FULL REPORT consolidation.docx

TITLE

1.0

: CONSOLIDATION TEST

INTRODUCTION

Consolidation is a process by which soils decrease in volume. According to Karl Terzaghi consolidation is any process which involves decrease in water content of a saturated soil without replacement of water by air. In general it is the process in which reduction in volume takes place by expulsion of water under long term static loads. It occurs when stress is applied to a soil that causes the soil particles to pack together more tightly, therefore reducing its bulk volume. When this occurs in a soil that is saturated with water, water will be squeezed out of the soil. The magnitude of consolidation can be predicted by many different methods. In the Classical Method, developed by Karl von Terzaghi, soils are tested with an odometer test to determine their compression index. This can be used to predict the amount of consolidation. When stress is removed from a consolidated soil, the soil will rebound, regaining some of the volume it had lost in the consolidation process. If the stress is reapplied, the soil will consolidate again along a recompression curve, defined by the recompression index. The soil which had its load removed is considered to be over consolidated. This is the case for soils which have previously had glaciers on them. The highest stress that it has been subjected to is termed the preconsolidation stress. The over consolidation ratio or OCR is defined as the highest stress experienced divided by the current stress. A soil which is currently experiencing its highest stress is said to be normally consolidated and to have an OCR of one. A soil could be considered under consolidated immediately after a new load is applied but before the excess pore water pressure has had time to dissipate.

Consolidation is defined as the reduction of the volume of a soil due to the expulsion of water. This will be accompanied by the dissipation of pore water pressures. A laboratory consolidation test is performed on an undisturbed sample of a cohesive soil to determine its compressibility characteristics. The soil sample is assumed to be representing a soil layer in the ground. A conventional consolidation test is conducted over a number of load increments. The number of load increments should cover the stress range from the initial stress state of the soil to the final stress state the soil layer is expected to experience due to the proposed construction. Increments in a conventional consolidation test are generally of 24 hr duration and the load is doubled in the successive increment. In this practical class one load increment of a multi increment consolidation test is conducted and the data will be analysis to obtain the compressibility characteristics of the soil. The compressibility characteristics of the soil are; (a). Parameters needed to estimate the amount of consolidation settlement (b). Parameters needed to estimate the rate of consolidation settlement in the field.

Using the data from a single load increment of the test, only the coefficient of volume compressibility mv can be estimated. Data from all the load increments should be combined to draw the e vs log σ graph and to obtain the compression index Cc - the other parameter used to estimate the consolidation settlement. The rate of consolidation settlement is estimated using the Coefficient of consolidation Cv. This parameter is determined for each load increment in the test. In this laboratory assignment, the coefficient of consolidation should be estimated using two methods the root time method (Taylor's method) and the log (time) method - Casagrande's method.

2.0

OBJECTIVE To determine the consolidation characteristics of soils of low permeability

3.0

THEORY

When a fully saturated soil is subjected to a compressive stress, its volume tends to decrease. The decreasing of its volume is due to 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), preconsolidation pressure (Pc) can be determined.

There are two methods for determining the coefficient of consolidation: (i) (ii)

Casagrande or log (time) or 50% consolidation Taylor or √ time or 90% consolidation

The coefficient of consolidation can be determined by this equation,

cv 

Tv H 2 t

Where, cv = coefficient of consolidation (m2/year) Tv = Time factor H = Maximum length of drainage path (m) t = Time to achieve 50% or 90% consolidation (year or minute)

Square Root Time (minute) 5 √t90 10

0

15

20

25

0

Settlement (mm)

5 10 15 20 x 25 30

1.15x

1

2

Figure : Settlement versus log Time

30

35

40

Time (minute) 0.1

1

t50

100

1000

δ δ

20 40

A

60

50% consolidation line

80

B = 4A

100 120

100% consolidation line

140 160

Figure : Settlement versus square root time

4.0

10000

0% consolidation line

0

Settlement (mm)

10

TEST EQUIPMENTS

1. Consolidation apparatus - Consolidation ring - Corrossion-resistant porous plate - Consolidation cell - Dial Gauge - Loading device

2. Balance readable to 0.1g 3. Vernier caliper 4. Stop-clock readable to 1 s

5.0

PROCEDURES

1. 2. 3. 4. 5. 6.

The internal diameter (D) and the height of the ring was measured by using internal vernier calipers. The ring was weighed to the nearest 0.01g (mR). The specimen was cut and was trim into ring. The initial moisture content from trimming soil are determined. The weight of ring and specimen (m1) are determined. The mass of bulk specimen (m) to the nearest 0.01 g was determined using this equation m = m 1 – mR

7. 8. 9. 10.

11. 12. 13. 14. 15. 16.

The consolidation ring and specimen (cutting edge uppermost) was placed centrally on the porous disc. Fit the ring retainer and cell body was fitted and then the upper porous disc was placed centrally on top of the specimen. The consolidation cell was placed centrally in position on the platform of the machine base. The end of the beam are lifted to allow the loading yoke to be raised to the 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 The compression dial gauge was attached to the arm on the support post. Weight (2.5 kg) was added carefully to the load hanger Water was added at room temperature to the cell and make sure that the specimen and upper porous disc are completely submerged. Wind down the beam support and at the same time start the clock. The compression gauge readings and the clock was observed, and the readings was recorded on a consolidation test form at the selected time intervals. The readings of the compression against time to a logarithmic scale and against square-root-time are plotted.

7.0

RESULTS AND CALCULATION

Date started: 17/2/2011

Sample No: 1

Soil Type: Clay

Load: 2.5kg / 50kN/m2

BEFORE TEST

Moisture content for trimming : 60.76 %

S.G (Assumed) : 2.7

Weight of ring: 121.0 g

Diameter of ring: 75.0 mm

Weight of sample + ring: 265.7 g

Area of ring: 4417mm2

Weight of sample: 144.7 g

Thickness of ring: 16 mm

Weight of dry sample: 102.6 g

Volume of ring : 70672 mm3

Weight of initial moisture: 42.1 g

Density, p : 2.05 Mg/m3 Dry density d: 1.45 Mg/m3

Initial void ratio, Gs -1 d

= 0.862

SETTLEMENT READINGS

Elapsed Time Hr

Min

Clock Sec

Time (min)

Gauge √time

Reading

Cumulative Compression

0

0

time (pm)

1

0

0

0

12.00

10

0.17

0.41

112

0.22

20

0.33

0.57

120

0.24

30

0.50

0.71

280

0.56

40

0.67

0.82

285

0.57

50

0.83

0.91

300

0.60

1

1

1

12.11

325

0.65

2

2

1.41

12.12

380

0.76

4

4

2

12.14

440

0.88

8

8

2.83

12.18

464

0.92

15

15

3.87

12.25

580

1.16

30

30

5.48

12.40

716

1.43

60

7.75

1.10

780

1.56

CALCULATION

Weight of sample = Weight of sample + ring - Weight of ring = 265.7g – 121.0g = 144.7g

Weight of initial moisture = Weight of sample - Weight of dry sample = 144.7g – 102.6 g = 42.1 g

Initial moisture contents = Weight of initial moisture / Weight of dry sample = 42.1/102.6 = 0,41 x 100% = 41% Area of ring = πD2/4 = π (75.0) 2/4 = 4417 mm2

Volume of ring = Area of ring x Thickness of ring = 4417 x 16 = 70672 mm3

Density,  = Weight of sample (ring) Volume of ring = 144.7 x 10 -6(Mg) 70672 x 10 -9(m3) = 2.05 (Mg/m3)

Dry density, d

= Weight of dry sample Volume of ring =

102.6 x 10 -6(Mg) 70672 x 10 -9 (m3)

= 1.45 (Mg/m3)

Date started: 17/2/2011

Sample No: 2

Soil Type: Clay

Load: 5.0 kg / 100 kN/m2

BEFORE TEST

Moisture content for trimming : 60.76 %

S.G (Assumed) : 2.7

Weight of ring: 108.6 g

Diameter of ring: 75.0 mm

Weight of sample + ring: 254.0 g

Area of ring: 4417mm2

Weight of sample: 145.4 g

Thickness of ring: 16 mm

Weight of dry sample: 109.2 g

Volume of ring : 70672 mm3

Weight of initial moisture: 36.2 g

Density, p : 2.06 Mg/m3 Dry density d: 1.55 Mg/m3

Initial void ratio, Gs -1 d

= 0.742

SETTLEMENT READINGS

Elapsed Time Hr

Min

Clock Sec

Time (min)

Gauge √time

Reading

Cumulative Compression

0

0

time (pm)

1

0

0

0

12.15

10

0.17

0.41

185

0.37

20

0.33

0.57

220

0.44

30

0.50

0.71

240

0.48

40

0.67

0.82

245

0.49

50

0.83

0.91

275

0.55

1

1

1

12.16

285

0.57

2

2

1.41

12.17

315

0.63

4

4

2

12.19

403

0.806

8

8

2.83

12.22

521

1.042

15

15

3.87

12.30

656

1.312

30

30

5.48

12.45

666

1.332

60

7.75

1.15

682

1.364

CALCULATION

Weight of sample = Weight of sample + ring - Weight of ring = 254.0g – 108.6g = 145.4 g

Weight of initial moisture = Weight of sample - Weight of dry sample = 145.4 g – 109.2 g = 36.2 g

Initial moisture contents = Weight of initial moisture / Weight of dry sample = 36.2 /109.2 = 0.331 x 100% = 33.1% Area of ring = πD2/4 = π (75.0) 2/4 = 4417 mm2

Volume of ring = Area of ring x Thickness of ring = 4417 x 16 = 70672 mm3

Density,  = Weight of sample (ring) Volume of ring = 145.4 x 10 -6(Mg) 70672 x 10 -9(m3) = 2.06 (Mg/m3)

Dry density, d

= Weight of dry sample Volume of ring =

109.2 x 10 -6(Mg) 70672 x 10 -9 (m3)

= 1.55 (Mg/m3)

OPEN ENDED QUESTIONS

QUESTIONS 1

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

Sample 1 : Load 2.5 kg (clay soil)

Graph settlement versus log time 1.8 1.6 1.4 Settlement (mm)

8.0

1.2 1 0.8 0.6 0.4 0.2 0 0.1

1

10

100 Time (minute)

Cv = 0.197 H² t50 = 0.197 (0.005)²mm 2min = 4.925 x 10-6 2 = 2.463 x 10-6 mm2/min

Cv = 2.463 x 10-13 (

= 1.294 x 10-6 m2/year

)

Sample 2 : Load 5.0 kg (clay soil)

Graph settlement versus log time 1.6 1.4

Settlement (mm)

1.2 1 0.8 0.6 0.4 0.2 0 0.1

Cv = 0.197 H² t50 = 0.197 (0.005)²mm 2.5 min = 4.925 x 10-6 2.5 = 1.97x 10-6 mm2/min

1

10

100 Time (minute)

Cv = 1.97 x 10-12 (

)

= 1.01 x 10-6 m2/year

2.) Clay samples collected from 5 metres deep in Batu Pahat has a unit weight () of 18 kN/m3. The following data were recorded during an oedometer test.

Effective Stress (kN/m2) 25

50

100

200

400

800 200 50

Void ratio (e)

0.82

0.71

0.57

0.43

0.3

0.85

0.4

(i) Plot the graph of void ratio against effective stress on semi-log graph and determine the compression index (Cc), Preconsolidation pressure (Pc) and coefficient of volume compressibility (mv)

void ratio

void ratio againts effective stress effective stress , kN/m2 1

10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

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

0.5

Cc = slope of the graph = 0.82 – 0.53 log(800/200) = 0.482

From graph, we obtained: Preconsolidation pressure, Pc=150kN/m2

Coefficient of volume compressibility, mv e 1 = '  1  eavg e  slope of the graph  '

eavg 

e1  e s 2

= (0.85 + 0.5 ) / 2 = 0.675

mv =

e 1 '  1  eavg

= (0.482) (1/ 1 + 0.675) = 0.288

(ii) Define whether the soil is normally consolidated or over consolidated. D = 10m P 0=   d = 18  10 = 180kN/m2

D = 10m

Pc P0 = 150/180 = 0.83 < 1

Overconsolidation, OCR=

The soil is over consolidated , OCR