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### Transcript of Elastic-Plastic Triaxial Tests

• GEO-SLOPE International Ltd, Calgary, Alberta, Canada www.geo-slope.com

SIGMA/W Example File: Elastic-Plastic triaxial tests.doc (pdf) Page 1 of 13

Verification of the Elastic-Plastic Soil Model by Triaxial Test Simulations

1 Introduction

This example simulates a series of triaxial tests, which can be used to verify that the elastic-plastic

constitutive model is functioning properly. The simulations include:

Consolidating the sample to an initial isotropic stress state

Drained strain-controlled tests

Extension tests

Consolidation with back-pressure

The verification includes comparisons with hand-calculated values and discussions relative to the elastic-

plastic theoretical framework.

2 Feature Highlights

GeoStudio feature highlights include:

Using the axisymmetric option to simulate a triaxial test

Displacement type boundary conditions to simulate a strain-controlled test

Use of total stress and effective stress soil parameters

3 General Methodology

All but one of the simulated shearing phases are preceded by the simulation of the consolidation phase of

a triaxial test. Consolidation is isotropic with the confining pressure equal to 100 kPa. The isotropic

stress state is simulated by applying a normal stress on the top and on the right side of the sample equal to

100 kPa. The consolidation stage is set as the Parent; that is, the initial condition, for the subsequent simulations involving shearing.

• GEO-SLOPE International Ltd, Calgary, Alberta, Canada www.geo-slope.com

SIGMA/W Example File: Elastic-Plastic triaxial tests.doc (pdf) Page 2 of 13

metres (x 0.001)

-10 0 10 20 30 40

metr

es (

x 0

.001)

-10

0

10

20

30

40

50

60

Figure 1 Triaxial test configuration for establishing initial stress state

The shearing phase of the analysis is simulated as a strain-rate controlled test. The definition of the

strain-rate involves defining the number of time steps and the displacement that occurs over each step. Although the time steps are being defined, it is more appropriate to think of the time steps as load steps. Absolute time has no meaning in the context of these analyses. The number of load steps defined in the

shear stage simulations is generally 50 or 75 and the incremental y-displacement (i.e. the boundary

condition) at the top of the specimen is defined as -0.0002 m (per load step), where the negative sign

indicates downward displacement. Consequently, 50 and 75 load steps multiplied by a y-displacement of

-0.0002 m per load step results in a total vertical displacement of 0.01 m and 0.015 m, respectively.

Symmetry is assumed about the vertical and horizontal centre-lines; consequently, only of the specimen

is simulated. The dimensions of the simulation portion of the specimen are 0.025 m by 0.05 m, which is

half of the width and height of a conventionally sized triaxial specimen. Total vertical y-displacements of

0.01 m and 0.015 m produce axial strains of 0.2 (or 20%) and 0.3 (30%), respectively.

Notice that Linear-Elastic parameters are used when setting up confining stresses; non-linear models are not

required for this and the value of E is not relevant

4 Analysis A-Cohesion only

An unconfined compression test is simulated in this example; therefore, the consolidation phase is not

simulated (there is no Parent analysis). The unconfined compressive strength Cu of the soil is 50 kPa and

the E is 1000 kPa.

Figure 2 shows the SIGMA/W simulated stress-strain curve. The unconfined compressive strength

defines the maximum shear stress that can be obtained, which is defined as:

• GEO-SLOPE International Ltd, Calgary, Alberta, Canada www.geo-slope.com

SIGMA/W Example File: Elastic-Plastic triaxial tests.doc (pdf) Page 3 of 13

max = (1 3) / 2

The vertical stress reaches 100 kPa and the horizontal stress 3 is zero because this is an unconfined test. Consequently, max is equal to the undrained strength of 50 kPa. The soil is linear elastic until the stress path reaches the failure line. Afterwards, the soil behavior becomes perfectly plastic. The (triaxial)

deviatoric stress (1 3) at failure is equal to the maximum vertical stress (Figure 3).

y-total stress: axial strain

Y-T

ota

l S

tress (

kP

a)

Y-Strain

0

20

40

60

80

100

120

0 0.05 0.1 0.15 0.2

Figure 2 Stress-strain curve for unconfined compression test

Deviatoric stress

Devia

tori

c S

tress (

q)

(kP

a)

Y-Strain

0

20

40

60

80

100

120

0 0.05 0.1 0.15 0.2

Figure 3 Deviatoric stress for unconfined compression test

5 Analysis 1 C only - confined

A confined triaxial compression test is simulated on a specimen with an undrained strength of 100 kPa.

Figure 4 and Figure 5 present the y-total stress and deviator stress versus axial strain. The maximum

vertical stress increases to 200 kPa because the initial confining stress was 100 kPa. The deviatoric stress

(1 3) remains the same as that simulated by unconfined compression test and max remains equal to 50 kPa = (200 100)/2.

• GEO-SLOPE International Ltd, Calgary, Alberta, Canada www.geo-slope.com

SIGMA/W Example File: Elastic-Plastic triaxial tests.doc (pdf) Page 4 of 13

y-total stress: axial strainY

-Tota

l S

tress (

kP

a)

Y-Strain

100

120

140

160

180

200

220

0 0.05 0.1 0.15 0.2

Figure 4 Stress-strain curve for confined compression test

Deviatoric stress

Devia

tori

c S

tress (

q)

(kP

a)

Y-Strain

0

20

40

60

80

100

120

0 0.05 0.1 0.15 0.2

Figure 5 Deviatoric stress versus strain for confined compression test

6 Analysis 2 Phi>0; c=0

A confined triaxial compression test is simulated on a specimen with = 30 and c = zero. Figure 6

presents the y-total stress versus axial strain. For a purely frictional soil, the principal stress ratio at the

point where the stress level reaches the soil strength is,

1

3

1 sin

1 sin which is equal to 3 if is 30 degrees.

The confining stress (3) is 100 kPa; therefore, 1 at failure should be 300 kPa. Note the starting stress is 100 kPa and the failure stress is 300 kPa. The maximum deviatoric stress is 200 kPa (300 100 = 200 kPa).

• GEO-SLOPE International Ltd, Calgary, Alberta, Canada www.geo-slope.com

SIGMA/W Example File: Elastic-Plastic triaxial tests.doc (pdf) Page 5 of 13

y-total stress: axial strainY

-Tota

l S

tress (

kP

a)

Y-Strain

100

150

200

250

300

350

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

Figure 6 Stress-strain curve for sand with no cohesion

7 Analysis: 3 Phi=30, C=20

Analysis 3 is a repeat of the previous case, but now the soil has some cohesive strength (c = 20 kPa).

Figure 7 presents the y-total stress versus axial strain. For a Mohr-Coulomb strength envelope in a

triaxial test the vertical stress 1 is,

1 3

1 sin 2 cos

1 sin 1 sin

c

The confining stress 3 is 100 kPa. The maximum 1 then will be 369.28 kPa. The deviatoric stress is 269.28 kPa.

y-total stress: axial strain

Y-T

ota

l S

tress (

kP

a)

Y-Strain

100

200

300

400

0 0.1 0.2 0.3

Figure 7 Stress-strain curve for = 30 and c = 20 kPa

• GEO-SLOPE International Ltd, Calgary, Alberta, Canada www.geo-slope.com

SIGMA/W Example File: Elastic-Plastic triaxial tests.doc (pdf) Page 6 of 13

8 Analysis 4 Coupled undrained

Analysis 4 simulates a consolidated undrained test on sand with pore-pressure measurements. This is

accomplished in SIGMA/W using a coupled analysis with the hydraulic boundary conditions set as no

flow (Q = 0) along the perimeter of the sample.

The sand properties are = 30 and c = zero.

A very important observation about the elastic-plastic model is that the effective stress path (q-p) is vertical during undrained loading. In other words, the mean effective stress remains a constant under triaxial loading conditions.

Figure 8 confirms this important characteristic of the elastic-plastic model under undrained loading

conditions.

p':q stress path

Devia

tori

c S

tress (

q)

(kP

a)

Mean Effective Stress (p') (kPa)

0

20

40

60

80

100

120

140

50 60 70 80 90 100 110

Figure 8 Effective q-p stress path under undrained conditions

The deviatoric stress is 120 kPa when the sample becomes plastic is 120 kPa, which corresponds to 1 must be 220 kPa with 3 being 100 kPa.

In this case the pore-pressure will increase while the soil is elastic and then remain constant when the soil

becomes plastic, as illustrated in Figure 9.

The maximum pore-pressure is 40 kPa. This makes 3 = 60 and 1 = 180. The ratio of 1 / 3 = 180 / 60 = 3.

• GEO-SLOPE International Ltd, Calgary, Alberta, Canada ww