CIVL 392 - ddddd2 - Earthmoving Materials and Operations

8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations

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EARTHMOVING MATERIALS

AND OPERATIONS


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Introduction to Earthmoving

Earthmoving is the process of moving soil or

rock from one location to another and

processing it so that it meets construction

requirements such as location, elevation,density, moisture content etc.

Activities of earthmoving: Excavation, loading,

hauling, damping, spreading, compacting,

grading and finishing.


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Equipment Selection

The major criterion in selecting an equipmentis the ability of equipment to do the job. Thiscan be specified as:

ability to maximise the profit

possible future use

availability of equipment, spare parts and

services the effects of equipment down-time on other

construction equipments


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Production of Earthmoving

Equipments

Production= Volume per cycle x Cycles per hour

Cost per unit of production =Equipment cost per hour

Equipment production per hour


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Production

In calculating the job efficiency for cycles per hourthere are two methods:

Using the number of effective working minutes in an

hour to calculate the cycles per hour.

To multiply the theoretical number of cycles per60 min (hour) by a numerical efficiency factor. The

following Table shows job efficiency factors forearthmoving operations.


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Job efficiency factors for

earthmoving operations

Job Conditions**

Management Conditions*

Excellent Good Fair Poor

Excellent 0.84 0.81 0.76 0.70

Good 0.78 0.75 0.71 0.65

Fair 0.72 0.69 0.65 0.60

Poor 0.63 0.61 0.57 0.52


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Management conditions include

Skill, training, and motivation of workers;

Selection, operation, and maintenance ofequipment;

Planning, job layout, supervision, andcoordination of work.


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Job conditions

Job conditions are the physical conditions of ajob that they affect production rate (notincluding the type of material involved). Theyinclude:

Topography and work dimensions.

Surface and weather conditions.

Specification requirements for work methodsor Sequence


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General Soil Characteristics

Trafficability : The ability of a soil to support

the weight of vehicles under repeated traffic.

Trafficability is measured qualitatively. It is

function of soil type and moisture.

Loadability : is the measure of the difficulty in

excavating and loading a soil. Loose granular

soils are highly loadable, whereas compacted

coehesive soils and rocks have low loadability.


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General Soil Characteristics

Unit weight : Soil unit weight is explained as kilogramper cubic meters. Unit weight depends on type of soil,moisture content and degree of compaction.

Moisture Content: In their natural state, all soilscontain some moisture. The moisture content isexpressed as percentage.

100Dry weight

Dry weight-htMoist weig=(%)contentMoisture


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Soil Volume Change Characteristics

Soil Conditions

There are three conditions that the soil may exist:

Bank (in place or in situ): Material in its natural statebefore disturbance (Bm3).

Loose : Material that has been excavated or loaded(Lm3).

Compacted : Material state after compaction (Cm3).


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Swell

Soil volume increases after excavation since the soil

grains are loosened during excavation and air fills the

void spaces created. As a result one unit volume of

soil in the bank condition will occupy more than one

unit volume after excavation. This is called as Swell.Swell can be calculated as:

100)1volumeseWeight/loo

k volumeWeight/ban

((%)Swell


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Shrinkage

When a soil is compacted, some of the air inside of the

soil is forced out. As a result the soil occupies less

volume than it does either in bank or loose conditions.

This is the reverse of swell and called as shrinkage.

100)

volumepactedWeight/Com

k volumeWeight/ban-(1=(%)Shrinkage


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Load and Shrinkage Factors

In earthmoving contracts, it is important to convert all

material volumes to a common unit of measure.

Haul unit or spoil bank volumes are commonlyexpressed in loose measure. Conversion factor to

convert from loose to bank volume is called as Load

Factor. Loose volume is multiplied by load factor to

find bank volume.


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Load and Shrinkage Factors

To convert a volume from bank volume to compacted volume,

shrinkage factoris used.

Load factor =Weight / loose unit volume

Weight / bank unit volume

Swell+1

1=FactorLoad

eunit volumpactedWeight/comeunit volumbankWeight/FactorShrinkage

Shrinkage1FactorShrinkage


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Example

Problem:A Soil Weighs 1163kg/Lm3,

1661kg/Bm3and 2077kg/Cm3,

a) Find load factor and shrinkage factor.

b) How many bank Bm3

and compactedCm3are contained in 593,300 Lm3of

this soil.


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Solution

a)

b)

Load Factor =1163

1661= 0 70.

Shrinkage Factor =

1661

2077=

080.

3Bm415,310=0.70593,300=VolumeBank

Cm332,248=0.8415,310=VolumeCompacted 3


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Example

A soil weighs 1471 kg/m3loose, 1839 kg/m3in placeand 2090 kg/m3compacted.

Find the swell and shrinkage.

Find load factor and shrinkage factor.

If an earth fill dam requires a compacted volume of

4.6*106 m3of the above soil. How many loose cubicmeter of this soil must be hauled to construct thedam.


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Solution

a- 100)1volumeseWeight/loo

k volumeWeight/ban((%)Swell

25%100)11471

1839((%)Swell

100)volumepactedWeight/Com

k volumeWeight/ban-(1=(%)Shrinkage

12%100)20901839-(1=(%)Shrinkage


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Solution

Swell+1

1=FactorLoad

80.00.25+1

1=FactorLoad

Shrinkage1FactorShrinkage

0.880.121FactorShrinkage


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Solution

Compacted volume = (bank volume x shrinkage factor)

Bank Volume =

36

522727288.0

10*6.4

_

_

BmfactorShrinkage

VolumeCompacted

3

653409180.0

5227272_ LmLoadFactor

BankVolume

VolumeLosse


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Triangular Spoil Bank

Where:

B: base width (m)

H: pile height (m)

L: pile length (m)R: angle of repose (deg)

V: pile volume (m3)

lengthareaSectionVolume

2

tanRB=H)

tanRL

4V(=B 1/2


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Material Angle of Repose (degrees)

Clay 35

Common earth, dry 32

Common earth, moist 37

Gravel 35

Sand, dry 25

Sand, moist 37

Typical Values of Angle of Repose

of Excavated Soils


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Example

Problem:

Find the base width and height of atriangular spoil bank containing 76.5 Bm3

if the pile is 9.14m, the soils angle of

repose is 37, and its swell is 25 %.


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Solution

Loose volume=76.5 x 1.25=95.6m3

7.45m=)37tan9.14

95.64(=widthBase 1/2

2.80m=37tan2

7.45=Height

2

tanRB=H)

tanRL

4V(=B 1/2


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Example

Problem:

Find the base diameter and height of aconical spoil pile that will contain 76.5

Bm3of excavation if the soils angle of

repose is 32and its swell is 12%.


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Example

A rectangular ditch having a cross-sectional

area of 2.4 m2is being excavated in a clay.

The soils angle of repose is 35oand its swellis 30%. Find the height and base width of the

triangular spoil bank that will result from the

trench excavation.


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Solution

mB 22.435tan

12.342/1

2/12/12/1 )tan

_sec4()

tan

_sec4()

tan

4(

R

areation

RL

Lareation

RL

VB

Considering section area as loose measure = 2.4 x 1.3 = 3.12 m2

mRB

H 478.12

)35(tan22.4

2

)(tan


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Soil Identification and classification

Fundamental Soil Types

Soil is considered to be composed of five fundamental soil types: gravel,sand, silt, clay, and organic material.

Gravel:is composed of individual particles larger than 6 mm but smaller

than 76 mm.

Sand: material smaller than gravel but larger than 0.07 mm (No. 200sieve).

Silt: particles passing No. 200 sieve but larger than 0.002 mm.

Clay: is composed of particles less than 0.002 mm in diameter.

Organic soils: contain partially decomposed vegetable matters.


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Soil Classification Systems

Liquid Limit(LL)of a soil is the water content(expressed in percentage of dry weight) at which thesoil will just start to flow when subjected to a standardshaking test.

Plastic limit(PL)of a soil is the moisture content inpercent at which the soil just begins to crumble whenrolled into a thread 0.3 cm in diameter.

Plasticity index (PI)is the numerical differencebetween the liquid and plastic limits and representsthe range in moisture content over which the soilremains plastic.

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CIVL 392 - ddddd2 - Earthmoving Materials and Operations

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