Non-linearity and small strain
plasticity of lacustrine clay
Sophie Messerklinger
Swiss Federal Institute of Technology
ETH Zurich
Content:
- Motivation
- Laboratory investigation
- Sampling
- Test data analysis
- Numerical modelling
- Conclusions
Motivation:
Switzerland:
Zürich
Soft clayey soils in
densly settled areas
Geological:
Lacustrine clay
Bern
Geneva
Source: hydrological map of Switzerland
Mechanical properties:
strength and stiffness ?
Wauwil @ 27 m
Fine sands, silts, clays
Alternating layers of
peat, clays, silts and
sands
State of the art:
On mechanical properties of
Swiss Lacustrine clays:
9 Post peak softening (Bucher, 1965)
9 Undrained shear strength & Insitu determination (Heil, 1999)
9 Strain rate dependency of stiffness
9 Initial shear stiffness
Trausch Giudici, 2004)
Non-linear, elasto-plastic stiffness response
Anisotropy & stress path dependency
Triaxial test equipment:
Back pressure
unit
Triaxial cell
7
Cell pressure
unit
Axial displacements:
local LVDT‘s :
membrane (Loctite 460)
• Measurement range: ± 5 mm
70 mm
• Fixities glued on the rubber
LVDT
top
fixity
• Accuracy: εa ± 0.05 %
bottom fixity
Triaxial test equipment:
Back pressure
unit
Triaxial cell
Cell pressure
unit
Laser scanning device
7
Measurement range:
± 5 mm
Distance to the sample:
Accuracy: εr: 0.2%
load frame
20 mm
laser in
housing
lead screw
connection plate
load frame
Axial velocity: 1 mm/s
lead screw
Radial displacements
with Lasers:
Soil investigated:
Sample tube:
• Inner diameter 196 mm
• Area ratio: 4 %
• Outer cutting edge angle 11°
196 mm
4 test specimen: à ø 50 mm
Soil properties:
block sample: ø 196 mm x 250 mm
• Plastic limit
wP: 14.4 %
USCS: CL
• Plasticity Index
IP: 12.3 %
• Insitu water content w: 26.5 %
• Specific gravity
ρs : 2.74 g/mm3
• Grain content < 2 µm : 20 %
• load rate: 1 kPa/hour
• break between stress paths: 24 hours
Analysis of test results:
q = σ 1 – σ3
Θ = 97° Θ = 90°
Θ = 42°
Θ = 37°
Θ = 31°
Θ = 180°
p‘= 150 kPa
q = 113 kPa
Θ = 217°
Θ = 0°
p‘= (σ1‘+ 2σ3‘)/3
Θ = 289°
2 x Θ = 270° Θ = 278°
q [kPa]
0
0.00
150
100
p‘= 300 kPa
q = 225 kPa
Y2 = 13 kPa
50
εs [%]
0.04
0.08
0.12
Y3 = 85 kPa
εs = 0.25 %
0
0.00
0.4
εs = 0.015 %
q [kPa]
13 Drained stress path tests:
20
0.30
εs [%]
Triaxial tests:
40
εs [%]
0.60
0.90
δεs/δεv = 1.67
0.2
δεs/δεv = 0.77
0.0
0.0
0.1
0.2
0.3
εv [%]
0.4
Test data analysis:
Y2 -–von
Y2
fromp'p‘ - εv plot
250
Y2 -–von
Y2
fromq q - εs plot
150
q [kPa]
Startpunkt der state
Consolidation
Y2
Wiederbelastung
50
-50 0
100
200
-150
-250
p' [kPa]
300
400
Test data analysis:
Y2 –
- von
Y2
fromp'p‘ - εv plot
q [kPa]
& δεs [-]
q [kPa]
250
Y2 –
- von
Y2
fromq q - εs plot
Startpunkt der state
Consolidation
Y2
150
Wiederbelastung
Dehnungsvektor
Strain
vector
50
-50 0
100
200
300
-150
-250
p' [kPa]
p‘ [kPa]
& δεv [-]
400
Test data analysis:
Y2 –
- von
Y2
fromp'p‘ - εv plot
q [kPa]
& δεs [-]
q [kPa]
250
Y2 –
- von
Y2
fromq q - εs plot
Startpunkt der state
Consolidation
Y2
150
Wiederbelastung
Y3 –
- von
Y3
fromp'p‘ – εv plot
Y3
50
Y3
fromq q – εs plot
Y3 –
- von
Strain vector
-50 0
100
200
300
-150
-250
p' [kPa]
p‘ [kPa]
& δεv [-]
400
Test data analysis:
Y2 –
- von
Y2
fromp'p‘ - εv plot
q [kPa]
& δεs [-]
q [kPa]
250
Y2 –
- von
Y2
fromq q - εs plot
Startpunkt der state
Consolidation
Y2
150
Y3
50
-50 0
Wiederbelastung
Y3 –
- von
Y3
fromp'p‘ – εv plot
100
200
300
-150
-250
p' [kPa]
p‘ [kPa]
& δεv [-]
Y3 –
- von
Y3
fromq q – εs plot
400
Dehnungsvektor
Strain
vector
Test data analysis:
q [kPa]
& δεs [-]
q [kPa]
250
Y2 –
- von
Y2
fromp'p‘ - εv plot
Mcomp = 1.25 → φ‘cv= 31°
Y2 –
- von
Y2
fromq q - εs plot
Startpunkt der state
Consolidation
Y2
150
Y3
50
-50 0
Wiederbelastung
Y3 –
- von
Y3
fromp'p‘ – εv plot
100
200
300
Y3 –
- von
Y3
fromq q – εs plot
400
Versagen
- von p'
Failure
state
Strain vector
-150
-250
Mext = 0.88 → φ‘cv= 31°
p' [kPa]
p‘ [kPa]
& δεv [-]
? Plastic strain vector
Elastic component ?
Elastic strain component:
Cross – anisotropic:
⎡ 1
⎡ε a ⎤ ⎢ Ea
⎢ε ⎥ = ⎢ ν
⎣ r ⎦ ⎢ − ar
⎢⎣ Ea
Kloten clay:
−2
ν ar ⎤
Ea ⎥ ⎡σ a '⎤
⎥⎢ ⎥
1
(1 − ν rr )⎥ ⎣σ r '⎦
⎥⎦
Er
ν ar = 0.52
b
Ea = 120 MPa
b = 8 .7 E − 6
? Ea > or < Er ?
Anisotropic elasticity:
Modified Cross – anisotropic matrix: (Houlsby & Graham, 1983)
*
*
*
⎤ ⎡ε1 ⎤
⎡
'
(
1
)
σ
ν
αν
αν
−
⎡ 1⎤
*
E
⎥⎢ ⎥
⎢
⎢σ '⎥ =
*
2
*
2 *
(
1
)
ε
αν
α
ν
α
ν
−
2
*
*
⎥
⎢
⎢ 2⎥
⎢ ⎥ (1 + ν )(1 − 2ν )
2 *
2
*
⎥ ⎢⎣ε 3 ⎥⎦
⎢ αν *
⎢⎣σ 3 '⎥⎦
(
1
)
α
ν
α
ν
−
⎦
⎣
Anisotropy factor α:
Er
α =
Ea
2
Kloten clay: α = 0.6 →
ν rr
α=
ν ar
Ea ≈ 2.8 Er
Plastic Potential:
p
σa' [kPa] & δεa [-]
600
400
Y3
Associated flow rule:
Y2
Y3 9
200
Y2 ~
0
0
200
400
p
σr' [kPa] & δεr [-]
Numerical modelling:
Constitutive models:
• Modified Cam Clay Model (Roscoe & Burland, 1968)
• 3 – SKH: Three surface kinematic hardening
Model (Stallebrass 1990)
• Soft Soil Model (Plaxis)
• S_CLAY1 Model (Wheeler 1997, Näätänen et al. 1999)
Numerical simulation:
300
200
q [kPa]
100
0
0
100
200
-100
-200
-300
p' [kPa]
300
400
Numerical simulation:
300
200
q [kPa]
100
0
0
100
200
-100
-200
-300
p' [kPa]
300
400
Numerical simulation:
300
Model Parameter:
Modified Cam Clay:
Mcomp: 1.25
Mext: 0.88
κ: 0.01
λ: 0.053
G‘: 11 MPa
Γ: 1.86
200
qq [kPa]
[kPa]
100
0
0
100
200
300
400
3-SKH: additional
-100
S: 0.18
T: 0.002
κ∗: 0.003
λ∗: 0.04
ψ: 2
-200
-300
p' [kPa]
Numerical simulation:
300
Model Parameter:
Modified Cam Clay:
Mcomp: 1.25
Mext: 0.88
κ: 0.01
λ: 0.053
G‘: 11 MPa
Γ: 1.86
200
q [kPa]
100
0
0
100
100
200
300
300
400
400
3-SKH: additional
-100
S: 0.18
T: 0.002
κ∗: 0.003
λ∗: 0.04
ψ: 2
-200
-300
p' [kPa]
Numerical simulation:
300
Model Parameter:
200
[kPa]
qq [kPa]
100
0
0
100
200
-100
-200
-300
p' [kPa]
300
400
Soft Soil:
ϕ‘: 31°
c‘: 0
κ∗: 0.006
λ∗: 0.033
ν‘: 0.3
M: 1.65
Numerical simulation:
300
Model Parameter:
200
[kPa]
qq [kPa]
100
0
0
100
200
-100
-200
-300
p' [kPa]
300
400
S_CLAY1:
Mcomp: 1.25
Mext: 0.88
κ: 0.01
λ: 0.053
ν‘: 0.3
e0: 1.86
pm‘: 80 kPa
αKo: 0.42
β: 1.31
µ: 50
Numerical simulation:
300
Model Parameter:
200
[kPa]
qq [kPa]
100
0
0
100
200
-100
-200
-300
p' [kPa]
300
400
S_CLAY1:
Mcomp: 1.25
Mext: 0.88
κ: 0.01
λ: 0.053
ν‘: 0.3
e0: 1.86
pm‘: 80 kPa
αKo: 0.42
β: 1.31
µ: 50
Numerical
simulation:
q = σ1 – σ3
p‘= 300 kPa
Θ
=
90°
Θ = 97°
q = 225 kPa
ΘΘ== 42°
42°
Θ = 37°
Θ = 31°
Θ = 180°
p‘= 150 kPa
q = 113 kPa
Θ = 217°
p‘= (σ1‘+ 2σ3‘)/3
Θ = 289°
0.1 m
2 x Θ = 270° Θ = 278°
• Axisymmetric
• 2 x 15 node elements
0.025 m
Θ = 0°
p' [kPa]
300
MCC Plaxis
Soft Soil
S_CLAY1
MCC Crisp
3-SKH
Test data
250
200
εv [-]
150
0.0E+00
5.0E-03
1.0E-02
q [kPa]
300
250
200
150
100
0.E+00
3.E-03
MCC Plaxis
Soft Soil
S_CLAY1 ε [-]
s
MCC Crisp
3-SKH
6.E-03
Test data
εs [-]
6.E-03
MCC Plaxis
Soft Soil
S_CLAY1
MCC Crisp
3-SKH
Test data
4.E-03
2.E-03
εv [-]
0.E+00
0.E+00
4.E-03
8.E-03
Numerical
simulation:
q = σ1 – σ3
Θ = 97° Θ = 90°
Θ = 42°
p‘= 300 kPa
q = 225 kPa
Θ = 37°
Θ = 31°
Θ = 180°
p‘= 150 kPa
q = 113 kPa
Θ = 217°
p‘= (σ1‘+ 2σ3‘)/3
Θ = 289°
0.1 m
Θ = 278°
2 x Θ = 270° Θ
= 278°
• Axisymmetric
• 2 x 15 node elements
0.025 m
Θ = 0°
q [kPa]
MCC Plaxis
Soft Soil
S_CLAY1
MCC Crisp
3-SKH
Test data
120
20
-2.E-03
2.E-03
-80
-180
-280
εv [-]
6.E-03
120
εs [-]
-1.E-02
-5.E-03
0.E+00
-80
MCC Plaxis
Soft Soil
S_CLAY1
MCC Crisp
3-SKH
Test data
-180
-280
q [kPa]
-2.E-02
20
-2.E-03
2.E-03
0.E+00
6.E-03
MCC Plaxis
Soft Soil
S_CLAY1
MCC Crisp
3-SKH
Test data
-2.E-02
εs [-]
-1.E-02
εv [-]
Summary and Conclusions:
-Triaxial test apparatus with
• Local LVDT‘s
• Laser scan device
- Block samples of lacustrine clay
- Triaxial stress path tests
• Y2 & Y3 yield surfaces,
• Anisotropic elastic stiffness,
• Plastic potential surface found
Summary and Conclusions:
- Numerical simulation with
• MCC, S_CLAY1, 3-SKH, Soft Soil Model
- It was found that:
• Shape of Y3 is crucial for deformation prediction
• Hardening rule has to be defined by
strain characteristics
• In extension all models which use a Drucker Prager
are on the unsafe side.
Non of the models are capable of simulating the
lacustrine clay behvaiour but the S_CLAY1 model
at least gives qualitatively correct results.
Non-linearity and small strain
plasticity of lacustrine clay
Sophie Messerklinger
Swiss Federal Institute of Technology
ETH Zurich