Compaction - Universiti Sains Malaysia CompactionNConsolidation.pdf · Compaction Densification of...
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Transcript of Compaction - Universiti Sains Malaysia CompactionNConsolidation.pdf · Compaction Densification of...
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