8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
1/34
EARTHMOVING MATERIALS
AND OPERATIONS
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
2/34
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.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
3/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
4/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
5/34
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.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
6/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
7/34
Management conditions include
Skill, training, and motivation of workers;
Selection, operation, and maintenance ofequipment;
Planning, job layout, supervision, andcoordination of work.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
8/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
9/34
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.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
10/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
11/34
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).
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
12/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
13/34
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
14/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
15/34
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.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
16/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
17/34
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.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
18/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
19/34
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.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
20/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
21/34
Solution
Swell+1
1=FactorLoad
80.00.25+1
1=FactorLoad
Shrinkage1FactorShrinkage
0.880.121FactorShrinkage
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
22/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
23/34
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
24/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
25/34
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
26/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
27/34
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 %.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
28/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
29/34
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%.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
30/34
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
31/34
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.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
32/34
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
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
33/34
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.
8/10/2019 CIVL 392 - ddddd2 - Earthmoving Materials and Operations
34/34
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.