U3 Consolidation test.docx

8/14/2019 U3 Consolidation test.docx

1/23

FACULTY OF CIVIL AND ENVIRONMENTAL

ENGINEERING

DEPARTMENT OF GEOTECHNICAL AND

TRANSPORTATION ENGINEERING

LAB GEOTECHNIC

FULL REPORT

Subject Code BFC 31901Code & Experiment Title U4 DIRECT SHEAR AND UNCONFINED COMPRESSION TESTCourse Code 3 BFFDate 19thNovember 2012Section / Group SECTION 10 / GROUP 3

Name Yong Hui Yen AF100227Members of Group Yeo Shi Wei AF100234Cheong Chin Lin AF100233

Nur Hanisa bt Hussin AF100093

Nurul Afizah bt Mahmud AF100084

Nur khairunnisa Mohd Nazrudin AF100067

Lecturer/Instructor/Tutor EN MOHD FIRDAUS B. MD DAN@ AZLAN

Received Date 10nd

December 2012

Comment by examiner Received


2/23

Title: Consolidation Test

1.0OBJECTIVE To determine the consolidation characteristics of the peat soils and clay of

low permeability.

2.0LEARNING OUTCOMEAt the end of this experiment, students are able to:

Conducted one dimensional consolidation test Identified the factors causes soil consolidation Determined the consolidation parameters ( cv, mv, Ccand Pc)

3.0THEORYWhen a fully saturated soil is exerted to the compressive stress, its volume

tends to decrease. The volume of the soil in the ring decrease is due to the

compression of the solid grains and escape of water from the voids. In a free

drainage soil such as saturated sand the escape of water can take place rapidly.

But in clay, due to low permeability, the movement of water occurs very much

slowly and therefore, considerable time may be required for excess water to be

squeezed out to permeable boundaries.

Settlement is the direct result of the decrease in soil volume and

consolidation is the rate of volume decrease with time. The consolidation test is

use to estimate the amount of settlement and time of consolidation. From this

test some consolidation parameters such as coefficient of consolidation( cv ),

coefficient of volume compressibility( mv ), compression index( Cc ),

pre-consolidation pressure ( Pc ), can be determined.


3/23

The methods which can be used for determining the coefficient of consolidation:

(i) Casagrande or log (time) or 50% consolidation

(ii) Taylor or time or 90% consolidation

The coefficient of consolidation can be determined by the equation:

Where :

= coefficient of consolidation (m2/year)

= Time factor

H = Maximum length of drainage path (m)

t = Time to achieve 50% or 90% consolidation ( year or minute )

The test was performed to determine the magnitude and rate of volume decrease

that a laterally confined soil specimen undergoes when subjected to different vertical

pressures. From the data, the consolidation curve (pressure-void ratio relationship)

can be plotted. The data obtained was useful in determined the compression index,

the recompression index and the pre-consolidation pressure (or maximum past

pressure) of the soil. In addition, the data obtained can also be used to determine

the coefficient of consolidation and the coefficient of secondary compression of the

soil.

The main purpose of consolidation tests was to obtain soil data which could be used

in predicting the rate and amount of settlement of structures founded on clay.

Although some of the settlement of a structure on clay may be caused by shear

strain, most of it was normally due to volumetric changes. It was particularly true if

the clay stratum was thin compared to the width of the loaded area or the stratum

which located at a significant depth below the structure.


4/23

The four most important soil properties furnished by a consolidation test are:

(i) The pre-consolidation stress, Sp,, which is the maximum stress that the soilhas felt in the past.

(ii) The compression index, Cc, which indicates the compressibility of anormally-consolidated soil.

(iii) The recompression index, Cr , which indicates the compressibility of anover-consolidated soil.

(iv) The coefficient of consolidation, Cv, which indicates the rate of compressionunder a load increment.


5/23

4.0 EQUIPMENTS & MATERIALS

1) Consolidation apparatus Consolidation ring Corrosionresistant porous plate Consolidation cell Dial Gauge Loading device

2) Balance readable to 0.1g

3) Vernier calliper

4) Stop-clock readable to 1s

5) Soil Type (Clay & Peat soil)
https://www.facebook.com/photo.php?fbid=4863749000144&set=o.211452272318378&type=1&relevant_count=1&ref=nfhttps://www.facebook.com/photo.php?fbid=4863753000244&set=o.211452272318378&type=1&relevant_count=1&ref=nfhttps://www.facebook.com/photo.php?fbid=4863749000144&set=o.211452272318378&type=1&relevant_count=1&ref=nfhttps://www.facebook.com/photo.php?fbid=4863753000244&set=o.211452272318378&type=1&relevant_count=1&ref=nfhttps://www.facebook.com/photo.php?fbid=4863749000144&set=o.211452272318378&type=1&relevant_count=1&ref=nfhttps://www.facebook.com/photo.php?fbid=4863753000244&set=o.211452272318378&type=1&relevant_count=1&ref=nfhttps://www.facebook.com/photo.php?fbid=4863749000144&set=o.211452272318378&type=1&relevant_count=1&ref=nfhttps://www.facebook.com/photo.php?fbid=4863753000244&set=o.211452272318378&type=1&relevant_count=1&ref=nf


6/23

5.0PROCEDURES

1. The internal diameter (D) and the height of the ring (H)were measuredby the internal vernier calipers.

2. The ring weighed to the neartest 0.01g (mR).3. The specimenswere cutted and trimmed into ring had been measured

dimensions.

4. The initial moisture content was determined from trimming soil.5. The weight of ring and specimen (m1) were determined.6. The mass of bulk specimen (m) to the nearest 0.01g was determined by

using the equation

m = m1mR

7. The consolidation ring and specimen (cutting edge uppermost) wereplaced centrally on the porous disc.

8. The ring retainer and cell body were fitted and the upper porous discwas placed centrally on top of the specimen.

9. The consolidation cell was placed centrally in position on the platform ofthe machine base.

10. The end of the beam was lifted to allow the loading yoke to be raised tothe vertical position and the loading stem was adjusted by screwing it

downwards until the end engages closely in the recess on the top of the

loading cap.
https://www.facebook.com/photo.php?fbid=4863766720587&set=o.211452272318378&type=1&relevant_count=1&ref=nf


7/23

11. The compression dial gauge was attached to the arm on the supportpost.

12. Weight (2.5 kg) was carefully added to the load hanger.13. Water was added at room temperature to the cell and the specimen and

upper porous disc had to submerged completely.

14. The beam support was wind down and at the same time the clock wasstarted.

15. The compression gauge readings and the clock were observed, and thereadings on a consolidation test form was recorded at the selected time

intervals.

16. The readings of the compression against time was plotted to alogarithmic scale and against square-root-of-time
https://www.facebook.com/photo.php?fbid=4863757360353&set=o.211452272318378&type=1&relevant_count=1&ref=nfhttps://www.facebook.com/photo.php?fbid=4863757360353&set=o.211452272318378&type=1&relevant_count=1&ref=nfhttps://www.facebook.com/photo.php?fbid=4863757360353&set=o.211452272318378&type=1&relevant_count=1&ref=nf


8/23

6.0 RESULTS

5.1 Data for Consolidation TestSpecimen 1( Peat Soil )

Date started: 5/11/2012 Sample No.: 1

Soil Type : Peat Soil Cell No: 2.5 kg

BEFORE TEST

Moisture content from trimming: - ( % ) S.G. (Assumed) : 2.7

Weight of ring : 106.40( g ) Diameter of ring : 74.70( mm )

Weight of sample + ring : 182.35 ( g ) Area of ring :4382.59( mm2)

Weight of sample :75.95( g ) Thickness of ring:19.15( mm )

Weight of dry sample : 53.40( g ) Volume of ring : 83926.60 ( mm3)

Weight of initial moisture : 22.55( g ) Density, :0.91( Mg/m3)Initial moisture content: 42.23( % ) Dry density, d: 0.64 ( Mg/m

3)

Initial void ratio, = 3.22

Table 6.1.1

Table 6.1.2

1d

sG

Elapse time

Time(min) time

Clock

time

Gauge

reading

Cumulative

compression, H (mm)hr min sec - - -

0 0 0 1205 0 0

10 0.167 0.409 441.0 0.882

20 0.333 0.577 569.0 1.138

30 0.500 0.707 620.0 1.240

40 0.667 0.817 664.0 1.328

50 0.833 0.913 746.0 1.492

1 1 1 1206 756.0 1.512

2 2 1.414 1207 827.0 1.6544 4 2 1209 883.0 1.766

8 8 2.828 1213 931.0 1.862

15 15 3.873 1220 964.0 1.928

30 30 5.477 1235 989.0 1.978

1 60 7.746 1305 1013.5 2.027

2 120 10.954 1405 1031.0 2.062

24 1440 37.947 1205 1117.0 2.234


9/23

6.2 Data for Consolidation TestSpecimen 2 ( Clay )

Date started : 5/11/2012 Sample No. : 1

Soil Type : Clay Cell No : 2.5 kg

BEFORE TEST

Moisture content from trimming : - ( % ) S.G. (Assumed) : 2.7

Weight of ring : 108.35( g ) Diameter of ring : 74.92( mm )

Weight of sample + ring :248.55 ( g ) Area of ring : 4408.45( mm2

)

Weight of sample : 140.20 ( g ) Thickness of ring :19.30( mm )

Weight of dry sample : 113.40( g ) Volume of ring :85083.09( mm3)

Weight of initial moisture : 26.80( g ) Density, : 1.65 ( Mg/m3

)

Initial moisture content : 23.63( % ) Dry density,d

:1.33 ( Mg/m3)

Initial void ratio, = 1.03

Table 6.2.1

Table 6.2.2

1d

sG

Elapse timeTime

(min)

time

Clock

time

Gauge

reading

Cumulative

compression, H (mm)

hr min sec - - -0 0 0 1215 0 0

10 0.167 0.409 167.5 0.335

20 0.333 0.577 235.0 0.470

30 0.500 0.707 250.0 0.500

40 0.667 0.817 320.0 0.640

50 0.833 0.913 336.0 0.672

1 1 1 1216 375.0 0.750

2 2 1.414 1217 422.5 0.845

4 4 2 1219 455.0 0.910

8 8 2.828 1223 575.0 1.150

15 15 3.873 1230 725.0 1.450

30 30 5.477 1245 950.0 1.900

1 60 7.746 1315 1199.0 2.398

2 120 10.954 1415 1363.0 2.726

24 1440 37.947 1215 1540.0 3.080


10/23

7.0 CALCULATION

7.1 Calculation for Specimen 1 ( Peat soil) :

7.1.1 Weight of sample

= (weight of sample + ring)(weight of ring)

= 182.35-106.40

= 75.95 g

7.1.2 Weight of initial moisture

= (weight of sample)(weight of dry sample)

= 75.9553.40

= 22.55 g

7.1.3 Initial moisture content = %100sampledryofweight

moistureinitialofweight

= %10040.53

55.22 g

g

= 42.23 %

7.1.4 Area of ring= 4

2D

= 4

)70.74( 2

= 4382.59 mm2


11/23

7.1.5 Volume of ring =4

2tD

=4

15.19)70.74( 2

= 83926.60 mm3

7.1.6 Density, = ringofvolume

sampleofweight

=39-

6

m1083926.60

Mg1075.95

= 0.91Mg/m3

7.1.7 Density, d= ringofvolume

sampledryofweight

=39-

6

m1083926.60

Mg1053.40

= 0.64 Mg/m3

7.1.8 Initial void ratio, 1d

sG

= 164.0

7.2

= 3.22


12/23

7.2 Calculation for Specimen 2( Clay) :

7.2.1 Weight of sample

= (weight of sample + ring)(weight of ring)

= 248.55-108.35

= 140.20 g

7.2.2 Weight of initial moisture

= (weight of sample)(weight of dry sample)

= 140.20-113.40

= 26.80 g

7.2.3 Initial moisture content = %100sampledryofweight

moistureinitialofweight

= %10040.11380.26

gg

= 23.63 %

7.2.4 Area of ring =4

2D

=4

)92.74( 2

= 4408.45 mm2


13/23

7.2.5 Volume of ring =4

2tD

=4

30.19)92.74( 2

= 85083.09 mm3

7.2.6 Density, = ringofvolume

sampleofweight

=39-

6

m1085083.09

Mg10140.2

= 1.65 Mg/m3

7.2.7 Density, d =ringofvolume

sampledryofweight

=39-

6

m1085083.09

Mg10113.40

= 1.33 Mg/m3

7.2.8 Initial void ratio, 1d

sG

= 133.1

7.2

= 1.03


14/23

SETTLEMENT READINGS

The gauge readings are multiplied with 0.002 mm to obtain cumulative compression,

H (mm). For example:

For, t = 10 sec, H = 441 0.002 mm = 0.882 mm

t = 20 sec, H = 569 0.002 mm = 1.138mm

t = 30 sec, H = 620 0.002 mm = 1.240 mm

(Refer to table 6.1.2)

The cumulative compression, H(mm)is divided with 0.002 mm to obtain gauge

readings. For example:

For, t = 10 sec, gauge reading = 0.335 0.002 mm = 167.5

t = 20 sec, gauge reading = 0.470 0.002 mm = 235.0

t = 30 sec, gauge reading = 0.500 0.002 mm = 250.0

(Refer to table 6.2.2)

8.0DATA ANALYSIS

Two graphs of consolidation curve are plotted from the results obtained. Using Taylors method, graph of settlement, H against square root of time,

t for peat soil and clay is plotted. Meanwhile, using Casagrandes method,

graph of settlement, H against time, t is plotted on semi-log graph paper.

The graphs can be viewed in the following page


15/23

9.0DISCUSSIONS

The aim of this experiment is to study the characteristic of the soil when

consolidation process takes place and obtain the sedimentation from this experiment.

Other than that, from this experiment we are able to determine the magnitude and

rate volume decrease that a laterally confined soil specimen undergoes.

When the saturated soil has been exposed to the same pressure, the volume will

be decrease and because the soil particle and water cannot compress the value of

compression unless if water exit from the spaces between the particle. This situation

will decrease the size of the spaces between the particle and soil particle become

more compact.

In this experiment show that consolidation process happened. It is because of

the movement horizontally as a volume rotation that call sediment. Two graphs had

been plotted for specimen 1 (peat soil) and specimen 2 (clay), which is graph of

settlement against log time with logarithmic scale and graph of settlement against

square root time.

As know, the values of the sediment increase as the time increase. This shows

that sediment is linearly proportional to the time. Consolidation takes most of the

sediment as a loading. After time t90the process continues slowly. This phenomenon

happens because of transformation moisture membrane that covers the solid

particle.


16/23

There are some errors maybe occur during the experiment and affect theaccuracy and precision of the recorded data, which are:

a) Disturbed soil specimen or excessive disturbance during trimmingb) Specimen not fitting into and filling the ringc) Permeability of the porous stones too lowd) Excessive friction between the specimen and ringe) Inappropriate load during inundationf) Improper specimen heightg) Parallax error

To reduce the value of the error, we must have methods for the step, it is:-

a.

the sample of the soil must be compacted

b. choose the best point of sampling on sitec. the value of water is not too muchd. using a larger diameter and thinner specimen


17/23

10.0 CONCLUSION

As a conclusion, there are 2 graphs which are graph of settlement against log

time with logarithmic scale and graph of settlement against square root time had

been plotted for each specimen of peat soil and clay.

For the specimen of peat soil, the t50 = 1.5 minutes has determine from

settlement against log time and thet90= 6.83 minutes has determine from graph of

settlement against square root time. By using taylor method, the calculated value of

coefficient of consolidation is Cv= yearm /98.46

2

while by using casagrande method,

the calculated value of coefficient of consolidation is Cv= yearm /328.62

.

For the specimen of clay, the t50= 2.2 minutes has determine from settlement

against log timeand the t90= 12.78 minutes has determine from graph of settlement

against square root time. By using taylor method, the calculated value of coefficient

of consolidation is Cv =

yearm /98.46 2

while by using casagrande method, the

calculated value of coefficient of consolidation is Cv= yearm /383.42

.

Overall, the characteristic of soil of low permeability has been determined.

This experiment has been done successfully.


18/23

11.0 QUESTIONS

QUESTION 1

1. From your experimental data, determine the coefficient of consolidation, cv(m2/year)using Casagrande method. Please comment your answer.

For Specimen 1Peat Soil:

t50 = 1.5, D = 19.15 mm, H = D/2

50

2)(197.0

t

HCv

)5.1(

)2/15.19(197.0 2

min/041.12 2mm

Cv= mm2

min

m

1000mm

m

1000mm

60 min

1 hour

24 hour

1 day

365 day

1 year

= 6.328 m2/year

For Specimen 2Clay:

t50 = 2.2 min, D = 19.30 mm, H = D/2

50

2)(197.0

t

HCv

)2.2(

)2/30.19(197.0 2

min/339.8 2mm

Cv= mm2

min

m

1000mm

m

1000mm

60 min

1 hour

24 hour

1 day

365 day

1 year

= 4.383 m2/year

From the calculation above, it shows that the coefficient of consolidation, c vfor peat soil is

bigger than clay per year.


19/23

2. A Clay sample collected from 5 metres deep in BatuPahat has a unit weight () of18kN/m3. The following data were recorded during an oedometer test.

i. Plot the graph of void ratio against effective stress on semi-log graph and determine thecompression index (Cc), pre-consolidation pressure (Pc) and coefficient of volume

compressibility (mv).

The compression index (Cc) is the slope of the graph

Cc = gradient of the graph

= 1

2

21

logP

P

ee

= 200

400log

43.057.0

= 0.465

From graph, we obtained: Pre-consolidation pressure, Pc= 68kN/m2

Coefficient of volume compressibility, mv= avge

e

1

1'

'

eslope of the graph

2

1 s

avg

eee

675.0

2

5.085.0

Effective Stress (kN/m2) 25 50 100 200 400 800 200 50

Void ratio (e) 0.85 0.82 0.71 0.57 0.43 0.3 0.4 0.5


20/23

mv=avge

e

1

1'

675.01

1)465.0(

278.0

ii. Define whether the soil is normally consolidated or over consolidated.

D = 5 m

P0 =d

= 18 5

= 90 kN/m2

Over consolidation, OCR = 0P

Pc

=90

68

= 0.76< 1

Since, OCR


21/23

QUESTION 2

1. From the experimental data, determine the coefficient of consolidation, cv(m2/year) using Taylors method. Please comment on your answer.

For Specimen 1peat soil:

t90 = 6.83, D = 19.15 mm, H = D/2

90

2)(848.0

t

HCv

)83.6(

)2/15.19(848.0 2

min/383.11 2mm

Cv= mm2

min

m

1000mm

m

1000mm

60 min

1 hour

24 hour

1 day

365 day

1 year

= 5.983 m2/year

For Specimen 2clay:

t90 = 12.78 D = 19.30 mm, H = D/2

90

2)(848.0

t

HCv

)78.12(

)2/30.19(848.0 2

min/179.6 2mm

Cv= mm2

min

m

1000mm

m

1000mm

60 min

1 hour

24 hour

1 day

365 day

1 year

= 3.247 m2/year

From the calculation above, it shows that the coefficient of consolidation, c vfor peat

soil is bigger than clay per year.


22/23

2. Clay samples collected form 10 metres deep in Parit Raja has a unit weight () of20 kN/m3. The following data were recorded during an oedometer test.

i. Plot the graph of void ration against effective stress on semi-log graph and

determine the compression index (Cc), preconsolidation (Pc) and coefficient of

volume compressibility (mv).

The compression index (Cc) is the slope of the graph

Cc = gradient of the graph

= 1

2

21

logP

P

ee

=

200

800log

3.057.0

= 0.448

From graph, we obtained: Pre-consolidation pressure, Pc= 106.5kN/m2

Coefficient of volume compressibility, mv= avge

e

1

1'

'

eslope of the graph

2

1 s

avg

eee

775.02

6.095.0

Effective Stress(kN/m2) 50 100 200 400 800 1600 400 100

Void ratio (e) 0.95 0.92 0.81 0.67 0.53 0.4 0.5 0.6


23/23

mv=avge

e

1

1'

775.01

1)448.0(

252.0

ii. Define whether the soil is normally consolidated or over consolidated.

D = 5 m

P0 =d

= 20x10

= 200 kN/m2

Over consolidation, OCR = 0P

Pc

=200

106.5

= 0.53< 1

Since, OCR

Documents

U3 Consolidation test.docx