loading and hauling - Civil Engineering Department

LOADING AND HAULING
LOADING AND HAULING
Cycle time (time required for a haul unit to make one complete
cycle) can be broken into two as:
Cycle time = Fixed time +Variable time
Fixed time: includes spot time (moving the unit into position
to begin loading), load time, manoeuvre time, and dump
time.
Variable time: represents travel time required for a unit to
haul material to the unloading site and return. Variable time
depends on weight and power of vehicle, haul road
conditions, the grades encountered, and the altitude above
sea level.
Total Resistance
To determine the maximum speed of a vehicle in a
specific situation, it is necessary to determine the total
resistance of movement of the vehicle.
The resistance has two components:
Total Resistance = Grade Resistance + Rolling Resistance
Resistance is expressed as either in kilograms per ton
of vehicle weight or in kilograms. (Sometimes
resistance factor is used to denote kg/ton).
Rolling Resistance
Rolling Resistance is the resistance encountered by a
vehicle in moving over a surface. For rubber tired vehicles,
rolling resistance depends on:


Size of tire: larger tire means larger rolling resistance (rr)
Pressure of tires: high pressure on hard surface means low
rr; high pressure on soft surface means high rr (it sinks).
 Tread design of tires: narrow tread gives low rr ; broad
treads give high rr.
 Condition of surface:
Hard surface : low rr
Soft surface : high rr
Muddy surface: high rr
Rolling Resistance

Rolling resistance factor for conventional tires is about 20 kg/t. and for
radial tires is 15 kg/t.

The rolling resistance factor increases about 15 kg/t for each 2.5 cm of
tire penetration. Therefore leading to the following formula:
Rolling Resistance Factor (kg/t) = 20 + 6 x penetration (cm)
(for conventional tires)

Rolling resistance is found by multiplying rolling resistance factor by
vehicle's weight. (Table 4-1 shows typical values of rolling resistance
factors.)

Crawler mounted vehicles has no rolling resistance to consider.

However, if crawler mounted tractor tows a tired vehicle, the rolling
resistance of the towed vehicle will be considered.
Typical values of rolling resistance factors
(Table 4-1 p.82)
Rolling Resistance Factors
lb /ton
kg/ton
Concrete or asphalt
40 (30*)
20 (15)
Firm, smooth, flexing slightly
under load
64 (52)
32 (26)
Rutted dirt roadway,
1-2 in. penetration
100
50
Soft, rutted dirt, 3-4 in.
penetration
150
75
Loose sand or gravel
200
100
300-400
150-200
Type of Surface
Soft, muddy, deeply rutted
* Values in parentheses are for radial tires.
Grade Resistance

Grade resistance is positive when the
vehicle is travelling up a grade, and
negative when it is travelling downhill.

On Construction site, grade resistance
factor is considered to be 10 kg/ton for
each 1% of grade.
Grade Resistance
Grade Resistance factor(kg/ton) = 10  grade(%)
Grade Resistance (kg) = Vehicle weight (t)  grade resistancefactor (kg/t)
Grade Resistance (kg) = Vehicle weight (kg)  grade
Effective Grade
As a simpler method for expressing the
total resistance, it can be stated as a
grade (%) which is called effective grade,
equivalent grade, or percent total
resistance.
Effective Grade = Grade (%) +
Rolling Resistance factor (kg / t)
10
Example
A wheel tractor-scraper weighing 91t is being operated
on a haul road with a tire penetration of 5 cm. What is the
total resistance (in kg) and effective grade when:
a. the scraper is ascending a slope of 5%?
b. the scraper is descending a slope of 5% ?
Solution
Rolling resistancefactor = 20 + (6  5) = 50 kg/ton
Rolling resitance = 50 (kg/ton)  91ton = 4550kg
Grade Resistance
= 91 t  1000 (kg/t)  0.05 = 4550 kg
a.
Total Resistance = Rolling Resistance + Grade Resistance
4550kg + 4550kg = 9100kg
Effective grade = 5 +
50
 10%
10
Solution
b.
Grade Resistance = 91ton x 1000 kg/ton x (-0.05) = -4550kg.
Total Resistance = Rolling Resistance + Grade Resistance =
4550kg + (-4550kg) = 0.0
Effective grade = (-5) + 50/10 = 0.0
Effect of Altitude


All internal combustion engines lose power as their
elevation above sea level increases because of the
decreased density of air at higher elevations.
Engine power decreases approximately 3% for each
305m increase in altitude above the maximum altitude
at which full rated power is delivered.
Altitude (m) - 915*
Derating factor (%) = [
]
102
*Substitute maximum altitude for rated performance if known.
The percentage of rated power available = 100 - Derating factor
Effect of Traction

The power available to move a vehicle and its load is expressed as
rimpull for wheel vehicles and drawbar pull for crawler tractors.

Rimpull is the pull available at the surface of the tire.

A limiting factor in using the power of a vehicle is the maximum traction
that can be developed between the driving wheels and the road surface.

Traction depends on the coefficient of traction at the road surface and
the weight of the drivers (wheels). This represents the maximum pull that
a vehicle can develop regardless of vehicle horse power.
Maximum usable pull = Coefficient of traction  weight on drivers

For crawler tractors and all wheel-drive rubber-tired equipments, the
weight on the drivers is the total vehicle weight. Typical coefficient of
traction values are given in Table 4-2.
Typical values of coefficient of traction
(Table 4-2 p.85)
Type of Surface
Rubber Tires
Tracks
Concrete, dry
0.90
0.45
Concrete, wet
0.80
0.45
Earth or clay loam, dry
0.60
0.90
Earth or clay loam, wet
0.45
0.70
Gravel, loose
0.35
0.50
Quarry pit
0.65
0.55
Sand, dry, loose
0.25
0.30
Sand, wet
0.40
0.50
Snow, packed
0.20
0.25
Ice
0.10
0.15
Example
A four-wheel-drive tractor weighs 20,000
kg and produces a maximum rimpull of
18,160 kg at the sea level. The tractor is
being operated at an altitude of 3050 m
on wet earth. A pull of 10,000 kg is
required to move the tractor and its load.
Can the tractor perform under these
conditions?
Solution
Derating factor =
 3050  915


21
%


102
Percent rated power available = 100 - 21 = 79%
Maximum available power = 18160 x 0.79 = 14346 kg 10000 kg
Therefore, sufficient rimpull is available to move the tractor.
Solution
Coefficient of traction = 0.45 (Table 4 - 2)
Maximum useable pull = 0.45 x 20000 = 9000kg
Any pull applied more than this causes slippage
of tires.
Since required pull to make it moving is 10000
kg  9000 kg maximum useable pull, tires will
slip and tractor cannot perform this job.
DOZERS

Tractor equipped with a front mounted
earthmoving blade is known as bulldozer or
dozer.

Blade is lowered to cut the soil and push the
soil in front of blade.

It unloads the soil by pushing it over a cliff or
into a hopper or by raising the blade.
Bulldozers are used for
- stripping topsoil and clearing vegetation
- shallow excavating
- maintaining haul roads
- opening up pilot roads
- spreading and grading
- pushing scraper
- ripping
- touring some equipment, such as
compactors
Dozers

Dozers are normally crawler mounted and rarely
used wheel mounted.

Crawler dozers can operate on steeper side slopes,
climb greater grades than can wheel mounted dozers,
and can operate in rough terrain. Apply low ground
pressure, 6-9 lb/in² so good in low trafficable areas.

On the other hand, wheel dozers can move faster
than crawler dozer and move on paved roads without
damaging its surface.
Dozer blades
Straight blades
The section perpendicular to line of push is
straight.
The curvature causes to roll the material
forward.
The driver controls the depth of cut during
pushing action by feel.
When rear of the dozer is felt to rise, the
blade is starting to dig in.
Dozer blades
Universal Blade (U-Blade)
The blade in cross section has a much
deeper curvature, almost approaching
a "U" shape.
In addition the outer edges are angled
slightly inwards.
U-Blades can carry larger volumes of
soil.
Dozer blades
Angle Blade
The blade in plan view is angled up to
25° and cast material to one side. This
blade is mostly used in backfilling
along a trench and operating a pilot
road in hilly terrain.
Dozer blades
Cushion Blade
Especially larger machine bulldozers
(300 hp or more) are used to push
scrapers while loading. The blade
used in this case is usually fitted with
a shock absorbers and the blade is
much more stronger than the other
blades.
Estimating Dozer Production
Production = Volume per cycle cycles per hour
Estimation of blade volume:
Doze a full blade load, then lift the blade while moving forward on a
level surface until an even pile is formed:
- Measure the width (W) of the pile
- Measure the height (H) of the pile
- Measure the length (L) on the pile
Blade Load(Lm3 )  H(m)  W(m)  L(m)
Dozer cycle time = fixed cycle time + variable cycle time
Estimating Dozer Production
Fixed cycle time represents time for
manoeuvring, changing gears, start
loading and dump.
 Variable cycle time is the time required
to doze and return.

Typical dozer fixed cycle times
Table 4-4 p.96
Operating Conditions
Time (min)
Power-shift transmission
0.05
Direct-drive transmission
0.10
Hard digging
0.15
Typical dozer operating speeds
Table 4-5 p.97
Operating Conditions
Speeds
Dozing
Hard materials, haul 100 ft (30 m)
or less
1.5 mi/hr (2.4 km/hr)
Hard materials, haul over 100 ft
(30 m)
2.0 mi/hr (3.2 km/hr)
Loose materials, haul 100 ft (30 m)
or less
2.0 mi/hr (3.2 km/hr)
Loose materials, haul over 100 ft
(30 m)
2.5 mi/hr (4.0 km/hr)
Returning
100 ft (30 m) or less
Maximum reverse speed in second range (power
shift) or reverse speed in gear used for dozing (direct
drive)
Over 100 ft (30 m)
Maximum reverse speed in third range (power shift)
or highest reverse speed (direct drive)
Example

A power-shift crawler tractor has a rated
blade capacity of 10 LCY (7.65 Lm3).
The dozer is excavating loose common
earth and pushing it a distance of 200 ft
(61 m). Maximum reverse speed in third
range is 5 mi/h.
Solution
Fixed time = 0.05 min
Dozing speed = 2.5 mi / hr (4.0 km / hr)
200
Dozing time =
= 0.91 min
2.5  88


61
=
= 0.91 min

 4  16.7


Note: 1 mi/hr = 88 ft/min; 1 km/hr = 16.7 m/min
Solution
Return time =
200
= 0.45 min
5  88


61
 = 8  16.7 = 0.45 min
Cycle time = 0.05 + 0.91+ 0.45 =1.41 min
50
Production = 10 
= 355 LCY / hr
1.41


50
3
 = 7.65  1.41 = 271 Lm / hr 
LOADERS

A tractor equipped with a front end
bucket, called a loader, front end loader
or bucket loader. Both wheel loaders and
track loaders are available.
Loaders are used:
- to excavate soft to medium hard material
- loading haul units and hopers
- stockpiling material
- backfilling ditches
- moving concrete and other construction materials
Wheel Loaders
- higher speed (25 mil/hr or more)
- good job mobility
- articulated (hinged between front and rear
axles to provide great manoeuvrability)
Track Loaders
- overcoming steeper grades
- operating in areas of higher side slopes
- low ground pressure and high tractive
effect
- lower speed than has wheel loaders
Estimating Loader Production
Production(m3 )  bucket load  cycles per hour
Basic loader cycle times (Table 4-6 p.103):
Basic Cycle Time (min)
Loading conditions
Articulated wheel loader
Track Loader
Loose materials
0.35
0.30
Average material
0.50
0.35
Hard materials
0.65
0.45
Example

Estimate the hourly production in loose
volume (LCY or Lm3) of a 3½ yd (2.68
m3) wheel loader excavating sand and
gravel (average material) from a pit and
moving it to a stockpile. The average
haul distance is 200 ft (61 m), the
effective grade is 6%, the bucket fill
factor is 1.00, and job efficiency is 50
min/hr.
Figure 4-14
p.104
Solution
Bucket volume = 3.5 1 = 3.5 LCY (2.68 Lm 3 )
Basic cycle time = 0.50 min. (Table 4.6)
Travel time = 0.30 min. (Figure 4.11)
Cycle time = 0.50 + 0.30 = 0.80
Production = 3.5 
[= 2.68
50
 219 LCY/hr
0.80
50
 168 Lm3 / hr]
0.80
SCRAPERS

Scrapers are capable of excavating,
loading, hauling and dumping material
over medium to long haul distances.

Scraper excavates (cuts) at a depth of
150 to 300 mm by lowering the front
edge of its bowl into the soil.
Scrapers can be classified as:

Towed scrapers
A towed scraper is towed by a bulldozer having a capacity of
300 hp or more up to a maximum distance of 300 m.

Motorised scrapers
Motorised scrapers frequently require pushing assistance
during loading. Pushing can be performed by a bulldozer or by
another scraper. Haul speed up to 60 km/hr is possible. Their
capacity could be from 15 to 50 m3 heaped capacity. There are
three types of motorised scrapers.
Motorised scrapers

i) Single-engine scrapers
The scraper comprises a bowl mounted on a single rear axle.
The front end is hitched to the drive axle by means of single
arm called swan neck. The bowl height is controlled through
a pivot attachment to the swan neck and hydraulic cylinder.

ii) Double-engine scraper
Second engine is placed on rear axle and provide four wheel
drive. However, it also may require a pusher during loading.

Iii) Self loading (elevating) scraper
Elevating scrapers utilise a ladder type elevator to assist in
cutting and lifting material into the scraper bowl. Elevating
scrapers are not designed to be push-loaded and may be
damaged by pushing.
Cutting Action

The bowl is lowered and the apron is
opened, the forward movement of the
machine directs the cutting edge into the
soil causing it to boil up works into the
bowl. Excavation is carried out in layers
of 150 to 300 mm in depth. After
completion of excavating, the apron is
closed, the bowl is raised and scraper
moves to discharging area.
Discharging

During discharging out the soil from the
bowl of the scraper, again the height of
bowl is set to spread the material in a
controlled layer and the soil is pushed
out of the bowl with the aid of the ejector
plate.
Two methods of working with
motorised scrapers are:
Push Loading
Except for elevating and push-pull scrapers, wheel scrapers
require the assistance of pusher tractors (bulldozers) to obtain
maximum production. The three basic methods of push loading
scrapers are:
i. The back track method is used in sloping short distance loading
areas where loading to be done only in downhill direction.
However it is slower since pusher requires longer cycle time.
ii. Chain loading is suitable for a long and sloping loading areas to
load downhill.
iii. Shuttle loading requires level surface.
Push-pull Loading
In push-pull loading two scrapers assist each other while loading.
The sequence of loading operation is:
a. The first scraper in the cut starts to self loading.
b. The second scraper makes contact, couples and pushes the
front scraper to assist it in loading.
c. When front scraper is loaded, the operator raises its bowl. The
second scraper begins to load with the front scraper pulling to
assist until it becomes full.
d. The two scrapers uncouple and separate for the haul.
Estimating Scraper Production

Scraper cycle time is the sum of fixed cycle
time and variable cycle time.

Fixed cycle time includes spot time, load
time, and manoeuvre and dump time. Spot
time represents the time for a unit to position
itself in the cut and begin loading, including
any waiting for a pusher.

Variable cycle time, or travel time includes
haul and return times.
Estimating Scraper Production

The haul route should be broken up into
sections having similar total resistance values.

For payload per scraper cycle, two items
should be checked and the smaller to be
selected:
- Rated weight payload
- Heaped volume capacity
Scraper Fixed Time (min)
Spot Time
Single Pusher
Tandem pusher
Favorable
0.2
0.1
Average
0.3
0.2
Unfavorable
0.5
0.5
Load Time
Single
Pusher
Tandem
pusher
Elevating
scraper
Auger
Push-Pull*
Favourable
0.5
0.4
0.8
0.7
0.7
Average
0.6
0.5
1.0
0.9
1.0
Unfavourable
1.0
0.9
1.5
1.3
1.4
Manoeuvre and Dump Time
Single Engine
Double Engine
Favourable
0.3
0.3
Average
0.7
0.6
Unfavourable
1.0
0.9
* Per pair of scrapers
Scraper Travel Time (Loaded)
Scraper Travel Time (Empty)
Example
Estimate the production of a single engine two-axle tractor scraper
whose travel time curves are shown in fig. 4-4 and fig. 4-5, based on the
following information.
Maximum heaped volume = 31 LCY (24 Lm3)
Maximum payload = 75000 lb (34020 kg)
Material: Sandy clay, 3200 lb/BCY (1898 kg/Bm3), 2650 lb/LCY (1571
kg/Lm3), rolling resistance 100 lb/ton (50 kg/t)
Job efficiency = 50 min/hr
Operating conditions = average
Single pusher
haul road:
Section 1. Level loading area
Section 2. Down a 4% grade, 2000 ft (610 m)
Section 3. Level dumping area
Section 4. Up a 4% grade, 2000 ft (610 m)
Section 5. Level turn around, 600 ft (183 m)
Solution
Load per cycle:
Weight of heaped capacity = 31 2650 = 82150lb
[= 24 1571 = 37794 kg]
Weight exceeds rated payload of 75000 lb (34020 kg), therefore, maximum
capacity is:
Load =
[=
75000
= 23.4 BCY / load
3200
34020
= 17.9 Bm3 / load]
1898
Solution

Effective grade:
Haul  -4.0 
100
 1%
20
[ -4.0 
50
 1%]
10
100
 9%
20
50
[ 4.0 
 9%]
10
Return  4.0 
Turnaround  0 
100
 5%
20
50
[ 0 
 5%]
10
Solution
Travel time:
Section 2 = 1.02 min
Section 3 = 1.60 min
Section 4 = 0.45 min
Total = 3.07 min
(Figure 4-4)
(Figure 4-5)
(Figure 4-5)
Fixed cycle (Table 7.7):
Spot time = 0.3 min
Load time = 0.6 min
Manoeuvre and dump time = 0.7 min
Total = 1.6 min
Total cycle time = 3.07 + 1.6 min =4.67 min
Solution
50
Estimated production  23.4 
 251 BCY/hr
4.67
50
[ 17.9 
 192 Bm3 / hr]
4.67
Example

A rubber tired single engine two axle tractor
pulled scraper is operated on sand and gravel
at an altitude of 3600m. The scraper produces
maximum rim pull of 60,000kg at sea level and
its gross loaded weight is 72,000kg which is
made up of 36,000kg self weight and 36,000kg
payload. The total load is distributed with 50%
on each axle and the coefficient of traction is
0.70 and a pull of 22,000kg is required to
move the scraper and its load.
Problem
a) Determine if the scraper can perform under these conditions.
b) If the scraper above operating under same altitude level and operation conditions
has:
Max heaped volume 21.2 Lm3
Max payload is 40,000kg
Material: sand and gravel 1898kg/m3, 1698 kg/m3
Rolling resistance factor is 60kg/t
Operating condition unfavourable.
Job efficiency 45 min/hr
The scraper will be loaded with the assistance of bulldozer. After loaded it will
travel up 4% haul road 1000 m then on a level road of distance 1400 m to a fill
where it will deposit its load and returns empty over the same route. Calculate:
i) Cycle time of the scraper.
ii) Production of the scraper in Lm3/hr.
Solution
a- Derating factor= 26.32 %
percent rated power available = 100-26.32 = 73.68%
maximum available power = 60000*0.7368 = 44208 kg
maximum useable pull = 0.70 * 36000= 25200 kg > 22000
kg required pull.
Therefore the scraper can perform under these conditions.
Solution
b0%
fill area
1400m
4%
1000m
Load area
Solution
Loaded:
Effective grade: 4% + 60/10 = 10 %
Effective grade: 0% + 60/10 = 6%
4.4 min
4.0 min
Empty:
Effective grade = 0%+ 60/10 = 6%
Effective grade = -4% +60/10 = 2%
Scraper fixed time = 0.5 +1 + 1 =
2.45 min
1.40 min
2.50 min
Cycle time =
14.75 min
Derating factor= (3600- 915) / (102) = 26.32 %
Adjusted time = (1+0.2632)*(14.75) = 18.6322min
Weight of heaped volume = 1698 kg/Lm3 x 21.2 Lm3 =36000kg < 40000kg
(max. pay load)
Solution
Volume per load =
36000
3

21
.
20
Lm
/ load
3
1698kg / Lm
Production (Lm3/hr) = 21.20 x (60/18.632) x (75/100) =51.21 Lm3/hr
Calculating the Number of Pushers Required
Number of scrapers served 
Number of pushers required 

Number of scraper
Number served by one pusher
When the number of pushers actually used is less
than the number required, production will be reduced.
Production 

Scraper cycle time
Pusher cycle time
Number of pushers
 Number of scrpers Production per scraper
Required number of pushers
Use the precise number of pushers required, not the integer
value)
Job Management
Some techniques to maximise scraper production include:
. Use downhill loading whenever possible.
. Use chain or shuttle methods if possible.
. Use ripper's to loosen soil.
. Have pushers.
. Provide adequate drainage in the cut to improve traffic ability.
. Make the haul road wide enough to permit high-speed hauling
without danger.
. Keep the fill surface smooth and compacted to minimise the
scraper time.
. Boost scrapers on the fill if spreading time is excessive.
RIPPING
Ripping is usually cheaper than other type of rock
excavating, drilling and blasting. Employing heavy
tractors make it possible to rip all but the toughest
rock.
Advantages of ripping against drilling and blasting are:




Cheaper than drilling and blasting.
Increased production.
Fewer safety hazards.
Reduced insurance cost
Ripping Equipment (Rippers)
The most modern rippers are parallelogram type.
This type of ripper maintains a constant angle with the
ground as it is raised or lowered.
Impact Rippers
Impact rippers utilise a hydraulic mechanism to impart
a hammering action to a single shank ripper. As a
result, impact rippers are able to effectively rip tougher
rock that can conventional rippers and its production is
higher.
Ripper Production
The seismic velocity of a rock formation
provides a good indication of the rock's
ripability.
Production(Bm3 ) 
Where:
60  D  W  L  E
T
D = average penetration in (m)
W = average width loosened (m)
L = length of pass (m)
E = Job efficiency factor
T = Time for one ripper pass including turn (min)
Considerations in ripping

Ripping speed, depth of ripping, spacing of rippers and the number of shanks to
be used depends on rock type and soundness and tractor power.

Ripping is usually performed downhill. However, when ripping laminated material,
it may be necessary to rip uphill to enable the ripper teeth to penetrate under the
layers.

Depth of the ripping depends also on the number of shanks.

Ripping speed should be kept low to reduce wear on ripper teeth and shanks.

The spacing of ripper passes depends on rock hardness and the degree of
fracture desired.

When ripping extremely hard rock it may be economical to loosen by light blasting
before ripping.
Example

A tractor mounted ripper will be used for excavating.
The site is 150m by 250m and must be excavated by
an average depth of 3.6m. Field tests indicate that the
ripper can obtain satisfactory rock fracturing to a depth
of 0.80m with 3 passes of a single ripper shank at
0.90m interval. Average ripper speed for each 250m
pass is 2km/hr. Maneouvre and turn time at the end of
each pass averages 0.30 min. Job efficiency is
estimated 50 min/hr. Machine cost including the
operator is $160/hr.

a. Estimate the hourly production and unit cost of excavation.
b. Calculate the total hours required to excavate the site.
c. Calculate the total cost of excavation.


Solution
a- T =
b-
250
1000
2
60
Production =
+ 0.30 min = 7.80 min
60x0.80 / 3x0.90x 250x50 / 60
 384.6m3 / hr
7.80 min
Total volume = 250 x 150 x 3.6 = 135000 m3
Total hours required for excavation = 135000/384.6 = 351 hrs
c- Total cost of excavation =351 x $160/hr = $56160