Chapter 6
Soil Compaction
In the construction of highway embankments, earth
dams, and many other engineering
structures, loose soils must be compacted to increase
their unit weights. Compaction increases the strength
characteristics of soils, which increase the bearing
capacity of foundations constructed over them.
Compaction also decreases the amount of undesirable
settlement of structures and increases the stability of
slopes of embankments.
The degree of compaction of a soil is measured in
terms of its dry unit weight. When water is added to
the soil during compaction, it acts as a softening agent
on the soil particles. The soil particles slip over each
other and move into a densely packed position. The
dry unit weight after compaction first increases as the
moisture content increases.
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2014
Standard Proctor Test
In the Proctor test, the soil is compacted in a mold
that has a volume of 944 cm3( ft3). The diameter of
the mold is 101.6 mm (4 in.). During the laboratory
test, the mold is attached to a baseplate at the bottom
and to an extension at the top (Figure 6.2a). The soil
is mixed with varying amounts of water and then
compacted in three equal layers by a hammer
(Figure 6.2b) that delivers 25 blows to each layer. The
hammer has a mass of 2.5 kg (6.5 lb) and has a drop
of 30.5 mm (12 in.).
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Where
W = weight of the compacted soil in the mold
Vm =volume of the mold
For a given moisture content, the theoretical
maximum dry unit weight is obtained when no air is
in the void spaces—that is, when the degree of
saturation equals 100%.
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Specifications for Field Compaction
In most specifications for earthwork, the contractor is
instructed to achieve a compacted field dry unit
weight of 90 to 95% of the maximum dry unit weight
determined in the laboratory by either the standard
or modified Proctor test. This is a specification for
relative compaction, which can be expressed as
Relative compaction
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2014
Determination of Field Unit Weight of
Compaction
(Sand cone method)
The combined weight of the jar, the cone, and the
sand filling the jar is determined (W1).
the weight of the moist soil excavated from the hole
(W2)and the moisture content of the excavated soil is
known, the dry weight of the soil can be obtained as
the combined weight of the jar, the cone, and the
remaining sand in the jar is determined (W4)
W5 = weight of sand to fill the hole and cone
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2014
The volume of the excavated hole can then be
determined as
Where
Wc = weight of sand to fill the cone only
Ɣd (sand) = dry unit weight of Ottawa sand used
The dry unit weight of compaction made in the field
then can be determined as follows:
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6.1 Given Gs = 2.75, calculate the zero-air-void unit
weight for a soil in lb/ft3 at w = 5%, 8%, 10%, 12%,
and 15%.
Solution
w%
5
8
10
12
15
w
0.05
0.08
0.1
0.12
0.15
Ɣzav
149.408
139.396
133.434
127.961
120.545
6.3 Calculate the variation of dry density (kg/m3) of a
soil (Gs = 2.67) at w = 10% and 20% for degree of
saturation (S) = 80%, 90%, and 100%.
Solution
ρd=
S
w=10%
w=20%
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ρd
0.8 2001.87
0.9 2059.13
1 2107.34
0.8 1601.2
0.9 1675.73
1 1740.55
2014
6.4 The results of a standard Proctor test are given
below. Determine the maximum dry
unit weight of compaction and the optimum moisture
content.
Solution
V (ft3)
0.03333
0.03333
0.03333
0.03333
0.03333
W
(Ib)
3.26
4.15
4.67
4.02
3.63
w%
8.4
10.2
12.3
14.6
16.8
Ɣ
97.8
124.5
140.1
120.6
108.9
Ɣd
90.2214
112.976
124.755
105.236
93.2363
140
120
100
80
60
40
20
0
5
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6.6 The results of a standard Proctor test are given in
the following table. Determine the maximum dry
density (kg/m3) of compaction and the optimum
moisture content.
Othman A. Tayeh
2014
Solution
ρ(kg/m3) ρd(kg/m3)
V(cm3) M(Kg)
w%
V(m3)
w
943.3
1.68
9.9 0.000943
0.099 1780.98 1620.547461
943.3
1.71
10.6 0.000943
0.106 1812.78 1639.046025
943.3
1.77
12.1 0.000943
0.121 1876.39 1673.854944
943.3
1.83
13.8 0.000943
0.138
1940 1704.743304
943.3
1.86
15.1 0.000943
0.151
1971.8 1713.120003
943.3
1.88
17.4 0.000943
0.174
1993 1697.617791
943.3
1.87
19.4 0.000943
0.194
1982.4 1660.303354
943.3
1.85
21.2 0.000943
0.212
1961.2
1618.15185
1720
1700
1680
1660
1640
1620
1600
5
7
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6.7 A field unit weight determination test for the soil
described in Problem 6.6 yielded the following data:
moisture content = 10.5% and moist density = 1705
kg/m3.
Determine the relative compaction.
Solution
ρd=
R=
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2014
6.9 A proposed embankment fill requires 8000 m3 of
compacted soil. The void ratio of the compacted fill is
specified as 0.7. Four borrow pits are available as
described in the following table, which lists the
respective void ratios of the soil and the cost per
cubic meter for moving the soil to the proposed
construction site. Make the necessary calculations to
select the pit from which the soil should be bought to
minimize the cost. Assume Gs to be the same at all
pits.
Solution
V=8000m3
e=0.7
0.7=
Vs=4705.88 m3
For A
e=0.82
Vt=8564.7016m3
Cost=Vt*Cost=$68517.6
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2014
For B
e=1.1
Vt=9882.348m3
Cost=$49.411.74
For C
e=0.9
Vt=8941.172m3
Cost=$80470.548
For D
e=0.78
Vt=8376.4664m3
Cost=$100517.597
Choose B
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2014
6.10 The maximum and minimum dry unit weights of
a sand were determined in the laboratory to be 104
lb/ft3 and 93 lb/ft3, respectively. What would be the
relative compaction in the field if the relative density
is 78%?
Solution
R=??
Dr=0.78=
Ɣd=101.3624Ib/ft3
R=
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6.13 Following are the results of a field unit weight
determination test on a soil with the sand cone
method:
• Calibrated dry density of Ottawa sand = 1667
kg/m3
• Calibrated mass of Ottawa sand to fill the cone =
0.117 kg
• Mass of jar + cone + sand (before use) = 5.99 kg
• Mass of jar + cone + sand (after use) = 2.81 kg
• Mass of moist soil from hole = 3.331 kg
• Moisture content of moist soil = 11.6%
Determine the dry unit weight of compaction in the
field.
Solution
Mass of dry soil excavated=
Mass of soil in the hole and cone=5.99-2.81=3.18Kg
Mass of soil in the hole= 3.18-.117=3.063Kg
(
)
V=1.8374*10-3m3
Ɣd=
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2014
Q3 (2011)
Solution
V=7500m3
Dr=94%
emax=0.73
emin=0.4
Gs=2.67
e=0.4198
Vs=5282.4341
For A
S.e=Gs.w
e=
Vt=8451.89456m3
Cost=Vt*cost=$84518.9456
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For B
S.e=Gs.w
e=
Vt=8715.488m3
Cost=Vt*cost=$43577.440
Choose B
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2014