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Hong Kong University of Science and Technology

CIVL 3720 Soil Mechanics

Lab 3 - Consolidated Drained/Undrained Triaxial Compression

Test (CD and CU test)

Experiment date : 20th

March, 2013

Report submission date : 10th

April, 2013

Group Members

Name SID Contribution

(%)

Signature

CHAN, Yik Hin 20035984

CHAU, Lai Bun 20029284

CHAU, Man Kit 20031134

CHONG, Sing Pui 20031225

CHONG, Wai Ho 20029375CHOW, Jun Kang 20020628

FUNG, Hoi Tai 20030489

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Introduction

A widely used apparatus to determine the shear strength parameters and the

stress-strain behavior of soils is the triaxial apparatus. The name is a misnomer since

two, not three, stresses can be controlled. In the triaxial test, a cylindrical sample of

soil, usually with a length to diameter ratio of 2, is subjected to either controlled

increases in axial stresses or axial displacements and radial stresses. The sample size

is kept in this ratio so that the stress is uniformly distributed or no buckling occurs.

The axial stresses are applied by loading a plunger. If the axial stress is greater than

the radial stress, the soil is compressed vertically and the test is called triaxial

compression. For another case, if radial stress is greater than axial stress, the soil is

compressed laterally and the test is called triaxial extension.

In this experiment, 2 tests are performed- consolidated drained (CD) compression test

and consolidated undrained (CU) compression test. The below table summarizes the

features of these two tests.

CD Test CU Test

Purpose Determine cs, p and c.

Effective elastic moduli for

drained condition E and Escan be obtained too.

Determine su, cs and p

Loading stages 1st stage: Isotropic

consolidation phase

- Consolidating soil sample

until excess pore water pressure

dissipates.

1st stage: Isotropic consolidation

phase

- Consolidating soil sample until

excess pore water pressure

dissipates.

2nd stage: Shearing phase

- Pressure in the cell is kept

constant and additional axial

loads or displacmentes are

added very slowly until the

soil. sample fails.

2nd stage: Shearing phase

- Axial load is increased under

undrained condition and the

excess pore water pressure is

measured.

Objective

To determine the stress-strain-strength behavior of a dry medium-fine sand by

consolidated drained/undrained triaxial compression test (CD and CU tests)

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Equipment

1 Triaxial device (WF machine)2 Pressure gauge3 Dial gauge4 Device for measuring volume changes5 PC installed with data acquisition systemProcedures

Sample preparation and setup:

Sand samples would be prepared and set up in the triaxial apparatus by lab technicians.

The dimension of the specimen and detailed explanation of the experimental setup

would be given by TAs.

Degree of saturation Checkingby B-value (Skempton pore pressure parameters):

B-value could be used as an indicator to check the degree of saturation of the

specimen. Procedures below should be followed:

1 Valve of back pressure source was closed.2 Valve of cell pressure source was opened.3 Cell pressure was adjusted slowly to a certain increment, for example 50kPA.4 The corresponding excessive pore pressure,u, was recorded.5

The B value was calculated by the definition: B =u / 36 Step 7 was processed for a B-value larger than or equal to 0.95. For B-value

smaller than 0.95, an increment of back pressure was applied to improve the

degree of saturation. Cell pressure to the same increment as back pressure was

adjusted and the effective confinement was kept unchanged. The back pressure

valve was opened until equilibrium was reached. Steps (1) to step (6) was then

repeated.

7. The pressure increment was released to check the B-value.

Consolidation

1. Cell pressure valve was opened and the pressure was adjusted to the designedvalue (effective confinement).

2. The valve connecting to the device of measuring volume changes was opened.The sample was allowed to be consolidated about 5 minutes and the water would

flow out from the sample to the device.

3. The volume change during consolidation, which was inferred from the amount ofwater flowing out, was recorded

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Drained/Undrained Shear Test

1. The loading ram (plunger) was brought in contact with the loading cap on the topof the sample.

2. The LVDT was connected to measure the axial displacement during shearing.3. The rate of vertical displacement was set to 0.5mm/min.4. For a drained test, the drainage valve had to be opened to ensure a drained

condition. Similarly, the closing of the drainage valve would create an undrained

condition.

5. The shearing of sample was started (vertical loading).6. The test was stopped until axial strain reaches 15%.Remarks

Group 1 2 3 4

Effective consolidation pressure

(confinement)

50 kPa 100 kPa 200 kPa 300 kPa

*However, data processing and discussion would be done with last years results

(effective consolidation pressure was 100 kPa, 200 kPa, 300 kPa and 400 kPa) due to

time constraint.

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Data Processing and Discussion

For the drained and undrained test performed during your lab session, plot

1. '1and 3 vs e (void ratio)2. q vs p and p3. q vs 1 (%)4. v(%) vs 1(%) for drained test and u vs 1 (%) for undrained test5.Identify peak and/or ultimate shear strength from your own tests.

For the all tests (including results from other groups)

6.In a p-q space, plot all the peak (for drained test only) and ultimate strengthpoints. Calculate the shear strength parameters of the soil.

7.DiscussionIn compression test, we will denote the radial stress r as 3and the axial stress z as

1. Besides, we will denote compression stress as positive. For volumetric strain,

positive value indicates compression, negative sign indicates expansion in order to be

consistent with analysis in the text book. (This is opposite with data have been

recorded in the machine)

Axial total stress: 1= Pz/A + 3 Deviatoric stress: 13 = Pz/A = q

Axial strain: 1 = z/H0 Radial strain: 3= 2= r/r0 Volumetric strain: p= V/V0= 1+ 2+ 3= 1+ 23 Deviatoric strain: q= (2/3)*( 13)

Where Pz = the load on the plunger

A = cross-sectional area

r0 = initial radius of the sample

r = change in radius

V0 = initial volume

V = change in volume

H0 = initial height

z = change in height

Correction of cross-sectional area AThe area of the samples change during loading at any given instance is

= = =

1 1

=

(1 )1 To get void ratio, e, we have to do some derivation

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= = 1 + =1 + 1

= 1

1

Where Gs = specific gravity of sample (assume as 2.70)

w = density of water

md = dry weight of the sample

In order to draw the stress path (we only consider the stage 2 shear phase in report),

we have to find the value of p, p and q. In triaxial test, we assume axisymmetric

condition, therefore 2 = 3, 2 = 3. Therefore

= + + 3 = + 23

= + + 3 = +23 = 12 [ + + ]/ =

12 [2 ]/

= In order to determine shear strength parameters of soils, a critical state model (CSM)

is used to interpret it. In this model, we transform Mohr-coulomb failure envelope

from - space into p-q space. Derivation is made under the axisymmetric condition.(z= 1, r= = 3)

For axisymmetric condition, we will keep 3as constant and increases 1. Then we

are able to derive a relationship between friction angle and Mc.

= =

+ 2

3

=3 1

+ 2

= 1 + sin 1 sin

Then, we are able to get the following equation:

= 6sin

3 sin , sin =36 +

Similarly for axisymmetric extension, we are able to get the relationship between

friction angle and Me. Everything remains constant except decreasing 3. Below are

the derivations obtained:

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= 6sin

3 + sin , sin =36

Consolidated Drained Test (CD Test)

The graphs below show the results obtained.

400

401

402

403

404

405

406

407

0.72 0.73 0.74 0.75 0.76 0.77 0.78

'3(kPa)

Void ratio, e

Graph of'3 vs e

0

200

400

600

800

1000

1200

1400

1600

1800

0.72 0.73 0.74 0.75 0.76 0.77 0.78

'1(

kPa)

Void ratio, e

Graph of'1 vs e

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-200

0

200

400

600

800

1000

1200

1400

0 200 400 600 800 1000 1200

q(kPa)

p,p' (kPa)

Graphof q vs p and p'

q vs p (TSP)

q vs p' (ESP)

-200

0

200

400

600

800

1000

1200

1400

0 5 10 15 20

q

(kPa)

1 (%)

Graph of q vs 1(%)

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From the graph of v vs 1, we could suggest that the soil sample is dense soil as

dilation occurs. Although the peak in the graph of q vs 1 is not obvious, we could

estimate that peak qeviatoric stress is around 1242.70 kPa. Therefore, peak shear

strength, peakis 0.5qpeak= 571.25 kPa.

We assume the last value in graph of q vs 1 is ultimate value although it has notreached a constant value yet. Therefore, ultimate shear strength, ultimate is 0.5 qultimate =

1181.8/2 = 590.90 kPa.

-2

-1.5

-1

-0.5

0

0.5

1

0 5 10 15 20

v

(%)

1 (%)

Graph ofv(%) vs 1(%)

y = 1.6042x

y = 1.478x

0

200

400

600

800

1000

1200

1400

0 200 400 600 800 1000

q(kPa)

p' (kPa)

Graph of p' vs q (CD Test)

q - p' (peak)

q - p' (ultimate)

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By using critical state model, we are to estimate the friction angle for peak shear

strength and ultimate shear strength.

= sin 31.642

6+ 1.642 = 40.1

= sin 3 1.4786 + 1.478 = 36.4Since the soil sample used is dry medium-fine sand, we can assume the cohesion,

c=0.

Discussion

Based on the graph of q vs p, p, we could observe that the test was carried out under

a back pressure of approximately 200 kPa as there is a constant gap of value about

200 kPa between effective stress path (ESP) and total stress path (TSP). From the datacollected. Besides, the gradient of ESP and TSP are 2.96 and 3.00 respectively, which

are consistent with theoretical value, which is 3. This is consistent with the setup of

experiment, whereby drainage is allowed and excess pore water pressure is allowed to

dissipate gradually and the specimen to consolidate.

Next, based on graph of q vs 1, we could suggest that the soil sample is dense sand

although the peak of the graph is not obvious. However, we are able to further suggest

that the soil sample is dense sand based on the graph of vvs 1. From this graph, wecan observe that the soil sample compress initially and then dilate, which is

phenomena of dense sand for CD Test.

Furthermore, soil friction angle determined by the laboratory tests is influenced by 2

major factors. The energy applied to a soil by the external load is used both to

overcome the frictional resistance between the soil particles and also to expand the

soil against the confining pressure. The soil grains are highly irregular in shape and

must be lifted over one another for sliding to occur. This behavior is called dilatency.

This soil friction angle is corresponding to peak. Therefore, peak= cs+ , where

is dilation angle. In this test, = 40.1 - 36.4 = 3.7.

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Consolidated Undrained Test (CU Test)

The graphs below show the results obtained.

0

100

200

300

400

500

600

700

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

'3(kPa)

Void ratio, e

Graph of'3 vs e

0

500

1000

1500

2000

2500

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

'1(kPa)

Void ratio, e

Graph of'1 vs e

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-200

0

200

400

600

800

1000

1200

1400

1600

1800

0 200 400 600 800 1000 1200 1400

q(kPa)

p,p' (kPa)

Graphof q vs p and p'

q vs p (TSP)

Series2

-200

0

200

400

600

800

1000

1200

1400

1600

1800

0 5 10 15 20

q(kPa)

1 (%)

Graph of q vs 1 (%)

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From the graph of u vs 1, we observe that the graph intersects x-axis from positive

value to negative value. Therefore, we could suggest that the soil sample is heavily

consolidated clay. Due to high consolidation pressure (high normal effective pressure),

we could observe that the graph of q vs 1 almost becomes relatively flat when 1

increases. Therefore, we could ignore the peak ultimate shear strength.

We assume the last value in graph of q vs 1 is ultimate value although it has notreached a constant value yet. Therefore, ultimate shear strength, ultimate is 0.5 qultimate =

1593.2/2 = 796.60 kPa.

-50

0

50

100

150

200

250

300

350

400

450

0 5 10 15 20

u(kPa)

1 (%)

Graph ofu vs 1 (%)

y = 1.438x

0

200

400

600

800

1000

1200

1400

1600

1800

0 200 400 600 800 1000 1200 1400

q(kPa)

p' (kPa)

Graph of q vs p' (CU Test)

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Similar as CD Test, we could estimate the value of friction angle of critical state.

= sin 3 1.4386 + 1.438 = 35.5Since the soil sample used is dry medium-fine sand, we can assume the cohesion,

c=0.

Discussion

Based on the graph of q vs p, p, we could observe that the test was carried out under

a back pressure of approximately 200 kPa as there is a constant gap of value about

200 kPa between effective stress path (ESP) and total stress path (TSP) initially.

Based on the graph of 3vs e and 1 vs e, we could observe that the void ratio does

not change with increasing of both 3 and 1. It is because we have fixed theexperimental setup in undrained condition, which means that no water is allowed to

drain out. Consequently, the soil sample is compressed and shortened in axial

direction but expands in the lateral direction to maintain zero volume change during

shearing by applying axial loading. Hence, there is no change in volume, indicating

void ratio remains constant.

Next, based on the graph ofu vs 1, we could suggest the sample is overconsolidated

soil sample as the graph intersects x-axis and turns into negative value, which meansthat dilation have occurred. overconsolidated soil behaves similarly as dense soil. The

phenomena of experiencing dilation during shear could be explained in the following

way. If shear is induced, the soil particles will begin to move relative to one another.

For dense soil sample, the soil particles will have to over-ride other particles in order

to have relative movement. This contributes to the increase in void ratio and as a

result, water or air will flow into the soil sample, void ratio increases.

Then, based on the graph of q vs p and p, we could suggest that the sample is lightly

consolidated soil based on modified Cam Clay Model. We could see that TSP

increases with a ratio of 3 which is same as the theoretical value. For ESP, we could

see that its value decreases initially then increases until its value is almost equal to the

last value of TSP at the end of experiment. This could be explained in the following

way. Since the condition is undrained, any increase in vertical stress will not

contribute to effective stress. Therefore, TSP increases with gradient of 3 while ESP

remains as a vertical straight line until it touches the yield surface. Once ESP touches

the yield surface, soil sample will become elasto-plastic. Then, the ESP moves along

the roof of the yield surfaces. The size of yield surface becomes larger and larger

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because of a strain-hardening response. After touching the yield surface, ESP is no

longer a straight line but bent leftwards due to positive excess pore water pressure

generated. This means that p gradually decreases during undrained shearing because

soil tends to contract. After touching the CSL line, q tends increase along the CSL line,

making the difference between ESP and TSP is getting smaller, indicating the excess

pore water pressure is decreasing and eventually turns negative.

We could also interpret this phenomena based on graph of u vs 1. Initially, the pore

water pressure increases as the shearing stress is increased. But shortly after that the

pore water pressure starts to decrease, and then becomes negative 3 (suction), as the

shearing stress increases. When the ultimate shear strength is approached, the curve

levels out and the pore water pressure reaches its maximum negative value. (However,

we could not ensure the end result here shows pore water pressure has reached itsmaximum negative value as it may take longer time to achieve it). In other words, we

say that soil sample initially contract, thus excess pore water pressure increases. When

sample starts to dilate, excess pore water pressure in the soil sample will eventually

decreases, then effective stress will increase.

Compare the cs of CD and CU tests, 36.4 and 35.5 respectively, the values

obtained are more or less the same. It is because cs is fundamental property of soil.

It is a constant value as long as same soil sample is being tested. The slight differencein value could be due to improper handling of samples when doing triaxial test

leading further disturbing of soil samples.

Besides, we are able to obtain the value of undrained shear strength, su. Each Mohrs

circle of total stress is associated with a particular value of su because each test has a

different initial void ratio resulting from different confining pressure. For this CU test,

the undrained shear strength obtained is (1-3)f/2 = (1667.1)/2 = 833.55 kPa.

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Discussion on sources of error in Triaxial Test

i) Drained Tests

a) Rate of loading too fast, thus u0.

b) Ineffective seals in volume change system.

c) Calibration errors in volume change system.

d) Load loss in axial load piston due to poor lubrication.

e) Insensitivity of measurements at low strains due to high early soil stiffness.

ii) Undrained Test

a) Disturbance during sampling and preparation.

b) Air bubbles trapped between the soil and the rubber membrane or end-caps.

c) Rubber membrane is excessively thick or is punctured.

d) Poor water seals at ends; air bubbles in porewater line.e) Lateral stress developed across end-caps(these should be greased).

f) Soil not saturated, i.e. contains air which is compressible.

Conclusion

From CD Test, we are able to determine the drained shear strength parameters, peak=

40.1, cs = 36.4 and c = 0. While for CU Test, drained shear strength parameter

determined is cs = 35.5 and undrained shear strength parameter determined is su =

833.55 kPa.