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CHAPTER 3

OPERATING CYCLE OFEARTHMOVING EQUIPMENT

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3.1 INTRODUCTION

The operating cycle of earthmoving machines basically affect themachines productivity and accordingly the project time plan.Productivity mainly depends upon the following factors: Machine loading capacity Soil friability Worker efficiency Operating cycle time periodIn the following these factors will be explained.

3.2 EARTHMOVING MACHINE LOADING CAPACITY

The loading capacity of earthmoving machine is calculated as the totalsoil volume that can be charged in its loading box, Fig.3.1. this can beestimated as follows:

LC = BV + RSVWhere:LC = loading capacity or heap capacityBV = machine loading box volumeRSV = additional volume of the soil reposed over the box plane

The angle of reposed soil is in general considered to be as 450. However,some types of soil such as sand may get angles of repose less than 45degrees, this depends on their angle of internal friction when being in dryloose state.

SRV 450

BV

Fig.3.1 Heap (loading) capacity of machine.

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3.3 SOIL FRIABILTY

Soil volume in its original site (natural state) in the terrain is always lessthan its volume after being cut and transported (displaced). Thedifference in volumes before and after soil cutting depends on:

Soil type Moisture content Soil cohesion

In general, the coarse granular soils and gravel get considerably smallerfriable volume after being cut than rocks, and the moist cohesive soils geta little bigger friable volume than their volume when being dry.

When calculating the actual machine productivity, the degree of soilfriability should be considered through a correction friability factor(CFF). This is calculated as:

CFF = soil volume before cutting (SVBC) / soil volume after cutting(SVAC)

Table 3.1 presents the correction factor of soil friability for some soiltypes.

Table 3.1

Soil CFF Weight of one meter cubeof soil, ton

Dry clay 0.85 1.4Moist clay 0.8 1.8Silt 0.83 1.6Dry Sand 0.89 1.95Moist Sand 0.88 2.15Mixed Sand and gravel 0.9 1.85Stones and Rocks 0.74 2.3

When the cut soil is to be transported by trailers or transport vehicles andthey vertically loaded by using loaders, the soil friable volume become

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bigger. Accordingly, another correction factor, in addition to the CFF,has to be used.

The total friability factor is TCFF is thus calculated as:

TCFF = CFF .VLFWhere:VLF = vertical loading factor

= 0.9 for sand= 0.7 for clay

3.4 MACHINE OPERATOR EFFICIENCY

Regarding the difficult work conditions in fields and work-sites, themachine is not actually working every minute in each hour, this meansthat there is a certain portion of time wasted. The actual work hour istherefore less than sixty minutes.

The actual time of machine operation depends mainly on the operatorefficiency, the type and work characteristics.In general, the actual work hour is calculated according to the TimeCorrection Factor (TCF) given in Table 3.2.

Table 3.2

Machine type Actual work time TCFOne hour, min.

Tracked 50 0.83Wheeled 45 0.75

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3.5 MACHINE OPERATING CYCLE

3.5.1 Phases of Machine Operating Cycle

The machine productivity depends for a specific earthmoving machine onthe phases of the operating cycle.For example, the scraper machine has to carry out the following workphases in one operating cycle:

Filling the bowl Displacing the soil to the evacuation site Discharging the soil Return to the work site

Some of these phases require time variable from cycle to another such asthe displacing and return phases. Some other operations such as filing thebowl and discharging the soil take fixed time period.Therefore, the time for one operating cycle is calculated as:

Operating cycle time = Fixed time + variable time

3.5.2 Machine Productivity per Hour

The machine operating cycle stands as an important factor in determiningits productivity per hour which is different from its productivity orcapacity.

Machine productivity per hour depends on:

Machine capacity (Loading Capacity LC, m3) Soil Friability (Correction Factor of soil Friability CFF) Type of soil loading operation (Vertical Loading correction FactorVLF) Actual work time (Time Correction Factor TCF)

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3.6 RESISTANCE TO THE EARTHMOVINGEQUIPMENT MOTION

During their motion with a relatively low speed, the earthmovingequipment has to overcome the following road resistance forces:

Rolling or motion resistance Grade resistance Inertia resistance

The air resistance or drag is negligible due to the low speed of motion.The motion resistance is generated as a result of soil compaction andbulldozing that occurs during motion.The rolling resistance coefficient varies according to the soil type andtraction mechanism from 0.02 up to 0.16 or even more.Table 3.3 shows average values of the rolling resistance coefficient onvarious types of soil.

Table 3.3 Average values of the rolling resistance coefficient.

Soil or ground Rolling resistance coefficientTyre Track

Concrete 0.02 -Compact clay & gravel 0.03 -Compact clay 0.05 0.2Low dense clay 0.08 0.04Loose sand & gravel 0.10 0.05Loose sand 0.16 0.07

Grade resistance depends on the weight of equipment and the gradeangle. Grade angle is usually calculated in percentage.

Inertia resistance depends on the mass of equipment, the engaged gearratio, and the motion acceleration.

The total resistance is calculated as:

Rtot = W cos α f + W sin α + W θ a /g

Where:W = weight of the equipment

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f = coefficient of rolling resistanceα = angle of gradea = accelerationθ = coefficient related to rotating masses inertia

= 1 + (0.055 + 0.22 it2) Wmax / Wactual

it = total gear ratio

3.7 TRACTION COEFFICIENT

The earthmoving equipment productivity is basically dependent on thetraction between its tyres or tracks and the soil. If the traction forcesavailable at the traction elements of the machine is not enough towithstand the soil cutting and motion resistance forces the slip willincrease and bogging in may occur.

Table 3.5 gives average values of the traction coefficient which is definedas the ratio between the traction force multiplied by the weight on thedriven traction elements. This is written as:

μ = Ft / Wadh

where:Wadh = weight on the driven traction elementsFt = traction force

Table 3.5

Soil or road Traction coefficientTyres Tracks

Concrete 0.90 0.45Compact gravel 0.60 0.70Compact clay & gravel 0.50 0.70Compact clay 0.45 0.75Low dense clay 0.40 0.70Loose sand & Gravel 0.25 0.35Loose sand 0.15 0.30

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3.8 AVERAGE SPEED OF EARTHMOVING MACHINES

The operating speeds of wheeled earthmoving equipment may differ thanthat found in the specification manuals; this is due to the ground surfaceroughness and resistance. Such machine usually operate at average speedsthat are lower those found in the catalogues.The average speeds of operation may be calculated by using a certainSpeed Correction Factor SCF whose values are found in Table 3.6.

Table 3.6.

Distance of Operating conditionsLoading, m heavy medium good

< 150 0.32-0.46 0.46-0.57 0.57-0.65150-300 0.46-0.57 0.57-0.65 0.65-0.71300-450 0.57-0.65 0.65-0.71 0.71-0.75450-600 0.65-0.71 0.71-0.75 0.75-0.79600-900 0.71-0.75 0.75-0.79 0.79-0.83>900 0.750.79 0.79-0.83 0.83-0.87

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3.9 CALCULATION OF EQUIPMENT PRODUCTIVITY

3.9.1 Productivity of Scrapers

The productivity of a scraper is defined as:

Productivity = Cycle load x Number of cycles per hour

It is important to firstly determine the cycle load as the scraper weight haseffect on haul speeds, after that it should be remembered that scraperswork in teams assisted by one or more pusher tractors.

a) Work data:Scraper data is usually given as:

Type of material to be worked out. Manufacturers` specifications. Details related to haul road, distances, grades, and surface

conditions. Hourly production required and anticipated overall job efficiency

expressed in working minutes per hour.

b) Cycle Load:From Manufacturer`s specifications take the heaped volume capacity ofthe scraper. This must be multiplied by a fill factor depending on materialto be loaded. Table (3.7) is a rough guide.

Table (3.7)Material Fill factor

Conventional scraper Elevating scraperSoft clay 0.75 – 1.05 0.90 – 1.00Sand & gravel 0.70 – 0.90 0.85 – 0.95Rock-well blasted 0.50 – 0.80 N.A

Cycle Load (loose volume) = Scraper heaped capacity x Fillfactor

Cycle Load (weight) = Loose volume x Loose density

Loaded weight = Scraper weight + Cycle load weight

Cycle Load (bank volume) = Loaded weight / Solid density

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c) Cycle Time:Cycle time is made up of two main parts:

Cycle Time = Fixed Time + Variable Time

Fixed timeIs the time required to load and subsequently unload the scraper, plus anyother non-traveling time, waiting for the push tractor, maneuvering etc.

Table (3.8) gives a rough estimation of the fixed time.

Table (3.8)Scrapertype

Fixed time, secondsLoad Dump Wait &

ManeuverTotal

Conventional 30 – 60 10 – 20 30 – 60 70 -140Elevated 55 – 80 15 – 25 10 -25 80 - 130

Variable timeThis is the time to travel from the cut (loading area) to the fill or dump(unloading area) and the return.The haul road is divided into sections having gradient and rollingresistance.If the job is done at high altitude (over 3000 m) the power loss may besignificant. Therefore, the total resistance is divided by an altitude factor.The altitude factor is calculated as:

Altitude factor = (100 – power loss factor) / 100

This will give the maximum theoretical travel speed for a given totalresistance.The theoretical travel speed should also be multiplied by a speed factor.Figure (3.2) shows the speed factor in relation to the haul section length.

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Fig.3.2 Speed factor.

Time = Distance (m) x 3.6 / speed (km/h)

Cycles per hour = Working minutes per hour x 60 / Cycle time (sec)

d) Number of scrapers required

Number of scrapers required to do the job is calculated as:

No of scrapers = Production required / Production per scraper

Production per scraper = Cycle load x Cycles per hour

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e) Number of Push tractorsIf the scrapers are to be push loaded, the number of push tractors must bedetermined.

Number of pushers = Number of scrapers required / Number ofscrapers per pusher

Number of scrapers per pusher = Scraper cycle time / pusher cycletime

Pusher cycle time = 1.4 x scraper load time + 15 seconds

e) Form of the output calculations

The forms shown in figs.(3.3, 3.4) can be used to calculate the scraperoutput.To complete the required calculations for scrapers output somerelationships showing the data of scraper engine, and pusher enginepower as related to heap capacity are required.These are given in Figs.

The characteristics of scraper engine enable selecting the travel speed asaffected by the road resistance.

The relationship of required pusher engine power versus heap capacityengine enables selecting the correct pusher.

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Fig. 3.3 Form for calculation of scraper output.

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Fig.3.4 Form for calculation of total cycle time.

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Fig. 3.5 Scraper engine characteristics.

Fig. 3.6 Pusher tractor power vs heap capacity.

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e) Solved example:

Calculation of scraper outputHow many scrapers and pushers are required to haul 350 m3 of commonearth soil with land topography as shown in the Figure using conventionalscrapers that have engine characteristics as given by the figures.Suppose that the actual working time is fifty minutes per hour, and thatthe return is by same route.Given data:

Bank soil density = 1700 kg/m3 , loose soil density = 1350 kg/m3

Anticipated job efficiency = 50 % Altitude = 100 m Heaped scraper capacity = 16.1 m3

Scraper rated load = 22860 kg, scraper empty weight is 26948 kg Fill factor = 0.9

Fig. (3.7) Land topography.

Solution:Procedure of calculation is as follows:-First calculate cycle load

estimated cycle load

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cycle load weight loaded scraper weight cycle load bank volume

Second calculate number of scrapers required cycle per hour output per scraper number of scrapers required

Third total cycle timeCalculate for each road section the following:

scraper weight distance traveled road resistance max speed speed factor actual speed

After that calculate the total cycle time

Fourth calculate number of push tractors required cycle time number of scraper per pusher number of pushers require

Figs. (3.8, 3.9) show the solved example calculations.

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Fig. 3.8 Solved example calculations sheet 1.

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Fig. 3.9 Solved example calculations, sheet 2.

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3.7.2 Productivity of Wheel Loaders

The productivity of a dozer is defined using same equation for a scraperproductivity.

a) Cycle Load:From Manufacturer`s specifications take the heaped volume capacity ofthe wheel loader. This must be multiplied by a fill factor depending onmaterial to be loaded. Table (3.9) is a rough guide.

Table (3.9)

Material Fill factorSoft clay, loam 1.00 – 1.10Sand & gravel 0.95 – 1.00Uniform aggregate (over 3 mm) 0.85 – 0.95Hard clay, cemented materials 0.85 – 0.95Rock-well blasted

-Poorly blasted0.80 – 0.850.60 – 0.65

Cycle Load (bank volume) = Cycle load weight / Bank density

b) Cycle Time:As a starting point for calculations, the basic cycle time for a wheelloader is taken as 25 seconds.This includes bucket filling and dumping.The time must be adjusted for varying conditions, as those shown byTables (3.10).

Digging conditions

Table 3.10

Conditions Adjustment, seconds Revised cycle,seconds

Loose, free flowing -2 23Medium digging None 25Hard digging frombank, large rocks

+10 and more +35

Dumping conditions

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Table 3.11

Conditions Adjustment,seconds

Common ownership of truck (one size &loaders

Up to -2

Independent owned or mixed sizes oftrucks

Up to +3

Small or fragile target Up to +4Dumping on ground from embankment Up to -2

Travel distanceTravel time is calculated as:

Travel time = Haul distance (m) / speed (km/h)

Total travel time is:

Total travel time = Haul time + Return Time

Haul distance is taken as the distance between the point of bucket fillingto the dumping point.Maximum speed in each gear can be found from machine specificationsas shown by Fig. (3.25).

Fig.3.10 Average speed – rolling resistance relationship.

Although higher speeds may be possible, there will be a tendency ofbucket shaking and a safety hazard may be created.

c) Hourly output:

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Cycles per working hour can be calculated as:

Cycles per working hour = working time per hour x 60 / Total cycletime (s)

Hourly output = Cycles per working hour x Cycle load

d) Truck size:The effective capacity of haul can be achieved if the loader can fill thetruck from four tom six bucketfuls.

e) Form of calculations:Fig.(3.11) shows a typical form of calculations.Procedure can be run as follows:

First calculate cycle load Second calculate cycle time Finally calculate hourly output.

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Fig. 3.11 Form of calculation of hourly output of loaders.

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f) Solved exampleA loader has to excavate and load stiff dry clay into a mixed fleet oftrucks. Assuming a 45 minute hour work efficiency, can an output of 200m3/ h be achieved?Given data:

Bucket heaped capacity 3.5 m3, Static tipping load = 10971 kg Bank density = 1800 kg/ m3, loose density = 1400 kg/ m3

Fill factor = 1.00Solution:Fig. 3.12 shows the results of calculation of hourly output of the wheelloader.

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Fig. 3.21 Results ofFig 3.12