lecture soil compaction
TRANSCRIPT
Soil Compaction
COSC 323: Soils in Construction
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?
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
content moisture
tunit weigh wet 1
w
wd
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.
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
Compaction Test
Dry out soil sample Add water to the soil sample Compact the soil sample in the compaction mold Measure wet unit weight () Measure moisture content (w) Calculate dry unit weight
Repeat 4 times with different moisture contents
wd
1
Proctor Curve
Example
ConditionThe 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?
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
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
Compaction
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
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
Field Compaction
Smooth wheel roller Provide a smooth finished grade Used for paving
Field Compaction
Sheepsfoot roller Greater compaction pressure Effective for compacting fine-grained soil
Field Compaction
Pneumatic roller Effective for compacting clayey soil and silty soils
Field Compaction
Vibratory rollerEffective for compacting granular materials:
clean sands and gravels
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.
Dynamic Compaction
Dynamic Compaction
How deep soil will be affected?Approximate depth of influence of dynamic
compaction (D)
(m)height :
tons)(metric Weight :
(m) compaction dynamic of influence ofDepth :
soil cohesive :
soil sscohesionle:5.0
h
W
D
WhD
WhD
Dynamic Compaction
How many drops do we need?Typically 5~10 dropsAdditional drops beyond 10 drops improves
littleWhat about craters?
Need to be backfilled and compacted by other means
In-Place Soil Unit Weight Test
Destructive testing methodSand-coneRubber-balloon methods
Nondestructive methodNuclear moisture-density apparatus
Soil unit weight is inversely proportional to the amount of radiation that reaches the detector.
Speedy Moisture Tester
In-Place Test: Sand-cone method
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%
Example 4-3
33
33
33
/8.117137.01
/9.133
1
/9.1330123.0
/6.453/747 place-in soil oft unit weighWet
0123.0/0.98
/6.453/548 hole test of Volume
548g319g-867g
funel fill tosand of Weight - funnel and hole test fill tosand ofWeight
holein test used sand ofWeight
ftlbftlb
w
ftlbft
lbgg
ftftlb
lbgg
d
In-Place Test: Rubber-balloon
In-Place Test: Nuclear Apparatus
Nuclear moisture-density apparatus
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
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
Example
Solution
%4.99100/5.118
/8.117
100tunit weighdry laboratory Maximum
tunit weighdry place-In
achieved compactionProctor Standard ofPercent
3
3
ftlb
ftlb
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.
Example
Solution
33
33
500,1617.18kN/m
283,500kN
pit borrow thefrom obtained be torequired soil of Volume
283,500kN))(15,000m(18.90kN/m
pit borrow thefrom excavated be torequred soil of dry weight Total
fill compacted efurnish th torequired dry weight Total
m
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
Example
Solution
3
3
3
042,7)(
000,10)(42.0)(
000,10)()(
)())(42.0(
)(
)(42.0
fill in the soil
mV
mVV
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fs
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s
v
Example
Solution
333
3
3
112,12042,7070,5)()()(
3070,5)(
)()042,7)(72.0(
042,7)()(
)())(72.0(
)(
)(72.0
pit borrow in the soil
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mV
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b
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