EARTHMOVING MATERIALS
AND OPERATIONS
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.
Equipment Selection
The major criterion in selecting an equipment
is the ability of equipment to do the job. This
can 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
Production of Earthmoving
Equipments

Production= Volume per cycle x Cycles per hour
Equipment cost per hour
Cost per unit of production =
Equipment production per hour
Production
In calculating the job efficiency for cycles per hour
there 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 per
60 min (hour) by a numerical efficiency factor. The
following Table shows job efficiency factors for
earthmoving operations.
Job efficiency factors for
earthmoving operations
Management Conditions*
Job 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
Management conditions include

Skill, training, and motivation of workers;

Selection, operation, and maintenance of
equipment;

Planning, job layout, supervision, and
coordination of work.
Job conditions
Job conditions are the physical conditions of a
job that they affect production rate (not
including the type of material involved). They
include:

Topography and work dimensions.

Surface and weather conditions.

Specification requirements for work methods
or Sequence
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.
General Soil Characteristics

Unit weight : Soil unit weight is explained as kilogram
per cubic meters. Unit weight depends on type of soil,
moisture content and degree of compaction.

Moisture Content: In their natural state, all soils
contain some moisture. The moisture content is
expressed as percentage.
Moist weight - Dry weight
Moisture content (%) =
 100
Dry weight
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 state
before disturbance (Bm3).

Loose : Material that has been excavated or loaded
(Lm3).

Compacted : Material state after compaction (Cm3).
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:
Weight/bank volume
Swell (%)  (
 1)  100
Weight/loose volume
Example

Problem : Find the swell of a soil that weighs
2800kg/m3 in its natural state and 2000kg/m3 after
excavation.

Solution :
Swell  (

2800
 1)  100 = 40%
2000
This means that 1 bank m3 of material will expand to
1.4 m3 loose volume after excavation.
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.
Weight/bank volume
Shrinkage (%) = (1 )  100
Weight/Compacted volume
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 commonly
expressed 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.
Load and Shrinkage Factors
Load factor =
Weight / loose unit volume
Weight / bank unit volume
1
Load Factor =
1 + Swell

To convert a volume from bank volume to compacted volume,
shrinkage factor is used.
Shrinkage Factor 
Weight/ bank unit volum e
Weight/compacted unit volum e
Shrinkage Factor  1  Shrinkage
Example
Problem: A Soil Weighs 1163kg/Lm3,
1661kg/Bm3 and 2077kg/Cm3,
a) Find load factor and shrinkage factor.
b) How many bank Bm3 and compacted
Cm3 are contained in 593,300
Lm3 of
this soil.
Solution
a)
Load Factor =
1163
= 0.70
1661
1661
Shrinkage Factor =
= 080
.
2077
b) Bank Volume = 593,300 0.70 = 415,310 Bm3
Compacted Volume = 415,310 0.8 = 332,248Cm 3
Example
A soil weighs 1471 kg/m3 loose, 1839 kg/m3 in place
and 2090 kg/m3 compacted.

Find the swell and shrinkage.

Find load factor and shrinkage factor.

If an earth fill dam requires a compacted volume of
4.6*106 m3 of the above soil. How many loose cubic
meter of this soil must be hauled to construct the
dam.
Solution
a-
Swell (%)  (
Weight/bank volume
 1)  100
Weight/loose volume
1839
Swell (%)  (
 1) 100  25%
1471
Shrinkage (%) = (1 -
Weight/bank volume
)  100
Weight/Compacted volume
Shrinkage (%) = (1-
1839
) 100  12%
2090
Solution
Load Factor =
Load Factor =
1
1 + Swell
1
 0.80
1 + 0.25
Shrinkage Factor  1  Shrinkage
Shrinkage Factor  1  0.12  0.88
Solution
Compacted volume = (bank volume x shrinkage factor)
Bank Volume =
Compacted _ Volume 4.6 *106

 5227272Bm 3
Shrinkage_ factor
0.88
BankVolume 5227272
Losse _ Volume 

 6534091Lm 3
LoadFactor
0.80
Spoil Banks
In earthwork planning it may be necessary to
determine the size of the pile of soil that will be
created by excavation. If the pile of material is long in
relation to its width, it is referred to as spoil bank. If
the material is dumped from a fixed position, a soil
pile is created which has a conical shape. To
determine the dimensions of spoil banks or piles, it is
first necessary to convert the volume of excavation
from in-place to loose conditions. Angle of repose of
soil is the angle that the sides of a spoil bank or pile
naturally form with the horizontal surface when the
excavated soil is dumped onto the pile.
Triangular Spoil Bank
Volume  Section area  length
4V
B=(
)1/2
L  tanR
B  tanR
H=
2
Where:
B: base width (m)
H: pile height (m)
L: pile length (m)
R: angle of repose (deg)
V: pile volume (m3)
Conical Spoil Pile
1
Volume =  Base area  height
3
7.64V 1/ 3
D=(
)
tan R
D
H =  tan R
2
Where D is the diameter of the pile (m).
Table 2.6 (p. 31) shows typical values of angle of
repose of soils.
Typical Values of Angle of Repose
of Excavated Soils
Material
Angle of Repose (degrees)
Clay
35
Common earth, dry
32
Common earth, moist
37
Gravel
35
Sand, dry
25
Sand, moist
37
Example
Problem:
Find the base width and height of a
triangular spoil bank containing 76.5 Bm3
if the pile is 9.14m, the soil’s angle of
repose is 37°, and its swell is 25 %.
Solution
Loose volume=76.5 x 1.25=95.6m3
4V
B=(
)1/2
L  tanR
B  tanR
H=
2
4  95.6
1/2
Base width = (
)
= 7.45m

9.14  tan 37
7.45
Height =
 tan 37 = 2.80m
2
Example
Problem:
Find the base diameter and height of a
conical spoil pile that will contain 76.5
Bm3 of excavation if the soils angle of
repose is 32° and its swell is 12%.
Solution
Loose volume=76.5 x 1.12=87.5m3
7.64V 1/ 3
D=(
)
tan R
D
H =  tan R
2
7.64  85.7 1 / 3
Base diameter = (
)  10.16m

tan 32
Height =
10.16
 tan 32  3.17m
2
Example
A rectangular ditch having a cross-sectional
area of 2.4 m2 is being excavated in a clay.
The soil’s angle of repose is 35o and its swell
is 30%. Find the height and base width of the
triangular spoil bank that will result from the
trench excavation.
Solution
B(
4V 1/ 2
4 section _ area L 1/ 2
4 section _ area 1/ 2
) (
) (
)
L tan R
L tan R
tan R
Considering section area as loose measure = 2.4 x 1.3 = 3.12 m2
 4  3.12 
B

 tan 35 
1/ 2
H
 4.22m
B(tan R) 4.22(tan 35)

 1.478m
2
2
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. 200
sieve).

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.
Soil Classification Systems

Liquid Limit (LL) of a soil is the water content
(expressed in percentage of dry weight) at which the
soil will just start to flow when subjected to a standard
shaking test.

Plastic limit (PL) of a soil is the moisture content in
percent at which the soil just begins to crumble when
rolled into a thread 0.3 cm in diameter.

Plasticity index (PI) is the numerical difference
between the liquid and plastic limits and represents
the range in moisture content over which the soil
remains plastic.