Elastic-Plastic Triaxial Tests

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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

Load-unload-reload 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

A displacement boundary function to simulate loading, unloading, and reloading

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.



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:



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.



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).




-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-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



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.



pore-pressure:axial strainP

ore

-Wate

r P

ressure

(kP

a)

Y-Strain

0

10

20

30

40

0 0.05 0.1 0.15 0.2

Figure 9 Pore-pressure response under undrained loading

9 Analysis 5 Coupled drained

This analysis repeats the previous case, but with drained conditions. Drained conditions can be simulated

by specifying the pore-pressure as a boundary condition at all the nodes. In this case the boundary

condition at each node is specified as u = zero or H = y.

metres (x 0.001)

-10 0 10 20 30 40

metr

es (

x 0

.001)

-10

0

10

20

30

40

50

60

Figure 10 Specification of pore-pressure for drained test

In this test the total stresses are the same as the effective stresses, since the pore-pressure is zero.



Once again the ratio of 1 / 3 must be 3 when the soil becomes plastic. Therefore, the maximum 1 will be 300 kPa.


Y-T

ota

l S

tress (

kP

a)

Y-Strain

100

150

200

250

300

350

0 0.1 0.2 0.3

Figure 11 Stress-strain curve for coupled analysis under drained conditions

The effective p-q stress path now is the same as the total stress path, and under triaxial condition will have a slope of 1h:3v. The change in p is 66.67 and the change in q is 200 or 3 times 66.67 kPa confirming the correct behavior.

p':q stress path

Devia

tori

c S

tress (

q)

(kP

a)


0

50

100

150

200

250

40 60 80 100 120 140 160 180

Figure 12 Total and effective stress path under drained conditions

10 Analysis 6 Coupled undrained C=20

This is a coupled undrained analysis, but with cohesion specified as 20 kPa.

As noted previously, the q-p effective stress path is a vertical straight line for the elastic-plastic constitutive model, which is confirmed in Figure 13. The maximum deviatoric stress is 160 kPa.



q - p'D

evia

tori

c S

tress (

q)

(kP

a)


0

20

40

60

80

100

120

140

160

180

0 20 40 60 80 100 120 140 160 180

Figure 13 Stress path q-p for soil with = 30 and c = 20 kPa

With qmax = 160, 1 must be 260.

Figure 14 shows that the maximum pore-pressure is 54 kPa. The effective minor principal stress then is

3 = 100 54 = 46 kPa.

pore-pressure:axial strain

Pore

-Wate

r P

ressure

(kP

a)

Y-Strain

0

10

20

30

40

50

60

0 0.05 0.1 0.15 0.2

Figure 14 Pore pressure response for soil with = 30 and c = 20 kPa

As discussed earlier, the major principal stress at the point when the soil becomes plastic must be,

1 3

1 sin 2 cos

1 sin 1 sin

c

With 3 = 46, = 30 and c = 20, 1 = 207.3 kPa as is correctly illustrated in Figure 15.



y-effective stress:starinY

-Effective S

tress (

kP

a)

Y-Strain

100

120

140

160

180

200

220

0 0.05 0.1 0.15 0.2

Figure 15 Stress-strain curve for soil with = 30 and c = 20 kPa

11 Analysis 7 Unload-reload

This analysis examines the behavior of a frictional soil under loading, unloading and then reloading.

This loading path can be simulated with a displacement function such as in Figure 16. The negative sign

means the top of the sample is pushed downwards. SIGMA/W is an incremental formulation and

therefore the applied displacement for each load step is the difference between the function value at the

current load step and the function value at the previous load step; that is, it is the incremental change

along the function. This means that where the function changes direction, the sign of the applied

displacement changes and consequently the sample is unloaded.

load-unload-reload

Y-D

ispla

cem

ent (m

)

Time (sec)

-0.005

-0.01

-0.015

-0.02

0

0 20 40 60 80 100

Figure 16 Loading, unloading and reloading boundary function



With the elastic-plastic constitutive soil model the soil returns to being perfectly elastic when it is

unloaded after having gone plastic. This response is illustrated in Figure 17.


Y-T

ota

l S

tress (

kP

a)

Y-Strain

100

150

200

250

300

350

0 0.1 0.2 0.3 0.4

Figure 17 Stress-strain during unloading and reloading

12 Analysis 8 Extension test

This analysis simulates an extension test.

This again is a drained test on sand that has a = 30 and c = zero. The confining stress again is 100 kPa.

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 extension test is simulated by pulling upward on the specimen (refer to the associated boundary

condition).

In this case, the confining stress is 100 kPa and is the major principal stress 1. The vertical stress is the minor principal stress 3. Therefore 3 at failure should approach 33.3 kPa. Figure 18 again confirms this. Note the starting stress is 100 and then drops to 33.3 as it should. The deviatoric stress at the end should

be 66.7 as is correctly shown in Figure 18.




-Tota

l S

tress (

kP

a)

Y-Strain

30

40

50

60

70

80

90

100

110

-0.05-0.1-0.15-0.2 0

Figure 18 Stress-strain curve for extension test

Deviatoric stress

Devia

tori

c S

tress (

q)

(kP

a)

Y-Strain

0

10

20

30

40

50

60

70

-0.05-0.1-0.15-0.2 0

Figure 19 Deviatoric stress for extension test

13 Analysis C Initial with pwp

In this analysis, the initial consolidated effective confining stress state is established by applying a normal

stress of 100 kPa on the top and on the right side of the sample. The initial back pressure of 50 kPa is

specified as an Activation PWP of 50 kPa. This makes the initial effective confining pressure equal to 50

kPa.

14 Analysis 9 Phi>0; C=0 backpressure

This analysis simulates a test with a backpressure but the pore-pressures remain at the backpressure

during the loading. This is done with a Load/Deformation type of analysis using Effective-Drained

strength parameters. The initial effective stress is used to establish the soil parameters, but the pore-

pressures do not change during the loading. This would be like loading a sand stratum that is under water.

Figure 20 shows both the total and effective stress-strain curves.



y-total stress:

axial strain :

(0.0131449,

0.0270671)

y-effective

stress:starin

: (0.0131802,

0.0258657)

kP

a

Y-Strain

40

60

80

100

120

140

160

180

200

220

0 0.05 0.1 0.15 0.2

Figure 20 Total and effective stress-strain curves for test with backpressure

As shown in Figure 20, 3 = 50 kPa. 1 must be 3 times 3, making it 150 kPa as correctly above.

15 Summary

These triaxial test simulations verify that the elastic-plastic constitutive relationship in SIGMA/W is

functioning and behaving correctly.

Elastic-Plastic Triaxial Tests

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