# Uniform Nonlinear Constitutive Model and Parameters for Clay in Different Consolidation Conditions Based on Regression Method

Authors: Tao Cheng; Keqin Yan; Huazhi Zhang; Xianfeng Luo; Shengfang Li
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

The nonlinear constitutive relations of clay are investigated considering different initial conditions. Highly compressible clay is selected as the test sample. Two groups of tri-axial compression tests are performed, respectively, afterK
0 consolidation and isotropic consolidation. On the basis of the framework of E
~
v model, a uniform nonlinear constitutive model is proposed by fitting the test data. With the average slope of the unloading-reloading curve selected as the unloading modulus, the unloading function is constructed as the loading-unloading criterion. Moreover, a comparison between the experimental stress-strain curves and the results predicted by the constitutive model is made. It is shown that the prediction is reasonable, which can reflect the stress-strain behavior of the soil under the K
0 consolidation and isotropic consolidation conditions. The maximum relative error of the two series of curves is not remarkable, less than 6%.


---

## Body

## 1. Introduction

In geotechnical engineering, the nonlinearity nature of soil not only is influenced by a stress path but also has a close relationship with a load and a loading method. Therefore, it is a key to choose a suitable mechanical model of the medium material in geotechnical calculation.Duncan [1, 2] presented an incremental nonlinear elastic model based on this relationship curve, which was generally called the Duncan-Chang model. A hyperbolic function was adopted to fit the tri-axial experiment stress-strain curve of soil [3]. The hyperbolic model has been a widely used nonlinear constitutive relationship for clay because of its simple model structure and accessible parameters [4]. This constitutive model was based on the data of tri-axial shear tests [5] under regular loading path. Thus, it had a simple framework and can raise the calculation efficiency.Subsequently, some researchers attempted to improve the Duncan-Chang model and studied different soil constitutive models and their influence on results [6–8]. They concluded that the modified Duncan-Chang model was a good compromise between prediction accuracy and availability of parameters from conventional tri-axial compression testing. It was developed for cohesionless soils that had two parts [9, 10]: an elastic nonlinear part and two mechanisms of plasticity. It was allowed to take into account the nonlinearity of the behavior at low stress level for overconsolidated materials [11]. The description of the model and its parameters are given [12]. To consider the effect of different stress paths, some scholars [13, 14] developed the Duncan-Chang model, derived a formula of tangent elastic modulus under different stress paths, and provided approximate elastic-plastic models for clay [15, 16]. Lode parameter was introduced into Duncan-Chang model to reflect the effect of intermediate principal stress [17]. To consider the soil structural damage during loading, the concept of damage rate was introduced to modify Duncan-Chang model [18, 19].In these modified models, the initial stress state was assumed to be isotropic. However, the initial stress state of naturally deposited clay is usually anisotropic, known asK
0 consolidation [20]. The anisotropic initial stress and strain states result in different strength and deformation characteristics, and the constitutive relationship was different from that of isotropic stress states [21]. To overcome these differences, a unified nonlinear constitutive model was proposed for clay under different initial conditions. Two groups of tri-axial compression tests for highly compressible clay are performed, as K
0 consolidation and isotropic consolidation. By fitting the test data, a factor of anisotropic consolidation is introduced into the framework of E
~
v model to construct a uniform nonlinear constitutive model. With the average slope of the unloading-reloading curve selected as the unloading modulus, the unloading function is constructed as the loading-unloading criterion. Furthermore, the rationality of this model is checked. A comparison is made between the experimental stress-strain curves and the computational results predicted by the constitutive model. It is shown that the model can reflect the stress-strain behavior of the soil under the K
0 consolidation and isotropic consolidation conditions.


## 2. Experiment Analysis

### 2.1. Sample Preparation

In this study, natural saturation clay sampled from Wuhan is studied. Considering the anisotropic initial stress and strain state, two parallel drain tri-axial tests were carried out. In these tests, the static lateral pressure coefficientK
0
=
σ
3
/
σ
1 was set to 0.562 [22]. Then, two parallel lateral confinement compression tests were carried out to acquire the natural compression coefficient a, and it equals 0.567 MP
a
-
1. Some other physical properties of the studied clay are shown in Table 1.Table 1
Physical properties of the studied clay.
Physical  property
w (%)
ρ/(g/cm3)
G
S
w
P (%)
w
L (%)
e
S
r (%)
Value
28.53
1.961
2.697
28.18
47.82
0.897
98.9Considering the inevitable influence of disturbance and the difference characteristics in mechanics and physics of undisturbed soil sampling from different place, definite divergence was inevitable. Therefore, the moisture content is decided by field investigation. The remodeling soil sample of test passed by a series process: the undisturbed soil samples are made into uniform saturation samples, then tamping in 5 layers in tramper, and then extraction air before test. After the sample is installed, counter-pressure saturation is carried out until the saturation reaches larger than 99%. Then tri-axial shear test can be carried out under specified stress path.


### 2.2. Tri-Axial Test Paths

Consolidated trained tri-axial compression tests (C
D) were carried out in two different stress paths, by using SJ-1AG conventional tri-axial testing machine (shown as Figure 1). One test path is drained tri-axial compression tests under K
0 consolidation conditions (C
K
0
D). As comparison, the other is chosen as isotropic consolidation (C
D). The stress paths of the two group tests are shown in Figure 2.Figure 1
Experiment testing machine and sample.Stress path of compression experiment.
(a)
C
K
0
D tri-axial compression test
(b)
C
D tri-axial compression testIn theC
D test, prophase average consolidation pressures p
a were varied from 50 to 300 kPa, and the interval between every single test was 50 kPa. The tri-axial compression test cannot be carried out unless the pore pressure dissipation reaches 95%; otherwise consolidation will be sustained. Similarly, C
K
0
D tests also consist of 6 series of parallel tests with different consolidation pressures. In consolidation, the axial pressure and lateral pressure are added step by step until initial consolidation pressure p
a reaches 50~300 kPa. Unlike the C
D test, the initial σ
3 is different from σ
1 in the C
K
0
D tests. As p
a
=
(
σ
1
+
2
σ
3
)
/
3, the initial stresses of C
K
0
D tests are determined and shown in Table 2.Table 2
Initial stresses ofC
K
0
D tests.
p
a
/
kPa
σ
1
/
kPa
σ
3
/
kPa
50
70.6
39.7
100
141.3
79.4
150
211.9
119.1
200
282.4
158.7
250
353.1
198.4
300
423.9
238.2


### 2.3. Result of Test

Figure3 is the relationship between generalized shear stress q and generalized shear strain ε
- as initial consolidation pressure is 200 kPa of two different stress paths. Similarly, Figure 4 shows the relationship between generalized shear stress q and generalized volumetric strain ε
v.q
~
ε
- curves of different compression experiments.
(a)
C
D path
(b)
C
K
0
D pathq
~
ε
v curves of different compression experiments.
(a)
C
D path
(b)
C
K
0
D pathThe two figures indicate that there are different stress-strain relationships for different stress paths. In the initial stage of load, the stress in isotropic pressure test (C
D test) increases slower than that of K
0 consolidation (C
K
0
D test). In the later stage, both ε
- and ε
v in C
D test increase faster. Finally, the peak values in the C
K
0
D test are slightly above those of C
D test. It is because the lateral deformation constraints in C
K
0
D test are less than C
D. Thus, it is also shown that the influence of initial anisotropy on the constitutive relationship cannot be neglected.


## 2.1. Sample Preparation

In this study, natural saturation clay sampled from Wuhan is studied. Considering the anisotropic initial stress and strain state, two parallel drain tri-axial tests were carried out. In these tests, the static lateral pressure coefficientK
0
=
σ
3
/
σ
1 was set to 0.562 [22]. Then, two parallel lateral confinement compression tests were carried out to acquire the natural compression coefficient a, and it equals 0.567 MP
a
-
1. Some other physical properties of the studied clay are shown in Table 1.Table 1
Physical properties of the studied clay.
Physical  property
w (%)
ρ/(g/cm3)
G
S
w
P (%)
w
L (%)
e
S
r (%)
Value
28.53
1.961
2.697
28.18
47.82
0.897
98.9Considering the inevitable influence of disturbance and the difference characteristics in mechanics and physics of undisturbed soil sampling from different place, definite divergence was inevitable. Therefore, the moisture content is decided by field investigation. The remodeling soil sample of test passed by a series process: the undisturbed soil samples are made into uniform saturation samples, then tamping in 5 layers in tramper, and then extraction air before test. After the sample is installed, counter-pressure saturation is carried out until the saturation reaches larger than 99%. Then tri-axial shear test can be carried out under specified stress path.


## 2.2. Tri-Axial Test Paths

Consolidated trained tri-axial compression tests (C
D) were carried out in two different stress paths, by using SJ-1AG conventional tri-axial testing machine (shown as Figure 1). One test path is drained tri-axial compression tests under K
0 consolidation conditions (C
K
0
D). As comparison, the other is chosen as isotropic consolidation (C
D). The stress paths of the two group tests are shown in Figure 2.Figure 1
Experiment testing machine and sample.Stress path of compression experiment.
(a)
C
K
0
D tri-axial compression test
(b)
C
D tri-axial compression testIn theC
D test, prophase average consolidation pressures p
a were varied from 50 to 300 kPa, and the interval between every single test was 50 kPa. The tri-axial compression test cannot be carried out unless the pore pressure dissipation reaches 95%; otherwise consolidation will be sustained. Similarly, C
K
0
D tests also consist of 6 series of parallel tests with different consolidation pressures. In consolidation, the axial pressure and lateral pressure are added step by step until initial consolidation pressure p
a reaches 50~300 kPa. Unlike the C
D test, the initial σ
3 is different from σ
1 in the C
K
0
D tests. As p
a
=
(
σ
1
+
2
σ
3
)
/
3, the initial stresses of C
K
0
D tests are determined and shown in Table 2.Table 2
Initial stresses ofC
K
0
D tests.
p
a
/
kPa
σ
1
/
kPa
σ
3
/
kPa
50
70.6
39.7
100
141.3
79.4
150
211.9
119.1
200
282.4
158.7
250
353.1
198.4
300
423.9
238.2


## 2.3. Result of Test

Figure3 is the relationship between generalized shear stress q and generalized shear strain ε
- as initial consolidation pressure is 200 kPa of two different stress paths. Similarly, Figure 4 shows the relationship between generalized shear stress q and generalized volumetric strain ε
v.q
~
ε
- curves of different compression experiments.
(a)
C
D path
(b)
C
K
0
D pathq
~
ε
v curves of different compression experiments.
(a)
C
D path
(b)
C
K
0
D pathThe two figures indicate that there are different stress-strain relationships for different stress paths. In the initial stage of load, the stress in isotropic pressure test (C
D test) increases slower than that of K
0 consolidation (C
K
0
D test). In the later stage, both ε
- and ε
v in C
D test increase faster. Finally, the peak values in the C
K
0
D test are slightly above those of C
D test. It is because the lateral deformation constraints in C
K
0
D test are less than C
D. Thus, it is also shown that the influence of initial anisotropy on the constitutive relationship cannot be neglected.


## 3. Framework of Model

### 3.1. Tangent Elastic ModulusE
t

Assuming that the elastic relationship of(
σ
1
-
σ
3
)
~
ε
1 is hyperbolic based on the generalized Hooke law and test data, we choose the tangent elastic model (E
~
v model) as the model framework. In view of initial anisotropy, we introduce the lateral pressure coefficient into this model. Thus, it can be expressed as
(1)
E
t
=
E
i
[
1
-
R
f
(
1
-
sin
φ
)
(
σ
1
-
σ
3
)
0.85
/
K
0
2
c
cos
⁡
φ
+
2
σ
3
sin
φ
]
2
,
where K
0 is the lateral pressure coefficient; E
t is the tangent elastic modulus; R
f is the damage ratio, R
f
=
(
σ
1
-
σ
3
)
f
/
(
σ
1
-
σ
3
)
ult; (
σ
1
-
σ
3
)
ult is the asymptotic value of curve (
σ
1
-
σ
3
)
~
ε
1; (
σ
1
-
σ
3
)
f is the breakdown strength and it can be determined by the Mohr Coulomb failure criterion:
(2)
(
σ
1
-
σ
3
)
f
=
(
2
c
cos
⁡
φ
+
2
σ
3
sin
φ
)
(
1
-
sin
φ
)
.
In the equation above, c denotes the cohesive strength of soil; φ denotes the internal friction angle of soil; σ
1 and σ
3 denote the axial and radial stresses, respectively, based on the nonlinear regression analysis method [23]; the initial modulus of the curve E
i can be determined by the following:
(3)
E
i
=
K
p
a
(
σ
3
p
a
)
n
,
where p
a is the atmospheric pressure; K and n are experiment parameters. In log-log coordinates, there is linear relationship between E
i and σ
3 and the slope is n. K is the intercept of the line when σ
3
=
p
a.


### 3.2. Tangent Poisson Ratioν
t

The relationship betweenε
1 and ε
3 is always hyperbolic. Equation (4) shows the expression of the Poisson ratio ν
t deduced as
(4)
ν
t
=
ν
i
(
1
-
D
ε
1
)
2
,
where D is a test parameter determined by the slope of the linear relationship between ε
r
/
ε
1 and ε
r, ε
r
=
(
3
ε
v
-
ε
1
)
/
2; ε
v
,
ε
1 represent the volumetric strain and the axial strain, respectively; ν
i is the Poisson ratio and can be determined by the following:
(5)
ν
i
=
G
-
F
log
⁡
(
σ
3
p
a
)
.
In the equation above, G and F are experiment parameters and can be determined according to the curve between ν
i and σ
3 in semilog coordinates. F is the slope of the ν
i
-
log
⁡
σ
3 line and G is the intercept of the line when σ
3
=
p
a. Axial strain ε
1 can be determined by the following:
(6)
ε
1
=
(
σ
1
-
σ
3
)
×
(
E
i
[
1
-
R
f
(
1
-
sin
φ
)
(
σ
1
-
σ
3
)
0.85
/
K
0
2
c
cos
⁡
φ
+
2
σ
3
sin
φ
]
)
.
-
1


### 3.3. Unloading ModulusE
u
r

Based on experimental data, Duncan assumed that the unloading modulusE
ur does not change with the change of σ
1
-
σ
3. It is only related to σ
3. According to experimental unloading-reloading curves, based on the nonlinear regression analysis method [23], the relationship between E
ur and σ
3 can be drawn. It is approximately linear in the logarithmic coordinates, as shown in in the following:
(7)
E
ur
=
K
ur
p
a
(
σ
3
p
a
)
n
,
where K
ur is an empirical constant and can be expressed as the intercept of the line between log
⁡
(
E
ur
/
p
a
) and log
⁡
(
σ
3
/
p
a
) when σ
3
=
p
a.


## 3.1. Tangent Elastic ModulusE
t

Assuming that the elastic relationship of(
σ
1
-
σ
3
)
~
ε
1 is hyperbolic based on the generalized Hooke law and test data, we choose the tangent elastic model (E
~
v model) as the model framework. In view of initial anisotropy, we introduce the lateral pressure coefficient into this model. Thus, it can be expressed as
(1)
E
t
=
E
i
[
1
-
R
f
(
1
-
sin
φ
)
(
σ
1
-
σ
3
)
0.85
/
K
0
2
c
cos
⁡
φ
+
2
σ
3
sin
φ
]
2
,
where K
0 is the lateral pressure coefficient; E
t is the tangent elastic modulus; R
f is the damage ratio, R
f
=
(
σ
1
-
σ
3
)
f
/
(
σ
1
-
σ
3
)
ult; (
σ
1
-
σ
3
)
ult is the asymptotic value of curve (
σ
1
-
σ
3
)
~
ε
1; (
σ
1
-
σ
3
)
f is the breakdown strength and it can be determined by the Mohr Coulomb failure criterion:
(2)
(
σ
1
-
σ
3
)
f
=
(
2
c
cos
⁡
φ
+
2
σ
3
sin
φ
)
(
1
-
sin
φ
)
.
In the equation above, c denotes the cohesive strength of soil; φ denotes the internal friction angle of soil; σ
1 and σ
3 denote the axial and radial stresses, respectively, based on the nonlinear regression analysis method [23]; the initial modulus of the curve E
i can be determined by the following:
(3)
E
i
=
K
p
a
(
σ
3
p
a
)
n
,
where p
a is the atmospheric pressure; K and n are experiment parameters. In log-log coordinates, there is linear relationship between E
i and σ
3 and the slope is n. K is the intercept of the line when σ
3
=
p
a.


## 3.2. Tangent Poisson Ratioν
t

The relationship betweenε
1 and ε
3 is always hyperbolic. Equation (4) shows the expression of the Poisson ratio ν
t deduced as
(4)
ν
t
=
ν
i
(
1
-
D
ε
1
)
2
,
where D is a test parameter determined by the slope of the linear relationship between ε
r
/
ε
1 and ε
r, ε
r
=
(
3
ε
v
-
ε
1
)
/
2; ε
v
,
ε
1 represent the volumetric strain and the axial strain, respectively; ν
i is the Poisson ratio and can be determined by the following:
(5)
ν
i
=
G
-
F
log
⁡
(
σ
3
p
a
)
.
In the equation above, G and F are experiment parameters and can be determined according to the curve between ν
i and σ
3 in semilog coordinates. F is the slope of the ν
i
-
log
⁡
σ
3 line and G is the intercept of the line when σ
3
=
p
a. Axial strain ε
1 can be determined by the following:
(6)
ε
1
=
(
σ
1
-
σ
3
)
×
(
E
i
[
1
-
R
f
(
1
-
sin
φ
)
(
σ
1
-
σ
3
)
0.85
/
K
0
2
c
cos
⁡
φ
+
2
σ
3
sin
φ
]
)
.
-
1


## 3.3. Unloading ModulusE
u
r

Based on experimental data, Duncan assumed that the unloading modulusE
ur does not change with the change of σ
1
-
σ
3. It is only related to σ
3. According to experimental unloading-reloading curves, based on the nonlinear regression analysis method [23], the relationship between E
ur and σ
3 can be drawn. It is approximately linear in the logarithmic coordinates, as shown in in the following:
(7)
E
ur
=
K
ur
p
a
(
σ
3
p
a
)
n
,
where K
ur is an empirical constant and can be expressed as the intercept of the line between log
⁡
(
E
ur
/
p
a
) and log
⁡
(
σ
3
/
p
a
) when σ
3
=
p
a.


## 4. Constitutive Model

### 4.1. Parameters for the Model

InK
0 consolidation, the clay appears to be anisotropic state. For p equals 100, 200, and 300 kPa, the lateral pressure σ
3 can be determined from K
0. The corresponding values are 79.4, 158.7, and 238.1 kPa. By the relationship between ε
1
/
(
σ
1
-
σ
3
) and ε
1 (shown in Figure 5), the intercept 1
/
E
i can be determined. The Initial elastic modulus E
i can be calculated correspondingly. It is shown in Table 3.Table 3
Initial tangent elastic modulusE
i.
σ
3
/
100 kPa
1
/
E
i
E
i
/
100 kPa
E
i
¯
/
100 kPa
0.794
0.0045
222.2
242.8
1.587
0.0040
250.0
2.381
0.0039
256.0Figure 5
ε
1
/
(
σ
1
-
σ
3
)


~


ε
1 curves of different compression experiments.In log-log coordinates, the relationship betweenE
i and σ
3 is almost linear, and it is shown in Figure 6.Figure 6
E
i
~
σ
3 curve.Figure6 indicates when σ
3
=
100 kPa, the corresponding vertical coordinate K
=
230.1, and the slope n of the line equals 0.139.Based on the result of test, axial strainε
1 and volumetric strain ε
v are determined; after conversion between strains, the relationship curve can be drawn between ε
r
/
ε
1 and ε
r. In view of that we can determine the parameter G in fitting line (Figure 8) only when σ
3
=
100 kPa, and we choose some test data which can cover 100 kPa. That is, σ
3
=
39.7 kPa, 79.4 kPa, and 158.7 kPa (as Table 1). Thus, the curve of ε
r
/
ε
1
~
ε
r is shown in Figure 7.Figure 7
ε
r
/
ε
1
~
ε
r curves.Figure 8
ν
i
~
σ
3 curve.Based on (5), the linear relationship between ν
i and σ
3 in semilog coordinates is shown in Figure 8. The figure indicates that corresponding straight slope and unit intercept are F
=
0.09 and G
=
0.0642.Mohr’s circle can be drawn based on parameters of limit states for different ambient pressures. It is shown in Figure9 that c equals 21 kPa and φ equals 31.3°. (
σ
1
-
σ
3
)
ult and R
f are determined by the curve relationship between σ
1
-
σ
3 and ε
1. The calculation result is shown in Table 4.Table 4
Limit state parameter.
σ
3
/
100 kPa
(
σ
1
-
σ
3
)
f
(
σ
1
-
σ
3
)
ult
R
f
R
-
f
0.794
264.1
286.0
0.9140
0.820
1.587
446.0
600.0
0.7433
2.381
630.7
786.0
0.8024Figure 9
Mohr circle of shearing limit state.Similar to the solution to elastic modulusE
t, the relationship between E
ur and σ
3 can be drawn in log-log coordinates. It is shown in Figure 10. When σ
3 equals 100 kPa, the corresponding vertical coordinate K
ur equals 295.8, and the slope n equals 0.362.Figure 10
E
ur
~
σ
3 curve.


### 4.2. Constitutive Model

Integrating the results of the above calculation, theE
~
v model can be obtained for K
0 consolidation. The tangent elastic modulus is shown in the following:
(8)
E
t
=
230.1
[
1
-
0.394
(
σ
1
-
σ
3
)
0.85
/
K
0
0.403
+
1.039
σ
3
]
2
p
a
(
σ
3
p
a
)
0.139
.The tangent Poisson ratio is shown in the following:(9)
ν
t
=
{
0.064
-
0.09
log
⁡
(
σ
3
p
a
)
}
×
{
×
p
a
(
σ
3
p
a
)
0.139
)
-
1
1
-
(
×
p
a
(
σ
3
p
a
)
0.139
[
1
-
0.394
(
σ
1
-
σ
3
)
0.85
/
K
0
0.403
+
1.039
σ
3
]
2
)
-
1
(
3.13
(
σ
1
-
σ
3
)
)
×
(
230.1
[
1
-
0.394
(
σ
1
-
σ
3
)
0.85
/
K
0
0.403
+
1.039
σ
3
]
2
×
p
a
(
σ
3
p
a
)
0.139
[
1
-
0.394
(
σ
1
-
σ
3
)
0.85
/
K
0
0.403
+
1.039
σ
3
]
2
)
-
1
)
}
-
2
.The coefficientK
0 is determined by stress values in (8) and (9). Correspondingly, the uniform relationship between the stress and strain increments is shown in the following:
(10)
{
d
σ
}
=
[
D
]
{
d
ε
}
,
where {
d
σ
}
=
{
d
σ
x
(
r
)
,
d
σ
y
(
θ
)
,
d
σ
Z
,
d
τ
x
y
(
r
z
)
}
T; {
d
ε
}
=
{
d
ε
x
(
r
)
,
d
ε
y
(
θ
)
,
d
ε
z
,
d
ε
x
y
(
r
z
)
}
T.In (10), [
D
] can be described as follows:
(11)
[
D
]
=
E
(
1
+
ν
t
)
(
1
-
2
ν
t
)
[
1
-
ν
t
ν
t
ν
t
0
ν
t
1
-
ν
t
ν
t
0
ν
t
ν
t
1
-
ν
t
0
0
0
0
1
-
2
ν
t
]
,
where E represents the elastic modulus of clay. In the equation above, loading and unloading states can be determined by (7) and (8). At loading stage, E
=
E
t; at unloading stage E
ur
=
295.8
p
a
(
σ
3
/
p
a
)
0.362; at reloading stage, E
=
E
ur. This modulus is used to separate elastic strain from plastic one.


## 4.1. Parameters for the Model

InK
0 consolidation, the clay appears to be anisotropic state. For p equals 100, 200, and 300 kPa, the lateral pressure σ
3 can be determined from K
0. The corresponding values are 79.4, 158.7, and 238.1 kPa. By the relationship between ε
1
/
(
σ
1
-
σ
3
) and ε
1 (shown in Figure 5), the intercept 1
/
E
i can be determined. The Initial elastic modulus E
i can be calculated correspondingly. It is shown in Table 3.Table 3
Initial tangent elastic modulusE
i.
σ
3
/
100 kPa
1
/
E
i
E
i
/
100 kPa
E
i
¯
/
100 kPa
0.794
0.0045
222.2
242.8
1.587
0.0040
250.0
2.381
0.0039
256.0Figure 5
ε
1
/
(
σ
1
-
σ
3
)


~


ε
1 curves of different compression experiments.In log-log coordinates, the relationship betweenE
i and σ
3 is almost linear, and it is shown in Figure 6.Figure 6
E
i
~
σ
3 curve.Figure6 indicates when σ
3
=
100 kPa, the corresponding vertical coordinate K
=
230.1, and the slope n of the line equals 0.139.Based on the result of test, axial strainε
1 and volumetric strain ε
v are determined; after conversion between strains, the relationship curve can be drawn between ε
r
/
ε
1 and ε
r. In view of that we can determine the parameter G in fitting line (Figure 8) only when σ
3
=
100 kPa, and we choose some test data which can cover 100 kPa. That is, σ
3
=
39.7 kPa, 79.4 kPa, and 158.7 kPa (as Table 1). Thus, the curve of ε
r
/
ε
1
~
ε
r is shown in Figure 7.Figure 7
ε
r
/
ε
1
~
ε
r curves.Figure 8
ν
i
~
σ
3 curve.Based on (5), the linear relationship between ν
i and σ
3 in semilog coordinates is shown in Figure 8. The figure indicates that corresponding straight slope and unit intercept are F
=
0.09 and G
=
0.0642.Mohr’s circle can be drawn based on parameters of limit states for different ambient pressures. It is shown in Figure9 that c equals 21 kPa and φ equals 31.3°. (
σ
1
-
σ
3
)
ult and R
f are determined by the curve relationship between σ
1
-
σ
3 and ε
1. The calculation result is shown in Table 4.Table 4
Limit state parameter.
σ
3
/
100 kPa
(
σ
1
-
σ
3
)
f
(
σ
1
-
σ
3
)
ult
R
f
R
-
f
0.794
264.1
286.0
0.9140
0.820
1.587
446.0
600.0
0.7433
2.381
630.7
786.0
0.8024Figure 9
Mohr circle of shearing limit state.Similar to the solution to elastic modulusE
t, the relationship between E
ur and σ
3 can be drawn in log-log coordinates. It is shown in Figure 10. When σ
3 equals 100 kPa, the corresponding vertical coordinate K
ur equals 295.8, and the slope n equals 0.362.Figure 10
E
ur
~
σ
3 curve.


## 4.2. Constitutive Model

Integrating the results of the above calculation, theE
~
v model can be obtained for K
0 consolidation. The tangent elastic modulus is shown in the following:
(8)
E
t
=
230.1
[
1
-
0.394
(
σ
1
-
σ
3
)
0.85
/
K
0
0.403
+
1.039
σ
3
]
2
p
a
(
σ
3
p
a
)
0.139
.The tangent Poisson ratio is shown in the following:(9)
ν
t
=
{
0.064
-
0.09
log
⁡
(
σ
3
p
a
)
}
×
{
×
p
a
(
σ
3
p
a
)
0.139
)
-
1
1
-
(
×
p
a
(
σ
3
p
a
)
0.139
[
1
-
0.394
(
σ
1
-
σ
3
)
0.85
/
K
0
0.403
+
1.039
σ
3
]
2
)
-
1
(
3.13
(
σ
1
-
σ
3
)
)
×
(
230.1
[
1
-
0.394
(
σ
1
-
σ
3
)
0.85
/
K
0
0.403
+
1.039
σ
3
]
2
×
p
a
(
σ
3
p
a
)
0.139
[
1
-
0.394
(
σ
1
-
σ
3
)
0.85
/
K
0
0.403
+
1.039
σ
3
]
2
)
-
1
)
}
-
2
.The coefficientK
0 is determined by stress values in (8) and (9). Correspondingly, the uniform relationship between the stress and strain increments is shown in the following:
(10)
{
d
σ
}
=
[
D
]
{
d
ε
}
,
where {
d
σ
}
=
{
d
σ
x
(
r
)
,
d
σ
y
(
θ
)
,
d
σ
Z
,
d
τ
x
y
(
r
z
)
}
T; {
d
ε
}
=
{
d
ε
x
(
r
)
,
d
ε
y
(
θ
)
,
d
ε
z
,
d
ε
x
y
(
r
z
)
}
T.In (10), [
D
] can be described as follows:
(11)
[
D
]
=
E
(
1
+
ν
t
)
(
1
-
2
ν
t
)
[
1
-
ν
t
ν
t
ν
t
0
ν
t
1
-
ν
t
ν
t
0
ν
t
ν
t
1
-
ν
t
0
0
0
0
1
-
2
ν
t
]
,
where E represents the elastic modulus of clay. In the equation above, loading and unloading states can be determined by (7) and (8). At loading stage, E
=
E
t; at unloading stage E
ur
=
295.8
p
a
(
σ
3
/
p
a
)
0.362; at reloading stage, E
=
E
ur. This modulus is used to separate elastic strain from plastic one.


## 5. Numerical Analysis

To verify the validity of the model proposed in this paper, the authors have implemented the constitutive relationship into a finite element program and made a numerical simulation of tri-axial tests. The height and diameter of the clay sample are 8 cm and 3.91 cm, respectively. Taking the symmetry of the sample into consideration, a half of the sample is adopted in the numerical simulation. In view of symmetry, horizontal displacement of nodes on the axes of symmetry is zero. Similarly, because of rigid fastening of test base, vertical constraints are applied to all the nodes of the sample. Moreover, the other two boundaries are force boundaries. The upper boundary is applied withσ
1 and the right is σ
3. The computation model is shown in Figure 11. There are 45 nodes and 64 3-node triangular elements in the simulation.Figure 11
Computation diagram of FEA.The relationship ofq
~
ε
1 is obtained, in which, q is the shear stress of the sample and ε
1 is the average axial strain of the sample. Comparison between the numerical results and that of the corresponding test has been made. The part of comparison is shown in Figure 12.Comparison between numerical simulation and test curves.
(a)
C
K
0
D test
(b)
C
D testFigure12 indicates that the nonlinear model proposed in this paper can reflect the relationship of tri-axial shear test for K
0 consolidation (in C
K
0
D test, K
0
=
0.562; in C
D test, K
0
=
1). Separation appears only at the later stage when the sample is close to failure. At this stage, the result from test is slightly larger than that of simulation, but the maximum error is only 6.0%.


## 6. Conclusion

The initial condition of natural clay is usually underK
0 stress states. To display the different mechanical characteristics, two series of initial consolidation condition tri-axial compress experiments are done. It is shown that there are some differences of a soil under isotropic conditions. In the initial stage of load, the stress in isotropic pressure test increases slower than that of K
0 consolidation. In the later stage, the shear strain in K
0 consolidation increases faster. It indicates that the influence of stress paths to the constitutive relationship cannot be neglected.According to the tests, a factor of lateral pressure coefficient is introduced into theE
~
ν model. A uniform nonlinear constitutive model is proposed by fitting of the stress-strain behavior of clay with different initial conditions. Based on nonlinear regression analysis, some model parameters are provided.With the average slope of the unloading-reloading curve selected as the unloading modulus, the constitutive relations can be promoted to an elastoplastic model. The proposed model can reflect the stress-strain relationship underK
0 conditions. Some discrepancy between the model prediction and the test data occurs at the stage close to failure. The maximum error is just 6.0%.



---

*Source: 804586-2014-04-06.xml*