6 Soil Compressibility_GB_Dr. Asmaa Moddather

PBW N302

Credit Hours

CEM/WEE/STE

Dr. Asmaa ModdatherSoil Mechanics and Foundations

Faculty of Engineering – Cairo University

FALL 2012

Definitions

• Compressibility: is the property through which

particles of soil are brought closer to each other, due

to escapage of air and/or water from voids under the

effect of an applied pressure.

Dr. Asmaa Moddather – PBW N302 – Fall 2012

Settlement of Cohesive Soils

• Coefficient of Compressibility (av): is the rate of change of

void ratio (e) with respect to the applied effective pressure

(p) during compression.

p

ea v

∆

∆=

e

e-p curve

e

eo = initial void ratio

po = initial effective stress


p

e∆

p∆

eo

e1

po p1

po = initial effective stress

e1 = final void ratio

p1 = final effective stress

∆e = eo – e1

∆p = p1 – po

• Coefficient of Volume Compressibility: (mv) is the volume decrease

of a unit volume of soil per unit increase of effective pressure during


of a unit volume of soil per unit increase of effective pressure during

compression

)e(1

a

Δp

Δe

)e(1

1

Δp

)Ve(1

ΔeV

Δp

V

Δv

mo

v

o

so

s

Ov

+=

+=

+==


• For a thin layer , ∆p at mid depth is considered as an average

stress within the layer.q


stress within the layer.

)e(1

Δe

H

ΔH

o+=

H)e(1

Δp

ΔP

ΔeH

)e(1

ΔeSΔH

oo +=

+==

ΔpH)e(1

aS v

+=

H∆p

Clay


1

eo

1+eo H

∆e ∆H

ΔpH)e(1

So+

=

∆pHmS v=

mv = coefficient of volume compressibility

∆p = increase in effective stresses at mid depth of layer

H = thickness of layer


• For a thick layer the layer may be divided to number of sub-

layers, each of thickness Hi. The stress at mid-depth of each sub-

H1∆p1

q

iiH∆pmS vi∑=

layers, each of thickness Hi. The stress at mid-depth of each sub-

layer is ∆pi.


H1

Clay

H2

∆p2

• Sand may be considered as elastic material with Young’s Modulus.

• Young’s Modulus (E): is the slope for the linear portion of the

Settlement of Cohesionless Soils

• Young’s Modulus (E): is the slope for the linear portion of the

stress-strain curve.

E

1

stress

strain=

E

1

Δp

H

ΔH

=

q


E

p

H

H ∆=

∆

E

p

H

S ∆=

pHE

1S ∆=

E

1mv =

2kg/cm dense) (v. 800-loose) (v. 100E =

EΔp

H∆p

Sand

Example

• Compute the settlement due to

compressibility of the sand layer and

consolidation of the clay layer at points (a)

and (b) of the building shown in Figure.

The building has a basement and is

founded on raft foundation. The stress

from the building at the foundation level

is as shown in the figure.


is as shown in the figure.

Sand: γ = 1.70 t/m3 , E = 500 kg/cm2

Clay: mv = 0.03 cm2/kg.

� Due to distributed load:

qnet = q – γh = 20 – 1.70x3.0 = 14.9 t/m2

Example

For sand layer

z = 5.0 m = 1” = 2.5 cm

L = 30 m � 30x2.5/5.0 = 15.0 cm, B = 20 m � 20x2.5/5.0 = 10.0 cm,

na = 191 ∆σa,sand = n/200 (qnet) = 191/200 x 14.9 = 14.2 t/m2

nb = 50 ∆σb, sand = n/200 (qnet) = 50/200 x 14.9 = 3.7 t/m2


ab

� Due to distributed load:

qnet = q – γh = 20 – 1.70x3.0 = 14.9 t/m2

Example

For clay layer

z = 11.0 m = 1” = 2.5 cm

L = 30 m � 30x2.5/11.0 = 6.82 cm, B = 20 m � 20x2.5/11.0 = 4.55 cm,

na = 144 ∆σa,clay = n/200 (qnet) = 144/200 x 14.9 = 10.7 t/m2

nb = 47.3 ∆σb,clay = n/200 (qnet) = 47.3/200 x 14.9 = 3.5 t/m2


ab

Sa = Sa,sand + Sa,clay

Example

Sa,sand = 1/Esand ∆σa,sand H

H = 10.0 m E = 500 kg/cm2 ∆σa,sand = 14.2 t/m2

Sa,sand = 1/5000 x 14.2 x 10.0 = 0.0284 m = 2.84 cm

Sa,clay = mv ∆σa,clay H

H = 2.0 m m = 0.03 cm2/kg = 3.7 t/m2


H = 2.0 m mv = 0.03 cm2/kg ∆σa,clay = 3.7 t/m2

Sa,clay = 0.03/10 x 3.7 x 2.0 = 0.0222 m = 2.22 cm

Sa = Sa,sand + Sa,clay = 2.84 + 2.22 = 5.06 cm

Example

Sb = Sb,sand + Sb,clay

Sb,sand = 1/Esand ∆σb,sand H

H = 10.0 m E = 500 kg/cm2 ∆σb,sand = 10.7 t/m2

Sb,sand = 1/5000 x 10.7 x 10.0 = 0.0214 m = 2.14 cm

Sb,clay = mv ∆σb,clay H

H = 2.0 m m = 0.03 cm2/kg = 3.5 t/m2


H = 2.0 m mv = 0.03 cm2/kg ∆σb,clay = 3.5 t/m2

Sb,clay = 0.03/10 x 3.5 x 2.0 = 0.021 m = 2.10 cm

Sb = Sb,sand + Sb,clay = 2.14 + 2.10 = 4.24 cm

Settlement

H∆p

q

∆pHmS =

(Sand)

(Clay)

ΔpHE

1S =


∆pHmS v= (Clay)

Time of Settlement Compared to Construction Time

� Loading diagram:

Construction period

� Settlement of sandpH

E

1S ∆=

Time

Time


� Settlement of clay∆pHmS v=

Time

Causes of Settlement

• Static loads

•• Dynamic loads

• Groundwater lowering

• Capillary forces

•


• Loads from adjacent structures

Theory of Consolidation

• Consolidation: is the process of squeezing water out of soil under the

effect loads, provided no lateral movement occurs.effect loads, provided no lateral movement occurs.

• In sandy soils, high permeability, drainage of water out of the soil

under the effect of loading happens immediately.

• In clay soils, low permeability, drainage of water out of the soil under


• In clay soils, low permeability, drainage of water out of the soil under

the effect of loading is time dependent.

• Therefore, parameters governing consolidation process include: soil

properties, stress (p), time (t), drainage conditions.

Analogy:

Confined saturated soil (solids Valve ~ soil Piston for

Mechanism of Consolidation ProcessMechanism of Consolidation Process

and voids filled with water)

Versus

Container filled with water that

has a spring and a valve

Valve ~ soil permeability

Piston for loading


Remember: water is incompressible

Spring ~ Solids

Water ~ pore-water

Container

• If we apply a certain load, and the valve is closed, then water

(incompressible material) carries the entire added load.


(incompressible material) carries the entire added load.

• If the valve is opened, water flows out, and thus its volume is

reduced.

• Since the total volume is reduced, the spring starts to be


• Since the total volume is reduced, the spring starts to be

compressed and carries part of the load. Meanwhile, additional

pressure in the water decreases.

• The process continues until sufficient amount of water escapes

from the valve, which will result in the compression of the


from the valve, which will result in the compression of the

spring that would allow it to carry the entire load.

• At this point, the additional pressure in water decreases to zero,

the spring carries the entire load, the system is in equilibrium.


For Sands: high permeability ~ valve is open, water gets out immediately

(t = 0), ∆u = 0


Stress = p


At t = 0 (immediately), ∆u = 0Load carried by spring

For Clays: low permeability ~ valve is partially open, water gets out

slowly (time-dependent), ∆u decreases with t

Stress = p


Stress = p

Stress = p


At t = 0 (immediately), ∆u = pSpring carry no load

At t = t1, 0 < ∆u < pLoad carried by water and spring

For Clays:


Stress = p

Stress = p


At t = long time, ∆u = 0Load carried by spring

At t = t2, 0 < ∆u < pLoad carried by water and spring

• Stress distribution before application of load

Application on SoilApplication on Soil

GSGWT

Sand

Clay


Gravel

tσ 'σu

• Short time (immediately) after application of q (t = 0)


qGSGWT

Sand

Clay

q

q

q


Gravel

tσ 'σu

q

q

• Long time after application of q (t = h)


qGSGWT

Sand

Clay

q

q


Gravel

tσ 'σu

q

q

Application on Soil

• Total stress:

• Pore water pressure:

qΣγhσ t +=

• Pore water pressure:

o t = 0 � (clay)

� (sand)

o t = h � (clay)

� (sand)

• Effective stress:

qhγu ww +=

wwhγu =

wwhγu =

wwhγu = At time t ???


• Effective stress:

o t = 0 � (clay)

� (sand)

o t = h � (clay)

� (sand)

qhΣγ'σ' +=

hΣγ'σ'=

qhΣγ'σ' +=

qhΣγ'σ' +=

Rate of Consolidation

qGSGWT

Sand

Clay

q

q

q


Gravel

tσ 'σu

q

q

Rate of ConsolidationRate of Consolidation

Clay

u

t = 0

t = t1

t = t2


q

t = t3

t = t2

t = ∞

Assumptions of Theory of Consolidation

• Clay is homogeneous, isotropic, and saturated.

•• Water and clay particles are incompressible.

• Darcy’s law is valid.

• The clay layer is laterally confined.

• One-dimensional compression, and one-dimensional flow.


• One-dimensional compression, and one-dimensional flow.

•Consolidation parameters from test applies to clay layer from

which sample for test was taken.

• Soil properties are constant with time.

Consolidation Test

• Objectives:

o Volume change-effective pressure relationship

o Stress history of soil

o Volume change – time – pore water dissipation relationship


Consolidation Test Load cell LoadDial gauge

Compression cell

Oedometer


Consolidation Test

Load cell

Dial gauge

Compression cell


Oedometer

Consolidation Test

Load cell

Dial gauge

Compression cell

Load


Oedometer

Consolidation Test

Dial gaugeLoad Head

Dial gauge

Porous stone

Saturated sampleMetal ring

Filter paper


Porous stone

Filter paper

Compression Cell/Oedometer

Consolidation Test

Dial gaugeLoad Head

Load (P )Dial gauge

Porous stone

Soil saturated sampleMetal ring

Filter paper

Load (P1)


Compression Cell/Oedometer Porous stone

Filter paper

Consolidation Test

• Equipments:

o Compression cell/Oedometero Compression cell/Oedometer

o Loading frame to apply weights (Po)

o Dial gauge to measure soil compression

Note:

Load acting on the cell (P1) is magnified by the lever arm ratio of the


Load acting on the cell (P1) is magnified by the lever arm ratio of the

loading frame (LAR), where: P1 = LAR x P

• Procedure:

1. Measure initial conditions of sample: ho, eo, wo, Gs

Consolidation Test

o o o s

2. Trim the sample into the metal ring and place filter paper on both

sides of sample.

3. Place the ring into the compression cell between the two porous

stones.

4. A metal cap (loading head) is placed over the top porous stone, on


4. A metal cap (loading head) is placed over the top porous stone, on

which the load (P1) is applied.

5. Set-up the dial gage to zero reading. Fill the compression cell with

water until the top porous stone is covered with water.

• Procedure:

6. Place the hanging load (Po) such that the pressure on the clay

Consolidation Test

o

sample is equal to the first loading step.

7. Record readings of dial gage (compression of sample) with time

until compression stops (usually within 24 hours).

8. Repeat steps 6 and 7 for subsequent loading and unloading

increments.


increments.

• Loading scheme, for example:

o 0.25, 0.5, 1.0, 2.0, 4.0, 8.0, 16.0 (kg/cm2) Loading

Consolidation Test

o 16.0, 4.0, 1.0, 0.25 (kg/cm2) Unloading

• Results: for each loading/unloading increment, we record:

dial gage (∆h) that measures sample compression versus


elapsed time (t).

Consolidation Test Data Reduction

• For each loading increment, plot ∆h-log t:

0.1 1 10 100 1000Time (min)

0

10

20

30

40

50

h (

x 1

0-2

mm

)


60

70

80

90

100

∆∆ ∆∆h

(x

• At the end of each loading increment:

1. Calculate effective pressure (p) = Load x LAR/sample area


2. Measure ∆h = sample compression during this loading increment

3. Calculate void ratio (e):

oo h

∆h

e1

∆e=

+ eo

∆e ∆h


e = eo – ∆e

• Plot e-p curve

1

1+eo ho

232116582914.570Load (kg)

Example: LAR = 3, sample area = 41.85 cm2, ho = 25.4 mm, eo = 0.636


232116582914.570Load (kg)

705550359223135890Dial Reading x 10-2 (mm)= ∆h

16.638.324.162.081.047 x 3/41.85

= 0.500

Stress (p) = Load x lever arm ratio/area (kg/cm2)


0.45410.35430.23120.14360.0870.89/25.4 (1+0.636)= 0.0573

0∆e = ∆h/ho (1+eo)

0.18190.28170.40480.49240.54900.636-0.0573

= 0.57870.636e = eo - ∆e

• Plot e-p Curve

0.7


0.3

0.4

0.5

0.6

0.7

e


0

0.1

0.2

0 2 4 6 8 10 12 14 16 18

p (kg/cm2)

• From e-p curve, for a certain stress range, calculate av and

mv, where:


mv, where:

p

ea v

∆

∆=

e

e∆

p∆

eo

e1


)e(1

am

o

vv

+=

ppo p1

• Plot e-log(p) curve:

0.7


0.3

0.4

0.5

0.6

e

e∆

eo

e1


0

0.1

0.2

0.1 1 10 100

p (kg/cm2)

p)log(∆

po p1

e1

• From e-log(p) curve, the slope of the linear portion is the

Compression Index (Cc):


c

• Determine preconsolidation pressure (pc): is the largest

∆log(p)

∆eCc =

o1

1o

logplogp

ee

−

−= 402.0

log4.8log12.0

22.00.38=

−

−=


• Determine preconsolidation pressure (pc): is the largest

effective pressure that has been applied on the soil in its

geological history.

• Determination of preconsolidation pressure:

0.7

5


0.3

0.4

0.5

0.6

e

1

2

3

4

5

ba


0

0.1

0.2

0.1 1 10 100

p (kg/cm2)

pc = 1.8 kg/cm2

1. Determine point of maximum curvature (a).

2. At (a), draw tangent to e-log(p) curve.

Determination of Preconsolidation Pressure

3. At (a), draw horizontal line.

4. Draw a bisector to the angle enclosed between the tangent and

horizontal lines.

5. Extend the linear portion of the curve until it intersects the bisector at

point (b).


point (b).

6. The preconsolidation pressure (pc) is the x-coordinate of the point (b).

1. Normally consolidated clay:

Clay that has never been loaded in the past by more than the existing

effective overburden pressure (p ) p ~ p

Types of Clay w.r.t. Preconsolidation

effective overburden pressure (po) pc ~ po

2. Overconsolidated (preconsolidated) clay:

Clay that has been loaded in the past by more than the existing effective

overburden pressure (po) pc > po

Causes: pre-existing structures, erosion of overburden

3. Underconsolidated clay:


3. Underconsolidated clay:

Clay that has been loaded partially by the existing effective overburden

pressure (po) pc < po

o

c

p

pOCR =Define: Overconsolidation Ratio (OCR):

� For normally consolidated clays, Cc can be calculated as follows:

� From e-log(p) curve:

Determination of Compression IndexDetermination of Compression Index

� From liquid limit:

∆log(p)

∆eCc =


where wL is in (%)

)100.009(wC Lc −≈

0.6

0.7

Recompression curve

DefinitionsDefinitions

0.2

0.3

0.4

0.5

e

Virgin compression curve


0

0.1

0.1 1 10 100

p (kg/cm2)

Unloading/reloading curve

pc

� From e-log(p) curve, the recompression index (Cr) is the slope of

the equivalent linear portion of the unloading/reloading curve.

DefinitionsDefinitions

0.3

0.4

0.5

0.6

0.7

e

)log( p

eCr

∆

∆=

Cr

Cr

Cc


0

0.1

0.2

0.1 1 10 100p (kg/cm2)

Cr

� Calculate settlement (S) of normally consolidated clay layer of

thickness (H) due to added stress (∆p) given Cc of clay:

∆e

SettlementSettlement

∆log(p)

∆eCc =

)log(p)log(p

eeC

01

10c

−

−=

)p

log(

∆eC

1c =

0.1

0.2

0.3

0.4

0.5

0.6

0.7

e

eo

e1

e Cc

∆p


)p

plog(

0

1

.......1..........).........p

plog(C∆e

0

1c=

......2..............................H

S

h

∆h

e1

∆e

oo

==+

0

0.1

0.1 1 10 100p (kg/cm2)

p1op~c p

� From 1 and 2:

)e(1S

)p

log(C 1 +=


)e(1H

S)

p

plog(C o

o

1c +=

)p

plog(H

e1

CS

o

1

o

c

+=


� For overconsolidated clay, (po and p1) < pc:

0.7e


0

0.1

0.2

0.3

0.4

0.5

0.6

e

eo

e1

Cr

∆p


)p

plog(H

e1

CS

o

1

o

r

+=

0

0.1 1 10 100p (kg/cm2)

p1 pcpo

� For overconsolidated clay, po < pc and and p1 > pc

0.7e


0

0.1

0.2

0.3

0.4

0.5

0.6

e

eo

e1

Cr

∆p

Cc


)p

plog(H

e1

C)

p

plog(H

e1

CS

c

1

o

c

o

c

o

r

++

+=

0

0.1 1 10 100p (kg/cm2)

p1pcpo


q

Short time after application of load “q”

GSGWT

Sand

Clay

q

q

q


Gravel

tσ 'σu

q

q

Rate of consolidationRate of consolidation

u

t = 0

u

t = 0

Isochrone

Sand

zu σ'

Clay

t = 0

t = t1

t = t3

t = t2

t = ∞

t = t2

t = ∞

umax σ‘minClay

Sand

2H

hDu σ'


∆p ∆p Sand

At time “t” and depth “z”:

∆p = ∆u + ∆σ’ = increase in total stress due to added load∆u = excess pore water pressure due to added load∆σ’ = increase in effective stress due to added load

Hydrostatic pore water pressure

(GWT)

u

t = 0

Sand

z∆u ∆σ'

2H

Rate of consolidationRate of consolidation

Degree of Consolidation (Utz) at time “t” and depth “z”:

∆p

t = tt = ∞

Clay

Sand

2H


Degree of Consolidation (Utz) at time “t” and depth “z”:

∆p

∆u1

∆p

∆u∆p

∆p

∆σU

'

tz −=−

==

� Average degree of consolidation (Ut) for the clay layer at time “t”:


dzU2UDh

tzt ∫=

u

t = 0Sand

zdzU2U

0

tzt ∫=

diagram∆pofarea

diagramσ'ofareaU t

∆=

∆p

∆u∆p

∆p

∆σU

'

tz

−==

t = 0

t = t

t = ∞

umax σ‘minClay

2H

z

hDu σ'

Where


diagram∆pofarea

diagram∆pofarea

diagramuofarea-diagram∆pofareaU t

∆=

2H∆pdiagram∆pofarea =

2Hu3

2diagramuofarea max=∆

∆p

t = ∞

Sand

Where

� Settlement of clay layer of thickness 2H at time t=∞:

Sultimate = mv∆p (2H)


� Settlement of clay layer of thickness 2H at time t=t:

St = mv∆σ’ (2H)

� Average degree of consolidation (U ) at time “t”:


� Average degree of consolidation (Ut) at time “t”:

Ut = St/Sultimate

The degree of consolidation at any time (t) is the percentage of

settlement that took place at this time w.r.t. to ultimate

consolidation settlement.

� Equation governing rate of consolidation

Differential Equation of ConsolidationDifferential Equation of Consolidation

� Consider a differential element of soil

Surcharge load

ground surface

Qin Area = dAVolume = dV

dx

dyx

z


Qout

dzy z

kidAQ −=

)dzz

Q(QQ

t

(dV)inout

∂

∂=−=

∂

∂−compression and flow 1-D

Darcy’s law is valid


kidAQ −=

dAz

hk

∂

∂−=

)dAγ

u(

zk

w∂

∂−=

dAz

u

γ

k

w ∂

∂−=

Darcy’s law is valid

u = excess p.w.pQin Area = dA

Volume = dV

dxdz

dy

xy

dA)dzz

u

γ

k(

zt

(dV)

w ∂

∂−

∂

∂=

∂

∂−

dVz

u

γ

k

t

(dV)2

2

w ∂

∂=

∂

∂Equation (1) soil is homogeneous

Qout

dz y

z

e1

e

dV

dVv

+= dV

e1

edVv

+=

dVv

de


e1

dV

t

edV

t

e

t

)(edV

t

dV)e1

e(

t

)(dV

t

(dV)s

sv

+∂

∂=

∂

∂=

∂

∂=

∂

+∂

=∂

∂=

∂

∂

dVs

dV

1e1

1

dV

dVs

+= dV

e1

1dVs

+=

dVe(dV) ∂=

∂Equation (2)

e1tt +∂=

∂

From equations 1 and 2,

e1

dV

t

edV

z

u

γ

k2

2

w +∂

∂=

∂

∂

Equation (2)

t

e

z

u

γ

e)k(12

2

w ∂

∂=

∂

∂+

σ'ee ∂∂=

∂(chain rule)


t

σ'

σ'

e

t

e

∂

∂

∂

∂=

∂

∂

t

ua)

t

u

t

σ(a

t

evv

∂

∂=

∂

∂−

∂

∂−=

∂

∂

t

σ'a

t

ev

∂

∂−=

∂

∂

(chain rule)

σ'

ea: v

∂

∂=

0t

σ: =

∂

∂ (load doesn’t vary with time)

t

ua

z

u

γ

e)k(1v2

2

∂

∂=

∂

∂+

tzγv2

w ∂∂

2

2

wv z

u

γa

e)k(1

t

u

∂

∂+=

∂

∂

2

2

vz

uc

t

u

∂

∂=

∂

∂

wvwv

vγm

k

γa

e)k(1c: =

+=

� Coefficient of Consolidation: cv

v

kc = cm2/sec


where:k = soil permeabilitymv = coefficient of volume compressibility

wv

vγm

c = cm2/sec


� Boundary conditions:

Example: a clay layer, of thickness 2H, with top and bottom boundaries freely draining Sand


� u(z=0,t) = 0

� u(z=2H,t) = 0

� u(z,t=0) = ui

Sand

Clay

u

t = t1

t = t2z

as time increases, u decreases2H

� Solution:

)TM((sinMZ)expM

2ut)u(z, v

2i −=∑∞

Solution of Differential Equation of ConsolidationSolution of Differential Equation of Consolidation

where:

Tv = Time factor

)TM((sinMZ)expM

t)u(z, v

0m

∑=

1)(2m2

πM +=

2

vv

H

tcT =

Tv = Time factor

� Solution in terms of degree of consolidation (U):

……………… Equation 3)TMexp(M

21U v

2

0m2

−−= ∑∞

=

Equation 3 was solved and the results were tabulated:

Tv = f(U)


2

d

vv

h

tcT =

where:T = time factor corresponding to degree of consolidation occurring at time t


Tv = time factor corresponding to degree of consolidation occurring at time tcv = coefficient of consolidation (determined from consolidation test)hd = longest drainage path within clay layer

The above equation holds for consolidation of sample in the lab, and for clay layer consolidating in the field

0

10

20

3030

40

50

60

70

80

90

U (

%)


90

100

0 0.2 0.4 0.6 0.8 1 1.2

Tv

10099959080706050403020100U (%)

∞1.7811.1290.8480.5670.4030.2870.1970.1260.0710.0310.0080.0Tv

� Drainage Path: hd

Sand Sand


2H Clay

Sand

2H Clay

Impervious Rockhd = H hd = 2H


Determination of cv

� cv is determined from the consolidation test data

� The is a c value that corresponds to each loading � The is a cv value that corresponds to each loading

increment, i.e. cv is stress dependent

� For the following test data corresponding to a certain

loading increment, find cv.


885240120603015106.2542.2510.50.250Time (min)

223221220219215209202193184173161155153135Dial Reading x 10-2 (mm)

888685848074675849382620153-135

=18135-135

= 0= ∆h x 10-2 (mm)

0.1 1 10 100 1000

Time (min)

t50 = 3.6 min0.25 1.00

� ∆h-log t Curve

Determination of cv

0

10

20

30

40

50

h (

x 1

0-2

mm

)

U = 0%

t50 = 3.6 min0.25 1.00

U = 50%


60

70

80

90

100

∆∆ ∆∆h

(x

U = 100%

1. Plot the (∆h) or (e) on the vertical axis (arithmetic scale), and the time (t) on

horizontal axis (logarithmic scale) S-shape

2. Determine (∆h) or (e) corresponding to 0% consolidation:

Determination of Determination of ccvv


� Consider (∆h1) corresponding to a small time (t1)

� Consider (∆h2) corresponding to (4t1)

� Determine the difference between (∆h1) and (∆h2) y

� Determine (∆h) corresponding to 0% consolidation at a distance

equivalent to y above (∆h1)


1


� Extrapolate the linear portion at the middle of the consolidation curve

� Extrapolate the linear portion at the end of the consolidation curve

� The two lines intersect at a point corresponding to (∆h) or (e) at 100%

consolidation

1. Two points on the first portion of curve are selected for which the values

of “t” are in the ratio of 4:1.

2. The vertical distance between the two points is measured.


2. The vertical distance between the two points is measured.

3. An equal distance set off above the first point fixes a line corresponding

to U = 0%.

4. The line corresponding to U = 100% passes through the intersection of

the two linear parts of the curve.

5. The line corresponding to U = 50% is located at the midway between

the two lines (U = 0% & U = 50%)

6. The intersection between the curve and the line (U = 50%) is


6. The intersection between the curve and the line (U = 50%) is

corresponding to t50.

2

D

vv

h

tcT =

t

hTc

2

Dvv =

50

2

Dv

t

0.197hc =

4. Determine (∆h) or (e) corresponding to 50% consolidation at mid

distance between 0% consolidation and 100% consolidation.

5. Determine t


5. Determine t50

6. Calculate cv:

where: T = 0.197, h = ½ sample height (sample drains from both

2

d

vv

h

tcT =


where: T50 = 0.197, hd = ½ sample height (sample drains from both

sides), and t = t50 from consolidation curve

Types of Compression

0

0.1 1 10 100 1000

Time (min)Immediate consolidation

∆h = 00

10

20

30

40

50

h (

x 1

0-2

mm

)

U = 0%

Primary consolidation


60

70

80

90

100

∆∆ ∆∆h

(x

U = 100%

Secondary consolidation

1. Immediate: due to compression of entraped air and

machine parts.

Types of CompressionTypes of Compression

machine parts.

2. Primary: due to consolidation by escapage of water

(relief of excess pore-water pressure) under the effect of

pressure (loading)


pressure (loading)

3. Secondary: compression at constant effective stress (no

excess pore-water pressure) [creep]

� A rectangular building 24 × 16 m is

shown in Figure and is founded on a

raft and has its foundation level at

ExampleExample

(a) (c) (b)

raft and has its foundation level at

2.0 m below ground surface. If the

building exerts a net stress of 2.0

kg/cm2 on soil at foundation

level, determine:

i. The ultimate settlement of

(a) (c)


points a, b and c.

ii. The time required for these

points to settle 5.0 cm and the

time required undergo 90%

consolidation.

i. The ultimate settlement of points a, b and c.

qnet = 20 t/m2

Example

z = 8.0 m

H = 5.0 m

Using Influence Chart:

ca & bPoints

1224L (m)

88B (m)


88B (m)

1.53m

11n

0.2030.193Iz

0.203x20x4 = 16.240.193x20x2 = 7.72∆σ (t/m2)

0.004x16.24x5 = 0.325 m0.004x7.72x5 = 0.154 mS = mv ∆σ H


points to settle 5.0 cm.

Example

(a) (c) (b)

cv = 0.006 cm2/sec

bcaPoints

=5/15.4 = 0.3255/32.5 = 0.154=5/15.4 = 0.325U = St/Sf

0.0830.0190.083From Chart


0.0830.0190.083From Chart(Tv)

40.09.210.0t (days)

2

D

vv

h

tcT =

2(500)

t0.006019.0 =

2(250)

t0.006083.0 = 2(500)

t0.006083.0 =


points to undergo 90%

Example

(a) (c) (b)

consolidation.

cv = 0.006 cm2/sec

bcaPoints

0.900.900.90U = St/Sf

0.8480.8480.848From Chart


0.8480.8480.848From Chart(Tv)

409.0409.0102.2t (days)

0.90 x 15.4 = 13.9 cm0.90 x 32.5 = 29.3 cm0.90 x 15.4 = 13.9 cmSt (cm)

2

D

vv

h

tcT =

2(500)

t0.006848.0 =

2(250)

t0.006848.0 =

2(500)

t0.006848.0 =

6 Soil Compressibility_GB_Dr. Asmaa Moddather

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