Kuliah Konsolidasi 1

Consolidation

Skeletal Material(incompressible)

Pore water(incompressible)

Voids

Solid

Initial State

The consolidation process

Skeletal Material(incompressible)

Pore water(incompressible)

Voids

Solid

Voids

Solid

Initial State Deformed State


Water

+

Deformation of saturated soil occurs by reduction of pore space & the squeezing out of pore water. The water can only escape through the pores which for fine-grained soils are very small


Deformation of saturated soil occurs by reduction of pore space & the squeezing out of pore water. The water can only escape through the pores which for fine-grained soils are very small


Effective soil skeleton “spring”

water

water squeezed out


water

Instantaneously no water can flow, and hence there can be no change in volume.


water


For 1-D conditions this means

zz = v = = 0 (1)


water


For 1-D conditions this means

zz = v = = 0 (1)

and hence ´ = 0 instantaneously


water

From the principle of effective stress we have

´ + u (2)

and thus instantaneously we must have

u

Region of highexcess water pressure

Region of lowexcess water pressure

Flow


The consolidation process is the process of the dissipation of the excess pore pressures that occur on load application because water cannot freely drain from the void space.

TotalStress

Time


TotalStress

Time

Time

ExcessPorePressure


EffectiveStress

Time


EffectiveStress

Time

Settlement

Time


vv

zzz

z

PlanArea A

Elevation

vz

z

Derivation of consolidation governing equation

1. Water flow (due to consolidation)

vv

zzz

z

PlanArea A

Elevation

vz

zRate at which waterleaves the element

v

zzA


1. Water flow (due to consolidation)

v

tzA

Rate of volume decrease


2. Deformation of soil element (due to change in effective stress)

PlanArea A

Elevationz

Rate at which waterleaves the element

Rate of volume decreaseof soil element =

v

zzA

v

tzA


Assume: Soil particles and water incompressible

Rate at which waterleaves the element

Rate of volume decreaseof soil element =

v

zzA

v

tzA

v

z v

t(3)Storage Equation


Assume: Soil particles and water incompressible

v kh

zv

Assume Darcy’s law

(4)


3. Flow of water (due to consolidation)

v kh

zv

Assume Darcy’s law

(4)


3. Flow of water (due to consolidation)

Note that because only flows due to consolidation are of interest the head is the excess head, and this is related to the excess pore pressure by

hu

w

(5)

Elastic response v v em (7)

Assume soil behaves elastically


4. Stress, strain relation for soil


Assume soil behaves elastically


4. Stress, strain relation for soil

Note that mv has to be chosen with care. It is not a universal soil constant. For 1-D conditions it can be shown that

(9)


5. Principle of effective stress

Note that these are changes in stress due to consolidation

(8)

v

z v

t (3)Storage Equation

v kh

zv Darcy’s law (4)


+

+


5. Principle of effective stress

Note that these are changes in stress due to consolidation

(8)

Equation of 1-D Consolidation

z

k u

zm

u

t tv

wv

e[ ] [ ] (10)


Very Permeable

Very Impermeable

At a very permeable boundary

u = 0

At a very impermeable boundarySaturated Clay

u

z 0

Solution of consolidation equation

1. Boundary conditions

At the instantof loading

u e


2. Initial conditions (1-D)

TotalStressChange

Time

Time

ExcessPorePressure

(10)


3. Homogeneous soil

z

k u

zm

u

t tv

wv

e[ ] [ ]

(13)

cv is called the coefficient of consolidation



cv has units L2/T and can be estimated from an oedometer test.The procedure will be explained in the laboratory sessions.




The coefficient of volume decrease mv can be measuredfrom the oedometer test.




The coefficient of volume decrease mv can be measuredfrom the oedometer test.

The value of kv is difficult to measure directly for clays butcan be inferred from the expression for cv.


Uniformly distributed surcharge q

2HZ Homogeneous Saturated Clay Layer freeto drain at Upper and Lower Boundaries

Solution of consolidation equation for 2 way drainage

Governing Equation

cu

z

u

tv

2

2 (14a)


Governing Equation

Boundary Conditions

cu

z

u

tv

2

2

u = 0 when z = 2H for t > 0

u = 0 when z = 0 for t > 0

(14a)

(14 b,c)


Governing Equation

Boundary Conditions

Initial Condition

cu

z

u

tv

2

2

u = 0 when z = 2H for t > 0

u = 0 when z = 0 for t > 0

u = q when t = 0 for 0 < z < 2H

(14a)

(14 b,c)

(14d)


u q Z

where

and

Zz

H

Tc t

H

nn

nTv

n

vv

21

1

2

0

2

2

sin( )e

(n )

Solution

(15)


T=0.8 0.5 0.3 0.2 0.1

0

1

2

0.0 0.5 1.0

Z=z/H

u/q

Variation of Excess pore pressure with depth


Calculation of settlement

S vdzH

0

2


S vdzH

mv e u dzH

0

2

0

2( )


S vdzH

mv e u dzH

fromwhich it can be shown

S

SU Tv

nTv

n

e

0

2

0

2

1 2

2

20

( )

( )

(16c)

10-3 10-2 10-1 1 10

Dimensionless Time Tv

0.00

0.25

0.50

0.75

1.00

U

Relation of degree ofsettlement and time

Approximate Expressions for Degree of Settlement

UT

T

U e T

vv

Tvv

40 2

18

0 22

2 4

( . )

( . )/

Uniformly distributed surcharge q

HZ Homogeneous saturated clay layerresting on an impermeable base

Impermeable


Impermeable

Governing Equation

cu

z

u

tv

2

2 (18a)


Governing Equation

Boundary Conditions

cu

z

u

tv

2

2

u=0 when z = H for t > 0

u = 0 when z = 0 for t > 0

(18a)

(18b,c)u

z 0


Governing Equation

Boundary Conditions

Initial Condition

cu

z

u

tv

2

2

u=0 when z = H for t > 0

u = 0 when z = 0 for t > 0

u = q when t = 0 for 0 < z < H

(18a)

(18b,c)

(18d)

u

z 0


T=0.8 0.5 0.3 0.2 0.1

0

1

20.0 0.5 1.0

Z=z/H

u/q



T=0.8 0.5 0.3 0.2 0.1

0

1

20.0 0.5 1.0

Z=z/H

u/q



Solution is identical to that for 2 way drainage. Note that the maximum drainage path length is identical.

Gravel

4mClay

Clay

Sand

5m

Impermeable

Clay

Final settlement=100mm cv=0.4m2/year

Soil Profile

Final settlement=40mm cv=0.5m2/year

Example 1: Calculation of settlement at a given time

For the upper layer

Now using Figure 5 with Tv = 0.1


T vc v t

H

20 1

2 20 1

.4.

10-3 10-2 10-1 1 10


0.00

0.25

0.50

0.75

1.00

U


For the upper layer


U = 0.36so

S = 100 0.36 = 36mm


T vc v t

H

20 1

2 20 1

.4.

For the lower layer



T vc v t

H

20 5 1

5 20 02

..

10-3 10-2 10-1 1 10


0.00

0.25

0.50

0.75

1.00

U


0.02 0.05

For the lower layer


U = 0.16so

S = 40 0.6 = 6.4 mm


T vc v t

H

20 5 1

5 20 02

..

Example 2: Scaling

Oedometer U=0.5 after 2 minutes. 2 way drainage, H = 5 mm

Calculate time for U= 0.5 for 10 m thick layer of the same clay, 1 way drainage

Example 2: Scaling



Oedometer Tc t

H

ccv

v vv

2 2

2

0 00580000

.

Example 2: Scaling



Oedometer

Soil layer

Tc t

H

ccv

v vv

2 2

2

0 00580000

.

Tc t

H

c t c tv

v v v

2 210 100

Example 2: Scaling



Oedometer

Soil layer

Tv (oedometer) = Tv (soil layer)

hence t = 80000000 mins = 15.2 years

Tc t

H

ccv

v vv

2 2

2

0 00580000

.

Tc t

H

c t c tv

v v v

2 210 100

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Kuliah Konsolidasi 1