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Othman A. Tayeh 2014

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.


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.).


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%.


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


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


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:


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% w Ɣzav 5 0.05 149.408

8 0.08 139.396

10 0.1 133.434

12 0.12 127.961

15 0.15 120.545

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 ρd

0.8 2001.87

w=10% 0.9 2059.13

1 2107.34

0.8 1601.2

w=20% 0.9 1675.73

1 1740.55


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) W

(Ib) w% Ɣ Ɣd

0.03333 3.26 8.4 97.8 90.2214

0.03333 4.15 10.2 124.5 112.976

0.03333 4.67 12.3 140.1 124.755

0.03333 4.02 14.6 120.6 105.236

0.03333 3.63 16.8 108.9 93.2363

0

20

40

60

80

100

120

140

5 10 15 20


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.


Solution

V(cm3) M(Kg) w% V(m3) w ρ(kg/m3) ρd(kg/m3)

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

1600

1620

1640

1660

1680

1700

1720

5 7 9 11 13 15 17 19 21 23


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=


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


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


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=


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=


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


For B S.e=Gs.w

e=

Vt=8715.488m3 Cost=Vt*cost=$43577.440 Choose B

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