaxis, np.newaxis, :]
C0 = C0[np.newaxis, np.newaxis, :]
Z = (z / H)[:, np.newaxis, np.newaxis]
por = (2 / M * np.sin(M * Z) * ((1+C0 * np.exp(-a3 * Th))/
(1 + C0))**(8 / (a3 * mu0)))
por = ui * np.sum(por, axis=2)
return por
if __name__ == "__main__":
print(repr(dengetal2013(z=np.array([0.05, 0.1, 0.2, 0.5, 0.8, 1.0])*20,
t=[11025., 110250.],
rw=0.035, re=0.525,
A1=1, A2=0, A3=0,
H=20,
rs=0.175,
ks=2e-8/1.8,
kw0=1e-3,
kh=2e-8,
mv=0.2e-3,
gamw=10,
ui=1)))
# Reproduce figure 3, from Deng et al 2013
if 0:
#reproduce deng et al 2013 fig 3a.
# Note deng has kh/ks=5, but I can only get a match with kh/ks=1.8
res = {}
res['Th=0.1 A2=0.1'] = np.array([[0.777489, 0.00142817],
[0.811606, 0.0504537],
[0.836125, 0.0994905],
[0.864221, 0.198757],
[0.884466, 0.2992],
[0.896856, 0.398485],
[0.905755, 0.497774],
[0.912911, 0.598233],
[0.917452, 0.699863],
[0.920243, 0.799159],
[0.921291, 0.899625],
[0.92146, 0.996587]])
res['Th=1.0 A2=0.1'] = np.array([[0.0820304, 0.00340884],
[0.124869, 0.0500878],
[0.165101, 0.102611],
[0.23857, 0.200656],
[0.295462, 0.299888],
[0.341883, 0.399133],
[0.377835, 0.499559],
[0.405057, 0.597658],
[0.424434, 0.700439],
[0.437696, 0.799722],
[0.445723, 0.899012],
[0.449386, 0.998307]])
res['Th=0.1 A2=1.0'] = np.array([[0.778362, 0.00142715],
[0.811602, 0.0481173],
[0.837872, 0.100657],
[0.871204, 0.199917],
[0.894067, 0.300357],
[0.909074, 0.399639],
[0.919719, 0.498926],
[0.930365, 0.59938],
[0.936649, 0.69984],
[0.943803, 0.799131],
[0.948341, 0.899593],
[0.952005, 0.998888]])
res['Th=1 A2=1.0'] = np.array([[0.0829009, 0.0022396],
[0.123995, 0.0489206],
[0.168587, 0.10027],
[0.252532, 0.20064],
[0.326005, 0.301021],
[0.387258, 0.39908],
[0.438917, 0.499487],
[0.48534, 0.5999],
[0.523035, 0.699155],
[0.55724, 0.798414],
[0.586213, 0.900016],
[0.610813, 0.995782]])
z = np.linspace(0, 20, 100)
Th = np.array([0.1, 1.0])
rw=0.035
rs = 0.175
re=0.525
gamw = 10
mv=0.2e-3
kh=2e-8
ks = kh/1.8
kw0 = 1e-3
ch = kh / mv / gamw
t = Th * 4 * re**2 / ch
mu0 = smear_zones.mu_constant(re/rw, rs/rw, kh/ks)
print('mu0', mu0)
print('et', 2/mu0/re**2)
for A2 in [0.1,1.0]:
por = dengetal2013(z, t, rw=rw, re=re, A1=1, A2=A2, A3=0, H=20,
rs=rs, ks=ks, kw0=kw0, kh=kh, mv=mv, gamw=10, ui=1)
for j, t_ in enumerate(t):
uz0 = np.exp(-8*Th[j]/mu0)
plt.plot(por[:,j], z, label="m Th={0:.3g}, A2={1:.3g}".format(Th[j], A2))
plt.plot(uz0,0, marker='o', ms=14)
for key, val in res.items():
x = val[:,0]
y = val[:,1]*20
plt.plot(x,y, label=key, marker='s', linestyle='None')
leg = plt.legend(loc=3 )
leg.draggable()
plt.gca().set_xlabel('Pore pressure')
plt.gca().set_ylabel('Depth')
plt.gca().invert_yaxis()
plt.gca().set_xlim(0,1)
plt.gca().grid()
plt.title('Deng et al. 2013, figure 3a, depth variation')
plt.show()
if 0:
#reproduce deng et al 2013 fig 3b.
# Note deng has kh/ks=5, but I can only get a match with kh/ks=1.8
res = {}
res['Th=1.0 a3=0.1'] = np.array([[0.0835118, 0.00577201],
[0.127409, 0.0519481],
[0.170236, 0.0995671],
[0.245182, 0.20202],
[0.301927, 0.300144],
[0.349036, 0.399711],
[0.383298, 0.499278],
[0.411135, 0.598846],
[0.430407, 0.702742],
[0.442184, 0.800866],
[0.448608, 0.89899],
[0.452891, 0.994228]])
res['Th=1.0 a3=0.5'] = np.array([[0.0856531, 0.00577201],
[0.139186, 0.0505051],
[0.191649, 0.10101],
[0.278373, 0.199134],
[0.343683, 0.301587],
[0.391863, 0.401154],
[0.429336, 0.499278],
[0.458244, 0.601732],
[0.478587, 0.699856],
[0.491435, 0.800866],
[0.497859, 0.900433],
[0.501071, 0.992785]])
res['Th=1.0 a3=1.0'] = np.array([[0.155246, 0.049062],
[0.219486, 0.0995671],
[0.324411, 0.200577],
[0.398287, 0.301587],
[0.448608, 0.399711],
[0.489293, 0.497835],
[0.514989, 0.598846],
[0.537473, 0.702742],
[0.54818, 0.800866],
[0.557816, 0.901876],
[0.559957, 0.992785]])
z = np.linspace(0, 20, 100)
Th = np.array([0.1, 1.0])
rw=0.035
rs = 0.175
re=0.525
gamw = 10
mv=0.2e-3
kh=2e-8
ks = kh/1.8
kw0 = 1e-3
ch = kh / mv / gamw
t = Th * 4 * re**2 / ch
mu0 = smear_zones.mu_constant(re/rw, rs/rw, kh/ks)
a3 = np.array([0.1, 0.5, 1.0])
A3 = a3 * ch / re**2 /4
for a3_, A3_ in zip(a3, A3):
por = dengetal2013(z, t, rw=rw, re=re, A1=1, A2=0, A3=A3_, H=20,
rs=rs, ks=ks, kw0=kw0, kh=kh, mv=mv, gamw=10, ui=1)
for j, t_ in enumerate(t):
uz0 = np.exp(-8*Th[j]/mu0)
plt.plot(por[:,j], z, label="m Th={0:.3g}, a3={1:.3g}".format(Th[j], a3_))
plt.plot(uz0,0, marker='o', ms=14)
for key, val in res.items():
x = val[:,0]
y = val[:,1]*20
plt.plot(x,y, label=key, marker='s', linestyle='None')
leg = plt.legend(loc=3 )
leg.draggable()
plt.gca().set_xlabel('Pore pressure')
plt.gca().set_ylabel('Depth')
plt.gca().invert_yaxis()
plt.gca().set_xlim(0,1)
plt.gca().grid()
plt.title('Deng et al. 2013, figure 3b, time variation')
plt.show()
if 0:
#reproduce deng et al 2013 fig 3c.
# Note deng has kh/ks=5, but I can only get a match with kh/ks=1.8
res = {}
res['A2=1.0 a3=1.0'] = np.array([[0.0847639, 0.00168277],
[0.157725, 0.0522524],
[0.228541, 0.101377],
[0.344421, 0.203869],
[0.435622, 0.300534],
[0.504292, 0.401451],
[0.55794, 0.500886],
[0.600858, 0.60029],
[0.639485, 0.699682],
[0.670601, 0.800491],
[0.696352, 0.899846],
[0.716738, 0.994868]])
z = np.linspace(0, 20, 100)
Th = np.array([0.1, 1.0])
rw=0.035
rs = 0.175
re=0.525
gamw = 10
mv=0.2e-3
kh=2e-8
ks = kh/1.8
kw0 = 1e-3
ch = kh / mv / gamw
t = Th * 4 * re**2 / ch
mu0 = smear_zones.mu_constant(re/rw, rs/rw, kh/ks)
a3 = np.array([1.0])
A3 = a3 * ch / re**2 /4
for a3_, A3_ in zip(a3, A3):
por = dengetal2013(z, t, rw=rw, re=re, A1=1, A2=1.0, A3=A3_, H=20,
rs=rs, ks=ks, kw0=kw0, kh=kh, mv=mv, gamw=10, ui=1)
for j, t_ in enumerate(t):
uz0 = np.exp(-8*Th[j]/mu0)
plt.plot(por[:,j], z, label="m Th={0:.3g}, a3={1:.3g}".format(Th[j], a3_))
plt.plot(uz0,0, marker='o', ms=14)
for key, val in res.items():
x = val[:,0]
y = val[:,1]*20
plt.plot(x,y, label=key, marker='s', linestyle='None')
leg = plt.legend(loc=3 )
leg.draggable()
plt.gca().set_xlabel('Pore pressure')
plt.gca().set_ylabel('Depth')
plt.gca().invert_yaxis()
plt.gca().set_xlim(0,1)
plt.gca().grid()
plt.title('Deng et al. 2013, figure 3c, depth and time variation')
plt.show()
if 1:
#reproduce deng et al 2014 fig 3a.
# this is to show that the eq5 should initially have higher pore pressure
# but as depth increases it shuld have lower excess pore pressure than
# equation 8 method.
res = {}
res['digitized Eq5 Th=2.5'] = np.array([[0.0834123, 0.00510204],
[0.108057, 0.0229592],
[0.127014, 0.0433673],
[0.145972, 0.067602],
[0.165877, 0.0854592],
[0.175355, 0.0994898],
[0.224645, 0.16199],
[0.254028, 0.19898],
[0.29763, 0.265306],
[0.323223, 0.298469],
[0.377251, 0.397959],
[0.390521, 0.429847],
[0.408531, 0.469388],
[0.423697, 0.501276],
[0.43981, 0.544643],
[0.454028, 0.580357],
[0.474882, 0.646684],
[0.488152, 0.700255],
[0.500474, 0.756378],
[0.508057, 0.799745],
[0.520379, 0.90051],
[0.525118, 0.996173]])
res['digitized Eq5, Th=1.0'] = np.array([[0.363033, 0.00127551],
[0.375355, 0.00892857],
[0.384834, 0.0242347],
[0.393365, 0.0369898],
[0.406635, 0.057398],
[0.419905, 0.0739796],
[0.430332, 0.0918367],
[0.441706, 0.112245],
[0.487204, 0.19898],
[0.50237, 0.232143],
[0.534597, 0.298469],
[0.561137, 0.371173],
[0.570616, 0.399235],
[0.588626, 0.456633],
[0.603791, 0.501276],
[0.62654, 0.596939],
[0.646445, 0.683673],
[0.655924, 0.753827],
[0.664455, 0.830357],
[0.668246, 0.90051],
[0.672038, 0.993622]])
res['digitized Eq5 Th=0.25'] = np.array([[0.776303, 0.00127551],
[0.788626, 0.0446429],
[0.796209, 0.0688776],
[0.800948, 0.0956633],
[0.806635, 0.128827],
[0.817062, 0.179847],
[0.823697, 0.220663],
[0.834123, 0.299745],
[0.843602, 0.354592],
[0.847393, 0.399235],
[0.858768, 0.489796],
[0.865403, 0.543367],
[0.872038, 0.630102],
[0.878673, 0.728316],
[0.883412, 0.809949],
[0.88436, 0.908163],
[0.885308, 0.97449],
[0.886256, 0.998724]])
res['digitized Eq8, Th=0.25'] = np.array([[0.785782, 0.00382653],
[0.806635, 0.0561224],
[0.834123, 0.19898],
[0.849289, 0.302296],
[0.85782, 0.401786],
[0.865403, 0.502551],
[0.87109, 0.600765],
[0.876777, 0.69898],
[0.878673, 0.80102],
[0.880569, 0.90051],])
res['digitized Eq8, Th=1.0'] = np.array([[0.383886, 0.00127551],
[0.429384, 0.0471939],
[0.466351, 0.100765],
[0.522275, 0.19898],
[0.563981, 0.298469],
[0.592417, 0.40051],
[0.616114, 0.5],
[0.633175, 0.600765],
[0.642654, 0.69898],
[0.652133, 0.799745],
[0.65782, 0.90051],
[0.658768, 0.998724]])
res['digitized Eq8, Th=2.5'] = np.array([[0.092891, 0.00127551],
[0.136493, 0.0280612],
[0.216114, 0.0982143],
[0.299526, 0.19898],
[0.361137, 0.299745],
[0.404739, 0.40051],
[0.43981, 0.501276],
[0.461611, 0.598214],
[0.478673, 0.701531],
[0.491943, 0.799745],
[0.498578, 0.903061],
[0.500474, 0.992347]])
z = np.linspace(0, 20, 100)
Th = np.array([0.25, 1.0, 2.5])
rw=0.035
rs = 0.175
re=0.525
gamw = 10
mv=0.2e-3
kh=2e-8
ks = kh/5#1.8
kw0 = 1e-3
ch = kh / mv / gamw
t = Th * 4 * re**2 / ch
mu0 = smear_zones.mu_constant(re/rw, rs/rw, kh/ks)
a3 = np.array([1.0])
A3 = a3 * ch / re**2 /4
for a3_, A3_ in zip(a3, A3):
por5 = dengetal2013(z, t, rw=rw, re=re, A1=1, A2=0.0, A3=A3_, H=20,
rs=rs, ks=ks, kw0=kw0, kh=kh, mv=mv, gamw=10, ui=1)
por8 = dengetal2014(z, t, rw=rw, re=re, A3=A3_, H=20,
rs=rs, ks=ks, kw0=kw0, kh=kh, mv=mv, gamw=10, ui=1, nterms=1000)
for j, t_ in enumerate(t):
uz0 = np.exp(-8*Th[j]/mu0)
plt.plot(por5[:,j], z, marker='o', label="eq5 Th={0:.3g}, a3={1:.3g}".format(Th[j], a3_))
plt.plot(por8[:,j], z, marker='o', label="eq8 Th={0:.3g}, a3={1:.3g}".format(Th[j], a3_))
plt.plot(uz0,0, marker='o', ms=14)
for key, val in res.items():
x = val[:,0]
y = val[:,1]*20
plt.plot(x,y, label=key, marker='s', linestyle='None', markersize=7)
leg = plt.legend(loc=3 )
leg.draggable()
plt.gca().set_xlabel('Pore pressure')
plt.gca().set_ylabel('Depth')
plt.gca().invert_yaxis()
plt.gca().set_xlim(0,1)
plt.gca().grid()
plt.title('Deng et al. 2014, figure 3a, time variation 2 methods')
plt.show()
