Compaction

Densification of soil by removal of air using mechanical energy

Compaction vs. ConsolidationCONSOLIDATION:

REDUCTION OF SOIL VOLUME UNDER STRESS/LOADING. STRESS OR LOADING CAUSES

1) DEFORMATION OF SOIL PARTICLE

2) RELOCATION OF SOIL PARTICLE

3) EXPULSION OF WATER OR AIR ( IN MOST PRACTICAL CASES, ITS WATER)

Effect of moisture content• Water added act as

softening agent• Soil slip onto each

other and into a more dense packed position

• Beyond certain m%, dry unit weight is reduced because the moist/water takes up spaces of soil !!!

Optimum Moisture Content

Water

Moisture content, m

Unit Weight

γ=γd

Proctor compaction test

MOULD

Extension

HAMMER

25 BLOWS

Standard & Modified Proctor

Original Ilustration by Prof. Bengt B. Broms inFoundation Engineering

Test Concept

• Soil is mixed with varying amount of water then compacted.

• Unit wt =

W = weight of compacted soil in the moldVm = volume of mold

• Moisture content w% is determined in lab.

mVW

=γ

Plotting dry density against w%

• Dry density is calculated as:

• Plot the various γd vs. w%

100%1 wd

+=

γγ

w%

γd

S=100 and 80 curve• S=100% is the ZAV

curve – zero air void• ZAV plot can be

calculated using:

• Both ZAV and S80 can also be calculated using:

s

wzav

Gw 1+

=γγ

%1001×⎟⎟⎠

⎞⎜⎜⎝

⎛−=

sd

w

GSw

γγ

w%

γd

ZavS80

Specs. For Field Compaction

• Typical requirements: 90-95% MDD

• Specs. normally in term of relative compaction Cr.

( ) %100max

×=labd

dRC

γγ

100%Dam

95%Upper 150mm of sub-grade below roadway

90%Building platform or road

Minimum CRType of works

Site Procedure

• Soil is normally compacted in layers of 200-300mm.

• Constant check of field density should be carried out to ensure compliance.

• Rate of construction in embankment work should be control to prevent build up of pore pressure

• Heavier roller gives better compaction.

Site Procedure (cont’d)• For embankment work,

the best procedure is to compact a trial area and measure the dry density.

• When relative compaction is satisfactory, the number of roller passes is used for the actual embankment.

How to determine field density?

• Field density tests– Sand Cone Test/Sand Replacement Method– Rubber balloon test– Nuclear density test– Water ring test– Drive cylinder test

Sand Replacement Method

• ASTM D1556 or BS 1377

100%1 w

WW field

dry

+=

Compacted fill

Test hole filled with std. sand

VWdry

dry =γ

Example 1: Proctor compaction test

209.33238.81241.14212.65231.32Mass of tray + dry soil

240.29267.01263.45227.03240.85Mass of tray + wet soil

20.9920.3019.8121.2420.11Mass of tray

40404091403439213762Compacted Soil + mold

54321

Mass of mold = 2031 g, Volume of mold = 9.44 E-4 m3

1) Compute γd and w% for each data point and plot results

2) Calculate S80 and S100 using Gs = 2.69

3) Determine MDD and wo

Calculate moisture content for each data points:

%100% ×−

−=

traydry

drywet

MMMM

w

%10011.2032.23132.23185.240

1 ×−−

=w

= 4.5 %

Moisture content

Dry density

mVW

=γ

3/22.17045.0199.17

100%1

mkNwd =+

=+

=γγ

Remember W = mass x gravity

Gravity = 9.81 m/s2

W1 = (3.762-2.031) x 9.81

= 16.981 kg.m/s2 or N

= 0.016981 kN

γ = 0.016981 / 9.44E-4

= 17.99 kN/m3

Summary of data points

17.9418.9618.9018.2717.22γd (kN/m3)

16.4%12.9%10.1%7.5%4.5%w

20.8821.4120.8119.6417.99γ (kN/m3)

54321

S80 and S100

s

wzav

Gw 1+

=γγ %1001

×⎟⎟⎠

⎞⎜⎜⎝

⎛−=

sd

w

GSw

γγ

Use γd = 16, 18 and 20 (i.e. within data points range)

%3.19%10069.21

1681.9%8080 =×⎥⎦

⎤⎢⎣⎡ −=w

S80 and S100 data points

11.9%9.5%20

17.3%13.9%18

24.1%19.3%16

S100S80γd

Chart and Report

15

16

17

18

19

20

21

0 5 10 15 20 25 30Moisture content (%)

Dry

Den

sity

(kN

/m3)

curves80s100

OMC

MDD

Example 2: Sand Replacement test

• Determination of Dry Density of soil on site

• BS 1377• Specs.

Requirement of Compaction ratio is 90% MDD of ex.#1

• Initial mass of sand: 5.912 kg• Remaining sand after

pouring: 2.378 kg• Mass of soil from hole: 2.383

kg• Moisture content w = 7.0 %• Density of sand: 1490 kg/m3• Volume of cone: 1.114E-3 m3• Calculate γd and Compaction

ratio.

Example 2: Solution

3

33

2

2

3333

33

)(

/36.1707.01

58.181

/58.1810258.110337.2

10337.21000

181.9383.2

10258.1)10114.1()10372.2(

10372.21490

534.3383.2

534.3378.2912.5

mkNw

mkNVW

kNgMW

mVVV

mV

M

M

d

hole

soil

soilsoil

coneholeconehole

holecone

wetsoil

holeconesand

=+

=+

=

=××

==

×=⎟⎠⎞

⎜⎝⎛××=×=

×=×−×=−=

×==

=

=−=

−

−

−

−−−+

−+

+

γγ

γ

Example 2:cont’d

• Relative compaction What if CR < 90%?

– Ripping, mixing and re-compacting

– Add water( )

OK

Clabd

dR

%36.91%1000.19

36.17

%100max

=×=

×=γγ

Relative density• In place density

compared to its laboratory max and minimum density

• Max – compacted in laboratory

• Min – loosely filled into a test container

• Then compacted and loose density including void ration can be calculated

%100%minmax

max ×⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

eeee

D fieldR

⎥⎥⎦

⎤

⎢⎢⎣

⎡×⎥⎥⎦

⎤

⎢⎢⎣

⎡

−

−=

)(

(max)

(min)(max)

(min))(%fieldd

d

dd

dfielddRD

γγ

γγγγ

or

Typical values

>20>85Very dense

17-2065-85Dense

14-1735-65Medium dense

<14<35Loose

Unit Weight (kN/m3)

Relative density (%)

Consolidation and Settlement

The weight of any structure on the earth will result in stresses being imposed on the soils below the level of the base or foundation of that structure. The deformation that develop in the soil because of these stresses cause dimensional changes in the soil volume, with the result that the structure undergoes settlement. The extent of foundation settlement that will actually occur is related to the bearing pressure (stresses) imposed on the soils and the stress-strain properties of the soil.

D.F. McCarthy

Normally consolidated and overconsolidated clay

NORMALly consolidated – the present Po is the maximum pressure that the soil was subjected since the past.

OVER consolidated – the present Po is LESS THAN the maximum past pressure.

Preconsolidation pressure (Pc) – the maximum effective past pressure

Po = Pc ………Normally consolidatedPo < Pc ……… Overconsolidated

Overburden pressureThe effective overburden pressure (Po) at any depth is

determined by accumulating the weights of all layers above that depth as follows:

1) Soil above water table – multiply the total unit wt by the thickness of each respective soil layer above the level.

2) Soils below the water table – reduce the total unit wt by the wt of water (1000 kg/m3) … i.e. use the effective unit wt.

Pc1) Locate point ‘a’ at the

minimum radius of the curve

2) Draw a horizontal line ‘ab’

3) Draw the line ‘ac’tangent at ‘a’

4) Draw the line ‘ad’ …bisector of angle ‘bac’

5) Project straight line portion of the curve to intersect ‘ad’ … the intersection is the Pc

[Das (1998) after Cassagrande]

Chart from IKRAM (2002)

Oedometer TestOedometer Test

• (change of) Height• Applied Load

• Void Ratio• Applied Stress

Particular Sample Measurements:

General Derived General Derived Relationship:Relationship:

hh