training slide

1. Soil Compaction
COSC 323: Soils in Construction

2. Question
 The contractor, during construction of the soil
embankment, conducted a sand-cone in-place density
test to determine whether the required compaction was
achieved. The following data were obtained during the
test:
 Weight of sand used to fill test hole and funnel of sand-cone
device = 845 g.
 Weight of sand to fill funnel = 323g.
 Unit weigh of sand = 100 lb/ft3
 Weigh of wet soil from test hole = 600g
 Moisture content of soil from test hole = 17%.
 Based on the contract, the contractor is supposed to
attain the 95% compaction. Will you approve the
contractor’s work?

3. Compaction
 Compaction – expelling air from the void space
 Consolidation – extrusion of water
 Effects of compaction
 Increase soil’s shear strength
 Decrease in future settlement of the soil
 Decrease in its permeability
 How to quantify – use dry unit weight of soil
contentmoisture
tunit weighwet
1
=
=
+
=
w
w
d
γ
γ
γ

4. Compaction
 What does water do for compaction?
 Lubricant
 Too much water  lesser density
 Optimum moisture content
(=maximum dry unit weight)  best compaction
 How to use maximum dry unit weight?
 Target unit weight at the job site
 Need to know how much the soil can be compacted.

5. Compaction Test
 ASTM D 698
 Standard Proctor Compaction Test
 Hammer: 5.5 lb
 Drop Height: 12in
 ASTM D 1557
 Modified Proctor Compaction Test
 Hammer: 10 lb
 Drop Height: 18in
 For higher value of dry unit weight

7. Compaction Test
 Dry out soil sample
 Add water to the soil sample
 Compact the soil sample in the compaction mold
 Measure wet unit weight (γ)
 Measure moisture content (w)
 Calculate dry unit weight
 Repeat 4 times with different moisture contents
w
d
+
=
1
γ
γ

8. Proctor Curve

9. Example
Condition
The combined weight of a mold and the
specimen of compacted soil it contains is 9.0
lb.
The mold’s volume is 1/35ft3
The mold’s weight is 4.35 lb.
The specimen’s water content is 12%.
What is dry unit weight of the specimen?

10. Example
 A set of laboratory compaction test data and
results is tabulated as follows. Determine the
soil’s maximum dry unit weight and optimum
moisture content.
Determination
Number 1 2 3 4 5
Dry unit weight
(lb/ft3
) 112 115 115 113 109
Moisture
content(%) 5 10 15 20 25

11. What affects Compaction?
 Moisture content
 Compaction effort
 Compaction energy per unit volume (function of
number of blows per layer)
 For the stand proctor test: 12,400 ft-lb/ft3
 For the modified proctor test: 56,000 ft-lb/ft3
 Type of soil
 Grain size distribution
 Specific gravity of solids
 Type and amount of clay materials

12. Compaction

13. Facts about Compaction
 Maximum dry unit weight
 Min: Organic soils (60lb/ft3
)
 Max: Well-graded granular material (145 lb/ft3
)
 Optimum moisture contents
 Min: Granular material (5%)
 Max: Elastic silts and clays (35%)
 Higher optimum moisture contents = Lower dry
unit weight

16. Field Compaction
 Compacted in layers
 8 in. loose horizontal layer  compacted to a
thickness of 6 in.
 Sprinkling or drying to control moisture content
 Scarifying to provide bonding between layers.
 Equipments
 Tempers – Limited in scope and compacting ability
 Rollers
 Smooth wheel roller, Sheepsfoot roller, Pneumatic roller,
Vibratory roller

17. Field Compaction
 Smooth wheel roller
 Provide a smooth finished grade
 Used for paving

18. Field Compaction
 Sheepsfoot roller
 Greater compaction pressure
 Effective for compacting fine-grained soil

19. Field Compaction
 Pneumatic roller
 Effective for compacting clayey soil and silty soils

20. Field Compaction
Vibratory roller
Effective for compacting granular materials:
clean sands and gravels

21. Dynamic Compaction
 When
 Existing surface or near-surface soil is poor with
regard to foundation support
 For which soil?
 Both cohesive and cohesionless soils
 How
 Drop a very heavy (2~20 tons) weight onto the soil
from a relatively great height (20 ~ 100 ft)
 Dropping weight randomly?  a closely spaced grid
pattern is selected.

22. Dynamic Compaction

23. Dynamic Compaction
How deep soil will be affected?
Approximate depth of influence of dynamic
compaction (D)
(m)height:
tons)(metricWeight:
(m)compactiondynamicofinfluenceofDepth:
soilcohesive:
soilsscohesionle:5.0
h
W
D
WhD
WhD
=
=

24. Dynamic Compaction
How many drops do we need?
Typically 5~10 drops
Additional drops beyond 10 drops improves
little
What about craters?
Need to be backfilled and compacted by other
means

25. In-Place Soil Unit Weight Test
Destructive testing method
Sand-cone
Rubber-balloon methods
Nondestructive method
Nuclear moisture-density apparatus
Soil unit weight is inversely proportional to the
amount of radiation that reaches the detector.
Speedy Moisture Tester

26. In-Place Test: Sand-cone
method

27. Example
 During construction of a soil embankment, a
sand-cone in-place unit weight test was
performed in the field.
 Weight of sand used to fill test hole and funnel of
sand-cone device = 867g
 Weigh of sand to fill funnel = 319g
 Unit weigh of sand = 98.0 lb/ft3
 Weigh of wet soil from the test hole = 747g
 Moisture content of soil from test hole = 13.7%

28. Example 4-3
3
3
3
3
3
3
/8.117
137.01
/9.133
1
/9.133
0123.0
/6.453/747
place-insoiloftunit weighWet
0123.0
/0.98
/6.453/548
holetestofVolume
548g319g-867g
funelfilltosandofWeight-funnelandholetestfilltosandofWeight
holein testusedsandofWeight
ftlb
ftlb
w
ftlb
ft
lbgg
ft
ftlb
lbgg
d =
+
=
+
=
==
==
==
=
γ
γ

29. In-Place Test: Rubber-balloon

30. In-Place Test: Nuclear
Apparatus
Nuclear moisture-
density apparatus

31. Field Control of Compaction
 Required percent of compaction
the required in-place dry unit weight
= ----------------------------------------------------- x 100%
the maximum laboratory dry unit weight
 Minimum number of field unit weight tests
required.
 Maximum thickness of loose lifts (layers)
 Methods to obtain maximum dry unit weight
 Methods to determine in-place unit weight

32. Example
 Given
 Soil from a borrow pit to be used for construction of
an embankment gave the following laboratory results
when subjected to the ASTM D 698 Standard Proctor
test
 Maximum dry unit weight = 118.5 lb/ft3
 Optimum moisture content = 12.5%
 The contractor, during construction of the soil
embankment, achieved the following
 Dry unit weight reached by field compaction = 117.8 lb/ft3
 Actual water content = 13.7%
 Required
 Percent of compaction achieved by the contractor

33. Example
 Solution
%4.99100
/5.118
/8.117
100
tunit weighdrylaboratoryMaximum
tunit weighdryplace-In
achievedcompactionProctorStandardofPercent
3
3
=×=
×=
ftlb
ftlb

34. Example
 Given
 A borrow pit’s soil is being used as earth fill at a construction
project.
 The in situ dry unit weight of the borrow pit soil was determined
to be 17.18 kN/m3
 The soil at the construction site is to be compacted to a dry unit
weight of 18.90 kN/m3
 The construction project requires 15,000m3
of compacted soil fill.
 Required
 Volume of soil required to be excavated from the borrow pit to
provide the necessary volume of compacted fill.

35. Example
 Solution
3
3
33
500,16
17.18kN/m
283,500kN
pitborrowthefromobtainedbetorequiredsoilofVolume
283,500kN))(15,000m(18.90kN/m
pitborrowthefromexcavatedbetorequredsoilofdry weightTotal
fillcompactedefurnish thtorequireddry weightTotal
m==
==
=

36. Example
 Given
 The in situ void ratio of a borrow pit’s oil is 0.72.
 The borrow pit soil is to be excavated and transported
to fill a construction site where it will be compacted to
a void ratio of 0.42.
 The construction project requires 10,000 m3
of
compacted soil fill.
 Required
 Volume of soil that must be excavated from the
borrow pit to provide the required volume of fill

37. Example
 Solution
3
3
3
042,7)(
000,10)(42.0)(
000,10)()(
)())(42.0(
)(
)(
42.0
fillin thesoil
mV
mVV
mVV
VfV
V
V
f
V
V
e
fs
fsfs
fvfs
fvs
fs
fv
s
v
=
=+
=+
=
=
=
=

38. Example
 Solution
333
3
3
112,12042,7070,5)()()(
3070,5)(
)()042,7)(72.0(
042,7)()(
)())(72.0(
)(
)(
72.0
pitborrowin thesoil
mmmVVV
mV
Vm
mVV
VV
V
V
b
bvbvb
bv
bv
fsbs
bvbs
bs
bv
=+=+=
=
=
==
=
=
=