1 Consolidation

SOIL CONSOLIDATION

Unsuitable ground

1.0 What Is Consolidation

What Is Consolidation

unlike

Effect of on

START Soil Particle

may be taken up by soil particles if they are allowed to move and rearrange. However,

was not taken up by soil particle as movement of solid particles are restraint by water in the pores (water is incompressible); hence pore water is the one that picking up the

Pore water

is taken up by pore water

ue = us +

where ue = excess pore water pressure

us = static pore water pressure (before P

is applied)(atmospheric pressure)

1.1 Consolidation Theory

AT TIME t

Soil Particles

Rearranging while pore water dissipating (flowing out)

is being transferred gradually to soil particle.

Pore Water

Pore water draining out due to existence of pressure difference (ue and us). This process is known as dissipation of pore water pressure and will continue until pore water pressure ue reduces to uss (steady state pore water pressure)

The soil at this state is said in drained condition. Before dissipation of pore water, the soil is said to be in undrained condition. The time required for pore water to dissipate depends on permeability of the soil.

Consolidation Theory

COMPLETE Soil Particles

New particle arrangement established.

All is taken up by soil particles

Pore water

All been completely transferred to soil particle 0 +

Dissipation of pore water pressure is completed where ue has reached uss

Soil is said to be in drained condition


Consolidation Analogy - 1-Dimensional

water pore water valve flow control (soil permeability) Load P pressure/external load spring soil solid particle

Spring and piston system

State of the System Spring Pore Water

Equilibrium 0 u0

Applied Load P () ; closed valve

0 u0 +

Applied Load P () ; opened valve ( = a + b)

0 + a u0 + b

Complete 0 + u0

For clay soil only


Consolidation process stages of , and u

Consolidation Analogy - 1-Dimensional


So.. Consolidation

Definition: Gradual reduction in volume of fully saturated soil of low permeability due to drainage of some of the pore water, the process continuing until the excess pore eater pressure set up by an increase in total stress has completely dissipated. End results: Consolidation settlement which can be measured by recording levels at reference points on the ground surface Consolidation progress measurement: Through measuring changes in pore water pressure using piezometer.

BS 1377-Part5: 1990

At the end of the test , draw curves of Consolidation settlement against time

Void ratio against stress

1-D Consolidation Test Oedometer Test

https://www.youtube.com/watch?v=ByER2wnnei4 https://www.youtube.com/watch?v=3bvevFBNYw0&list=PLiD_qZ1L2hL4vdbNFbRLLgIb06Wzn8VNd

https://www.youtube.com/watch?v=bZTN8z0EIrM

Calculation Formulae Water content measured at the end of the test = w1 Void ratio at the end of the test = e1 = w1 Gs (assuming Sr = 1) Thickness of the specimen at start of test = H0 Change in the thickness during the test = H Void ratio at start of test = e0 = e1 + e where

0

01

H

e

H

e

1.2 Oedometer Test

Initial compression Expansion recompression

During compression, changes in soil structure continuously take place and the clay does not revert to the original structure during expansion.

1.2.1 Compressibility Characteristics

Compressibility of Clay can be represented by Coefficient of volume compressibility (mv) Compression Index (Cc)

Coefficient of compressibility (av) Coefficient of compressibility = change in void ratio change in effective stress

''' 01

10

eeeav

Compressibility of Clay

00 1 e

e

V

V

Coefficient of volume compressibility (mv)

therefore '

1.

1 0

e

emv

or

01 e

am vv

Definition : the volume change per unit volume per unit increase in effective stress (m2/MN)

(volume strain)


volume change per unit volume =

Compressibility Index (Cc )

Compressibility index is the slope of the linear portion of the plot e -log

'

'log

0

1

1

eeC oc

The slope on the expansion part is known as Expansion index Ce. The value of Cc for normally consolidated soil can be related to liquid limit (wL) where

%)10(0009.0 Lc wC

For low plasticity soil or porous rock , Cc is given by

)(75.0 aeCc

Where e = void ratio of undisturbed sample a = constant depending on soil type


Reading from oedometer test on saturated clay sample are as below

Presure (kN/m2) 0 50 100 200 400 0

Layer thickness (mm) 18.5 18.02 17.86 17.35 17.22 17.75

At the end of the loading, load has been removed and the sample has been allowed to expand for 24 hours and the thickness of the sample at the end is 17.75mm. Moisture content and specific gravity are 28.5% and 2.65 respectively. Plot e - curve Tfind the coefficient of volume compressibility, mv for the stress range between 90 kN/m2 an d260 kN/m2

Example 1

Overconsolidated Soil Soil that has experienced vertical effective stress es greater than its existing vertical effective stress Overconsolidated soil may occur in a situation such as after melting of ice cap, erosion of overburden and an increase in water table. The ratio by which the current vertical effective stress in the soil was exceeded in the past is called overconsolidation ratio,

'1

'0 /OCR

'

1

'

0 /OCR

Overconsolidated Clay

'1

'0 /OCR

Preconsolidated Pressure (c) is the maximum vertical effective stress that a soil was subjected to in the past It can be measured through e log curve for oversonsolidated clay

1.2.2 Preconsolidated Pressure

'1

'0 /OCR

Procedure of Determination 1. Determine and lable points A, B and C. Point B being the meeting

point between the straight and the curve part of the whole curve. 2. Extend back the straight line BC of the curve. 3. Determine the point D on curve AB; where point D is the

maximum point on curve AB. 4. Draw the tangent through D. 5. Draw horizontal line through D. 6. Draw a line that divide the angle between horizontal line and

tangent into two equal parts (the bisector line). 7. Find the intersection point between the bisector line and the

extended line BC. 8. Read the pressure as an approximation of preconsolidated

pressure c . Usage c In construction, intended applied pressure on

overconsolidated clay must be less than c in order to avoid large settlement.

Determination of Preconsolidated Pressure

'1

'0 /OCR

The slope of virgin compression line for in-situ soil is slightly greater than the slope of virgin compression line obtained from the lab test due to the fact that the soil has been disturbed during sample preparation. However, the void ratio at the beginning of the lab test (e0) is the same for both in-situ and lab samples . Lab virgin line may be expected to intersect the in-situ virgin line at ~0.42 e0

In- situ e log curve

'1

'0 /OCR

Draw horizontal line representing e0 and 0.42e0

Draw vertical line c until it crosses the horizontal line e0 (lable this intersection point as E)

Extend downwards the straight line of the lab curve until it crosses the horizontal line 0.42e0 (lable this intersection point as F)

Connect EF which represent the straight part of the in-situ curve (in-situ virgin consolidation line)

Construction of In-situ consolidation line

In-situ virgin consolidation line

'1

'0 /OCR

Lable 0 as current effective stress

Draw vertical line 0 until it crosses horzontal line e0 (lable this intersection point as G)

Draw a parallel line with the average slope of the lab curve part (GH) from G

In-situ overconsolidated line

Construction of In-situ consolidation line

'1

'0 /OCR

The following compression readings were obtained in an oedometer test on a specimen of saturate clay (Gs = 2.73)

Pressure (kN/m3) 0 54 107 214 429 858 1716 3432 0

Dial gauge after 24 hrs (mm)

5.000 4.747 4.493 4.108 3.449 2.608 1.676 0.737 1.480

The thickness of the specimen was 19.0 mm and at the end of the test the water content was 19.8%. Plot e - log curve Determine the preconsolidation pressure Determine the values of mv for stresses increment 100-200 kN/m2 and 1000-1500 kN/m2. What is the value of Cc for the latter increment?

Example 2

'1

'0 /OCR

1-Dimensional

The term consolidation settlement can be derived from mv or Cc

For overconsolidated clay Given

'1

1

0

e

emv

or 00

'1 H

Hm

e

ev

0' HmH v

or Hms vc '

e e0

1

H

H0

1.3 Consolidation Settlement

'1

'0 /OCR

For overconsolidated clay

e e0

1

H

H0

If mv and do not change with depth so Sc can be written as

He

eesc

0

10

1

If changed linearly, therefore the the change in Sc can be written as

dzmds

H

vc '

0

Where dz = thickness of one element of soil

Consolidation Settlement

1-Dimensional

'1

'0 /OCR

Given

'

'log

'

'log

0

1

0

1

10

eeeCc

'

'log

0

0

cCe

He

eesc

0

10

1

Insert in e into Sc

He

Cs cc

'0

'1

0

log1


1-Dimensional For normally consolidated clay

'1

'0 /OCR

For normally consolidated soil

If the change in mv and Sc with depth need to be considered , Sc can be calculated using graphical procedure as shown below


1-Dimensional

'1

'0 /OCR

A building is supported on a raft 45m x 30m, the net foundation pressure (assumed to be uniformly distributed) being 125 kN/m2. the soil profile is as shown below. The value of mv for clay is 0.35 m2/MN. Determine the final settlement under the centre of the raft due to consolidation of the clay.

Example 3

'1

'0 /OCR

A large embankment will be constructed on a site with the soil profile and characteristics as shown in diagram below. Water table is static and at the ground level. Embankment cosntruction will induce change in stress uniformly at 150 kN/m2. based on the given data, calculate the final settlement of the embankment. Given unit weight of water w is 9.81 kN/m

3. Assume the changes occur instantaneously.

Example 4

'1

'0 /OCR

The progress of consolidation process can be known from the following expression:

10

0

ee

eeU z

where Uz = degree of consolidation at a perticular instant of time at depth z e0 = void ratio before consolidation started e1 = void ratio at the end of consolidation e = void ratio at the time in question ; during consolidation

How to measure e in-situ?

1.4 Degree of Consolidation

'1

'0 /OCR

If e - curve is assumed to be linear, therefore, degree of consolidation can be written as

'0

'1

'0'

zU

During consolidation = -u and relationship between and u can be written as

uui ''

0'1

therefore

ii

iz

u

u

u

uuU

1

Linear relationship e-

Degree of Consolidation

'1

'0 /OCR

The theory relates three quantities: u excess pore water pressure z depth below the top of the clay layer t time from the instantaneous application of a a total stress

increment

Assumptions: The soil is homogeous and fully saturated Soil particles and water are incompressible Compression and flow are one-dimensional (vertical) Strains are small Darcys law is valid at all hydraulic gradients The value of k and mv remains constant through out the process There is a unique relationship, independent of time, between e and .*

1.5 Terzaghi 1-D Consolidation Theory

'1

'0 /OCR

The differential consolidation equation can be mathematically written as :

2

2

z

uk

t

um

wv

or

2

2

z

uc

t

uv

where

vwwv

va

ek

m

kc

)1(

and k = coeff. of permeability e = void ratio w = unit weight of water

(coefficient of consolidation; m2/year)

Terzaghi 1-D Consolidation Theory

'1

'0 /OCR

Excess pore water pressure at depth z and time t is given by

vm

m

i TMd

Mz

M

uu 2

1

expsin2

where

2d

tcT vv (time factor)

nM2

.

and n is odd d half of clay layer thickness ui is the pore water pressure at t=0

1.5.1 Solution of Consolidation Equation

Isochrones

'1

'0 /OCR

vm

m

i TMd

Mz

M

uu 2

1

expsin2

Plot u at any time t garisan semasa

The shape is determine by

Initial distribution of u

Drainage conditions at the boundaries of the clay layer (open or half closed)

Degree of consolidation at depth z and time t is given by

vm

m

iz TM

d

Mz

M

uU 2

1

expsin2

1

In practical problems, the average degree of consolidation (U) over the depth of the layer for constant ui is

vm

m

TMM

U 2

02

exp2

1

The above equation can be represented almost exactly by the following empirical equation For U < 0.60 or 60%,

For U > 0.60 or 60%, 085.0)1log(9333.0 UTv

2

4UTv

Degree of consolidation

Relationship between U and Tv

In an oedometer test on clay sample of 20mm thick, 90% consolidation been achieved in 30 minutes. Based on this test data, calculate the coefficient of consolidation of the clay tested. On-site, the thickness of the clay layer is 3.5 m. Calculate the time required for the clay layer to consolidate to 50% and 90% if the stratum of the clay layer free to allow the flow of water through: Top and bottom layers Bottom layers only.

Example 5

'

'log

0

1

1

eeC oc

'

1.

1 0

e

emv

Void ratio - stress

Oedometer Test

Dial gauge reading - time

50

2196.0

t

dcv

90

2848.0

t

dcv

1.5 Determination of Coefficient of Consolidation

Curve Fitting

50

2196.0

t

dcv

The figure shown curves obtained from oedometer test and from theory. Theoretical curve consist parabolic part (1) followed by a linear part (2) and a final curve which the horizontal axis is an asymtote (3) at U = 1.0 (100%)

1.5.1The Log Time Method (Casagrande)

Procedures to find cv from experimental curve:

Lable initial dial gauge reading as a0 Select two points A and B which the

t ratio is 4:1. Measure vertical distance between A

and B . Extend same vertical distance from

point A upwards. Lable dail gauge reading from this

extension as as. Draw tangent on the linear part (2)

and another tangent on the asymptote curve part (3)

Lable dial gauge reading on the intersection as a100

Take the midpoint between as and a100 as a50.

Determine the corresponding t value for a50 as t50.

The Log Time Method (Casagrande)

Curve Fitting

If U = 50% then

196.05.04

2

vT

Therefore

50

2196.0

t

dcv

Where d = average thickness of the sample for a particular stress increment

The Log Time Method (Casagrande)

Curve Fitting

Theoretical curve is linear for almost 60% of consolidation and at 90% consolidation. Straight line AC=1.15AB. This characteristic is used for determination of point on experimental curve corresponding to U=90%.

Curve Fitting

1.5.2 The Root Time Method (Taylor)

Procedures for determining cv from experimental curve:

Draw tangent on the linear part of the curve and extend the tangent line until it crosses the vertical axis.

Mark the intersection point between the tangent and vertical axis as D.

Draw line DE where E crosses the experimental curve and the horizontal distance from this point to the vertical axis is 1.15xFG

Label point E as a90 and the corresponding value for t90

F G

Curve Fitting


F G

If U = 90% therefore

848.0085.0)9.01log(933.0 vT

Hence

90

2848.0

t

dcv


Curve Fitting

Where d = average thickness of the sample for a particular stress increment

Data in the table below is obtained from an oedometer test on fully saturated clay when the pressure is doubled from 800 to 1600 kN/m2.

After 1440 minutes (24 jam) passed, the final thickness of the speciment is 22.15 mm. Specific gravity is 2.72. calculate Coeff. consolidation (cv) using Root Time and Log time methods. Coeff. of volume compressibility mv Coeff. Of permeability k

Time (t) Dial gauge reading (mm) Time (t) Dial gauge reading (mm)

0.00 2.000 9.00 2.501

0.25 2.132 16.00 2.632

0.50 2.162 25.00 2.740

0.75 2.185 36.00 2.812

1.00 2.206 49.00 2.860

1.50 2.240 64.00 2.891

2.25 2.281 81.00 2.913

3.00 2.310 100.00 2.930

4.00 2.355 144.00 2.951

5.00 2.392 256.00 2.978

6.25 2.430 1100.00 3.017

7.50 2.461 1440.00 3.024

Example 6

Hydraulic Oedometer

Soil sample with diameter 250mm with thickness 100mm is considered adequate to give constant value of cv .

The settlement rate for full scale structure often bigger than those estimated using cv obtained from oedometer test where the sample size is smaller. This is due to effects of soil fabric for example existence of root and layers of silt and fine sand.

1.6 Hydraulic Oedometer

Compression does not cease when the excess pore pore eater pressure has dissipated to zero but continues at a gradually decreasing rate under constant effective stress

This compression is known as secondary compression and is due to thye gradual readjustment of the clay particles into a more stable configuration caused by decrease in void ratio.

The rate of secondary compression is determined by the viscosity of the adsorp water around clay mineral particles.

During secondary compression, the flow of adsorp water is very slow. The viscosity of the adsorp water increases when clay particles move

closer together resulting in the secondary compression reduces further. Normally, secondary compression is assumed to had occued concurrently

with the main compression.

Secondary Compression

C

)/log( pscs ttdCS

Where d = initial thickness of the clay layer, ts = time when compression is to be determined, tp = time to complete the premier compression.

Secondary Compression

The following compression readings are taken while oedometer test being carried out on fully saturated clay sample (Gs = 2.73) when the applied pressure was increases from 214 to 429 kN/m2.

Time (min) 0 0.25 0.5 1 2.25 4 9 16 25

Dial gauge reading (mm)

5.00 4.67 4.62 4.53 4.41 4.28 4.01 3.75 3.49

Time (min) 36 49 64 81 100 200 400 1440

Dial gauge reading (mm)

3.28 3.15 3.06 3.00 2.96 2.84 2.76 2.61

After 1440 min. the thickness of the sample became 13.60mm and water content 35.9%. Determine the coeff. Of consolidation from log time and root time

methods. Determine the coeff. of permeability.

Example 7

During construction total load are not applied instantaneously

Foundation- excavation there is a reduction in the net pressure

Structure structural load applied exceed the weight of soil excavated consolidation settlement

To take into account changes in the applied load during construction period, correction method on time-settlement curve need to be carried out.

1.7 Correction for Construction Period

Correction procedures 1. Mark the construction period tc; a point on time axis when net load P is

initially applied on the on soil layer experiencing consolidation. 2. Mark a point tc /2 on instantaneous curve and lable this point as A. 3. From point A, draw horizontal line and lable it as B where B is the

corresponding point on tc. 4. Select time t where t< tc on instantaneous curve . 5. Mark point C on instantaneous curve where t at that time is t/2 6. Draw horizontal line from C till it crosses vertical line tc and lable it as

point D. 7. Connect point D to O and lable the intersection between OD and

vertical line at time t as point E. 8. Repeat procedure 4,5,6 and 7 to obtained corrected curve before time

tc 9. The corrected curve after tc (beyond point B), can be constructed

through projecting points on instantaneous curve horizontally as much as tc /2

Correction for Construction Period

Correction for Construction Period

An 8m depth of sand overlies a 6m layer of clay, below which is an impermeable stratum. (as shown in the diagram). The water table is 2m below the surface of the sand. Over a period of 1 year a 3m depth of fill (unit weight 20 kN/m3) is to be dumped on the surface over an extensive area. The saturated unit weight of the sand is 19 kN/m3 and that of the clay is 20 kN/m3. Above the water table the unit weight of the sand is 17 kN/m3 . Given that Cc is 0.32. For the clay, the relationship between void ratio and effective stress (units kN/m2) can be represented by the following equation

100

'log32.088.0

e

and the coefficient of consolidation is 1.26m2/year Calculate the final settlement of the area due to consolidation of the clay and the settlement after a period of 3 years from the start of dumping. If a very thin layer of sand, freely draining, existed 1.5m above the bottom of the clay layer, what would be the values of the final and 3 year settlement.

Example 8

Example 8

Method to speed up the flow of water from the soil mass so that maximum settlement can be accelerated for highly compressible soil (increase the rate of settlement).

Involved provision of bore holes filled with sand (traditional method)

Requires position of boreholes to be designed to ensure high flow rate normally square or triangle arrangement.

The flow distance must be smaller than the thickness of the clay layer

If the clay layer is thin, the use of vertical drain is uneconomical

1.7 Vertical Drain

requires Ch and Cv,values, horizontal and vertical and coefficient of consolidation

Normally Ch/Cv ~ 1 and 2 or the higher the better.

Disadvantages : act as weak piles

It has been shown that the vertical drains is not successful / effective if the soil is having high ratio of secondary settlement for example high plasticity clay or peat; where the secondary settlement cannot be controlled with vertical drains.

Vertical Drain Design

Equations for degree of consolidation can be written as follows

)( vv TfU and )( rr TfU

with Uv is the average of degree of consolidation due to vertical drainage only and Ur is the average of degree of consolidation due to horizontal drainage (radial) only.


Therefore , the time factor is

2d

tcT vv (time factor due to vertical drainage)

24R

tcT hh (time factor due to horizontal drainage)


The expression for Tr confirms the fact that the closer he spacing of the drains the quicker the consolidation process due to radial drainage proceeds. The solution for radial, due to Barron, is given in the next, the Ur / Tr relationship depending on the ratio n=R / rd, where R is the radius of the equivalent cylindrical block and rd iis the radius of the drain. It can be shown that

)1)(1(1 rv UUU

where U is the average degree of consolidation under combined vertical and radial drainage..


Solution for radial consolidation


Vertical Drain

1 Consolidation

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