PRESENTED BY :
BHIM DHOBLE (701016)
SWAPNIL DIVEKAR (701017)
SAISHRAVAN DUBE (701018)
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Dozers are self contained units equipped with a blade.
They are designed to provide tractive power for drawbar work.
The amount of material the dozer moves is dependent on the quantity that will remain in front of the blade during the push.
They are used for dozing (pushing material),land clearing, ripping, assisting scrapers in loading and towing other pieces of construction equipments.
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Moving earth or rock for short distances. Spreading earth or rock fills. Back-filling trenches. Opening up pilot roads through mountains
or rocky terrain. Clearing the floors of borrow and quarry
pits. Clearing land of timber , stumps and root
mat.
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Dozers are classified as 1. Crawler Dozer 2. Wheel
Dozer
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Sr.No.
Wheel Dozer Crawler Dozer
1. Good on firm soils, concrete and abrasive soils that have no sharp edged pieces.
Can work on variety of soils
2. Best for level and downhill work. Can work over almost any terrain.
3. Wet weather causing soft and slick surface conditions will slow or stop operation.
Can work on soft ground and over mud slick surfaces.
4. Good for long travel distances. Good for short work distances.
5. Best in handling loose soils. Can handle tight soils.
6. Fast return speeds, 8-26 mph. Slow return speeds, 5-10 mph.
7. Can only handle moderate blade loads.
Can push large blade loads.
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The usable force available to perform work is often limited by traction, which depends on
1. Coefficient of traction of the surface being traversed. 2. Weight carried by the drive wheels.
Coefficient of traction= Usable force for traction Weight of running gear
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Manufacturers provide dozers with variety of transmissions but primarily available options are
1. Direct drive 2. Torque converter and power shift transmission Some less than 100 hp dozers are
available with hydrostatic powertrains. Larger dozers are always equipped with
power shift transmissions.7
Power is transmitted straight through the transmission as if there is a single shaft.
This usually happens when the transmission is in its highest gear.
They are superior when the work involves constant loading conditions.
Example : A job where full blade loads must be pushed at long distances
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Transmissions that can be shifted while transmitting full engine power are known as power shift.
These transmissions are teamed with torque converters to absorb drive train shock loads caused by changes in gear ratios.
It gives superior performance in applications involving variable load conditions.
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Reference : Figure 6.4 Construction planning, Equipment and Methods – Peurifoy 6th edition
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Most of the wheel dozers are equipped with torque converters and power-shift transmissions.
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Reference : Figure 6.5 Construction planning, Equipment and Methods – Peurifoy 6th edition
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Data : Working surface is Dry clay loam Coefficient of traction factors Rubber tires = 0.5 – 0.7 Track = 0.9 Dozers having power shift transmission.
The usable drawbar pull is = Weight x COT
Crawler = 20666 kg x 0.9 = 18599.4 kg Wheel = 20580 kg x 0.6 = 12348 kg 13
The two machines have approx. same operating weight and power, yet because of effect of traction the crawler can supply one and a half times the usable power of Wheel type dozer.
Therefore the wheel type dozer must be heavier (approx.50% ) than crawler type to develop same amount of usable force.
But if the weight is increased then a larger engine will be required to maintain the weight to horsepower ratio.
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A dozer blade consists of a moldboard with replaceable cutting edges and side bits.
Push arms and tilt cylinders connect the blade to dozer.
Blades vary in size and design based on specific work applications.
The hardened steel cutting edges and side bits are bolted for easy replacement.
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Basically only five blades are common to earthwork
1. The straight “S” blade 2. The angle “A” blade 3. The universal “U” blade 4. The semi-U “SU” blade 5. The cushion “C” blade
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Straight “S” Blade Angle “A” Blade
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Universal “U” Blade Semi – U “SU” Blade
Cushion “C” Blade
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Tilt – This movement is within the vertical plane of the blade. It enables the concentration of dozer driving power on limited portion of blades length.
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Pitch – This is a pivotal movement about the point of connection between the dozer and blade. It increases the angle of cutting edge attack.
Angling – Turning the blade so that it is not
perpendicular to the direction of the dozer’s travel is known as angling. It causes the pushed material to roll off the trailing end of the blade.
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A dozer’s pushing potential is measured by two standard ratios:
1. Horsepower per foot of cutting edge- It provides indication of the ability of the blade to penetrate and obtain a load. The higher this ratio the more aggressive the blade.
2. Horsepower per loose cubic yard of material retained in front of the blade- It gives blade’s ability to push a load. Higher ratio means dozer can push a load at greater speed.
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Stripping Sidehill cuts Ditching Backfilling Shifting of rocks or frozen ground Spreading Slot dozing Blade to Blade dozing
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A dozer has no set volumetric capacity. The amount of material dozer moves is
dependent on the quantity that will remain in front of the blade during the push.
The factors that control dozer production rates are
1. Blade type 2. Type and condition of material 3. Cycle time
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Straight blades roll material in front of the blade.
Universal and Semi–U blades control side spillage holding the material to move to center.
In U and SU blades quantity of loose material will be greater than that of S blade.
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It affects the shape of the pushed mass in front of the blade.
Cohesive materials will boil and heap. Materials with slippery quality or having
high mica content will ride over ground and swell out.
Cohesionless materials will not exhibit heap or swell properties.
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The sum of the time required to push, backtrack and maneuver into position to push represents the complete dozer cycle.
The time required to push and backtrack can be calculated by considering the travel distance and obtaining a speed from the machine’s performance chart.
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Methods to calculate Blade Load are 1. Manufacturer’s blade rating 2. Previous experience 3. Field measurements
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Vs = 0.8 WH2
Vu = Vs + ZH(W - Z) tanx0
Where, Vs = capacity of straight or angle blade in lcy
Vu = capacity of universal blade in lcy
W = blade width in yards H = effective blade height in yards Z = wing length measured parallel to blade
width x = wing angle
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Obtain a normal blade load The dozer pushes a normal blade load
onto level area. Stop dozers forward motion and reverse
it. Measurement Measure height(H) and width(W) of the
pile at the inside edge of each track. Measure the greatest length of pile.
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Computation Calculate average height and width. If the measurements are in feet the
Blade load in lcy is given by, Blade load (lcy) = 0.0139HWL
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The dozer production in lcy per 60 min is given as below,
Production = 60 min x Blade load(lcy/hr) Push time(min) + Return
time(min) + maneuver
time(min)
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Data : Track type dozer with power shift push an avg. blade load of 6.15 lcy the material is silty sand and avg. push distance is 90 ft.Calculate production in loose cubic yards. Solution : Push time: 2 mph avg. speed (sandy material) Push time = 90ft x 1 x 60 min/hr 5290ft/mile 2mph = 0.51 min
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Return time: second gear because less than 100ftMaximum speed: 4 mphReturn time = 90ft x 1 x 60 min/hr 5290ft/mile 4mph = 0.26 minThe chart provides information based on steady
state velocity. But the dozer must accelerate to attain
that velocity. Therefore allowance of 0.05 min is made
for acceleration time. Return time = 0.26 + 0.05 = 0.31 min
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Maneuver time = 0.05 minProduction = 60 min x 6.5 lcy 0.51 min + 0.31 min + 0.05 min = 424 lcy/hrThis is based on ideal condition of working for 60min. But on field job efficiency is affected by conditions of equipment or difficulty of work.So, efficiency factor is expressed as working
minutesper hour.Example: 50 min/ hr or 0.83 efficiency factor
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To calculate the production in bank cubic yards , percent of swell is required.
So in previous example assume % of swell 0.25 for the silty sand and efficiency equals to 50min/hr
So, actual production = 424 lcy x 50 min 1.25 60
min = 283 bcy/hr
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Final step is to calculate the unit cost for pushing the material.
The ratio of cost to operate to amount of material moved gives most economical machine for job.
In the previous example assume an owning and operating cost of $40.5/hr and a wage of $15.5/hr.
So, unit cost = $40.5/hr + $15.5/hr 283bcy/hr = $0.197/ bcy 37
Given by equipment manufacturers International Harvester.
Production(lcy/60min) = net hp x 330 (D + 50)Where , net hp = net horsepower of dozer D = one way push distance in feet
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Production curves are based on following set of ideal
conditions A 60 min hour (100% efficiency) Power shift machines with 0.05 min fixed
time The machines cuts for 50 ft A soil density of 2300 lb/ lcy Coefficient of traction The use of hydraulic controlled blades
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Reference :Figure 6.12 Construction planning, Equipment and Methods – Peurifoy 6th edition
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Reference : Figure 6.4 Construction planning, Equipment and Methods – Peurifoy 6th edition
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Equipments used : Crawler dozer with earthmoving blades Crawler dozer with special clearing
blades Crawler dozer with clearing rakes
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Constant speed clearingProduction = width of cut (ft) x speed (mph)
x (acre/hr) 5280 (ft/mile) x efficiency 43560 sf/acre American society of Agricultural Engineer’s It is based on 0.825 efficiencyProduction = width of cut (ft) x speed (mph) (acre/hr) 10
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Only trained workers should be allowed to operate the dozers.
It should be kept far away from overhead power lines.
Do not use dozers on materials having more height than that of dozer.
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A loader is one of the most commonly used heavy machinery in the construction industry, used to handle and transport bulk material , such as earth and rock.
The main purpose of using a loader is to excavate and upload sand , debris, dirt and mud into other vehicles. 46
Loaders are also used in the farming, mining, urban engineering projects, small earthmoving works and logging industry.
The hydraulic-activated lifting system exert maximum breakout force with an upward motion of bucket.
It does not required any other equipment to level, smooth or clean up the area in which it is working.
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Loader, is a machine usually wheeled, that uses a wide tilting bucket on the end of movable arms to lift and move materials. Often the bucket can be replaced with other devices depending on the nature of the work.
The loader assembly can be removed or permanently mounted. Usually, the bucket can be replaced with mount forks to lift heavy pallets
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Basic feature The loader is versatile piece of equipment designed to excavate at or above wheel /track level
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1) Crawler–tractor mounted
(track mounted )
2) Wheel –tractor mounted
(wheel mounted)
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Track mounted
1. Due to track, speed and mobility is low.
2. They are mainly used to load sharp edged materials, debris in construction sites
3. There is no rubber tyres.
4. Stability can be increased using long and wide track
Wheel mounted
1. Wheels provide better mobility and speed.
2. used in the areas where the ground is soft and muddy.
3. there is the chance of rubber tyres being destroyed .
4. Stability can not be increased
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1 .Skid Steer Loader 2 .Telescopic Loader
3. Backhoe Loader 4 .Wheel Loader
Depending on the sizes and the nature of work to be performed, loaders can be classified in to various types
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The capacity of a bucket in a loader can be anywhere from 0.5 to 36 m³.
5 .Scraper Loader 6.crawler Loader
7 .Skip Loader
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General purpose•General purpose (one-piece)bucket is made up of heavy duty , all welded steel .•The replaceable cutting edge are bolted onto the bucket proper .•Replaceable teeth are bolted onto cutting edge.
Multipurpose•Multipurpose (two-piece) hinged-jaw bucket is made of heavy duty, all welded steel.•This is four in one bucket used to dig like 1)normal bucket , 2)blade , 3)clam , 4)grapple
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Types of Bucket / attachments 1. Rock bucket :- having V shape cutting edge used for loosening rock.
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2. Side –dump bucket:- used in confined areas , along roads in trffic , for filling truck .
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3. Forklit:- forklift can be attached to loader in place of bucket
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4. other:- demolition bucket, plow blades for snow removal ,brush racks for clearing appication
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Struck capacity:- it is volume actually enclosed by bucket with no allowance for bucket teeth.
Heaped capacity:-it is net section bucket volume depending on angle of repose. (i.e. PCSA &SAE use 1:1 angle of repose , CECE specifies 2:1 angle of repose)
Fill factor :- it is percentage that ,when multiplied by a rated-heaped capacity, adjust the volume by accounting for how the specific material will load into bucket
it is depending upon type of material & voids present between material particles
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The rated capacity of loader bucket is expressed in cubic yard for all sizes ¾ cy or over.It is measured in cubic feet for all sizes under ¾ cy.
Rated capacities are stated in intervals of 1 cf for buckets under ¾ cy,
Rated capacities are stated in intervals of 1/8 cf for buckets under ¾ cy to 3 cy,
Rated capacities are stated in intervals of ¼ cf for buckets over 3 cy,
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Table fill factor 9.6
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•Unlike shovel or hoe , to position bucket to dump , a loader must maneuver and travel with the load .
•A shovel or hoe simply swing about its center pin and does not required travel movement when moving the bucket from loading to dump position.
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•SAE has established operating load wt limits for loaders
1)wheel loader is limited to an operating load , by weight i.e. less than 50% of rated full turn static tipping load considering the combined weight of bucket & the load , measured from the center of gravity of extended bucket at its maximum reach ,with standard counterweights and non ballasted tires.
2) For track loader operating load is limited to less than35% of static tipping load
Most buckets are sized based on 3,000 lb/lcy material
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•Two critical factors to be considered in choosing a loader are :- 1) The type of material,
2) Volume of material to be handled.
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•Wheel loaders are excellent machines for soft to medium-hard material.
•In medium to hard material production rate of wheel loader decrease rapidly.
• At the time of loading Height that the material is to be lifted is also important factor.
•Wheel loader attains its highest production rate when working on smooth surface and sufficient space to maneuver.
Wheel loader production rate
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•Wheel loader work in repetitive cycles, constantly reversing direction, loading , turning, dumping
•The production rate for wheel loader will depend on:-
1. fixed cycle time required to load the bucket2.Time required to travel from loading to dumping position3.Time required to return to loading position4.Volume of material hauled each cycle
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Loader size wheel loader track loader Cy sec sec 1-3.75 27-30 15-21 4-5.5 30-33 -- 6.0-7.0 33-36 -- 14-23 36-42 ---
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Calculation of wheel loader production
A 4-cy wheel loader will be used to load trucks from a quarry stockpile of processed aggregate having maximum size of 1 ¼ in. the haul distance is negligible . The agg has a loose unit weight of 3100lb/cy estimate the loader production in tons based on a 50-min hour efficiency factor. Use conservative fill factor.
Step 1:- size of bucket 4cy,
Step 2:-bucket fill factor (for agg. over 1 in) is 85-90%;using 85%
load weight=4cy x .85=3.4 lcy
3.4 lcy x 3100 lb/lcy(loose unit weight) =10,540 lb
From table, 4-cy machine static tipping load at full turn is 25000 lb
Operating load is 50% of static tipping load,
Operating load=0.5 x 25000 lb=12500 lb72
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As the actual load (10,540 lb) <operating load(12,500 lb)
………………….o.k.
Step 3 :-Fixed cycle time for 4cy loader =30 to 33 sec
Using 30 sec,
Step 4 :-Efficiency factor=50 min hour
Step 5 :-Class of material , aggregate 3100 lb/lcy
step 6 :-Probable production:-
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A 2-cy track loader having the following specification used to load truck from a bank of moist loam. This operation require the loader to travel 30 ft for both the haul and return .Estimate loader production based on 50 min hour efficiency factor Travel speed by gear for 2-cy track loader Gear mph fpmForward first 1.9 167 second 2.9 255 third 4.0 352Reverse first 2.3 202 second 3.6 317 third 5,0 440
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Assume that the loader will travel at an average of 80% of the specified speeds in second gear , forward and reverse.
Step 1 :- size of bucket = 2cy ,
Step 2 :-bucket fill factor (for moist loam)=100-120%, using 110%, load weight=2 cy x 1.10 =2.2 cy , unit weight moist loam from table=2,580 lb lcy 2.2 lcy x 2,580 lb/ lcy =5,676lb from table 2 cy machine static tipping load=19000lb
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Operating load (35% tipping load) 0.35 x 19000 lb =6,650 lb 5,676 (actual load) < 6,650 (operating load)……….okStep 3:- Fixed time cycle time for 2-cy track loader=15-21sec taking 21 sec, Travel loaded:30 ft , use 80% first gear max speed
Return empty: 30 ft, use 60% of second gear max sped
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1.Fixed time 30 sec 2 cy track loader 2.Travel with load 13sec 30 ft ,80% first gear 3.Return travel 12 sec 30ft ,60%second gear Cycle time 55 sec
Step 4:- efficiency factor, 50-min hour
Step 5:- moist loam , swell 25 type
Step 6 :- probable production=
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High travel speed, low hauling cost, high degree of flexibility.
Main classification:-1.off-highway trucks2.Highway trucks.
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OFF-HIGHWAY TRUCKS HIGHWAY TRUCKS
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1.The method of dumping the load- rear dump, bottom dump, side dump.
2. The type of frame- rigid or articulated.
3. The size and type of engine- gasoline, diesel ,etc.
4. The kind of drive- two wheel, four wheel or six wheel.
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5. The no. of wheels and axles and their arrangement
6. The class of material hauled- earth, rock, coal etc
7. The capacity- gravimetric or volumetric
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1. suitable for carrying any type of material .
2 the shapes used are flat or ’v’ shaped bottoms
3. The body is made of high tensile steel.
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ADT is specifically designed to operate over rough or soft ground.
An articulated joint & oscillating ring between tractor & dump body permit all wheels to maintain contact with ground.
ADT can climb steeper grades up to about 35%
High hydraulic dumping pressures with steeper dump angles
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An economic hauler when material moved is free flowing as sand, gravels etc.
Reduces the time required for unloading the material
Time advantage over rear dump trucks These trailers should be considered when:-1. Material to be hauled is free flowing2. Unrestricted loadings and dump sites3. Haul route grades are less than about 5%
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1. Gravimetric – expressed as a weight2. Struck capacity – if the load was water
level in the body3. Heaped volume – if the load was
heaped on 2:1 slope above the body The heaped volume:- The volume of material that truck can
haul when load is heaped above sides
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Check to ensure that volumetric load does not cause a condition where load weight exceeds gravimetric load.
If possible Overloading should be avoided.
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Advantages of small trucks to large trucks
1. More flexible In maneuvering2. Achieve higher hauls and return speed3. Loss in production is less when truck
fails down4. Easier to balance between no. of trucks
& output of excavator
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1. Difficult to load & small target for depositing the bucket load
2. More time is lost in spotting the trucks as large no. of trucks are required
3. More drivers are required to haul the truck
4. Due to large no. there may be bunching at the pit, along the haul road or at dump
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Data:- 3-cy shovel,90° swing ,no delays for
haulers 20 sec cycle time The time for travel cycle includes
travelling, dumping and returning to shovel times will be
6 min. for several sizes of trucks.
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If 12-cy trucks are used (12/3)=4 no. of buckets to fill truckTime required to fill the truck will be (20*4)=80 sec. or 1.33 min.The roundtrip cycle for truck 6+1.33=7.33 min.Min no. of trucks required to keep shovel busy (7.33/1.33)=5.51
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6 trucks:The time required to load the truck (6 x 1.33)=7.98minThus shovel will lose(7.98-7.33)=0.65
minThe percentage time lost will be (0.65/7.98) x 100=8.2%Operating factor 91.8%
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If 24-cy truck usedIt will require(24/3)=8 trucksTime required to fill the truck will be (8 x 20)=160 sec or 2.66
min.Min. roundtrip required=8.66 min.The min no. of trucks required to keep
shovel busy (8.66/2.66)=3.26If 4 trucks are used them loading time (2.66 x 4)=10.64min.
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Loss of time will be (10.64-8.66)=1.98 min.
Operating factor (8.66/10.64)x 100=81.4%
It will be advisable to use 6 trucks of 12-cy with greater operating factor
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balanced number of bucket = truck capacity(lcy)
bucket capacity(lcy)
It will determine the number of excavator bucket loads to load the truck.
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Load time=number of bucket swings x bucket cycle
time Truckload(volumetric)= no. of bucket swings x volume of
bucket
Truckload(gravimetric)= volumetric(lcy) x unit weight(loose vol.
lb/cy)101
Haul time(min)= Haul distance(ft) 88fpm/mph x haul
speed(mph)
Hauling should be at highest safe speed & in proper gear to increase the efficiency.
Haul speed can be determined by manufacturers performance chart
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Return time(min)= Return distance(ft)
88(fpm/mph) x return speed(mph)
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It will depend on type of hauling unit & congestion in dump area
For rear dump it is 0.7 min in favorable & 1.5 in worse conditions
For bottom dump it is 0.3 min in favorable & 1.5 in worse conditions
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Truck cycle time=Load time+ Haul time + Dump time +
Return time
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Number of trucks= Truck cycle time(min) Loader cycle
time(min)
The no. of trucks must be an integer number
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Production(lcy/hr)= truck load(lcy) x no.of trucks x
60 min. truck cycle
time(min) If there is not sufficient no. of trucks
then there will be loss in production
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The production calculated above is based on 60 min. this can be adjusted by efficiency factor
Longer hauling distances result in better driver efficiency
It can be increase up to 8000ft and then remains constant
Adjusted production= production(lcy) x working
time(min/hr) 60min 108
Data: Hauling sandy clay Capacities struck 14.7cy, heaped
2:1,18.3cy Net empty load = 36860lb payload = 44000lb gross vehicle wt. = 80860lb Hydraulic hoe of 3 cy l(haul road)= 3-mile, downhill grade 1% Earth is poorly maintained 109
Dump time = 2 min Hoe cycle = 20 sec Loose unit wt of sand=2150 lb/cy realistic efficiency 50 min-hr
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Therefore bucket fill factor is 110%
Bucket volume=3x1.1=3.3 cy
Reference: Table 8.4,construction planning, equipments & methods-Peurifoy, 6th edition 111
Balanced no. of buckets=(18.3/3.3) = 5.5 ≈ 6 bucketsWe will investigate for both 5 and 6 bucketsLoad time=5 x 20/60=1.66minLoad volume= 5 x 3.3= 16.5 lcyCheck with load wt.= 16.5 x 2150= 35045lb < 44000ok
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6 buckets load time=6 x (20/60)=2 min
load volume=(6 x 3.3)=19.3 this will equals truck capacity & excess will
spill off check for load wt.= 18.3 x 2150= 39345 lb <44000 lb so ok
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Reference: Table 5.1,construction planning, equipments & methods-Peurifoy, 6th edition 114
Rolling resistance 100-140 lb per ton using avg. value as 120
Grade resistance(%)= rolling res. per ton/20 lb per ton
= (120/20)=6% Grade resistance = -1%
Total resistance = 5%115
Description 5 buckets 6 buckets
Empty truck net wt. 36860 lb 36860 lb
Load wt. 35045 lb 39345 lb
Gross wt. 71905 lb 76205 lb
speed 16 mph 13 mph
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Reference: Fig.10.9,construction planning, equipments & methods-Peurifoy, 6th edition 117
Haul time 5 trucks= ( 3mile x 5280ft/mile)
88 x 16 = 11.25 min Haul time for 6 trucks = 13.85 min
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Now for return journey grade resistance will be 1%
Total resistance 7% Empty truck wt= 36860 lb Speed = 22 mph Return time = 8.18 min
Dump time is 2 min
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Description 5 bucket 6 bucket
Load time 1.66 2
Haul time 11.25 13.85
Dump time 2 2
Return time 8.18 8.18
Truck cycle time 23.09 26.03
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Description 5 bucket 6 bucket
Truck cycle time 23.09 min 26.03 min
Loader cycle time 1.66 min 2 min
No. of trucks 13.9 13
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For 5 bucket loader if 13 trucks are used production = 16.5 x 13 x 60/23.09= 557
lcy per hr if 14 buckets are used production = 16.5 x 23.09/1.66= 596
lcy per hr For 5 bucket loader if 13 trucks are used production = 18.3 x 13 x 60/26.03= 548
lcy per hr122
Adjusted production = 596 lcy per hr x 50/60
= 497 lcy per hr
Adjusted production = 497 x 2150/2000 = 534 tons
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The selection of proper tire size & correct air pressure in it will reduce the rolling resistance
For rigid road such as concrete low diameter & high pressure gives low rolling resistance
For soft soils large diameters with low pressures gives low rolling resistance
Tires are about 35% of truck’s operating cost Tire manufacturers provide ton-mile-per-hour (TMPH) limit for there tires
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TMPH job rate= (avg. tire load x avg. speed during day’s
operation)
Avg. tire load= empty tire load+ loaded tire load 2Avg. speed(mph)=round trip distance (miles) x
no. of trips total hrs worked 125
Data empty truck wt=70000 lb loaded truck wt = 150000 lb wt distribution for empty truck=50% rear & 50% front wt distribution for loaded truck=67% rear
& 33% front Truck has 2 tires in front & 4 tires at rear end truck works for 8 hrs with 14 trips per day &
5.5 miles one way haul distance126
Description Front ,2 tires Rear, 4 tires
Empty wt 70000 lb 35000 lb 35000 lb
Loaded wt 150,000 lb 50,000 lb 100,000 lb
Empty wt per tire 17500 lb 8750 lb
Loaded wt per tire 25000 lb 25000 lb
Avg. tire load 10.6 tons 8.4 tons
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avg. speed= (2 x 5.5) x14 8hr =19.25 mph TMPH job rate = 10.6 x 19.25 mph = 204 while calculating the TMPH job rate
select max. highest load
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Thank you!!!
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