soil mechanics chapter 8 consolidation• terzaghi, k. v. (1924) die theorie der hydrodynamischen...

Universitat Politècnica de Catalunya · BARCELONATECH Escola Tècnica Superior d’Enginyers de Camins, Canals i Ports

Soil Mechanics

Chapter 8

Consolidation

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Chapter 6

1. One-dimensional consolidation theory

2. Consolidation with radial flow

Exercises

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6.1

One-dimensional consolidation theory

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Introduction: coupling between the hydraulic and

mechanical problems

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Introduction

• We have seen that to study failure in soils:

– if it is short term → work with total stress

because we don’t know how to calculate pw

– if it is long term → work with effective stress

because we know how to calculate pw: either it is

the hydrostatic pressure or it can be obtained from

the flow conditions – steady state in both cases (no

time-dependent)

– that is why we have not dealt directly with

calculating pw in the failure analysis

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Equations so far:

(a) Equilibrium equations

total stress: effective stress:

(b) Compatibility equations

(c) Constitutive law

(d) Boundary conditions

(0,0, )

ij ij wi i

j j i

nat

pb b

x x x

b

water pressure:

hydrostatic

flow

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Porewater pressure

• To understand what happens with the

porewater pressure, we need more

equations

• Add the continuity (mass conservation)

equation and Darcy’s law (chapter 2):

– Conservation:

– Darcy:

• Assume saturated soil; h = piezometric level

div 0

grad

F

F

n

t

q K h

q

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Equations:

(a) Equilibrium

(b) Compatibility

(c) Constitutive law

(d) Boundary

conditions

(e) Continuity of

mass of water

(f) Darcy’s law +

Let’s study in detail equations (e) and (f),

combining them, but assuming now that the

porosity n is not constant (as was done in

chapter 2)

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Continuity + Darcy

• In general, the mass conservation

equation, with one phase only, would be:

¶r

¶t+ div rv( ) = 0 1( )

Þ¶r

¶t+ v ×gradr + r div v = 0

Þdr

dt+ r div v = 0

Mass conservation equation in

spatial form, where v is the

absolute velocity (taking a control

volume fixed in space)

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Continuity + Darcy

• But in soils there are actually two phases

• It is easy to check that

• Although ρfluid and ρsolid may be constant

(undeformable water and soil particles), the

porosity n is not constant (soil is deformable)

• It is easier to write the conservation equation

separately for each of the phases

fluid solid 1n n

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Continuity + Darcy

• Water:

• Solid:

• Where:

– vfluid = vsolid + vrel

– vrel is the relative velocity of the fluid with

respect to the solid particles → q

fluid

fluid fluid

solid

solid solid

div 0

1div 1 0

t

nn

t

nn

v

v

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Continuity + Darcy

• Elimination of ρfluid and ρsolid (because they

are constant in time and space) leads to:

• Adding the two equations results in:

solid rel

sol

fluid

id

div div 0

1di 1 0

t

0

v

nn

t

n

n

n

nt

v v

v

v

rel soliddiv div 0n v v

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Continuity + Darcy

• On the other hand:

vsolid: absolute velocity of solid (spatial derivative)

usolid: displacement vector of solid particles

• Therefore:

solidsolid

t

uv

solidsolid soliddiv div div vol

t t t

uv u

changing the order

of the differential

operators

definition of volumetric

deformation and Soil Mechanics

sign criterion

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Continuity + Darcy

• Resulting in:

• Darcy’s law provides the relationship

between nvrel and the piezometric head:

reldiv 0volnt

v

rel ; w

w

pn h h z

v K grad

unit flow vector q defined in

chapter 2, but now we consider

also the particle’s motion

piezometric head

(potential)

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Continuity + Darcy

• Finally:

• that is a general form of the continuity equation

that includes the motion of the solid skeleton,

which is more strict

• If K = ct. →

• and if there is no deformation, we recover the

flow equation 2h = 0 (chapter 2)

div volht

K grad (e) + (f)

2 volK ht

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Summary of equations (1)

(a) Equilibrium, written in terms of effective

stresses because now we do want to know the

porewater pressure:

(b) Compatibility: in general not imposed “a priori”,

but need to check. Often automatically fulfilled.

(c) Constitutive law, written in vector form for

simplicity:

ij wi

j i

pb

x x

d d σ D ε

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Summary of equations (2)

(d) Boundary conditions: include also initial

conditions because the variable time also

appears in the equations.

(e) Continuity

(f) Darcy’s law

• We know that (a), (b) and (c) represent 6 PDE

with 6 unknown functions σx, σz, τxz, εx, εz, εxz

• Now we have one more equation, and one

more unknown: the porewater pressure

div w vol

w

pz

t

K grad

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• These equations represent a coupling

between a mechanical problem (stress –

strain) and a hydraulic problem (flow):

– Mechanical problem (a) + (c):

– Hydraulic problem (e) + (f):

Coupling

ij wi

j ixd

pb d

x

(σ )D 1εwith

div w vol

w

pz

t

K grad (2)

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• Equations (1) and (2) are coupled: in

general they cannot be solved separately

• Only if εvol does not change with time (2) is

the classical (chapter 2) flow equation and

in this case it can be solved independently

from (1)

• This happens, e.g., for long term analysis

when the problem is time-independent

(∂εvol/∂t = 0)

Coupling

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• Therefore, at long term (what we have

called drained conditions), equations (1)

and (2) are uncoupled:

• Equation (2) can be solved directly to

obtain pw, and then substitute it in (1)

working with effective stresses

This is how we have worked so far...

Coupling

ij wi

j i

pb

x x

(1) div 0w

w

pz

K grad (2)

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• But in any other circumstance (1) and (2)

must be solved simultaneously.

• If the constitutive model (c) is complicated

(plasticity, Cam-clay, ...) it is not possible

in general to obtain an analytical solution

and we must resort to numerical methods

(i.e. finite elements)

• In some very simple cases there are

analytical solutions

Coupling

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One-dimensional consolidation equation

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• Assume that the problem is one-dimensional

and that the soil is an elastic material

there is only vertical deformation

lateral deformation is zero

• This is the case in the oedometer:

1-D consolidation theory

z

x

z

H

WT

▼

z

uniformly distributed load Δσ

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• The relationship between and is the

one seen for the oedometric test

• Because we assume the soil to be elastic,

there is no need to write this relationship in

incremental form:


z z

1 1 2

1

zz z

mE E

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• In this case, also: εvol = εz

• If K is constant, equation (2) becomes:

• where:


2

2

1 1 2

1z w

hK p

z t E

,hydrostatic; ww

w

w

ph z p up

excess porewater

pressure (over

hydrostatic pressure) the hydrostatic pressure is

constant with time

linear with depth, z

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• After derivation, the hydrostatic

component of the pressure vanishes from

the equations:

• And so equation (2) becomes:


2

,hydrostatic ,hydrostatic

20

w wp p

z t

2

2

1z

w m

K uu

z E t

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• If the external load is constant with time,

σz = ct.:

• This is the one-dimensional consolidation equation

usually attributed to Terzaghi. The general theory is

from Biot (1941)


2

2

2

2

10z

w m

m

w

KE

K u u

t

u

z E t

u

z t

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• Definition:

– coefficient of consolidation:

– can be obtained from tests with the

oedometer

– combines:

permeability (flow): K

deformability: Em

– cv for clays: between 10-2 cm2/s and 10-4 cm2/s


mv

w

KEc

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• Terzaghi, K. v. (1923) Die Berechnung der Durchlässigkeitsziffer des

Tones aus dem Verlauf der hydrodynamischen Spannungserscheinungen.

Akad. Wiss. Wien. Math-naturw. Klasse 132, No. 3/4, 125-128.

• Terzaghi, K. v. (1924) Die Theorie der hydrodynamischen

Spannungserscheinungen und ihr erdbautechnisches Anwendungsgebiet.

Proceedings of the International Congress for Applied Mechanics, 288-294.

Delft.

• Terzaghi, K. v. (1925) Erdbaumechanik auf bodenphysikalischer Grundlage.

Leipzig and Vienna: Deuticke. (Principles of soil mechanics)

• Terzaghi, K. v. & Fröhlich, O.K. (1936) Theorie der Setzung von

Tonschichten: eine Einführung in die Analytische Tonmechanik, Leipzig:

Deuticke. (Theory of the settlement of clay layers: an introduction to the

analytical mechanics of clay)

• Biot, M.A. (1941) General Theory of Three-Dimensional Consolidation. J.

Appl. Physics, 12, 155-164

Original references

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• The equation

• is a parabolic PDE with analytical solution

• After a change of variables, we obtain a

dimensionless form of the equation:

• Δσ = external load; H = thickness of layer; τ = time (to be defined)


2

2v

u uc

z t

u z tW Z T

H

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• Substituting these equations into the

original differential equation:

• Therefore:


2 2

2 2 2;

u W u W

z H Z t T

2

2 2v

W Wc

H Z T

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• Defining now

• results finally in

• where


2

v

H

c

2

2

W W

Z T

2

vcT t

H

T = dimensionless problem time

t = real (physical) time

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Boundary conditions

H

WT

▼

z

Δσ

z = 0; Z = 0

z = H; Z = 1

0 0

1 0

1

0 0

0

0 0

z u

uz

Z W

WZ

Z

T W

Hz

t u

Initial condition :

z= H Þ q= 0 Þ K¶j

¶z= 0

j = H - z+p

w

gw

= H - z+p

w

hydrostatic

gw

+u

gw

¶j

¶z=

¶

¶zH - z+

pw

hydrostatic

gw

é

ëêê

ù

ûúú+

1

gw

¶u

¶z=

=¶

¶zH - z+

gw

× z

gw

é

ëê

ù

ûú

=0

+1

gw

¶u

¶z=

1

gw

¶u

¶z

¶j

¶z= 0 Þ

¶u

¶z= 0

2

2

0 0; 1 0

0 1

W W

Z T

WZ W Z

Z

T W

Therefore we must solve :

with conditions :

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Solution

• Using separation of variables, the

following solution is obtained:

W(Z,T) =4

(2n+1)pn=0

¥

å exp -p 2 2n+1( )

2

4T

é

ë

êêê

ù

û

úúú

f (T )

sin2n+1( )p

2Z

é

ë

êê

ù

û

úú

f (Z )

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Isochrones

22

0

2 1 2 14( , ) exp sin

(2 1) 4 2n

n nW Z T T Z

n

Z

Z = 0

Z = 1

W = 0 W = 1

Δσ

T = 0+ → W = 1

T = ∞ → W = 0

porewater pressure

over the hydrostatic

isochrones

T1

T2

T3

T4

•Porewater pressure dissipation process in time

•Remember that u = Δσ·W is the excess porewater pressure over the hydrostatic

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Isochrones

1 Z

W

T

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Isochrones

z

z = 0

z = H

t = 0+ → u = Δσ

wH

isochrones

t1

t2

t3

t4 hydrostatic

pressure

Δσ

in terms of the “physical” variables (z, u, t)

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P

0 uuS

Pt = 0

C0

ut

't

P

0

tt

tt

uu

uS

P

t

Ct

tt

tt

u

uS

P

0

t = ∞

' P

C∞

Terzaghi’s hydraulic analogy P

ere

Pra

t. E

ngin

yeria

del

Ter

reny

. UP

C.


Double-sided drainage

z

Δσ

z = 0; Z = 0

z = 2H; Z = 2

• Double-sided drainage is equivalent to single-side drainage of a

layer of half the thickness: the middle plane of the layer behaves as

an impermeable boundary.

• The same set of formulas are used, but the layer thickness is

designated as 2H

2H

isochrones WT

▼

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Degree of consolidation

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• The degree of consolidation at a point is defined as:

• where:

Degree of consolidation at a point

,( , )

,P

z tU z t

z

0 0 0

0

,1 1 1

, ,1

v v

t

v

a aez

e e e

az t z t

e

at the end for

av = coefficient of

compressibility (chapter 3)

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Degree of consolidation at a point

u Δσ'

Δσ z

u

( , )

, ,,

,1

W Z

P

T

z t u z tU z t

u z t

, 1 ,PU Z T W Z T

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• The degree of consolidation of a layer is defined as:

• where:

Degree of consolidation of a layer

( ) tsU t

s

s¥

= e z,¥( )dz0

H

ò =a

v

1+ e0

Ds × H

st= e z,t( )dz

0

H

ò =a

v

1+ e0

D ¢s z,t( )dz0

H

ò =

=a

v

1+ e0

Ds -u z,t( )éë

ùûdz

0

H

ò

surface settlement at time t

final surface settlement

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W 1-W

Z

W

U t( ) =s

t

s¥

=1-u z,t( )Ds

dz0

H

ò

U T( ) =1- W Z,T( )dZ0

1

ò

1 0

1

0

• If the layer drains on both sides and its thickness is 2H, then the

degree of consolidation is

U T( ) =1-1

2W Z,T( )dZ

0

2

ò

area ≡ U(T) : degree of consolidation

of the layer at time T

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U (T) =1- W(Z,T)dZ0

1

òT = 0®U = 0

T =¥®U =1 (100% of consolidation)

Solving the integral results in :

U (T) =1-8

p 2

1

2n+1( )2

n=0

¥

å exp -p 2 2n+1( )

2

T

4

é

ë

êêê

ù

û

úúú

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2( ) v

t

cs U T s T t

H

• Settlement at time t:

• U(T) is tabulated:

– T = 0.100 → U = 0.356 (35.6%)

– T = 0.500 → U = 0.764 (76.4%)

– T = 1.129 → U = 0.950 (95%)

consolidation is

practically finished

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2

2

80.2 : ( ) 1 exp

4

20.2 : ( )

T U T T

TT U T

• It can be demonstrated that U(T) can be

approximated as:

first term

of the

series

parabola

U

T 0%

100%

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Example

10 m

WT

▼

clay

cv = 10-3 cm2/s

Clay layer, 10 m thick

How long does it take to reach 95%

consolidation?

2 2 29

3 2

10000.95 1.129 1.129 1.13 10

10v

H cmU T t T s

c cm s 36 years

If the bottom boundary is permeable and the layer drains on both sides, 2H = 10 m

2 2 2

3 2

5001.129

10v

H cmt T

c cm s 9 years

It takes a long time: consolidation settlements may last many years

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Consolidation caused by changes in the hydraulic

conditions

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Examples

• The differential equation is the same, only

boundary conditions change.

• Two examples:

– Example 1: decreasing piezometric level at

the layer’s bottom boundary

– Example 2: raising the water table

• Generalization to changes of water

pressure at the layer’s top and bottom

boundaries

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Example 1

WT

▼

sand

sand

clay

initial pw (γwz)

final pw decrease of pw

well

At the sand layer the changes

of water pressure are fast; at

the clay layer they are slow

At the clay layer settlements

occur as pressure dissipates:

t→∞ isochrones

Δpw < 0 → Δσ' = Δσ – Δpw > 0

→ Δe < 0 → settlement

e

Δσ'

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Example 2

sand

sand

clay

initial pw (γwz)

At the sand layer the changes

of water pressure are fast; at

the clay layer they are slow

At the clay layer swelling

occurs as pressure increases:

t→∞

isochrones

Δpw > 0 → Δσ' = Δσ – Δpw < 0

→ Δe > 0 → swelling

e

Δσ'

WTfin

▼ WTini

▼

final pw

increase of pw

assume that at this point the

initial water head is

maintained

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General case – linear increment

2H

z

steady state

conditions

initial

conditions

pw1

pw2

pw1

pw2

p'w1

p'w2

Differential equation:

2

2

w wv

p pc

z t

Boundary conditions:

2

1

0

2

w w

w w

z p p

z H p p

Initial conditions:

2 11

0

22

w ww w

t

p pp p H z

H

Steady state conditions:

2 11 2

2

w ww w

t

p pp p H z

H

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• Change of variable u = pw – pw,steady state; then:

• Changing to a dimensionless form with

2

2

2 1, , 1

0 0; 2 0

0 22

v

w wi w ini w sst w

u uc

z t

z u z H u

p pt u p p p H z

H

1 202

0

; ; ;2

v w wz c u p pZ T t W u

H H u

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• We obtain:

• And the solution is:

2

2

1

2

0 0; 2 0

2 ( 1)0 ;

1

w

w

W W

Z t

Z W Z W

Z pT W

p

1 2 2

1

1 1 4, sin exp

1 2 4

n

n

n Z n TW Z T

n

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• Calculating the degree of consolidation with this function

W(Z,T), we obtain:

U (t) =s

t

s¥

=D ¢s

t0

2H

ò dz

D ¢s¥0

2H

ò dz

D ¢st= ¢s

t- ¢s

ini= s

total- p

w( )¢st

- stotal

- pw,ini( )

¢sini

= pw,ini

- pw

u+pw,fin

= pw,ini

- pw,fin( ) - u

D ¢s¥

= pw,ini

- pw,fin

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• Therefore:

U (t) =1+udz

0

2H

ò

pw,ini

- pw,fin( )dz

0

2H

ò=1-

udz0

2H

ò2H ×u

0

U (T) =1-1

2W(Z,T)dZ

0

2

ò

1wp

2wp

2H 1 202 2

2

w wp pH H u

Area :

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• Introducing the values of the obtained W(Z,T)

results in a degree of consolidation U(T) equal

to the one obtained for the case of a uniform load

Δσ

• Therefore: any problem with a linear increment of

porewater pressure results in the same degree of

consolidation U(T)

• There are also tables for non-linear increments

of porewater pressure

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U-T Relations

u0 u0 u0 u0 u0 2H

h

Case I

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U-T Relations

u0 2H

h

u2

2H

h

u3

u0

h

u4

u0

0 2sin

4hu uH

0 3sin

2hu uH

4

4

de 0 a

(2 )de a 2

0

0

u hh h H

H

u H hh H h H

H

u

u

Case II Case III Case IV

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U

Case I

T

Case II

T

Case III

T

Case IV

T

0.05 0.0017 0.0021 0.0208 0.0247

0.10 0.0077 0.0114 0.0427 0.0500

0.15 0.0177 0.0238 0.0659 0.0750

0.20 0.0314 0.0403 0.0904 0.102

0.25 0.0491 0.0608 0.117 0.128

0.30 0.0707 0.0845 0.145 0.157

0.35 0.0962 0.112 0.175 0.188

0.40 0.126 0.143 0.207 0.221

0.45 0.159 0.177 0.242 0.257

0.50 0.196 0.215 0.281 0.294

0.55 0.238 0.257 0.324 0.336

0.60 0.286 0.304 0.371 0.384

0.65 0.342 0.358 0.425 0.438

0.70 0.403 0.421 0.488 0.501

0.75 0.477 0.494 0.562 0.575

0.80 0.567 0.586 0.652 0.665

0.85 0.684 0.700 0.769 0.782

0.90 0.848 0.862 0.933 0.946

0.95 1.129 1.163 1.214 1.227

1.00 Infinite in all cases

Table of T for several values of U P

ere

Pra

t. E

ngin

yeria

del

Ter

reny

. UP

C.


T

Case I

U

Case II

U

Case III

U

Case IV

U

0.004 0.0735 0.0649 0.0098 0.0085

0.008 0.1038 0.0862 0.0195 0.0162

0.012 0.1248 0.1049 0.0292 0.0241

0.020 0.1598 0.1367 0.0481 0.0400

0.028 0.1889 0.1638 0.0667 0.0560

0.036 0.2141 0.1876 0.0850 0.0720

0.048 0.2464 0.2196 0.1117 0.0950

0.060 0.2764 0.2481 0.1376 0.1198

0.072 0.3028 0.2743 0.1628 0.1436

0.083 0.3233 0.2967 0.1852 0.1646

0.100 0.3562 0.3288 0.2187 0.1976

0.125 0.3989 0.3719 0.2654 0.2442

0.150 0.4370 0.4112 0.3093 0.2886

T

Case I

U

Case II

U

Case III

U

Case IV

U

0.175 0.4718 0.4473 0.3507 0.3306

0.200 0.5041 0.4809 0.3895 0.3704

0.250 0.5622 0.5417 0.4603 0.4432

0.300 0.6132 0.5950 0.5230 0.5078

0.350 0.6582 0.6421 0.5783 0.5649

0.40 0.6973 0.6836 0.6273 0.6154

0.50 0.7640 0.7528 0.7088 0.6994

0.60 0.8156 0.8069 0.7725 0.7652

0.70 0.8559 0.8491 0.8222 0.8165

0.80 0.8874 0.8821 0.8611 0.8566

0.90 0.9119 0.9079 0.8915 0.8880

1.00 0.9313 0.9280 0.9152 0.9125

2.00 0.9942 — — —

1.0000 1.0000 1.0000 1.0000

Table of U for several values of T

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6.2

Consolidation with radial flow

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Introduction

• Clay layers are “slow” to consolidate

• The preload technique allows improving the

ground

• Problem: all loading/unloading processes are

slow

• Why? The main variable controlling the

problem is the maximum distance the water

must travel to exit the layer, so that pressure

may dissipate:

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Introduction

U

T 0%

95%

1.129

2

2

v w

m

c HT t t T

H K Et t H

• The time depends essentially on the distance:

• To decrease the distance we may introduce vertical drains:

Vertical and radial flow (small vertical flow)

Vertical deformation → vertical

settlement

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Radial consolidation

• Consolidation equation in cylindrical coordinates:

¶2u

¶r 2+

1

r

¶u

¶r+

1

r 2

¶2u

¶q 2+

¶2u

¶z2

»0,symmetry radial part

=¶u

¶t

cvr

¶2u

¶r 2+

1

r

¶u

¶r

é

ëê

ù

ûú=

¶u

¶t; c

vr=

Kr× E

m

gw

cvr

1

r

¶

¶rr

¶u

¶r

æ

èç

ö

ø÷

é

ëê

ù

ûú=

¶u

¶tK

r: radial permeability

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• It can be proved that

• With

– radial + vertical flows

– only vertical deformation

• The vertical deformation (settlement) is:

• The vertical part is the classical solution already seen:

1 1 1rz r zU U U

t rzs U s

2; ;vz z m

z z z z vz

w

c K EU U T T t c

H

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• Drain pattern

– rw : drain radius

– rs : radius of remoulded zone, with

permeability Ks

– re : radius of the drain influence zone,

with permeability Kr

wr

sr

er

er

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• Radial degree of consolidation Ur:

2

2 2 2

2 2 2

13 1 1ln

1 4

p

;

1 ex 2r r

vrr

e

r

s

e s

w w

cT t

r

K sn n nm n

n n K n

r r

m

n sr r

U T

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6.3

Time-dependent load

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Time-dependent load

0

0T

0T 0 0 2

vcT t

H

u

integration variable

has dimensions (kPa)

2 2

2 2 00 0

01,3,5,... 0

4 1, sin exp 4 exp

2 4

T

n

n Z n T du Z T n T dT

n dT

Per

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soil mechanics chapter 8 consolidation• terzaghi, k. v. (1924) die theorie der hydrodynamischen...

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