LAB 5 Consolidation

UNIVERSITY PUTRA MALAYSIA

FACULTY OF ENGINEERING

CIVIL ENGINEERING DEPARTMENT

SOIL MECHANICS 2

(ECV3303)

LABORATORY WORK 5:

ONE DIMENSIONAL CONSOLIDATION TEST – OEDOMETER TEST

GROUP 8

NAME MATRIC NUMBERNOR SUHAIZA BINTI ABDUL RAHMAN 152191YUSOF AMANAH BIN MARINSAH 152864WAN MOHD HELMIE BIN WAN MEZAH 153712SITI AISYAH BINTI IBRAHIM 154612ROSMALIANA BINTI ZUBER 154796NURHAFIZA BINTI KAMARUDDIN 154936

COURSE : BACHELOR OF ENGINEERING (CIVIL)LECTURER : PROF. DR. BUJANG B.K HUATLAB DEMO : MR. HOSSEIN MOAYEDIDUE DATE : 22nd NOVEMBER 2011

1

5.1 INTRODUCTION

In soil mechanics, the settlement of cohesive soil comprise of three components:1. Immediate settlement2. Consolidation settlement3. Secondary compression (creep)

a) Immediate settlementIt occurs when saturated clay is loaded instantaneously and it is resulted in vertical deformation. The clay will deform and its pore water pressure will increase. Deformation will occur without any change in soil volume due to low permeability of the soil.

b) Consolidation settlementWhen the saturated soil is loaded, its volume will be reduced due to:

Compression of solid particles Compression of water in the soil void Drainage of water from soil voids

c) Secondary compression (creep)It occurs due to the reorientation of soil particles, creep or decay of organic matters. This settlement is not dependent on dissipation of pore water pressure.

Consolidation is the process of gradual transfer of an applied load from the pore water to the soil structure as pore water is squeezed out of the voids. The amount of water that escapes depends on the size of the load and compressibility of the soil, the rate at which it escapes depends on the coefficient of permeability, thickness, and compressibility of the soil. The rate and amount of consolidation with load are usually determined in the laboratory by the one-dimensional consolidation test. In this test, a laterally confined soil is subjected to successively increase vertical pressure, allowing free drainage from the top and bottom surfaces.

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.

Terzaghi’s theory of 1-D consolidation makes the following simplifying assumptions:

o The soil is homogeneous.o The soil is fully saturated.o The solid particles and the pore water are incompressible.o The flow of water and compression of soil are one-dimensional (vertical).o Strains are small.o Darcy’s law is valid at all hydraulic gradients.o The coefficient of permeability and the coefficient of volume compressibility remain

constant throughout the consolidation process.o There is a unique relationship, independent of time, between void ratio and

effective stress.

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5.2 THEORETICAL BACKGROUND

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 Terzaghi, soils are tested with an oedometer 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 pre-consolidation 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.

A cylindrical specimen of soil enclosed in a metal ring is subjected to a series of increasing static loads, while changes in thickness are recorded against time. From the changes in thickness at the end of each load stage the compressibility of the soil may be observed, and parameters measured such as Compression Index (Cc) and Coefficient of Volume Compressibility (mv). From the changes in thickness recorded against time during a load stage the rate of consolidation may be observed and the coefficient of consolidation (cv) measured. In this experiment, the sample is in disc shape, constraint on its side and applied with vertical load. Free drainage is allowed through top and bottom surface of the sample.

Time factor, Tv = cv.t

d2 at degree of consolidation, U = 90% and 50% and time factor,

Tv = 0.848 and 0.197 respectively.

Using Taylor & Merchant method (square root method) and Casagrande method (log method), we can find soil coefficient of consolidation, cv.

cv=T v d

2

t where, d = drainage path length = sample height (thickness) H/2

Soil coefficient of volume compressibility, mv

mv=∆e∆ p

11+eo

= ∆ HH og∆ P

where, ∆ e = average void ratio

eo = initial void ratio ∆ H= change in thickness

H o = initial length (thickness) of soil sample

3

http://en.wikipedia.org/wiki/Pore_water_pressure

http://en.wikipedia.org/wiki/Glaciers

http://en.wikipedia.org/wiki/Oedometer_tests

http://en.wikipedia.org/wiki/Stress_(physics)

http://en.wikipedia.org/wiki/Karl_Terzaghi

http://en.wikipedia.org/wiki/Karl_Terzaghi

http://en.wikipedia.org/wiki/Volume

http://en.wikipedia.org/wiki/Soil

∆ P= load (pressure) increment (kN/m2)

Coefficient of permeability, K=cvm vγw

5.3 OBJECTIVE

To determine the magnitude and rate of consolidation for saturated soil samples.

5.4 EQUIPMENTS

1. Metal cutting ring

2. Perforated plates (porous disc) – to be placed at top and bottom of soil sample3. Consolidation cell (consolidometer)4. Dial gauge with accuracy of 0.002mm and maximum travel of at least 6mm or equivalent

displacement transducer5. Loading apparatus6. Palette knife, wire saw, steel edge/ruler7. Moisture content apparatus8. Filter paper, silicone grease, evaporating disc

9. Stop watch10. Tools for determining soil density

4

5.5 PROCEDURES

Sample preparation

1. Ring and glass plates are cleaned and dried. They were weighed and recorded. A small amount of silicon grease is applied to the cutting ring.

2. The sample is placed on the glass plate. Some distilled water is added to the soil and they are mixed thoroughly using palette knives.

Test procedure

1. Bottom perforated plate (porous disc) is placed at the centre of consolidation cell. A filter paper is put on the plate then the cutting ring is placed with the sample in it.

2. The plate must first be saturated in water.3. They are placed on the load hanger on the consolidation cell. The arm is ensured to

be levelled.4. Load is gently placed on the hanger. Consolidation cell is filled with water after 2

minutes. Dial gauge is read for compression intervals of 6s, 15s, 1min, 2,25min, 4min, 6.25min, 9min, 12.25min, 16min, 20.25min, 20min, 25min, 36min,100min, and 24 hours. First applied load is 250kN/m2. After 24 hours, the load is increased to 500kN/m2

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5.6 RESULTS AND CALCULATION

a) Sample 1 (organic soil)

Table 1a: Dial gauge reading for organic soil

Time, t (min)

Time1/2 log time

Consolidation Pressure0.25kg (12.5 kPa) 0.5kg (25 kPa)

Dial gauge

reading (mm)

ΔH (mm)

Sample height (mm)

Strain, %

Dial gauge

reading (mm)

ΔH (mm)

sample height (mm)

Strain, %

0 0 - 0 0 20 0 2.3 0 17.845 0¼ 0.5 -0.6 0.8 0.8 19.2 4 2.42 0.12 17.725 0.672½ 0.707 -0.3 0.92 0.12 19.08 0.6 2.43 0.01 17.715 0.0561 1 0 1.13 0.21 18.87 1.05 2.44 0.01 17.705 0.056

1 ½ 1.225 0.176 1.22 0.09 18.78 0.45 2.45 0.01 17.695 0.0562 1.414 0.301 1.26 0.04 18.74 0.2 2.46 0.01 17.685 0.0563 1.732 0.477 1.27 0.01 18.73 0.05 2.47 0.01 17.675 0.0564 2 0.602 1.28 0.01 18.72 0.05 2.48 0.01 17.665 0.0565 2.236 0.699 1.29 0.01 18.71 0.05 2.5 0.02 17.645 0.1127 2.646 0.845 1.29 0 18.71 0 2.5 0 17.645 09 3 0.301 1.29 0 18.71 0 2.515 0.02 17.63 0.084

11 3.317 0.477 1.29 0 18.71 0 2.52 0.01 17.625 0.02813 3.606 0.602 1.3 0.01 18.7 0.05 2.525 0.01 17.62 0.02815 3.873 0.699 1.31 0.01 18.69 0.05 2.53 0.01 17.615 0.02820 4.472 0.845 1.31 0 18.69 0 2.545 0.02 17.6 0.08425 5 0.954 1.31 0 18.69 0 2.56 0.02 17.585 0.08430 5.477 1.041 1.31 0 18.69 0 2.57 0.01 17.575 0.05635 5.916 1.114 1.31 0 18.69 0 2.575 0.01 17.57 0.02840 6.325 1.176 1.31 0 18.69 0 2.582 0.01 17.563 0.03950 7.071 1.301 1.31 0 18.69 0 2.595 0.01 17.55 0.07360 7.746 1.398 1.311 0 18.689 0.005 2.62 0.03 17.525 0.1490 9.487 1.477 1.312 0 18.688 0.005 2.62 0 17.525 0

120 10.954 1.544 1.312 0 18.688 0 2.62 0 17.525 0180 13.416 1.602 1.312 0 18.688 0 2.621 0 17.524 0.006

1440 37.947 1.699 2.155 0.84 17.845 4.215 3.511 0.89 16.634 4.987

6

t90

0 5 10 15 20 25 30 35 4016.5

17

17.5

18

18.5

19

19.5

20

20.5

Thickness vs Time1/2

Time1/2 (min)

Thick

ness

(mm

)

Graph 1: Thickness vs Time1/2 for 12.5 kPa

0 5 10 15 20 25 30 35 4016

16.2

16.4

16.6

16.8

17

17.2

17.4

17.6

17.8

18


Time1/2 (min)

Thick

ness

(mm

)


7

t90

t50

t50

0 5 10 15 20 25 3016.5

17

17.5

18

18.5

19

19.5

20

20.5

Thickness vs log time

log time, t (min)

Thick

ness

(mm

)

Graph 3: Thickness vs log time for 12.5 kPa

0 5 10 15 20 25 3016

16.2

16.4

16.6

16.8

17

17.2

17.4

17.6

17.8

18


log time (min)

Thick

ness

(mm

)


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Density/Moisture Content Determination

Table 2a: Moisture content determination

Density Moisture Content(1) Weight of ring & soil 211.506 g Container number(2) Weight of ring 82.759 g Weight of container + moist soil 38.80 gWeight of initial moist sample (M) 129 g Weight of container 14.38 gInitial weight density, ρ

ρ=MV o

= (1 )−(2)π r2h

= 0.212−0.083

π (0.03752 ) (0.035 )

= 0.129

1.5463−4

834.28 kg/m3

Weight of moisture 24.42 g

Initial dry density

ρd=ρ

1+wo

= 834.281+0.1656

= 715.75 kg/m3

715.75 kg/m3

Weight of dry soil 20.95 g

Moisture content (w)

w=www s

= 24.42−20.9520.95

= 16.56 %

16.56 %

Soil particle specific gravity, Gs = assume to be 1.3

Weight of solid, W s=M1+w

= 0.129

1+0.1656

= 0.1107 kg

Volume of soil, V s=W s

G sρw

= 0.11071.3(1000)

= 8.52 x 10-5 m3

Volume of water, Vw = Vo - Vs

= = (1.5463×10¿¿−4)−(8.52×10¿¿−5)¿¿

9

= 6.493×10−5 m3

Initial void ratio, eo=V wV s

= 6.493×10−5

8.52x 10−5

= 0.76

Degree of saturation, Sr = eowGs

= 0.76(0.1656) (1.3)= 0.1636 ×100= 16.36 %

Table 3a: Coefficient of volume compressibility, mv determination for Organic soil

Pressure, P (kPa)

Change in thickness, ∆H (mm)

Initial thickness, Ho

(mm)

Pressure increment, ∆P (kPa)

Coefficient of volume compressibility,

mv=∆ HH o

1∆P

12.5 2.155 20.000 12.5 8.62 x 10-3

25.0 1.211 17.845 12.5 5.43 x 10-3

Table 4a: Coefficient of consolidation,cv determination for organic soil

Pressure range (Pa)

Average thickness, Ho

(mm)

Drainage path length,

d = Ho/2

Square root method Log time method

t90 (min)

Coefficient of consolidation,

cv=0.848d2

t 90(mm2/min)

t50 (min)


cv=0.197 d2

t50(mm2/min)

0 - 12.5 20.000 10.00 31.5 2.69 4.0 4.92512.5 -25.0 17.845 8.92 34.3 1.97 24.5 0.640

- Coefficient of permeability, k (for square root method)

*Taking for pressure = 0.0025 kPaK=cvm vγw = 2.69 (8.62 x 10-3)(9.81) = 0.227 mm/min

*Taking for pressure = 0.0050 kPaK=cvm vγw = 1.97(5.43 x 10-3)(9.81) = 0.105 mm/min

- Equivalent solid height, H s=V s

A

H s=8.52 x10−54.26×10−3

¿0.02m¿20.0mm

- Void ratio after test

e f=H s−H f

H f

¿ 20−18.3818.38

¿0.062¿6.2%

- Final degree of saturationS f=e f wGs¿0.062 (0.1854 ) (1.3 )¿0.0133¿1.33 %

10

For 12.5 kPa:

Δe = mv∆ P(1+e0) = 8.62 x 10-3 (12.5)(1+0.76) = 0.19

Δe = e0 – e1

e1 = 0.76 – 0.19 = 0.57

For 25 kPa:

Δe = mv∆ P(1+e0) = 5.43 x 10-3 (25)(1+0.76) = 0.24

Δe = e0 – e1

e1 = 0.76 – 0.24 = 0.52

Pressure, σ’ (kPa) Log σ’ Void ratio, e0 - 0.76

12.5 1.097 0.5725 1.398 0.52

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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Void ratio vs log σ'

log σ'

Void

ratio

, e

Graph 9: Void ratio, e vs log σ’

b) Sample B (clay)

Table 1b: Dial gauge reading for clay

Time, t (min) log time

Consolidation Pressure

0.25kg (12.5 kPa) 0.5kg (25 kPa)Dial

gauge reading (mm)

ΔH (mm)

Sample height (mm)

Strain, %

Dial gauge

reading (mm)

ΔH (mm)

sample height (mm)

Strain, %

0 0 - 0 0 20 0 1.8 0 18.38 0

¼ 0.5 -0.60206 0.34 0.34 19.66 1.7 2.14 0.34 18.04 1.849837

½ 0.707 -0.30103 0.5 0.16 19.5 0.8 2.25 0.11 17.93 0.598477

1 1 0 0.65 0.15 19.35 0.75 2.3 0.05 17.88 0.272035

1 ½ 1.225 0.17609 0.8 0.15 19.2 0.75 2.31 0.01 17.87 0.054407

2 1.414 0.30103 0.88 0.08 19.12 0.4 2.32 0.01 17.86 0.054407

3 1.732 0.477121 0.98 0.1 19.02 0.5 2.39 0.07 17.79 0.380849

4 2 0.60206 1.028 0.048 18.972 0.24 2.41 0.02 17.77 0.108814

5 2.236 0.69897 1.05 0.022 18.95 0.11 2.43 0.02 17.75 0.108814

7 2.646 0.845098 1.07 0.02 18.93 0.1 2.44 0.01 17.74 0.054407

9 3 0.30103 1.095 0.025 18.905 0.125 2.45 0.01 17.73 0.054407

11 3.317 0.477121 1.11 0.015 18.89 0.075 2.46 0.01 17.72 0.054407

13 3.606 0.60206 1.12 0.01 18.88 0.05 2.46 0 17.72 0

15 3.873 0.69897 1.125 0.005 18.875 0.025 2.47 0.01 17.71 0.054407

20 4.472 0.845098 1.13 0.005 18.87 0.025 2.48 0.01 17.7 0.054407

12

25 5 0.954243 1.14 0.01 18.86 0.05 2.485 0.005 17.695 0.027203

30 5.477 1.041393 1.145 0.005 18.855 0.025 2.49 0.005 17.69 0.027203

35 5.916 1.113943 1.148 0.003 18.852 0.015 2.49 0 17.69 0

40 6.325 1.176091 1.15 0.002 18.85 0.01 2.49 0 17.69 0

50 7.071 1.30103 1.15 0 18.85 0 2.491 0.001 17.689 0.005441

60 7.746 1.39794 1.152 0.002 18.848 0.01 2.498 0.007 17.682 0.038085

90 9.487 1.477121 1.157 0.005 18.843 0.025 2.51 0.012 17.67 0.065288

120 10.954 1.544068 1.159 0.002 18.841 0.01 2.511 0.001 17.669 0.005441

180 13.416 1.60206 1.159 0 18.841 0 2.511 0 17.669 024

hours 37.947 1.69897 1.62 0.461 18.38 2.305 3.132 0.621 17.048 3.378672

0 5 10 15 20 25 30 35 4017.5

18

18.5

19

19.5

20

20.5


Time1/2 (min)

Thick

ness

(mm

)


13

t90

0 5 10 15 20 25 30 35 4016

16.5

17

17.5

18

18.5


Time1/2

Thick

ness

(mm

)


0 5 10 15 20 25 3017.5

18

18.5

19

19.5

20

20.5


log time, t (min)

Thick

ness

(mm

)


14

t90

t50

0 5 10 15 20 25 3016

16.5

17

17.5

18

18.5


log time, t (min)

Thick

ness

(mm

)


Density/Moisture Content Determination

Table 2b: Moisture content determination

Density Moisture ContentWeight of ring & soil 167.400 g Container numberWeight of ring 83.361 g Weight of container + moist soil 50.24 gWeight of initial moist sample (M) 84.039 g Weight of container 14.44 gInitial weight density, ρ

ρ=MV o

= (1 )−(2)π r2h

= 0.1674−0.08336π (0.03752 ) (0.035 )

= 0.0840

1.5463−4

543.48 kg/m3Weight of moisture 35.80 g

Initial dry density456.71 kg/m3 Weight of dry soil 30.20 g

15

t50

ρd=ρ

1+wo

= 543.481+0.19

= 456.71 kg/m3

Moisture content (w)

w=www s

= 35.80−30.2030.20

= 18.54 %

18.54 %

Soil particle specific gravity, Gs = assume to be 1.3

Weight of solid, W s=M1+w

= 0.08401+0.1854

= 0.0709 kg

Volume of soil, V s=W s

G sρw

= 0.07091.3(1000)

= 5.45 x 10-5 m3

Volume of water,

Vw = Vo - Vs

= (1.5463×10¿¿−4)−(5.45×10¿¿−5)¿¿ = 1.0013×10−4 m3

Initial void ratio, eo=V wV s

=1.0013×10−4

5.45×10−5

= 1.837

Degree of saturation, Sr = eowGs

= 1.837(0.1854) (1.3)= 0.4428×100 = 44.28 %

Table 3b: Coefficient of volume compressibility, mv determination for Clay

Pressure, P (Pa) Change in thickness, ∆H (mm)

Initial thickness, Ho

(mm)Pressure

increment, ∆P (Pa)

Coefficient of volume

compressibility,

mv=∆ HH o

1∆P

12.5 1.620 20.00 12.5 6.48 x 10-3

25.0 1.332 18.38 12.5 5.80 x 10-3

Table 4b: Coefficient of consolidation,cv determination for Clay

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Pressure range (kPa)

Average thickness, Ho

(mm)

Drainage path length,

d = Ho/2

Square root method Log time method

t90 (min)


cv=0.848d2

t 90(mm2/min)

t50 (min)


cv=0.197 d2

t50(mm2/min)

0 - 12.5 20 10.00 28.5 2.975 5.00 3.9412.5 - 20.0 18.38 9.19 32.5 2.204 11.5 1.45

- Coefficient of permeability, k (for log time method)



- Equivalent solid height, H s=V s

A

H s=5.45×10−5

2.76×10−3

¿0.0197m¿19.7mm

- Void ratio after test

e f=H s−H f

H f

¿ 19.7−18.3818.38

¿0.07¿7%

- Final degree of saturationS f=e f wGs¿0.007 (0.1854 ) (1.3 )¿1.687×10−3

¿0.16 %

For 12.5 kPa:

Δe = mv∆ P(1+e0) = 6.48 x 10-3(12.5)(1+1.837) = 0.230

Δe = e0 – e1

e1 = 1.837 – 0.230 = 1.607

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For 25 kPa:

Δe = mv∆ P(1+e0) = 5.80 x 10-3(25)(1+1.837) = 0.411

Δe = e0 – e1

e1 = 1.837 – 0.411 = 1.426

Pressure, σ’ (kPa) Log σ’ Void ratio, e0 - 1.837

12.5 1.097 1.60725 1.398 1.426

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Void ratio vs log σ'

log σ'

Void

ratio

, e

Graph 10: Void ratio,e vs log σ'

5.7 DISCUSSION

In this test, we found out that we lack some information. Because of that, some data demand cannot be fulfilled. This led to error in calculation and thus, affects the result for the overall test.

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From the result we get, we found out that consolidation and settlement is increase evenly with time. From time to time, the sample is consolidated and settled. The further details can refer to the graph (Graph 1 to Graph 10).

For determination of soil coefficient of consolidation, cv, we are using Using Taylor & Merchant method (square root method) and Casagrande method (log method). While soil coefficient of volume compressibility, mv can be found by using equation of:

mv=∆e∆ p

11+eo

= ∆ HH og∆ P

And Coefficient of permeability, K=cvm vγw

All of the calculation and result are shown in Table 2a/b, Table 3a/b and Table 4a/b.

5.8 CONCLUSION AND RECOMMENDATION

We can conclude that this experiment is successful since we are able to determine the magnitude and rate of consolidation for saturated soil samples. But there are some error occurs since the value we get is not realistic.

A summary of the result is:

- Organic soil

Pressure (kPa)

Coefficient of consolidation, cv for time1/2

(mm/min)


cv for log time(mm/min)


compressibility, mv

Coefficient of permeability, K

for time1/2

(mm/min)


for log time (mm/min)

12.5 2.69 4.925 8.62 x 10-3 0.227 0.41625.0 1.97 0.640 5.43 x 10-3 0.105 0.003

- Clay

Pressure (kPa)

Coefficient of consolidation, cv for time1/2

(mm/min)


cv for log time(mm/min)


compressibility, mv


for time1/2

(mm/min)


for log time (mm/min)

12.5 2.975 3.94 6.48 x 10-3 0.189 0.2525.0 2.204 1.45 5.80 x 10-3 0.125 0.08

Precaution that need to take care of:1. Weighed the weight of sample correctly2. Make sure we read the lab manual carefully and noted what we have to take

note in this test.3. List down data needed properly

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5.8 REFERENCE1. http://en.wikipedia.org/wiki/Consolidation_(soil)2. Bujang B.K Huat, Faisal Hj. Ali. (2008). Essential Soil Mechanics for Engineers. Universiti

Putra Malaysia, Serdang: Malaysia.3. C. Venkatramaiah. (2006). Geotechnical Engineering. 3rd Ed. New Age International

Publishers

5.9 APPENDICES

Figure 1: Sample preparation

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http://en.wikipedia.org/wiki/Consolidation_(soil)

LAB 5 Consolidation

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