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Geologic repository of
high-level nuclear waste
-Constitutive modeling and boundary
value problems-
Ho CHO, Nagoya Institute of Technology
Taiwan-Japan joint Workshop, May 22-23, 2016, Taiwan
Background
Disposal of high-level radioactive waste (HLRW) becomes one of
the largest environmental issues in 21st century
Nuclear waste contains radioactive substances that are dangerous to human being
Radioactivity usually
decays with long time
During the decaying, it can
generate huge amount of heat
How to deal with HLRW, safely, certainly and
permanently without care of human being ?
Moreover
Background
Technological barrier &engineered
barrier
Natural barrier
Geologic repository
Geologic repository of HLRW is
usually executed at the place below
300m in stable geologic environment.
One of the most important factors in
geological disposal of HLRW, is to understand
the thermo-hydraulic-mechanical-air (THMA)
behavior of natural barrier or host rock
Multiple barrier
THMA coupling process
mechanical
heat
hydraulic
conv
ectio
n
visco
sity
pore pressure
porosity
diffusiontherm
al stress
I. Thermal elastoplastic model
for saturated/unsaturated
geomaterials
State parameters of unsaturated geomaterials
1. Net stress
Basic Barcelona model: BBM 2. Suction
3. Void ratio
State variables:
1. Skeleton stress
2. Degree of saturation
3. Void ratio
Smooth transition between saturated and unsaturated state
Easy to incorporate with other physical states, such as density, structure, and anisotropy
Simple framework with a small number of parameters with definite physical meaning
Compared with BBM
" t
ij ij ijU
(1 )r w r aU S u S u
" ( )t n
ij ij a ij r a w ij ij r iju S u u S s
n t
ij ij a iju a ws u u
U : mean pore pressure
s : suction
Sr : degree of saturation
: net stress
: skeleton stress
Skeleton stress tensor here after:
n
ij"
ij
"
ij ij
0.0
0.2
0.4
0.6
0.8
1.0
0 20 40 60 80 100
1
2
3
4
5
6
7
c = S r
Co
effi
cien
t c
Degree of saturation Sr [%]
1: compacted boulder clay (Bishop et al, 1960)
2: compacted sale (Bishop et al, 1960)
3: breahead silt (Bishop & Donald, 1961)
4: silt (Jennings & Burland, 1962)
5: silty clay (Jennings & Burland, 1962)
6: sterrebeek silt (Zerhouni, 1991)
7: white clay (Zerhouni, 1991)
⇒Bishop’s effective stress(1959):
" = - ua+c (ua - uw) = net+c s
Sr=c
,
Triaxial compression device for unsaturated/saturated
sample with double-cell type volume meter
(Axial translation method)
① Sample
② Pedestal
③ Load cap
④ Membrane
⑤ Inner cell
⑥ Load cell
⑦ Pore water pressure
meter
⑧ Axial displacement
transducer
⑨ Dual-tube bullet
⑩ Pressure difference
meter for volumetric
stain
⑪ Axial actuator (Up)
⑫ Axial actuator (Down)
⑬ Air supplying tank
⑭ Reference water level
for inner cell
⑮ Air pressure meter
0.40
0.60
0.80
1.0
1.2
1.4
1.6
1 10 102 103 104
Initial Sr=15.7%Initial Sr=21.3%Initial Sr=25.9%Saturated
Void
rat
io e
Mean skelton stress p (kPa)
0.40
0.60
0.80
1.0
1.2
1.4
1.6
1 10 102 103 104
Initial Sr=15.7%Initial Sr=21.3%Initial Sr=25.9%saturated
Voi
d ra
tio e
Mean skelton stress p (kPa)
(a) s=73.5 kPa (b) s=294 kPa
Influence of degree of saturation on e-ln p relation
(Results of oedometer test for unsaturated soils
extracted from the work by Honda, 2000)
N.C.L.
C.S.L.
Mean skeleton stress p
Void
rat
io
e
N
G
pr
eN1e
N.C.L.SC.S.L.S
p0
e0
pN 1 pN1e
rs
r
ln p
( )rSG
( )rS
e-lnp relation considering moving up of
N.C.L. and C.S.L. due to instauration
( ) ln ( 0)r
r
p qe N S
p p
( ) ln ( )r
r
p qe S M
p pG
( , ) lnr
r
pe S
pc
2 2
2
( ) ( )( ) ln ln
ln 2
r rr
r
N S S M pe N S
M p
G
2 2
2
0 0 0 0
( ) ( ) 1 1 1ln ln 0
(1 ) ln 2 1 1
ps er rv
p p p p
N S Sp Mf
p C e M e C e C C
r rG
0 ( ) ( )=( )ln2p
v r r
M
fd N S S
p
G
2 2
2
0 0 0
1 1 1ln ln 0
1 1
p
vs e
p p p
p Mf
p M e C e C C
r r
Associated flow rule:
Hardening parameter
p
ij
ij
fd
0
( ) ,1
ee sd a b
e p
r rr r r
s rd QdSr r
s r
r r
N NQ
S S
Evolution of density:
Evolution of saturation :
( )p
ij ijkl ijkl ij ijkl
kl
fd E E d AE
/p
ijkl ijqr mnkl
mn qr
f fE E E D
σ
0
1 1
1p
r
QA dS
C e D
p
p
mnkl
mn kl
h f fD E
C
p
mm mm
fh
r
On condition that elastic strain caused bychange of temperature obeys Hooke’s theory,equivalent stress is then defined as;
Thermal effect described with the concept of equivalent stress (Zhang & Zhang, 2009)
Change of temperature may also cause elastic and plastic strain
0mp
0mp
mp
mp
T0T
0mp
0mp
mp
mp
Temperature change
(pm0=const.)
Stress change
(T=const.)equivalent
stress
Concept of equivalent stress is
used to consider the influence of
temperature and is expressed as,pm0: real stress
K: bulk modulus
s: thermal expansion coefficient
T: current temperature
T0: reference temperature (15C)
Only one parameter, thermal
expansion coefficient , T , is
added to the constitutive model!
1 1( ) ln[( OCR) / ]e N Np pr
New void ratio difference,
considering the influence
of temperature, is given.
0 0
0
( ) ln[( OCR) / ]
( ) ,1
e N
e
N
e s
p
d a be
p
r
rr r
rr
p
Associate flow rule:
( ) ( ); ( )
( ) ( );
s rrr r r r rs r
r r
s rs r r r s r
r r
s r
N NN S N S S N N S
S S
N NN S N Q S S Q
S S
d QdS
r
r
p
ij
ij
fd
r
rS s
rSand
2 2
2
0 0 0
1 1 1ln ln 0
1 1
psv
p p p
ep Mf
p M e C e C C
r
r
Elastoplastic model for saturated/unsaturated geomaterials
(i) Evolution of is dependent on:
Saturation , Overconsolidation and
temperature
(ii) Evolution of rs is dependent on:
Saturation
Overconsolidation termSaturation term
: residual saturation 100% saturation
Compared with Cam-clay model for saturated soils, only saturation is added
Equivalent stress
er
Mean skeleton stress p
Dd
evia
tory
str
ess
q
p0 pN 1 pN1e p
(p* , q*)
(p , q)
Normal yielding surface
Subloading yielding surface
Extension of subloading concept (Hashiguchi and Ueno,
1977) to unsaturated soil in skeleton stress space
(1) Initial drying curve110 0 12
( ) tan (( 1) / )dc sc ss s r
r r r rS S S S e e
(2) Main drying curve
1112( ) tan (( 1) / )dc sc ss s r
r r r rS S S S e e
2212( ) tan (( 1) / )wc sc ss s r
r r r rS S S S e e
Skeleton curve
Scanning curve
1 1 1
0 1s s sk k k
1 1 3
1(1 )s
s s
rk k c
r
2
1
/ 0
/ 0
dsr
ds
Parameters:
Sd: drying AEV; Sw: wetting AEV
Srr: residual Sr; Sr
s; saturated Sr
c1 c2 c3: scanning parameters
ks0 gradient at r =0
rdS ds - 1sk
MCC
Degree of saturation Sr
Suc
tion
s
Srr Sr
s Sr
s 0=1
Drying A.E.V.
Primary drying curve
Secondary drying curve
Wetting curve
W.E.V.
Main wetting curve
Initial drying curve
Main drying curve
Current 1
Moisture characteristic curve (MCC)
Tangential stiffness of suction-
saturation relation
(3) Main wetting curve
Parameters involved in constitutive model
Compression index 0.050 Swelling index 0.010 Critical state parameter Μ 1.0 Void ratio N (p’=98 kPa on N.C.L.) 1.14 Poisson’s ratio ν 0.30
Parameter of overconsolidation a 5.00 Parameter of suction b 5.50 Parameter of overconsolidation 1.0 Void ratio Nr (p’=98 kPa on N.C.L.S.) 1.28
Parameters involved in MCC
Saturated degrees of saturation s
rS 0.82
Residual degrees of saturation r
rS 0.64
Parameter corresponding to drying AEV (kPa) Sd 550
Parameter corresponding to wetting AEV (kPa) Sw 320
Initial stiffness of scanning curve (kPa)e
spk 200000
Parameter of shape function c1 0.008
Parameter of shape function c2 0.013
Parameter of shape function c3 10.0
0
200
400
600
800
1000
1200
60 65 70 75 80 85 90 95 100
Secondary drying curve
Wetting curve
Primary drying curve
Scanning curve
Suction (kPa)
Degree of saturation (%)
Starting point (0.715,450)
Simulated moisture characteristics curve
of unsaturated fictional silt
Test and simulated moisture characteristics
curve of unsaturated rockfill (test data from the
work by Kohgo et al, 2007)
Verification of proposed model
Verification of the model by drained and exhausted triaxial
compression tests for a rockfill with submergence process (test
data from the work by Kohgo et al, 2007)
Verification of proposed model
Verification of the model by drained and exhausted
triaxial compression tests for a rockfill with submergence
process (test data from the work by Kohgo et al, 2007)
Verification of proposed model
Verification of proposed model
1. Consolidation tests
2. Triaxial compression tests
under different temperature .(Test data from Uchaipichat and Khalili , 2009)
Material parameters of silt
Saturated degrees of saturation 1.00
Residual degrees of saturation 0.30
Parameter corresponding to drying AEV(kPa) Sd
18.0
Parameter corresponding to wetting AEV(kPa) Sw
5.0
Initial stiffness of scanning curve (kPa) 90.0
Parameter of shape function c1 0.009
Parameter of shape function c2 0.013
Parameter of shape function c3 3.0
s
rSr
rS
e
spk
Parameters of silt in MCC
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1 10 100 1000
Drying curve
Wetting curve
Suction s (kPa)
Deg
ree
of
satu
rati
on
Sr
Line: simulation
Dot: experiment
MCC
Compression index 0.09
Swelling index 0.006
Critical state parameter Μ 2.45
Void ratio N (p’=98 kPa on N.C.L.) 0.638
Poisson’s ratio ν 0.30
Parameter of overconsolidation a 60.00
Parameter of suction b 0.00
Parameter of overconsolidation 2.0
Void ratio Nr (p’=98 kPa on N.C.L.S.) 0.665
0.48
0.50
0.52
0.54
0.56
0.58
0.60
10 100 1000
T=25o
T=40o
T=60o
Vo
id r
atio
e
Mean effective stress '' (kPa)
Line: simulation
Dot: experiment
0.48
0.50
0.52
0.54
0.56
0.58
0.60
10 100 1000
T=25o
T=40o
T=60o
Void
rat
io e
Mean effective stress '' (kPa)
Line: simulation
Dot: experiment
0.48
0.50
0.52
0.54
0.56
0.58
0.60
10 100 1000
T=25o
T=40o
T=60o
Vo
id r
atio
e
Mean effective stress '' (kPa)
Line: simulation
Dot: experiment
0.50
0.52
0.54
0.56
0.58
10 100 1000
T=25o
T=40o
T=60o
Vo
id r
atio
e
Mean effective stress '' (kPa)
Line: simulation
Dot: experiment
s=0 kPa s=10 kPa
s=50 kPa s=100 kPa
Consolidation tests under different temperature
Verification of proposed model (Test data from Uchaipichat and Khalili , 2009)
0.0
1.0
2.0
3.0
4.00 5 10 15 20 25
Axial strain (%)
Volu
metr
ic s
train
(%
)
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
s=300kPa, T=25o
s=300kPa, T=40o
s=300kPa, T=60o
Calculated result
0.00
50.0
100
150
200
250
0 5 10 15 20 25
Dev
iato
r st
ress
(k
Pa)
Axial strain (%)
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
s=300kPa, T=25o
s=300kPa, T=40o
s=300kPa, T=60o
Experimental result
(a)
(b)
0.00
50.0
100
150
200
250
0 5 10 15 20 25
Dev
iato
r st
ress
(k
Pa)
Axial strain (%)
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
s=300kPa, T=25o
s=300kPa, T=40o
s=300kPa, T=60o
Calculated result
0.0
1.0
2.0
3.0
4.00 5 10 15 20 25
Vo
lum
etr
ic s
train
(%
)
Axial strain (%)
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
s=300kPa, T=25o
s=300kPa, T=40o
s=300kPa, T=60o
Experimental result
Conventional triaxial compression tests under different constant suction
and temperature conditions (initial mean net stress= 50 kPa)
(a) experimental results (Uchaipichat and Khalili, 2009)
(b) calculated results
s=50kPa, T=25o
s=50kPa, T=40o
s=50kPa, T=60o
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
Conventional triaxial compression tests under different constant suction
and temperature conditions (initial mean net stress= 100 kPa)
0.00
50.0
100
150
200
250
300
350
0 5 10 15 20 25
Axial strain (%)
Dev
iato
r st
ress
(k
Pa)
0.0
2.0
4.0
6.0
8.00 5 10 15 20 25
Volu
metr
ic s
train
(%
)
Axial strain (%)
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
s=300kPa, T=25o
s=300kPa, T=40o
s=300kPa, T=60o
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
s=300kPa, T=25o
s=300kPa, T=40o
s=300kPa, T=60o
Experimental result
Calculated result
(a)
(b)
0
50
100
150
200
250
300
350
0 5 10 15 20 25
Dev
iato
r st
ress
(k
Pa)
Axial strain (%)
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
s=300kPa, T=25o
s=300kPa, T=40o
s=300kPa, T=60o
Calculated result
0.0
2.0
4.0
6.0
8.00 5 10 15 20 25
Vo
lum
etr
ic s
train
(%
)
Axial strain (%)
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
s=300kPa, T=25o
s=300kPa, T=40o
s=300kPa, T=60o
Experimental result
(a) experimental results (Uchaipichat and Khalili, 2009)
(b) calculated results
s=100kPa, T=25o
s=100kPa, T=40o
s=100kPa, T=60o
s=300kPa, T=25o
s=300kPa, T=40o
s=300kPa, T=60o
-2
0
2
4
6
8
10
12 -6
-5
-4
-3
-2
-1
0
10 1 2 3 4 5
dev
iato
r st
ress
a-
r (
MP
a) volu
metric strain
v (%)
axial strain a (%)
○ 20℃△ 40℃□ 60℃× 80℃
-2
0
2
4
6
8
10
12 -6
-5
-4
-3
-2
-1
0
10 1 2 3 4 5
dev
iato
r st
ress
a-
r (
MP
a) volu
metric strain
v (%)
axial strain a (%)
○ 20℃△ 40℃□ 60℃× 80℃
-2
0
2
4
6
8
10
12 -6
-5
-4
-3
-2
-1
0
10 1 2 3 4 5
dev
iato
r st
ress
a-
r (
MP
a) volu
metric strain
v (%)
axial strain a (%)
○ 20℃△ 40℃□ 60℃× 80℃
3=0.49MPa 3=0.98MPa 3=1.47MPa
Stress-strain-dilatancy relation
Triaxial compression test results
Soft sedimentary rock (saturated state)
(Test data from Zhang, Nishimura and Kageyama, 2013 & 2014)
soft sedimentary rock (saturated state, 3=0.49MPa)
-2
0
2
4
6
8
10
12 -6
-5
-4
-3
-2
-1
0
1
0 1 2 3 4 5
dev
iato
r st
ress
a-
r (M
Pa) v
olu
metric strain
v (%)
axial strain a (%)
○ 20℃△ 40℃□ 60℃× 80℃
-2
0
2
4
6
8
10
12 -6
-5
-4
-3
-2
-1
0
10 1 2 3 4 5
dev
iato
r st
ress
a-
r (
MP
a) volu
metric strain
v (%)
axial strain a (%)
○ 20℃△ 40℃□ 60℃× 80℃
Tests Simulation
Verification of proposed model
(Test data from Zhang, Nishimura and Kageyama, 2013 & 2014)
There is a basic and intrinsic relationship among the state
variables, void ratio e, degree of saturation Sr and gravimetric
water content w ,
r
s
S ew
G
s r rG dw edS S de
Which means that even if dw=0 (constant water content), if
void ratio e changed, degree of saturation will also change!
A simple finite deformation scheme for MCC
Simulated behavior of soil at the process of drying-isotropic loading to full
saturation-suction reduction
0
50
100
150
200
250
300
350
1 10 100 1000 10000
Suct
ion (
kP
a)
Mean net stress (kPa)
A
B C
D
(a)
Dry
ing
Net stress loading
Wet
tin
g0.2
0.4
0.6
0.8
1.0
1.2
1 10 100 1000 10000
Mean net stress (kPa)
Void
rat
io
A
B
C. D
(b)
NCL---saturated soil
Dry
ing
Net stress loading
Saturation point
0.2
0.4
0.6
0.8
1.0
1 10 100
Suction (kPa)
Vo
id r
atio
(c)A
B
CD
Drying
Net
str
ess
load
ing
Wetting
0.2
0.4
0.6
0.8
1.0
1.2
1 10 100
Suction (kPa)
Deg
ree
of
satu
rati
on A
D
B
C
(e)
Drying
Net
str
ess
load
ing
Wetting
0.2
0.4
0.6
0.8
1.0
1.2
1 10 100 1000 10000
Vo
id r
atio
Mean skeleton stress (kPa)
(f)A
B
C.D
Saturation point
Drying
Net stress loading
NCL---saturated soil
0.2
0.4
0.6
0.8
1.01 10 100 1000 10000
Deg
ree
of s
atur
atio
n
Mean skeleton stress (kPa)
A
B
C. D
(g)
Dry
ing
Net stress loading
Saturation point
0.2
0.4
0.6
0.8
1.0
1.2
1 10 100 1000 10000
Deg
ree
of
satu
rati
on
Mean net stress (kPa)
A
B
C
D
(d)
Dry
ing
Net stress loading
Saturation point
0.00
0.01
0.02
0.03
0.04
0.05
0.06
1 10 100 1000 10000
void
rat
io o
f po
ssib
le c
olla
pse
Mean net stress (kPa)
(h)
Simulated behavior of soil at the process of drying-isotropic
loading-wetting collapse
0.4
0.6
0.8
1.0
1.2
1 10 100 1000
Mean net stress (kPa)
Void
rat
io
(a)
A
B
C
D
Dry
ing
Net stress loading
NCL---saturated soil
Wet
tin
g
0.4
0.6
0.8
1.0
1.2
1 10 100 1000
Mean skeleton stress (kPa)
Void
rat
io
(b)
A
B
C
D
Drying
Net stress loading
Wet
tin
g
0.4
0.6
0.8
1.0
1 10 100
Suction (kPa)
Vo
id r
atio
(c)A
B
CD
Drying
Net
str
ess
load
ing
Wetting collapse
0.2
0.4
0.6
0.8
1.0
1.2
1 10 100 1000
Deg
ree
of
satu
rati
on
Mean net stress (kPa)
(d)
A
B
C
D
Dry
ing
Net stress loading
Wet
tin
g
0.2
0.4
0.6
0.8
1.0
1.2
1 10 100
Deg
ree
of
satu
rati
on
Suction (kPa)
(e)
A
B
C
D
Drying
Lo
adin
g
Wetting
Wetting
0.2
0.4
0.6
0.8
1.01 10 100 1000
Deg
ree
of
satu
rati
on
Mean skeleton stress (kPa)
(f)
A
B
C
D
Net stress loading
Wet
tin
g
II. Field equations for THMA
analysis
1. Stress2. Saturation3. Water head4. Air pressure5. Temperature
( )0
s ss
i
i
u
t x
rr
( )0
w ww
i
i
u
t x
rr
( )0
a aa
i
i
u
t x
rr
Conservative equations for all phases
0ij
i
j
bx
r
1. Equilibrium equation
2. Continuum equations (solid + liquid)
2 10
s wwwii r
w w
i i r
p Skp
n x x K S
3. Continuum equations (solid + air)
2 10
1
s aaaii r
a a
i i r
p Skp
n x x K S
Boundary condition
1. Stress2. Displacement3. Water head4. Air pressure5. Temperature
Discretiza
tion
in sp
ace
an
d tim
e d
om
ain
4. Energy conservation equation
2
( ) (1 )( )w w a a
r i r i t
i i i i
T T T Tc nS c v n S c v k Q
t x x x xr r r
Field
equ
atio
ns
Soil-water-
air coupling
Neglecting energy from plastic deformation
Static loading
Initial condition
Discretization of governing equations
Equilibrium equation
| | | | | | |
w w a a w w a a
N t t Sat dE t t Sat dE t t t t T N t t Sat dE t Sat dE tK u I p I p F F T I p I p
| |
1
| |[ ]
mT w a w
r t v sr dE sr dE t t i idEN t t t t t t
i
w a
sr dE sr dE t WT N t tt
S K u A F p F p p
A F p F p K T
| |
1
| |
(1 )
[ ]
mT w a a
r t v sr dE t t sr dE i idEN t t t t t t
i
w a
sr dE t sr dE AT N t tt
S K u F p B F p p
F p B F p K T
Continuum equations(solid + liquid)
Displacement, temperature:4 node
Pressures of Water and air: center of gravity
2D isoparametric element
Continuum equations(solid + air)
t t t
N t N tC T K T f
Discretization in space
(Galerkin)
N t t N t N t N t t N tT T t T t T T
1t t t t
N t t N t N tC t K T F K T t T
Tt
V
C c N N dVr
( )
(1 S )( )
T
Tt w
t r
i i iV V
Ta a
r i
iV
N N NK k dV nS c N dV
x x x
Nn c v N dV
x
r
r
w
iv
T Tt
V S
f N EdV q N dS
Discretization in time
(Newmark )
En
ergy
con
serva
tion
equ
atio
n
Discretization of governing equations
Overall stiffness matrix
| |
1
|
|
1
||
0
(1 ) [ ]r
w w aN t tSat Sat
mT w w w w
r v sr sr dE t t i idE t t
iaT wdEt t mv sr sr
a
i idE t t
i
w w
t t ST N t t at d
uK H H
S K A F F h h
pS K F B F
p
F H hF T
| |
| |
|
|
( )
( )
w a a
E t Sat dE t
w w a
sr dE sr dE WT N tt t
w w a
sr dE t sr dE t
t
AT N t t
H p
A K T
K
F h F p
F h F TB p
| | |1t t t t
N t t N t N tC t K T f K T tT
Displacement, water head, air pressure and
temperature as unknown variables
Discretization of governing equations
III. Slope failure in unsaturated
Sirasu ground
THMA coupling analysis in FE-FD
scheme (Temperature=Const.)
Triaxial compression test of silty clay under undrained
and unvented condition (Test data from Oka et al., 2011)
MCC
One element mesh and boundary conditions
30= 450kPa
pa0= 250kPa
S0 =10kPa, 30kPa
50kPa, 100kPa
0= 0.5%/min
Temp. = 20oC
Parameters of MCC of silty clay
Material parameters of silty clay
0
20
40
60
80
100
70 75 80 85 90 95 100
Suct
ion (
kP
a)
Degree of saturation (%)
.
Comparison between test and calculation
Test
Calcu
latio
n
Pore water pressureAir pressureStress-strain relation
Test data from Oka et al., 2011
Model test (Kitamura et al., 2007)
180cm180cm
90cm
Test box
Rain falling device
90cm
Water tank
Water injection (Width:
60cm, Depth: 180cm)
Flow meter
10cm
Detail of rain fall tank
Sirasu
Drainage
Water h
ead
Water tank
Test conditions
Physical properties of Sirasu
Position of Water injectionCase 1
Case 2
Case 3
Drained
condition is the
same as test
FEM mesh and loading condition
Location of tensiometers and
water injection pattern
Drainage conditions
No.1
No.2
No.3
No.4
No.5
No.6
No.7
No.8
No.9
No.10
No.11
No.12
No.13
No.14
No.15
Range of water injection (Case 3)
Range of water injection (Case 1)
Ran
ge
of
wat
er
inje
ctio
n (
Ca
se 2
)
40 cm
25
cm
60 cm
45o
80
cm
140cm
10
cm
10
cm
10
cm
10
cm
10
cm
20cm 20cm 20cm
Case1 Case2 Case3
Top drained drained drained
Bottom undrained undrained undrained
Back side undrained undrained undrained
Slope face drained drained drained
Case1 Case2 Case3
Density of soil
particle (g/cm3)2.45 2.40 2.45
Water content
in nature (%) 25.6 23.3 23.1
Void ratio 1.57 1.47 1.57
Total density of
soil (g/cm3)1.20 1.20 1.17
Material parameters
Element behavior of Sirasu
10-7
10-6
10-5
10-4
10-3
10-7
10-6
10-5
10-4
10-3
0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Water permeability
Air permeability
Wat
er p
erm
eabi
lity
(m/s
) Air perm
eability (m/s)
Degree of saturation (%)
Relationship between permeability of
water & air with saturation
Gravitation stress field+5kPa
-10
-5
0
5
10 -2
-1
0
1
20 2 4 6 8 10
q (suction=8kPa)
q (suction=0kPa)
v (suction=8kPa)
v (suction=0kPa)
1
3 (
kP
a)
v (%
)
a (%)
3=10kPa
Calculation conditions
Compression index 0.055
Swelling index 0.010
Critical state parameter Μ 1.0
Void ratio N (p’=98 kPa on N.C.L.) 1.55
Poisson’s ratio ν 0.30
Parameter of overconsolidation a 2.00
Parameter of suction b 0.50
Parameter of overconsolidation 1.0
Void ratio Nr (p’=98 kPa on N.C.L.S.) 1.57
Case 1 Case 2 Case 3
Saturated degrees of saturation 0.87 0.89 0.95
Residual degrees of saturation 0.20 0.25 0.20
Parameter corresponding to drying AEV(kPa) Sd
12.0 15.0 12.0
Parameter corresponding to wetting AEV(kPa) Sw
0.07 0.10 0.17
Initial stiffness of scanning curve (kPa) 90.0 20.0 50.0
Parameter of shape function c1 0.30 0.30 0.33
Parameter of shape function c2 0.60 0.40 0.18
Parameter of shape function c3 30.0 50.0 5.0
s
rSr
rS
e
spk
Parameters of MCC
Soil water characteristic curve (MCC)
Case 1 Case 2 Case 3
0
5
10
15
20
25
30
20 30 40 50 60 70 80 90 100
ExperimentSimulation
Degree of saturation (%)
Su
ctio
n (
kP
a)
0
5
10
15
20
25
30
20 30 40 50 60 70 80 90 100
ExperimentSimulation
Degree of saturation (%)
Suct
ion (
kP
a)
0
5
10
15
20
25
30
20 30 40 50 60 70 80 90 100
ExperimentSimulation
Su
ctio
n (
kP
a)
Degree of saturation (%)
Initial condition
Suction: 8kPa
Saturation: 38%
Case 1 (底面注水)
calculated experimental
No.6
No.7
No.8
No.6
No.7
No.8
No.10No.9
No.10
No.9
-12
-10
-8
-6
-4
-2
0
0 50 100 150 200 250 300
Elapsed Time (min)
Neg
ativ
e P
WP
(kP
a)
-12
-10
-8
-6
-4
-2
0
0 50 100 150 200 250 300
Elapsed Time (min)
Neg
ativ
e P
WP
(kP
a)
calculated experimental
No.11
No.12
No.13
No.11
No.12
No.13
No.15No.14
No.15
No.14
-12
-10
-8
-6
-4
-2
0
0 50 100 150 200 250 300
Elapsed Time (min)
Neg
ativ
e P
WP
(kP
a)
calculated experimental
No.1
No.2
No.3
No.1
No.2
No.3
No.4
No.5No.4
No.5
No.1
No.2
No.3
No.4
No.5
No.6
No.7
No.8
No.9
No.10
No.11
No.12
No.13
No.14
No.15
45o
Comparison of EPWP (Case 1)
Position where
water is injected
Calculation is on the whole coincident
with test. The failure time is also the same
in calculation and test. In slope surface,
however, the lose of suction in calculation
is much slower than that at test
Case 2 (背面注水)
calculated experimental
No.6
No.7
No.8
No.6
No.7
No.8
No.10No.9
No.10
No.9
-12
-10
-8
-6
-4
-2
0
0 50 100 150 200 250 300
Elapsed Time (min)
Neg
ativ
e P
WP
(kP
a)
calculated experimental
No.11
No.12
No.13
No.11
No.12
No.13
No.15No.14
No.15
No.14
-12
-10
-8
-6
-4
-2
0
0 50 100 150 200 250 300
Elapsed Time (min)
Neg
ativ
e P
WP
(kP
a)
calculated experimental
No.1
No.2
No.3
No.1
No.2
No.3
No.4
No.5No.4
No.5
-12
-10
-8
-6
-4
-2
0
0 50 100 150 200 250 300
Elapsed Time (min)
Neg
ativ
e P
WP
(kP
a)
No.1
No.2
No.3
No.4
No.5
No.6
No.7
No.8
No.9
No.10
No.11
No.12
No.13
No.14
No.15
45o
Comparison of EPWP (Case 2)
Position where
water is injected
Calculation is on the whole coincident
with test. In slope surface, however, the
lose of suction in calculation is much
slower than that at test
Case 3 (上面注水)
calculated experimental
No.11
No.12
No.13
No.11
No.12
No.13
No.15No.14
No.15
No.14
calculated experimental
No.6
No.7
No.8
No.6
No.7
No.8
No.10No.9
No.10
No.9
calculated experimental
No.1
No.2
No.3
No.1
No.2
No.3
No.4
No.5No.4
No.5
-10
-8
-6
-4
-2
0
0 20 40 60 80 100 120Neg
ativ
e P
WP
(kP
a)
Elapsed Time (min)
-10
-8
-6
-4
-2
0
0 20 40 60 80 100 120
Elapsed Time (min)
Neg
ativ
e P
WP
(kP
a)
-10
-8
-6
-4
-2
0
0 20 40 60 80 100 120
Elapsed Time (min)
Neg
ativ
e P
WP
(kP
a)
No.1
No.2
No.3
No.4
No.5
No.6
No.7
No.8
No.9
No.10
No.11
No.12
No.13
No.14
No.15
45o
Comparison of EPWP (Case 3)
Case 3 上面降雨Position where
water is injected
Calculation is on the whole coincident with test. The
failure time is also the same in calculation and test.
Change of degree of saturation
Case 1 Case 2
T=100 min
T=200 min
T=300 min
T=100 min
T=200 min
T=300 min
Case 3
T=13.3min
T=53.3min
T=93.3min
T=26.6min
T=66.6min
T=106.6min
T=40.0min
T=80.0min
T=120.0min
Change of degree of saturation
T=300 min T=300 min T=120 min
T=300 min T=300 min T=120 min
22 pI ( : 2nd invariant of deviatoric plastic strain)
The calculations are totally the same as the tests, that is, Case 1 and
Case 3 are much easier to fail than Case 2
Shear strain2
pI
Distribution of deviatoric plastic strain
Distribution of displacement vector
Case 1 Case 2 Case 3
IV. THMA coupling analysis
of heating test for saturated
geomaterials (pa=constant, Sr=1.0)
Layout of test site
Stage Start:
day
End:
day
Duration
: day
Activity
1 0 90 90 Heating
(325W/heater)
2 91 339 248 Heating
(975W/heater)
3 340 518 180 Cooling
Multi stages of the heating tests
Heating test for saturated rock (Gens et al.,2007)
Heater (1m)
Separation:0.8m
3D FEM mesh
Initial condition:
Water pressure: 1 MPa
Mean stress: 5 MPa
Temperature: 15oC
Boundary conditions:
Up, back and right planes:
undrained and heat insulated
Other planes:
Drained and fixed temperature (15oC)
3D THMA analysis conditions
x
zy
5
10
15
20
-1 -0.5 0 0.5 1 1.5 2
Model
Model
Experiment
Experiment
σ1-σ3(MPa)
ε3 (%) ε
1(%)
10-13
10-12
10-11
10-10
10-9
10-8
10-7
10-6
0 20 40 60 80 100 120
Per
mea
bil
ity
(m
/s)
Temperature (o
C)
Element simulation of triaxial
compression test (38 MPa)
Relation between permeability
and temperature
Thermal expansion coefficient αt(K-1) 8.0e-06
Thermal expansion coefficient αw(K-1) 2.1e-04
Thermal conductivity kt (kJ m-1 K-1Min-1) 0.18
Specific heat C (kJ Mg-1 K-1) 840.
Heat transfer coefficient of air boundary ac
((kJ m-2 K-1 Min-1)236.
Specific heat of water Cw (kJ Mg-1 K-1) 4184.
Physical parameters
Material parameters
Material parameters and physical parameters
Compression index 0.0026
Swelling index 0.00064
Critical state parameter Μ 1.0
Void ratio N (p’=98 kPa on N.C.L.) 0.85
Poisson’s ratio ν 0.30
Parameter of overconsolidation a 2.00
Parameter of suction b 0.50
Parameter of overconsolidation 1.0
Void ratio Nr (p’=98 kPa on N.C.L.S.) 0.85
Comparison between test and calculation (temperature)
Change of temperature at different positions
D: distant of measuring point
to heater
Z: z coordinate of measuring
point
y
xz
0
20
40
60
80
100
120
0 100 200 300 400 500
Model
Experiment
Tem
peratu
re (
o
C)
Time (days)
X
Y
D=0m
Z=8m
heate
r
0
10
20
30
40
50
60
0 100 200 300 400 500
Model
Experiment
Time (days) T
em
peratu
re (
o
C)
X
Y
heate
r
D=0.872m
Z=8.000m
0
10
20
30
40
50
60
0 100 200 300 400 500
Model
Experiment
Time (days)
Tem
peratu
re (
o
C)
X
Y
heate
rD=1.068m
Z=8.000m
0
10
20
30
40
50
60
0 100 200 300 400 500
Model
Experiment
Time (days)
Tem
per
atu
re (
o
C)
X
Y
heaterD=1.200m
Z=6.580m
0
10
20
30
40
50
60
0 100 200 300 400 500
Model
Experiment
Time (days) T
emp
era
ture
(o
C)
heater
X
Y
D=2.200m
Z=5.040m
ヒータ
ヒータ
ヒータ
ヒータ
ヒータ
-2
-1
0
1
2
3
4
0 100 200 300 400 500
Model
Experiment
Ex
cess
Po
re W
ate
r P
ress
ure
(M
Pa
)
Time (days)
D=0.4m
Z=8m
heater
X
Y
-2
-1
0
1
2
3
0 100 200 300 400 500
Model
Experiment
Ex
cess
Po
re W
ate
r P
ress
ure
(M
Pa
)Time (days)
heater
D=0.00m
Z=5.04m
X
Y
-2
-1
0
1
2
3
0 100 200 300 400 500
Model
Experiment
Ex
cess
Po
re W
ate
r P
ress
ure
(M
Pa
)Time (days)
-2
-1
0
1
2
3
0 100 200 300 400 500
Model
Experiment
Time (days)
Exce
ss P
ore
Wate
r P
ress
ure
(M
Pa)
D=0.83m
Z=6.89m
heater
X
Y
-2
-1
0
1
2
3
0 100 200 300 400 500
Model
Experiment
Exce
ss P
ore
Wate
r P
ress
ure
(M
Pa)
Time (days)
heaterD=0.21m
Z=7.225m
X
Y
heater
X
Y
D=0.36m
Z=8m
y
xz
ヒータ
ヒータ
ヒータ
ヒータ
ヒータ
Comparison between test and calculation (EPWP)
D: distant of measuring point
to heater
Z: z coordinate of measuring
point
Change of EPWP at different positions
-0.04
-0.03
-0.02
-0.01
0
0.01
100 200 300 400 500
Model
Experiment
Time (days)S
train
(%
)
X
Y
D=1.8m
Z=8m
heater
-0.04
-0.03
-0.02
-0.01
0
0.01
100 200 300 400 500
Model
Experiment
Time (days)
Str
ain
(%
)
X
Y
D=1.15m
Z=8m
heater
-0.04
-0.03
-0.02
-0.01
0
0.01
100 200 300 400 500
Str
ain
(%
)Time (days)
X
Y
D=0.0m
Z=8m
heater
Model
Experiment
heater
-0.04
-0.03
-0.02
-0.01
0
0.01
100 200 300 400 500
Str
ain
(%
)
Time (days)X
Y
D=0.18m
Z=8m
Model
Experiment
-0.04
-0.03
-0.02
-0.01
0
0.01
100 200 300 400 500
Model
Experiment
Str
ain
(%
)
Time (days)
X
Y
D=2.14m
Z=8m
heater
y
xz
ヒータ
ヒータ
ヒータ
ヒータ
ヒータ
Comparison between test and calculation (Strain)
D: distant of measuring point
to heater
Z: z coordinate of measuring
point
Change of Deviatory strain at different positions
0
20
40
60
80
100
120
0 1 2 3 4 5
At the end of 1st phase
At the end of 2nd
phase
initial temperature
Tem
pera
ture
(o
C)
Distance From Heater (m)
Distribution of temperature (only calculation)
Distribution of temperature
Temperature only changed
in limited area (D<5m)
Reason : Small thermal
conductivity of rock
V. THMA coupling analysis of
heating test for saturated/
unsaturated geomaterial (pa=0, Sr≤1.0)
Heating tests (Munoz, 2006)
Layout of heating test
Boring hole
(f 0.3m; Depth: 7m)
Rock
Bentonite buffer
Heater
(f75mm)
Stage Start (day) End (day) Time (day) Action
1 0 982 982 hydration phase
2 982 1522 540 heating phase
3 1523 1800 278 Cooling phase
Testing stages
Analytical conditions
heater
bentonite
5m
2m
20 m40
m
FEM mesh and boundary condition
MCC
Initial conditions
Unsaturated bentonite: s=136MPa; Sr=70%
Saturated soft rock: water head=40m
102
103
104
105
106
40 50 60 70 80 90 100
Degree of saturation (%)
Suct
ion (
MP
a)
Nature rock
Bentonite
Line: calculation
Dot: experiment
20m40m
Dが大きいほど飽和に至るのが早い
約400日に渡って孔内は全部飽和した
Result: 1st stage
50 days 150 days 250 days 600 days
Distribution of saturation
Change of saturation in bentonite
and neighboring rock
D:distance
from heater
heater
bentonite
70
80
90
100
0 200 400 600 800 1000
D=0.05m
D=0.10m
D=0.15m
Sat
ura
tion (
%)
Time (days)
90
90
100
0 200 400 600 800 1000
D=2.00m
D=0.65m
D=0.20m
Sat
ura
tio
n (
%)
Time (days)
0.0
20.0
40.0
60.0
80.0
100.0
120.0
600 800 1000 1200 1400 1600
D=0.05 mD=0.65 mD=5.00 m
Tem
per
atu
re (
o
C)
Time (days)
Line: modelling
Dot : experimental data
0
20
40
60
80
100
0 1 2 3 4 5
990 days
1030 days
1200 days
Tem
per
atu
re (
o
C)
Distance from heater (m)
Line: modelling
Dot: experimental data
D=0.65m
D=1.65m
-100
0.00
100
200
300
400
500
800 1000 1200 1400 1600 1800
Ex
cess
ive
po
re w
ater
pre
ssu
re (
kP
a)
Time (days)
Line: modelling
Dot : experimental data
D:Distance
from heater
Distribution of temperatureChange of temperature
Change EPWP
Quick change of temperature can be
clearly simulated
No prominent change of temperature at
the place 5m away from the heat source
Change of EPWP due to temperature can
be clearly simulated
Result: 2nd and 3rd stages
VI. THMA coupling analysis
for geologic repository of
HLRW(pa=0, Sr≤1.0)
520m
210m
300m
20
m
7.6
0 [MPa]
20m520m
300m
x y
z
7.6
x
y
2D meshInitial stress field
THMA coupling analysis for geologic repository of HLRW
FEM mesh
Thermal, boundary and initial conditions• Boundary condition:fixed in side and bottom surfaces
• Hydraulic condition:only drained at top surface
• Thermal condition:all the surfaces far from the
heating source are at constant temperature of 15ºC
• Material: heating source is elastic
• Initial saturation: soft rock is saturated but heating
source and bentonite are unsaturated
3D meshInitial stress field
0
[MPa]
Time (year)
Q(t
)/Q
0
Case1(10~310 years)
Case2(100~410 years)
Background→ Constitutive model→FEM→ Conclusions
32
0 1 1( ) / (1 )tt
Q t Q e e
The relation of decay rate and time
Heat released from nuclear waste is given as follow
proposed by Thunvik and Braester (1991) :
The initial heat Q0 is given 28.28 kJ/m3 /min.
Heat emission as input in FEM
0
5
10
15
20
25
30
0 50 100 150 200 250 300
CASE 1 (10-310 years)
CASE 2 (100-400 years)
Heat
Q (
kJ/m
3/m
in)
Time (year)
100%
FEM---Setting of heat source
Background→ Constitutive model→FEM→ Conclusions
FEM---Setting of heat source
The schematic diagram of
the field project
The heat emission as input in FEM
1. Volume of HLRW only occupies about 1.6%
of total tunnel volume, correspondingly the heat
emission as input in FEM is given 1.6%.
2. Heat emission decreases greatly between 10
years and 100 years from the decay rate-time
relation. Two cases are considered starting from
10 years and 100 years after emplacing into
repository respectively. The calculated times are
300 years in two cases.0
0.1
0.2
0.3
0.4
0.5
0 50 100 150 200 250 300
CASE 1 (10-310 years)
CASE 2 (100-400years)
Hea
t Q
(k
J/m
3/m
in)
Time (year)
1.6%
3.7m
4.5
m
30m
Waste repository
R=0.4m
7m
4.5
m
3m
MCC and initial condition
MCC(Munoz et al, 2006)
0.1
1
10
100
1000
40 50 60 70 80 90 100
RockBentonite
Sucti
on(M
Pa)
Degree of saturation (%)
10-14
10-13
10-12
10-11
10-10
10-9
10-8
40 50 60 70 80 90 100
BentoniteRock
Perm
eab
ilit
y o
f w
ate
r(m
/s)
Degree of saturation(%)
Rock Bentonite
Saturation [%] 100 86
Suction [MPa] 0 62
Total head [m] 520 -6945
Initial condition
Material parameters for constitutive model
Bentonite Rock
Parameter of overconsolidation a 5.000 2.000
Parameter of suction b 0.5 0.5
Parameter of overconsolidation β 1.00 1.10
Void ratio (p’=98 kPa on N.C.L) N 1.040 0.450
Void ratio (p’=98 kPa on N.C.L.S) Nr 1.060 0.530
Thermal expansion coefficient (1/K) αT -1.0×10-6 -2.5×10-5
Thermal conductivity (kJ m−1 K−1 Min−1) KtS 0.060 0.200
Specific heat (kJ Mg-1 K-1) CS 723 840
Thermal expansion coefficient of water (1/K) αT 2.1×10-4 2.1×10-4
Bentonite Rock
Compression index λ 0.050 0.010
Swelling index κ 0.010 0.001
Void ratio (p’=98 kPa on N.C.L) e0 1.040 0.500
Poisson’s ratio ν 0.300 0.120
Saturated degrees of saturation 𝑆𝑟𝑠 1.00 1.00
Residual degrees of saturation 𝑆𝑟𝑟 0.40 0.40
Parameter corresponding to drying AEV (kPa) 𝑆𝑑 11000 21000
Parameter corresponding to wetting AEV (kPa) 𝑆𝑤 800 1000
Initial stiffness of scanning curve (kPa) 𝑘𝑠𝑝𝑒 25000 90000
Parameter of shape function c1 0.000001 0.00003
Parameter of shape function c2 0.000005 0.00006
Parameter of shape function c3 30.0 50.0
Parameters for
MCC
Temperature change 2 (3D)
Case1
300year15year
0 [℃] 100
Case1
300year15year
0
50
100
150
200
0 50 100 150 200 250 300
0m(ベントナイト)2m10m200m
tem
per
ature
(℃
)
time (year)
熱源からの水平距離Case1
0
50
100
150
200
0 50 100 150 200 250 300
0m2m10m200m
tem
per
ature
(℃
)
time (year)
.
Case1 熱源からの水平距離
Saturated condition(without bentonite)
Unsaturated condition
(with bentonite)
Bentonite
Distance form heat source
Distance form heat source
2D analysis
-10
0
10
20
30
40
50
60
70
0 10 20 30 40 50
0m(ベントナイト)2m10m200m
su
ctio
n(M
Pa)
time (year)
Case1 熱源からの水平距離
-4
-3
-2
-1
0
0 10 20 30 40 50
su
ctio
n(M
Pa)
time (year)
Change of saturation Change of suction
-10
0
10
20
30
40
50
60
70
0 50 100 150 200 250 300
0m(ベントナイト)2m10m200m
su
ctio
n(M
Pa)
time (year)
.
熱源からの水平距離Case1
80
85
90
95
100
0 50 100 150 200 250 300
0m(ベントナイト)2m10m200m
deg
ree
of
satu
rati
on(%
)
time (year)
.
熱源からの水平距離
Case1
80
85
90
95
100
0 10 20 30 40 50
0m(ベントナイト)2m10m200m
deg
ree
of
satu
rati
on(%
)
time (year)
Case1熱源からの水平距離
Changes of saturation and suction
2D analysis
Bentonite → absorption
Rock → change from drainage to absorption
• Initial suction is -3MPa (300m beneath water level)
• Bentonite is much less in 3D than in 2D →water supplying from rock to bentonite is much less in 3D
3D analysis
Bentonite → absorption
Rock → always saturated
Bentonite
Distance form heat source
Bentonite
Distance form heat source
Distance form heat source
Distance form heat source
Bentonite
Bentonite
3D analysis
Bentonite→absorption
Changes of volumetric and deviatory strains (2D)
Saturated condition (without bentonite)
Unsaturated condition (with bentonite)
-25
-20
-15
-10
-5
0
5
0 50 100 150 200 250 300
0m(ベントナイト)2m10m
volu
met
ric
stra
in
v (
%)
time (year)
. Case1熱源からの水平距離
-0.10
-0.05
0.00
0.05
0.10
0 50 100 150 200 250 300
0m2m10m
volu
met
ric
stra
in
v (
%)
time (year)
Case1 熱源からの水平距離
0
1
2
3
0 50 100 150 200 250 300
2m10m
time (year)
Case1 熱源からの水平距離
0
0.005
0.01
0.015
0.02
0 50 100 150 200 250 300
2m10m
time (year)
Case1 熱源からの水平距離
Volumetric strain Deviatory strain
Volumetric expansion of bentonite is far larger than that of rock
mainly coursed by huge decrease of suction
Volumetric expansion near heating source
Thermal effect is confirmed
Distance form heat source
Bentonite
Distance form heat source
Distance form heat source
Distance form heat source
Substantial shear strain happened in rock when bentonite is involved !
Comparison of temperature between Case 1 and 2 (2D)
0
50
100
150
200
250
300
0 50 100 150 200 250 300
0m(ベントナイト)2m10m200m
tem
per
ature
(℃
)
time (year)
.
熱源からの水平距離Case1
0
50
100
150
200
250
300
0 50 100 150 200 250 300
0m(ベントナイト)2m
10m
200m
tem
per
ature
(℃
)
time (year)
.
Case2 熱源からの水平距離
Case1
Case2
300year15year
300year15year
0 [ºC] 180
Distribution of temperatureChange of temperature
Heat emission input in FEM
0
0.1
0.2
0.3
0.4
0.5
0 50 100 150 200 250 300
CASE 1 (10-310 years)
CASE 2 (100-400years)
Hea
t Q
(k
J/m
3/m
in)
Time (year)
1.6%
Distance form heat source
Distance form heat source
Bentonite
Bentonite
Change of saturation (2D)
80
85
90
95
100
0 50 100 150 200 250 300
0m(ベントナイト)2m10m200m
deg
ree
of
satu
rati
on(%
)
time (year)
.
熱源からの水平距離
Case1
Bentonite saturated 70years later
80
85
90
95
100
0 50 100 150 200 250 300
0m(ベントナイト)2m10m200m
deg
ree
of
satu
rati
on(%
)
time (year)
.
Case2
熱源からの水平距離
Bentonite saturated 80years later
Reason why saturated quickly when heating is strong
-7000
-6000
-5000
-4000
-3000
-2000
-1000
0
1000
0 20 40 60 80 100
Case1Case2
wat
er h
ead (
m)
time (year)
.
Change of total water head
strong heating results in quick increase of excess pore water pressure
quick decrease of suction (s=ua-uw)
Saturated quickly
Comparison of saturation between Case 1 and 2 (2D)
Distance form heat source
Distance form heat source
Bentonite
Bentonite
-30
-25
-20
-15
-10
-5
0
5
0 50 100 150 200 250 300
Case1Case2
volu
met
ric
stra
in
v (
%)
time (year)
.
In Case 2 (relative weak heating source), expansive of bentonite is larger, why?
Change of volumetric
strain of bentonite (2D)
s rd QdS r
Trace in absorption process of bentonite e-lnp
Taking more time to reach saturated state →Sr developed slower→also developed slowersr
sr
lnpMean skelton stress p
Void
rati
o e
Δe
Due to slower develop of , extra Δe happened in Case 2
Comparison of volumetric strain between Case 1 and 2 (2D)
sr
-30
-25
-20
-15
-10
-5
0
5
0 50 100 150 200 250 300
Case1Case2
volu
met
ric
stra
in
v (
%)
time (year)
.
0
0.1
0.2
0.3
0.4
0.5
0.6
0 50 100 150 200 250 300
2m10m
time (year)
熱源からの水平距離Case1
0
0.1
0.2
0.3
0.4
0.5
0.6
0 50 100 150 200 250 300
2m10m
time (year)
熱源からの水平距離Case2
• Deviatory strain in surrounding rock due to huge swelling of bentonite
• In case 2, volumetric strain in bentonite is much larger→larger deviatory stain
Comparison of deviatory strain between Case 1 and 2 (2D)
Change of volumetric
strain of bentonite (2D)
Change of deviatory strain of rock (2D)
dev
iato
ry s
tra
in (
%)
Distance form heat source Distance form heat source
A thermo-elastoplastic model for unsaturated/saturatedgeomaterials has been introduced based on theconcept of equivalent stress and subloading surface,which can not only describe properly thethermodynamic behavior, but also overconsolidatedbehavior of soft sedimentary rocks.
Verification by laboratory tests and field observationproved the validity of the proposed numerical method
Field equations in THMA FE-FD analysis has beenintroduced to treat thermo-soil-water-air couplingproblem. As a application, a field problem of geologicalrepository of high-level nuclear waste is simulated inunsaturated/saturated condition.
In 2D simulation, the calculated results shows that thetemperature has a great influence on THMA couplingbehavior of natural soft rock.
CONCLUSIONS
Thank you for your attention!
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