Construction Planning, Equipment, CHAPTER GEOTECHNICAL … · 2018-09-02 · COMPACTION, AND STABILIZATION By Dr. Ibrahim Assakkaf ENCE 420 – Construction Equipment and Methods

1

• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering

Sixth EditionCHAPTER

4b

Construction Planning, Equipment, and Methods

GEOTECHNICAL MATERIALS, GEOTECHNICAL MATERIALS,

COMPACTION, AND STABILIZATIONCOMPACTION, AND STABILIZATION

ByByDr. Ibrahim AssakkafDr. Ibrahim Assakkaf

ENCEENCE 420 420 –– Construction Equipment and MethodsConstruction Equipment and MethodsSpring 2003 Spring 2003

Department of Civil and Environmental EngineeringDepartment of Civil and Environmental EngineeringUniversity of Maryland, College ParkUniversity of Maryland, College Park

CHAPTER 4b. GEOTECHNICAL MATERIALS & COMPACTIONENCE 420 ©Assakkaf

Slide No. 75

Same weight but different volume.

MATERIAL PROPERTIESMATERIAL PROPERTIES

2


Slide No. 76

Example 4Example 4

The soil borrow material to be used to The soil borrow material to be used to construct a highway embankment has a construct a highway embankment has a mass unit weight 96.0 lb per cu ft (mass unit weight 96.0 lb per cu ft (pcfpcf) ) and water content of 8%, and specific and water content of 8%, and specific gravity of soil solids is 2.66. The gravity of soil solids is 2.66. The specifications require that the soil be specifications require that the soil be compacted to dry unit weight of 112 compacted to dry unit weight of 112 pcfpcfand that the water content be held to and that the water content be held to 13%.13%.


Slide No. 77

Example 4 (cont’d)Example 4 (cont’d)

a)a) How many cubic yards of borrow are How many cubic yards of borrow are required to construct an embankment required to construct an embankment having a 250,000having a 250,000--cucu--yd net section yd net section volume?volume?

b)b) How many gallons of water must be How many gallons of water must be added per cubic yard of borrow material added per cubic yard of borrow material assuming no loss by evaporation and one assuming no loss by evaporation and one gallon of water equals 8.34 lb?gallon of water equals 8.34 lb?

c)c) If the compacted fill becomes saturated at If the compacted fill becomes saturated at a constant volume, what will be the water a constant volume, what will be the water content and mass unit weight of the soil?content and mass unit weight of the soil?

3


Slide No. 78


Borrow:Borrow:

Embankment:Embankment:

3

3

ftlb89.88

08.0196

1

66.2 8.0%, ,ftlb 0.96

=+

=+

=

===

ωγγ

γ

d

sGω

( ) ( ) 3

3

ftlb56.12613.011121

13.0%, ,ftlb 0.112

=+=+=

==

ωγγ

γ

d

d ω


Slide No. 79


(a):(a):

(b):(b):

ydcu 315,000yd)cu 001.26(250,0Required Borrow of Volume

26.189.88

112eightDry Unit WBank

eightDry Unit W CompactedFactor Shrinkage

==

===

( ) lb 000,280,98yard

ft 27yard 000,250ftlb56.14 needed Water ofWeight

ftlb56.1411256.126

:embankmentin neededWater

3

33

3

3

=

=

=−=− dγγ

4


Slide No. 80



Slide No. 81


(b) continued:(b) continued:

( )

borrow ydcu gal 39.14

lb/gal 34.81

ydcu lb120

ydcu lb120

yard 315,000lb 450,809,37 WaterdReq' of Gallons

lb 450,809,37550,470,60000,280,98Required Water Additional ofWeight

lb 550,470,60yard

ft 27yard 000,315ftlb11.7 Borrow from Water ofWeight

ftlb11.789.8896

:borrow fromWater

3

3

33

3

3

==

==

=−=

=

=

=−=− dγγ

5


Slide No. 82


(c):(c): If the fill becomes saturated all voids between the solid soilIf the fill becomes saturated all voids between the solid soilparticles are filled with water. Therefore, the total weparticles are filled with water. Therefore, the total weight is ight is increased by the added weight of water:increased by the added weight of water:

Water

AIR

Soil

Weight air = 0

Weight water = Ww

Total weight W

Weight soil solids

Ws

Volume air Va

Volumesoil solids

Vs

Volumewater

Vw

Volumevoids

Vv

Totalvolume

V

AdditionalWater to replace Air


Slide No. 83


Total volume includesTotal volume includes

AirAirWaterWaterSolids Solids

6


Slide No. 84


(c): (c): continuedcontinued

Weight of extra waterWeight of extra water::

3

3

3

ft 092.0233.0675.0000.1

ft 233.04.62

56.14

ft 675.0)4.62(66.2

112

=−−=

===

===⇒=

v

w

ww

ws

ss

ws

ss

V

WV

GWV

VWG

γ

γγ

3ftlb3.13211274.556.14

%1.18181.0112

)74.556.14(lb 74.5)4.62(092.0

=++=

==+

==

===

γ

ω

γ

s

w

www

WWVW


Slide No. 85

COMPACTION SPECIFICATION COMPACTION SPECIFICATION AND CONTROLAND CONTROL

7


Slide No. 86


The engineering properties of most soils can be improved by compaction.Compaction is the art of mechanically densifyingmaterials.


Slide No. 87

NON-COHESIVE

SMALL GRAINED < #200 MESH SIEVECOHESIVE

SOIL TYPESSOIL TYPES

8


Slide No. 88

ORGANIC SOILS

Will usually have to remove before building.

SOIL TYPESSOIL TYPES


Slide No. 89


Before the specifications for a Before the specifications for a project are prepared project are prepared representative representative soil samplessoil samples are are usually collected and tested in usually collected and tested in the laboratory to determine the laboratory to determine material properties. material properties.

9


Slide No. 90


SOIL CLASSIFICATION SOIL CLASSIFICATION ((AtterburgAtterburg Limits)Limits)LL - Liquid limitPL - Plastic limitPI - Plasticity Index


Slide No. 91


SOIL LIMITSSOIL LIMITS

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Slide No. 92

LL - Liquid limitis the water content of a soil when it passes from the plastic to liquid state.

SOIL CLASSIFICATIONSOIL CLASSIFICATION


Slide No. 93

LL - Liquid limitNon-cohesive or sandy soils have low LLs -- less than 20.Clay soils have LLs ranging from 20 to 100.


11


Slide No. 94

PL - Liquid limitis the lowest water content at which a soil remains plastic.

1/8 inch diameter thread



Slide No. 95

PI - Plastic Index

PI = LL - PL

The higher the PI the more clay that is present in the soil.


12


Slide No. 96


Normal testing would include Normal testing would include graingrain--size analysissize analysis because the because the size of the grains and the size of the grains and the distribution of those sizes are distribution of those sizes are important properties, which important properties, which affect a soil's suitability.affect a soil's suitability.


Slide No. 97

Soil gradation is the distribution, in percent (%) by weight, of individual particle sizes.

COMPACTIONCOMPACTION

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Slide No. 98


Soil Gradation (ParticleSoil Gradation (Particle--size Distribution)size Distribution)


Slide No. 99


Maximum Dry Density/Optimum Maximum Dry Density/Optimum MoistureMoisture

Critical test is the construction of a compaction curve. From compaction curves the maximum dry unit weight (density) and the percent water required to achieve maximum density can be determined.

14


Slide No. 100



Slide No. 101


Maximum Dry Density/Optimum Maximum Dry Density/Optimum Moisture (cont’d)Moisture (cont’d)

This percent of water, which corresponds to the maximum dry density (for a given compactiveeffort), is known as the optimum water content.

15


Slide No. 102

COMPACTION TESTSCOMPACTION TESTS

The standard laboratory tests that are used for evaluation of maximum dry unit weights (γd’s) and optimum moisture contents for various soils are:

1. The Standard Proctor Test (ASTM D-698 and AASHTO T-99).

2. The Modified Proctor Test (ASTM, D-1557 and AASHTO T-180)


Slide No. 103


Standard Proctor TestStandard Proctor TestThe soil is compacted in a mold that has a volume of 1/30 ft3 (943.3 cm3).The diameter of the mold is 4 in. (101.6 mm)During the laboratory test, the mold is attached to a base plate at the bottom and to an extension at the top (see Figure 1).

16


Slide No. 104


Standard Proctor TestStandard Proctor TestThe soil is mixed with varying amounts of water and then compacted in three equal layers by a hammer (Figure 2) that deliver 25 blows to each layer.


Slide No. 105


Figure 1. Standard Proctor Test Equipment: (a) mold; (b) hammer

17


Slide No. 106

PROCTOR TESTStandard ProctororAASHTO T-99

Soil sample 1/30 cubic foot3 layers



Slide No. 107


Figure 2Figure 2. Compaction . Compaction of Soil using Standard of Soil using Standard Proctor HammerProctor Hammer( courtesy of John ( courtesy of John Hester, Carterville, IL)Hester, Carterville, IL)

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Slide No. 108


Standard Proctor Test (continued)Standard Proctor Test (continued)The hammer weighs 5.5 lb (mass = 2.5 kg) and has a drop of 12 in. (304.8 mm).For each test, the moist unit weight

of compaction γ can be calculated as

mVW

=γ (28)


Slide No. 109


Standard Proctor Test (continued)Standard Proctor Test (continued)WhereW = weight of compacted soil in moldVm = volume of mold (1/30 ft3)

For each test, the moisture content w of the compacted soil is determined in the laboratory.With known moisture content, the dry unit weight γd can be calculated as

19


Slide No. 110


Standard Proctor Test (continued)Standard Proctor Test (continued)

ωγγ+

=1d

(29)


Slide No. 111


Standard Proctor Test (continued)Standard Proctor Test (continued)Where w = moisture contentThe values of γd determined from the above equation can be plotted against the corresponding moisture contents for the soil as shown the following figure (Fig. 3), which is a compaction for silty clay.

20


Slide No. 112

COMPACTION TESTSCOMPACTION TESTSFigure 3. Standard Proctor Compaction Test Results for a Silty Clay

Figure 3Figure 3


Slide No. 113


Standard Proctor Test (continued)Standard Proctor Test (continued)For a given moisture content ω, the theoretical maximum dry unit weight is obtained when there is no air in the void spaces, that is, when the degree of saturation S equal 100%. Thus, the maximum dry unit weight at a given moisture content with zero air voids can computed from

21


Slide No. 114


Standard Proctor Test (continued)Standard Proctor Test (continued)

s

ws

s

ws

GG

Gee

G

ωγγ

ω

γγ

+=

=+

=

1

so , ,saturation 100%For 1

zav

zav(30)

(31)


Slide No. 115


Modified Proctor TestModified Proctor TestThe soil is compacted in a mold that has a volume of 1/30 ft3 (943.3 cm3).The diameter of the mold is 4 in. (101.6 mm)During the laboratory test, the mold is attached to a base plate at the bottom and to an extension at the top (see Figure 4).

22


Slide No. 116


Modified Proctor Test (cont’d)Modified Proctor Test (cont’d)The soil is mixed with varying amounts of water and then compacted in five equal layers by a hammer (Figure 2) that deliver 25blows to each layer.


Slide No. 117

COMPACTION TESTSCOMPACTION TESTSFigure 4. Modified Proctor Test Equipment: (a) mold; (b) hammer

23


Slide No. 118

Soil sample 1/30 cubic foot5 layers


PROCTOR TESTModified ProctororAASHTO T-180


Slide No. 119


Modified Proctor Test (cont’d)Modified Proctor Test (cont’d)The hammer weighs 10 lb (mass = 4.54 kg) and has a drop of 18 in. (457.2 mm).For each test, the moist unit weight

of compaction g can be calculated as

mVW

=γ (32)

24


Slide No. 120


Modified Proctor Test (cont’d)Modified Proctor Test (cont’d)W = weight of compacted soil in moldVm = volume of mold (1/30 ft3)

For each test, the moisture content ω of the compacted soil is determined in the laboratory.With known moisture content, the dry unit weight γd can be calculated (Eq.29)


Slide No. 121


Modified Proctor Test (cont’d)Modified Proctor Test (cont’d)The values of γd determined from the above equation can be plotted against the corresponding moisture contents for the soil as shown the following figure (Fig. 5).

25


Slide No. 122

COMPACTION TESTSCOMPACTION TESTSFigure 5. Standard and Modified Compaction Curves


Slide No. 123

Comparison between Standard Comparison between Standard & Modified Proctor Tests& Modified Proctor Tests

Figure 5 shows compaction Figure 5 shows compaction curves which illustrate the effect curves which illustrate the effect of varying amounts of moisture of varying amounts of moisture on the density of a soil subjected on the density of a soil subjected to given to given compactivecompactive efforts.efforts.The two energy levels depicted The two energy levels depicted are known as standard and are known as standard and modified Proctor tests. modified Proctor tests.

26


Slide No. 124

It should be noted that the It should be noted that the modified Proctor, which is a modified Proctor, which is a higher energy level, gives a higher energy level, gives a higher density at a lower higher density at a lower moisture content than the moisture content than the standard Proctor, as shown standard Proctor, as shown in Figure 5.in Figure 5.



Slide No. 125

In this figure, the In this figure, the optimum moisture for optimum moisture for the standard Proctor is the standard Proctor is 16%, versus 12% for the 16%, versus 12% for the modified Proctor. modified Proctor.


27


Slide No. 126

Example 5Example 5

The laboratory test data for a The laboratory test data for a standard Proctor test are given standard Proctor test are given as shown in Table 3. Find the as shown in Table 3. Find the maximum dry unit weight and maximum dry unit weight and the optimum moisture content.the optimum moisture content.


Slide No. 127


Table 3. Test Data for Example 5Table 3. Test Data for Example 5Volume of Mold (ft3) Weight of Wet Soil in the

Mold (lb) Moisture Content ω (%) 1/30 3.88 12 1/30 4.09 14 1/30 4.23 16 1/30 4.28 18 1/30 4.24 20 1/30 4.19 22

28


Slide No. 128


Volume, V (ft3) Weight, W (lb) γ (lb/ft3) ω (%) γ d (lb/ft3) 1/30 3.88 116.4 12 103.9 1/30 4.09 122.7 14 107.6 1/30 4.23 126.9 16 109.4 1/30 4.28 128.4 18 108.8 1/30 4.24 127.2 20 106.0 1/30 4.19 125.7 22 103.0

( ) 33 ftlb6.107

14.017.122

1 ,

ftlb7.122

30/109.4

Hence 12%, 4.09, Soil Wet ofWeight :nCalculatio Sample

=+

=+

====

==

ωγ

ω

dγVWγ


Slide No. 129


Plot the dry unit weight gd against the moisture content w as shown in the following figure(Figure 6). From the figure find the maximum γd and optimum ω.

Maximum dry unit weight = 109.5 lb/ftMaximum dry unit weight = 109.5 lb/ft33

Optimum moisture content = 16.5%Optimum moisture content = 16.5%

29


Slide No. 130


Figure 6. Compaction Curve for the Data of Example 5Figure 6. Compaction Curve for the Data of Example 5

102

103

104

105

106

107

108

109

110

10 12 14 16 18 20 22 24

Moisture Content, ω (% )

Dry

Uni

t Wei

ght,

(lb/ft

^3)


Slide No. 131

COMPACTION COMPACTION SPECIFICATIONSSPECIFICATIONS

Typically specifications give an acceptable range of water content, OMC ± 2% for example.

30


Slide No. 132

In most specifications for earth work, In most specifications for earth work, the contractor is required to achieve a the contractor is required to achieve a compacted field dry unit weight of 90% compacted field dry unit weight of 90% to 95% of the maximum dry unit weight.to 95% of the maximum dry unit weight.The maximum dry unit weight is the The maximum dry unit weight is the maximum unit weight that is determined maximum unit weight that is determined by either the by either the standardstandard or or modifiedmodifiedProctor test.Proctor test.

SPECIFICATIONS FOR FIELD SPECIFICATIONS FOR FIELD COMPACTIONCOMPACTION


Slide No. 133

COMPACTION COMPACTION SPECIFICATIONSSPECIFICATIONS

The specification also sets a minimum density, 90% or95% of max. dry density for a specific test.

123.5

31


Slide No. 134

CO

MPAC

TIO

N

CO

MPAC

TIO

N

SPEC

IFIC

ATIO

NS

SPEC

IFIC

ATIO

NS 123.

5Must work in the box.


Slide No. 135

COMPACTION COMPACTION SPECIFICATIONSSPECIFICATIONSLift.Lift. A layer of soil placed on top of soil previously placed in an embankment. The term can be used in reference to material as spread or as compacted.

32


Slide No. 136



Slide No. 137

SPECIFICATIONS FOR SPECIFICATIONS FOR FIELD COMPACTIONFIELD COMPACTION

The specification for field The specification for field compaction can be based compaction can be based either oneither on

(1) relative compaction RC or(2) relative density Dr

33


Slide No. 138


The relative compaction (RC), is therefore, defined as the ratio of the dry unit weight of the soil in the field to the maximum dry unit weight of the same soil determined in the laboratory

100(%) lab) (max,

(field) ×=d

dRCγγ

(33)


Slide No. 139

The relative density Dr is given by

(34)

( )

−

−=

RCR

RDr 0

0

11

1

(max)

(min)0

d

dRγγ

=

whereγd(min) = dry unit weight in the loosest condition

(at a void ratio of emax)γd(max) = dry unit weight in the densest condition

(at a void ratio of emin)

SPECIFICATIONS FOR FIELD SPECIFICATIONS FOR FIELD COMPACTIONCOMPACTION

34


Slide No. 140


ASTM Test Designation DASTM Test Designation D--2049 2049 provides a procedure for the provides a procedure for the determination of the minimum and determination of the minimum and maximum dry unit weights of maximum dry unit weights of granular soils.granular soils.For sands, this done by using a For sands, this done by using a mold with a volume of 0.1 ftmold with a volume of 0.1 ft33 (2830 (2830 cmcm33).).


Slide No. 141


For determination of the minimum For determination of the minimum dry unit weight, sand is loosely dry unit weight, sand is loosely poured into the mold from a funnel poured into the mold from a funnel with a 1/2with a 1/2--in (12.7in (12.7--mm) diameter mm) diameter spout.spout.The average height of the fall of The average height of the fall of sand into the mold is kept at about sand into the mold is kept at about 1 in (25.4 mm)1 in (25.4 mm)

35


Slide No. 142


The value of The value of γγdd(min(min)) can then be can then be determined asdetermined as

m

sd V

W=(min)γ

whereWs = weight of sand required to fill the moldVm = volume of the mold (0.1 ft3)

(35)


Slide No. 143


The maximum dry unit weight is determined by vibrating sand in the mold for 8 min.A surcharge of 2 lb/in2 (13.8 kN/m2) is added to the top of the sand in the mold.The mold is placed on a table that vibrates at a frequency of 3600 cycles/min and that has an amplitude of vibration of 0.025 in (0.635 mm).

36


Slide No. 144


The value of The value of γγdd(max(max)) can then be can then be determined at the end of the determined at the end of the vibrating period with the knowledge vibrating period with the knowledge of the weight and volume of sand.of the weight and volume of sand.An empirical formula has been An empirical formula has been developed by Lee and Singh (1971) developed by Lee and Singh (1971) to give a relationship between to give a relationship between RCRCand and DDrr..


Slide No. 145


For granular soils, the relationship For granular soils, the relationship is given asis given as

RC (%) = 80 +0.2 Dr

According to Lee and Singh (1971), the correlation between RC and Dr was based on the observation of 47 soil samples.

(36)

Documents

Construction Planning, Equipment, CHAPTER GEOTECHNICAL … · 2018-09-02 · COMPACTION, AND STABILIZATION By Dr. Ibrahim Assakkaf ENCE 420 – Construction Equipment and Methods