Consolidation & Settlements

Consolidation &

Consolidation &

Settlements

Settlements

updated April 23, 2007

Terms and Definition

Terms and Definition

z

Settlement

z

Settlement

}

total vertical deformation at soil surface resulting from the load

z

Consolidation (volume change velocity)

z

Consolidation (volume change velocity)

}

rate of decrease in volume with respect to time

z

Compressibility (volume change flexibility)

z

Compressibility (volume change flexibility)

}

volume decrease due to a unit load

z

Contraction (temperature expansion)

z

Contraction (temperature expansion)

}

change in volume of soil due to a change in temperature

z

Swelling

z

Swelling

}

volume expansion of soil due to

increase

in water content

z

Shrinkage

z

Shrinkage

Introduction

Introduction

z

Soils

z

Soils

}

Considered elastic materials

}

Visco-elastic materials (time dependent in

stress-t i

)

strain response)

}

But, visco-elastic only applicable to material that are

linear

}

Soil is highly nonlinear materials

}

Soil have a ‘memory’ Æ non conservative material

}

Present theory can’t handle that.

}

Simplification

Introduction

Introduction

(cont’d)

z

When stressed Æ soil deform

z

When stressed Æ soil deform

z

Stressed released Æ deformation

remains

z

Soil deformation :

z

Soil deformation :

}

Distortion (change in shape)

}

Compression (change in volume)

Component of Setlement

Component of Setlement

z

Settlement

z

Settlement

total vertical deformation at soil surface resulting from the load

z

Soil Movement:

z

Downward

ƒ

load increase or lowering water table

z

Up ard

z

Upward

ƒ

temporary or permanent excavation

z

Points of interest:

z

Points of interest:

z

How much

z

How fast

]

settlement occurs

Component of Setlement

Component of Setlement

(cont’d)

z

Total Setlement : S

_t

= S

_i

+ S

_c

+ S

_s

}

S

_i

: immediate / distortion settlement

Initial compression

}

S

_i

: immediate / distortion settlement

z elastic theory, analog to deformation of column.

z 3D loading Æ distortion in soil.

z compression modulus & volume of stressed soil k

Initial compression

unknown

z design for shallow foundation

}

Sc : consolidation setlement

Primary consolidation

o

rmation

}

Sc : consolidation setlement

z time dependent process

z occurs in saturated fine-grained soil

z low coefficient of permeability

z S ttl t t d d f t Secondary consolidation

def

o

z Settlement rate depend of pore water pressure

}

Ss : secondary compression (time dependent)

z time dependent

Secondary consolidation

Time (log scale) z time dependent

z occurs at constant effective stress

Compressibility of Soil

Compressibility of Soil

z Compressibility (volume change flexibility) is the volume decrease due to a unit load

z Assumption in settlement : 100% saturated and 1D (vertical) soil deformation

z When soil is loaded it will compress because of:

} Deformation of soil grains (small, can be neglected) } Compression of air and water in the voidsp

} Squeezing out of water & air from the voids

z Compressible soil mostly found below water table Æ considered fully saturated

z As pore fluid squeezed out:

} Soil grain rearrange themselves Æ stable & denser configuration } Decrease in volume Æ surface setlement resulted

z How fast? Æ depend on permeability of soil

z How much rearrangement & compression? Æ depend on the rigidity of soil skeleton

C i f d i t tl

} Compression of sand occurs instantly

} Consolidation of cohesive soil is very time depend process

Consolidation of Clay

Consolidation of Clay

at equilibrium (t = 0)

z

System is analog to soil layer at

equilibrium with weight of all soil

layer (overburden) above it.

X

valve closed

p

o (overburden load)

z

In equilibrium, valve is closed.

z

Piston is loaded, compresses a

Piston

spring in chamber.

z

Hydrostatic pressure = u

_o

spring

u

o

water

spring ≈ soil skeleton water ≈ water in poresp

Consolidation of Clay

y

under load Δp (0 < t < ∞)

z

Soil is loaded by increment

Δp

z

Soil is loaded by increment Δp.

z

Valve initially closed.

valve closed initially

p

o +

Δp

X

z

Pressure (Δp) is transferred to

the water.

Piston

z

As water is incompressible and

valve still closed, no water is out,

no deformation of piston.

u

o +

Δu

spring

z

Pressure gauge read : Δu = Δp

where Δu is excess hydrostatic

pressure.

u

o +

Δu

water

z

To simulate a fine grained

cohesive soil, where permeability

is low valve can be opened

spring ≈ soil skeleton

water ≈ water in soil void

_{is low, valve can be opened.}

z

Water slowly leave chamber.

valve ≈ pore sizes in soil

Consolidation of Clay

y

at Equilibrium (t = ∞)

z

To simulate a fine grained

cohesive soil, where permeability

is low valve can be opened

=

valve open

p

o +

Δp

is low, valve can be opened.

z

Water slowly leave chamber

p

o +

Δp

S

z

As water flows out, load (Δp) is

transferred to the spring.

Piston

spring

z

At equlibrium, no further water

squeezed out, pore water

pressure back to its hydrostatic

condition.

water

u

o

condition.

z

Spring is in equilibrium with load

p

_o

+ Δp

spring ≈ soil skeleton water ≈ water in soil void

Δu = 0

z

Settlement ‘s’ exist

Setlement process:

Setlement process:

z

Initially all external load is transferred into excess pore

y

p

water (excess hydrostatic pressure)

}

No change in th effective stress in the soil

z

Gradually, as water squeezed out under pressure

gradient, the soil skeleton compress, take up the load,

and the effective stress increase

and the effective stress increase.

z

Eventually, excess hydrostatic pressure becomes zero

y,

y

p

and the pore water pressure is the same as hydrostatic

pressure prior to loading.

Wh il i l d dt t l l t th it

When soil is loadedto a stress level greaterthan it ever ‘experienced’ in the past, the soil structure is no longer able to sustain the increased load, and start to breakdown.

zP lid ti P P Void ratio

e

zPreconsolidation Pressure - Pc:

} Maximum pressure experienced by soil in the past

zNormal Consolidation: OCR = 1 Effective Consolidation Stress p’o

Pc

} when the preconsolidation pressure is equal to the existing effective

vertical overburden pressure P_c = p’_o

} present effective overburden pressure is the maximum pressure that

soil has been subjected in the past p’o

Pc

zOver Consolidation: OCR > 1

} when the preconsolidation pressure is greater than the existing

effective vertical overburden pressure P_c > p’_o

} present effective overburden pressure is less than that which soil

h b bj t d i th t

Pc= p’o

Pc

has been subjected in the past

} It also said soil is in preconsolidated condition } OCR (over consolidation ratio) =

_'

c o

P

p

Pc> p’o

p’o

zUnder Consolidation: OCR < 1

} when the preconsolidation pressure is less than the existing effective

vertical overburden pressure P_c < p’_o,

} tl d it d il l i ll ll

o

p

Pc

} e.g : recently deposited soil geologically or manually. _p’

o

Mechanism causing preconsolidation

Mechanism causing preconsolidation

Brumund, Jonas, and Ladd (1976)

z Change in Total Stress due to

R l f b d

} Removal of overburden } Past Structures

} Glaciation

z Ch i

z Change in pore water pressure

} Change in water table elevation } Artesian pressure

} Deep pumping; flow into tunnel } Dessication due to surface drying } Dessication due to plant life

z Environmental changes such as pH, temperature and salt concentration

z Chemical alteration due to ‘weathering’, precipitation, cementing agents, ion exchange

Consolidation Test data Plots

Arithmetic scale Log scale

(a) (a) train ε (% ) (a) train ε (% ) (a) Vertic al S mv= coefficient of volume change Vertic al S Cce= modified compression index

Effective consolidation stress p’o(kPa) Effective consolidation stress p’o(kPa)

o (e) (b) o (e) Cc= compression (b) Void Rati o av =coefficient of compressibility Void Rati o Cc compression index

Stress-strain history of a sedimentary clay during deposition sampling and reloading in the

1

Field virgin

O _{deposition, sampling and reloading in the}

laboratory by the consolidation test:

}OA represents the relationship between void ratio and the

log effective stress of a particular element in the ground

0.9 Field virgin compression curve in situ Rebound due to sampling A B

C _{during deposition. The process consolidates the element} g p g

to point A. This point represents the in situ e vs log p’o

coordinates of the normally consolidated clay element.

}When the boring is made and soil is sampled, overburden

C C’

Increasing sample

disturbance }When the boring is made and soil is sampled, overburden

stressed are removed by the sampling operation and the samples rebounds or swells along curved AB.

}When the sample is transferred from sampling tube into

consolidometer ring and then reloaded in the

0.8 rat io , e Laboratory consolidation test curve

consolidometer ring and then reloaded in the consolidation test, the curve BC is obtained.

}About point C, the soil structure start to break down and if

the loading continuos the laboratory virgin compression curve CD is obtained 0.7 Vo id r curve CD is obtained

}Eventually, the field curve OAD and lab curve BCD will

converge beyond point D (approximately 0.4e_o according to Terzaghi and Peck, 1967)

0.6

Reconsolidation

E

}If the sampling operation was poor quality and

mechanical disturbance to the soil dtructure occurred, curve BC’D would result upon reloading of the sample in the consolidometer.

Rebound D

F

}The proconsolidation pressure is much more difficult to

define when sample disturbance has occurred.

0.5

1 10 100

Pressure, p (log scale)

F

How to determine P ?

How to determine P

_c

?

(Cassagrande, 1936)

1. Choose point with minimum

radius Æ point. A

2 Dra hori ontal line from point

2 6 2.8

Pcpossibility range

E

D

2. Draw horizontal line from point A

3. Draw line tangent to the curve

t i t A 2 2 2.4 2.6 1 2 5 6

A B

C

at point A

4. Bisect the angle made by step

2 and 3 1 8 2 2.2 d ra ti o , e 1 3 4 α α

5. Extend the straight line portion of the virgin compression curve up to where it meets the

1 4 1.6 1.8 Vo id 3 p

bisector line obtained in step 4

6. Point of intersection step 4 and 5 is the (most probable)

1 1.2 1.4 ( p ) presonsolidation stress Æpoint B 1 1 10 100

Pressure, p (log scale)

Settlement Calculation:

Normally consolidated clay

voids ΔH = s Δe eo soil + water solids Hf voids solids ef 1 Ho 1 H

₌

2 4 2.6 2.8

1

v o o o o

L

H

s

e

or

L

H

H

e

ε

=

Δ

Δ

=

=

Δ

+

2 2.2 2.4 ra tio, e Cc

1

_o o v o

e

s

H

H

e

e

e

e

e

e

ε

Δ

=

=

+

Δ

1.4 1.6 1.8 Vo id r Cc 1 1 2 1 2 2 2 1 1

'

log

'

log

'

log

'

_log

'

c o

e

e

e

e

e

C

p

p

p

p

p

−

−

−Δ

=

=

=

Δ

−

1 1.2 1 10 100 P (l l ) Effective Consolidation Stress p’

Pc 2 1

'

log

1

'

o c c o

H

p

S

C

e

p

=

+

Pressure, p (log scale) Effective Consolidation Stress p’o

Settlement Calculation

(cont’d):

z

For normally consolidated clay

' H p + Δp log 1 ' ' log ' o o c c o o o c ce o H p p S C or e p p p S C H p + Δ = + + Δ =

p’1= p’o, and p’2include the additional

stress Δp applied by the structure when computing settlement using percentage

vertical strain vs log effective pressure

}

For layered normally consolidated clay:

'

H

p

p

⎡

+Δ

⎤

o p

z

In overconsolidated clay

log

1

'

o o c c o o

H

p

p

S

C

e

p

⎡

+Δ

⎤

= ⎢

₊

⎥

⎣

⎦

∑

z

In overconsolidated clay

'

log

1

'

o o c r o o

H

p

p

S

C

e

p

+Δ

=

+

p’1= p’o, and p’2=po+Δp < Pc o

p

o

'

log

log

1

'

1

'

o c o o c r c o o o o

H

P

H

p

p

S

C

C

e

p

e

p

⎛

⎞ ⎛

+ Δ

⎞

=

_⎜

_{⎟ ⎜}

+

_⎟

+

+

⎝

⎠ ⎝

⎠

p’1= p’o, and p’2=po+Δp > Pc

} Cris the slope of rebound curve (swell index); Cr ≈ 20% to 10% Cc

o

p

o o

p

o

Example:

a. Perform Cassagrande construction and

find Pc= 121 kPa

Example:

The void ratio vs log effective pressure data shown in Fig Ex. 8.9.

Determine:

(a) the preconsolidation pressure P

c

b. Using point a and b, ea= 0.870, eb=0.655, p’_a=100kPa, and p’_b=300kPa.

(a) the preconsolidation pressure P_c (b) the compression index C_c

(c) the modified compression index C_ce

2 2 1 1 0.870 0.655 0.451 ' ' 300 log log log 100 ' ' a b e e e Cc p p p p − Δ − = = = =

Another way is to find Δe over one ‘log cycle’; for example log (1000/100) = log 10 = 1. Therefore Cc= Δe.

I th fi th ti l l i t

1 1

p p

In the figure the vertical scale is not sufficient for finding Δp = 1 log cycle, therefore it will be done in 2 steps:

„ Extend e_{aa b}e_bto one full log cycle on the g y

same graph, chose ecat the same pressure as eb.

„ Draw the line e_ce_dparallel to e_ae_b.

Δe = C_c = (e_a-e_b)+(e_c-e_d)

(0.870-0.655)+(0.9-0.664)= 0.451

c. The modified compression index Cce

c. The modified compression index Cce

0.451 0.242 1 1 0.865 c ce o C C e = = = + +

Example

Example

z

Prior to placement of a fill covering a large area at a site the

z

Prior to placement of a fill covering a large area at a site, the

thickness of a compressible soil layer was 10m. Its original in

situ void ratio was 1.0. Some time after the fill was constructed,

measurements indicated that the average void ratio was 0 8

measurements indicated that the average void ratio was 0.8.

Estimate the settlement of the soil layer.

1 0 0 8

e

Δ

1.0 0.8

−

10m

1.0m

1

_o

o

1 1.0

e

s

H

e

Δ

=

=

=

+

_o

+

„

Factors affecting the determination of Pc from laboratory test:

g

y

1.

Sample disturbance

2.

Load increment ratio (LIR)

3

Load increment duration (LID)

3.

Load increment duration (LID)

1.

Increasing sampe disturbance:

}

Decreases the void ratio at

any given value of

any given value of

consolidation stress

}

Lowers the estimated value of

Pc from the Cassagrande

g

method

}

Increases the compressibility

at stresses less than Pc

}

Decreases the compressibility

at stresses greater than Pc

2.

Load Increment Ratio (LIR) denotes the

(

)

changes in consolidation stress divided by

the initial consolidation stress.

}

LIR =

p

p

Δ

o

p

3.

Load increment duration denotes the total

time t

_f

allowed for consolidation prior to

the application of the next load increment

the application of the next load increment.

Standard consolidation test often use a

duration of 1 day for a each increment.

Plot preferences

Plot preferences

z

The use of average vertical strain (

ε

) than void ratio (e) versus log effective

t

( ’ ) i

d d b

stress (p’

_o

) is recommended because:

} Strains are easier to compute than void ratio

} Differences in initial void ratio may cause samples to exhibit quite different plots } Differences in initial void ratio may cause samples to exhibit quite different plots

of void ratio versus stress but almost identical plots of strain versus stress

} Settlements are directly proportional to strain, but use of Δe data also requires a

knowledge of (1+e ) which introduces 2 variables Δe and (1+e ) This can only knowledge of (1+eo) which introduces 2 variables, Δe and (1+eo). This can only be determined at the end of test, not during the settlement test. The e vs log p’_o curve cannot be plootted during the test.

} Strain plot are easier to standardize than void ratio plots } Strain plot are easier to standardize than void ratio plots.

} Estimating field settlement is simple, percent compression can read directly

from the graph,once a good estimate of in situ overburden pressure.

Field Consolidation Curve

Field Consolidation Curve

Schertman (1955) Procedure

z Normally consolidated Soil } Find Pc using Cassagrande

} Calculate e_o(initial void ratio)

} Draw a horizontal line from e_oto Pc Æ Point 1

} Draw a horizontal line from 0.42 e_o to the extension

of laboratory virgin compression curve (L) Æ Point of laboratory virgin compression curve (L) Æ Point 2

} Draw a line from Point 1 to point 2 Æ Field virgin

consolidation curve (F)( )

z Overconsolidated Soil

} Calculate e_o and draw a line from e_o to the existing

overburden pressure p’_oÆPoint 1 overburden pressure p_o ÆPoint 1

} Find Pc using Cassagrande.From point 1 draw a

line paralel to rebound-reload curve to Pc Æ Point 2

Example 8 16

Example 8.16

Holtz & Kovacs

z

The void ratio vs pressure data shown below. The initial void

ratio is 0.725 and the existing vertical efective overburden

i 30 kP

pressure is 30 kPa.

Void ratio 0.708 0.691 0.670 0.632 0.635 0.650 0.642 0.623 0.574 0.510 0.445 0.460 0.492 0.530 Pressure (kPa) 25 50 100 200 100 25 50 200 400 800 1600 400 100 25

z

Required

q

1.

Plot the data as e vs log p’

_o 2.

Evaluate overconsolidation ratio

3

Determine the field compression index using Schertmann procedure

3.

Determine the field compression index using Schertmann procedure

4.

If this consolidation test is representation of a 12m thick clay layer,

compute the settlement of this layer if an additional stress of 220 kPa

were added

Solution

1. The data is plotted

Solution

2. The given value of p’ois plotted on the graph, and from Cassagrande construction a value for P_cc= 190 kPa is found.

3. OCR = Pc/p’o= 190/130 = 1.46. The soil is slightly overconsolidated.g y

4. Using Schmertmann procedure for

overconsolidated clay the values of Cr and Cc are 0.022 and 0.262

5. The settlement is:

' log log 1 ' 1 ' 12 190 12 130 220 0 022 log 0 262 log ⎛ ⎞ ⎛ + Δ ⎞ =_{⎜ ⎟ ⎜}+ _⎟ + + ⎝ ⎠ ⎝ ⎠ + ⎛ ⎞ ⎛ ⎞ =_{⎜ ⎟ ⎜}+ _⎟ o c o o c s c o o o o H P H p p S C C e p e p m m 0.022 log 0.262 log 1 0.725 130 1 0.725 190 0.025 0.484 0.509 0.5 + ⎜ ₊ ⎟ ⎜ ₊ ⎟ ⎝ ⎠ ⎝ ⎠ = m+ m= m≈ m

Time Rate of Consolidation

Time Rate of Consolidation

S

2

%

t t

S

U

S

U

π

=

⎛

%

⎞

0 to 60%,

4

100

60%,

1.781 0.933log(100

%)

t t v t v

U

for U

T

for U

T

U

π

⎛

⎞

=

_{= ⎜}

_⎟

⎝

⎠

>

=

−

−

2 2

or

v v dr v v v dr v

t

T H

T

c

t

H

c

=

=

z

U

_t

= average degree of consolidation (%)

z

S

_t

= settlement of the layer at time t

z

S = ultimate settlement of the layer due to primary consolidation

z

T

_v

= time factor

z

H

_dr

= average longest drainage path during consolidation

z

ffi i t f

lid ti

z

c

_v

= coefficient of consolidation

z

t

_v

= time for consolidation

Coefficient of consolidation (c )

Coefficient of consolidation (c

_v

)

E

z

Log-of-time

(Cassagrande) d₀

B

C

D

E

x

x

1. Extend the straight line portion of primary and secondary consolidation curve to intersect at A. A is d₁₀₀, the deformation at the end of consolidation

ncreasing)

_F

0.5(d₀ + d₁₀₀)

2. Select times t₁and t₂on the curve such that t2= 4t1. Let the difference is equal to

x

3. Draw a horizontal line (DE) such that the _{ormation (i}

n

d₅₀

F

3. Draw a horizontal line (DE) such that the vertical distance BD is equal to x. The deformation of DE is equal to d0.

4. The ordinate of point F represents the deformation at 50% primary consolidation

Def

o

d

A

deformation at 50% primary consolidation, and it abscissa represents t₅₀

2 50 0.197 dr v H c t = d₁₀₀ 50

Time (log scale) t₅₀

t₂ t₁

Coefficient of consolidation (c )

Coefficient of consolidation (c

_v

)

z

Square-root-of-time

(Taylor)

A

1. Draw a line AB through the early portion of the curve

2. Draw a line AC such that OC = 1.15 OB. The abscissa of D which is the

ncreasing)

intersection of AC and the consolidation curve, gives the square-root-of-time for 90% consolidation. 2 ormation (i n

D

2 90

0.848

_dr v

H

c

t

=

Def o

_D

Time (square root)

B C

O

t

₉₀

Secondary Consolidation

Secondary Consolidation

e

e

C

Δ

Δ

2 2 1 1

log

log

_log

e

e

C

t

t

t

t

α

Δ

Δ

=

=

−

' 2 ' 1

log

where

1

s p

C

t

S

C H

C

t

e

α α α

=

=

+

ratio e

void ratio at the end of primary consolidation

p

e

=

Vo id

Δe

e

_p t t Time (log scale) t1 t2

Example

Example

z

Calculate the settlement

z

Calculate the settlement

due to primary

con-solidation for 5m clay

Surcharge = 50kPa

layer due to a surcharge

of 50kPa applied at the

d l

l Th

l

i

2m

S d Sand

50% saturation Ground water table

ground level. The clay is

normally consolidated.

z

Calculate the time rate of

5m Sand

Gs=2.65, e=0.7

z

Calculate the time rate of

settlement when c

_v

is

given as 0.85m

2

_/yr

5m Clay_C c=0.45, eo=0.9

g

y

Rock c , o γsat=15kPa Rock

Solution

} Submerged unit weight of

Solution

z

Calculation of Average effective

} Submerged unit weight of

clay

z

Calculation of Average effective

Overburden Pressure (p

_o

)

} The moist unit weight of sand

b th d t t bl

( )

'_clay '_{sat clay w}

γ =γ −γ

} So

above the ground water table

(

2.65 0.5 0.7 9.81

)

1 1 0.7 s w w sand G Sr e e γ γ γ = ⋅ + ⋅ ⋅ = + ⋅ ⋅

+ + o sand sand clay

15 9.81 5.19 kPa 5 p' 2 +3 ' + ' 2 γ γ γ = − = = ⋅ ⋅ ⋅ 22.21kPa = y 2 5 2 22.21+3 9.516+ 5.19 85.94 kPa 2 = ⋅ ⋅ ⋅ =

z

Calculation of Settlement

} Submerged unit weight of sand

below the ground water table

' ' γ =γ −γ _{H ' + Δ}

(

)

(

)

( ) 1 1 1 2 65 1 9 81 sand sat sand w

s w s w w w G G e e e γ γ γ γ γ γ _γ = − ⋅ + ⋅ = − = + + − ⋅ ' log 1 ' 2.5 85.94 50 0.45 log 1 0 9 85 94 o o c c o o H p p S C e p + Δ = + + = +

(

2.65 1 9.81

)

9.516 kPa 1 0.7 = = + 1 0.9 85.94 0.592m 0.199m 0.792m 0.8m + = + =

Time rate of settlement

Time rate of settlement

( )

( )

10% 2 2 2 ; 0.8 10% 0.08m 2 5 0 008 2 5 t avg t c avg c s U s S U s S T T H = = × = × = U_avg Tv St (m) t (yr)

( )

( )

2 10% 2.5 0.008 2.5 ; 0.06 0.85 0.85 v dr v v Tv T H t yr t yr yr c ⋅ ⋅ = = = = Time (yr) ( ) 10 0.008 0.08 0.06 20 0.031 0.16 0.23 30 0 071 0 24 0 52 0 0.1 0 1 2 3 4 5 6 7 8 9 Time (yr) 30 0.071 0.24 0.52 40 0.126 0.32 0.93 50 0.197 0.40 1.45 0.2 0.3 t (m ) 60 0.287 0.48 2.11 70 0.403 0.56 2.96 80 0.567 0.64 4.17 0.4 0.5 S et tel em en t 90 0.848 0.72 6.24 95 1.163 0.76 8.55 100 ∞ 0.80 ∞ 0.6 0.7 0.8 100 0.80

Reference

Reference

z

Holtz, R.D and Kovacs, W.D. (1981) An Introduction to

z

Holtz, R.D and Kovacs, W.D. (1981) An Introduction to

Geotechnical Engineering

z

Das, B.M. (1985) Principles of Geotechnical

Engineering

z

Transportation Research Board Commision on

Sociotechnical System (1976) Estimation of

Sociotechnical System (1976) Estimation of

Consolidation Settlement