Equipment Cost Estimating - Otuonye - 2000

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Name

OperationReq'd

Unit Modellbs.

MaterialYd. lbs.Pay Load per Cycle: Loose Cu. Yds. BankCu.Yds lbs.

CEAPTER lPsrt2SUMMARY HAI]LING PRODUCTION AND COST ESTIMATION

Address

ProductionLocation

Heaped Cap Yds. Capacity

Bank Yd. lbs. Swell Factor Loose

Type ofloading UnitLoad

Bucket Size

Loading Production

No. OfPasses to

Tons orLoading ConditionsBank Cu. Yds.{.

A. LOADINGTIME Mn.

3

Loaded Haul-Total

B. TOTAL HAULING TIME Min.

Return

o Ft. Elevation Ft.

Section

Length

in Ft.

RollingResist.

Per CentGrade

Trans.

Gear

Max.

Speed

SpeedFactor

Hauling

TimeMin.

-Total Ft.Road

Section

Length

in Ft.

RollingResist.

Per CentGrade

Trans.Gear

Max.Speed

SpeedFactor

AverageSpeed

ReturnTime

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C. TOTAL RETURN TIME Min.

D. Turning and Dumping-ConditionsTime Min.E. Spot at Loading Machine-ConditionsTime Mn.F. TOTAL TIME pER COMPLETE HAULING CYCLE (A+B+C+D+E)

Min.

Turning and Dumping

Spotting

G. Average Trips per Hour: :,Yh.,Pr=od.,tr:_ =(F) Total Cycle Time

H. Hour Production : (G) Trips per Hour x Pay Load :Yds. Per Hour

J. Number. of Units Req'd : _!ur!L Production Req'd :- (H) Bank Yds. or Tons per Unit per Hour

ESTIMATING COSTS

Types of Costs

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Trips per Hour

Tons or Bank

Units

Fleet Production per Hour : J x H:Bank Yds.

Hauling Cost per Bank Yd. Or Ton

Hour Cost of Owning and Operating Units@_ each:

Hour Cost of Owning Spare each:

K. Hour Cost for Fleet ofUnits Total

ESTIMATED HAULING COST PER YD. OR TON:Fleet Production

orK-Req'd Production

@

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Ownership Costs

Those costs which are incidental to the purchase and possession of a piece of equipment before anyuse or income may be derived from that unit.

A fixed cost set up as a sinking fund or reserve to cover the economic life of equipment.

Operating Costs

Those costs which are incurred in using the unit to perform its designated tasks.

Ownership Costs

Depreciation

A reserve established to offset the effects of wear and obsolescence on a piece of equipment so that,theoreticall sufficient funds will be available to replace the existing unit when it has lost its value. Analternate definition is that depreciation is the loss in value of equipment resulting from use or age. Theowner of equipment must recover the loss in value of equipment during its useful life or he will sustainan equipment loss on those projects where the equipment is used.

In estimates, the straight line method of depreciation is generally used and it is assumed that theequipment possesses no scrap value at the end of the period.

Total Cost of Equipment: Purchase Price + Cost of Transportation + Cost of Unloading &Erecting '

Normal Depreciation Schedules:

a. Scrapers - 5 yrs at2,000 hrs/yr: 10,000 hrs

b. 75 ton trucks and smaller - 5 yrs at 4,000 hrs/yr:20,000 hrs

c. 85 ton trucks and larger - 5 yrs at 5,000 hrs/yr:25,000 hs

In practice, the owner will choose a depreciation method and schedule which will benefit him most.

Depreciation value is determined on the delivered price less the cost of tires since tires are consideredan operating expense.

It is convenient to express costs in an hourly basis. To obtain hourly depreciation rates, divide thedelivered price less tires by the total hourly depreciation schedule.

fnterest, fnsurance and Taxes

These items are generally taken together and considered as a percentage of the average annualinvestment.

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Average Annual Investment = Normal Interest Rate + fnsurance + Taxes

(Use interest, insurance and tor rates in effect at the time.)

Average Annual fnvestment : Ddived Price

(Where n: years of depreciation value.)

The formula is derived by taking the average value remaining in the unit at the beginning of eachsucceeding year of its useful life (refer to example I and 2).

Hour cost of Interest, fnsurance and Tax Rate

(Idcrest+ Isrznce+ TaxXArg Annral hnestmentxDelivelpd Price)Hors qreraEd pEr yeat

Operating Costs

Fuel cost per hour: (consumption, gals/hr) (fuel, cosUgal)Lubrication costs per hour are estimated from required service intervals as listed in vehiclemanufacturers service manuals.

Maintenance and repair costs per hour can be estimated by taking a constant times the fuelconsumption.

Tire costs are based on an average discounted price available to most fleet owners divided by tire life.

Operator's hour wage varies locally, and current labor rates should be consulted.Supervision and overhead costs should be scheduled in this estimate.

Other figures of importance are:

a) Scrapers, cost per yard

b) Trucks, cost per ton

n+12"

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hourlJr ownlp + qlrating cmtsprudwtion,tonsperlnur

Each manufacturer has detailed guides for calculating ownership and operating costs.

EXAMPLE

Original cost of equipment: $25,000

Estimated useful life:5 years

Average annual cost of depreciation: $25,000/5 or $5,000

Beginning of Year Cumulative Depreciation Value of Equipment

I $0 $25,000

2 5,000 20,000

3 10,000 15,000

4 15,000 10,000

5 20,000 5,000

6 25,000 0

Averagevalue: !$$ffi ry = $$,oooAverage value as % of original cost 6 : 1l'999 - fOO Er7o

25,000

Average yearly investment as o/o of purchase price : * - It o/o

EXAMPLE

Original cost of equipment: $12,000

Estimated useful life:4 years

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Average annual cost of depreciation = $12,000/4 or $3,000

Beginning of Year Cumulative Depreciation Value of Equipment

1 $0 $12,000

2 3,000 9,000

3 6,000 6,000

4 9,000 3,000

5 12,000 0

Average value: ryqq : $?Foo

Average value as % of original cost: ffi. tOO = 625to/o

o+l x looo/oAverage ygarly investment as o/o of purchase price

= 4+l

= 1*roo = frL.ilol88

COST ESTIMATION

Haulage Unit (Truck)A. Ownership Costs

1. Depreciation

a) Purchase Price: estimate as $7650/ton of truck capacity (live load).

b) Salvage Value: recoverable, estimate as l5Yo of purchase price.

c) Freight: estimate truck weight (dead load) as 1400lb/ton of truck capacity (live load). Use freightcharge as $4.00/cwt (cwt: l00lb)d) Unloading and moving cost: estimate as l}Yo of freight cost.

e) Tire cost: estimate as 5Yo of truck purchase price.

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f) Delivered price: u_rchase price) + (freight) + (unloading and moving cost) - (salvage value) - (tirecost)[a+c+d-b-e].g) operating Period: assume 200ohr/yr for I shifl/day of operation.8 hrlshift x I shifl/day x 5 days/week x 50 weeks/yr :2000 hrlyrh) Economic life: based on IRS guidelines:

Conditions Truck Life

Favorable 25,000 hrs

Average 20,000 hrs

Unfavorable 15,000 hrs

I) Depreciation: (delivered price) + (economic life).2. Fixed Charge

a) Rate: interest * taxes f other.

b) Average annual investment rate. n + I2n

where n: the economic life of equipment or unit in years.

c) Average annual investment: : (delivered price)(average annual investment).d) Annual fixed charge: : (average annual investment)(rate).

e) Fixed charge: (annual fixed charge) + (operating period).

3. Total ownership costs: (depreciation) + (fixed charge).B. Operating Costs

1. Tire replacement cost: (tire cost) + (tire life)Use the following guidelines for tire life:

Conditions Truck Tire Life

favorable 4000 hrs.

average 3200 hrs.

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unfavorable 2500 hrs.

2. Tire repair cost : (tire replacement cost)(tire repair factor)

Use the following guidelines for tire repair factor:

Conditions Tire Repair Factor

favorable l2Yo

average l5Yo

unfavorable lTYo

3. Repairs & maintennce cost: (depreciation)(repair factor)@conomic life) + 10,000 hrs

Use the following guidelines for repair factor:

Conditions Truck Repair Factor

favorable3TYo

average 45Yo

unfavorable 60Yo

4. Fuel or power cost: (truck hp)(fuel consumption, galhp-hr)(fuel cost, $/gal)Estimate haulage unit power as l0 hp/ton of truck capac.Select fuel consumption in gallhp-hr based on operating conditions.

Conditions Fuel Consumption Gallhp-hr)favorable 0.014

average 0.020

unfavorable 0.026

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5. Lubrication : (fuel, cost/hn)(lubrication factor)Select lubrication factor based on the following:

Conditions Lubrication Factor

favorable l/5

average l/3

unfavorable %

6. Labor = (wage rate) + (35%wage rate)Assume one (1) operator per truck and35yo of wage rate as benefits.7. Total operating costs : I + 2 +3 + 4 + 5 + 6

C. Total Ownership and Operating Costs: A3 + B7

D. Unit Cost

1. (Cost of owning and operating)(Units in-run)Dl : (x truck unitsxA3 + 87)Dl : (x truck unit$(C)2. Cost of owning y spare units

D2: (y truckunitsxA3)3. Total cost of owning and operating x units in-run and owning y spare units

D3:Dl +D2

4. Unit cost:

Hourly production: shift# of hosofqeration

NorE: Replace all estimated values with known varues if possible.

=$ton

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bN*shift

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EXAMPLE

Determine the total hourly and unit cost of owning and operating five 85-ton trucks and one spare 85-ton truck ifthe trucks operate on an 8-hour shift and the following information is provided.

Operating conditions: average

Truck and tire life: unfavorable

Diesel fuel cost: $1.00/gal

Operator \iage: $ 12.00/hr

Total output per shift: $10,000 tons

Interest : l4%o, taxes: 2Yo, other:2Yo

(see cost estimation from on the next page)

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Cost Estimation Form - 85-ton Truck Mining Equipment Unit

A. Ownership Costs

1. DEPRECIATION

a. Purchase price: $650.250

b. - Salvage value (UZ: - 97.538c. Freht 119,000 lb @ $4.Oq/crrr : 4.760d. Unloading and moving cost:476

e. Delivered price :557.948; say $525.000

f. Operating period 2000 tr/yr - 32,512 tires

g. Economic life 15,000 tr:7.5 yr (n)h. Depreciation: $525.000 (del. price less tire cost) : $_3j.0q /hr

U*88q hr

2. INTEREST, TAXES, INSURANCE AND STORAGE

a. Rate : interest U% + taxes lYo * other 2%o: lSYo

b. Average annual investment rate:

c. Average annual investment:$525.000 x 56.7Yo: $Difi

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n*l S.5- 56,79tn l5

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d. Annual fixed charge: 9ry.675x ISyo: $53J82

e. Fixed charge: $5_3J82

2000 hr/yr : S 26.79 lhr

TOTAL OWNERSHIP COSTS : $ 61.79 /hr

B. Operating Costs

1. TIRE REPLACEMENT COST

Purchase price I set of tires : W212Tire life 2500 hr

Tire cost : V2_I2

25OOtrr: $ 13.00 /hr

2. TIRE REPAIR COST: lTYo x $13.0L tire cost : $ 2.2L {hr

3. REPAIRS & MAINTENA{CE: 45Yo x $35-QQ x 15/10 deprec. : $ 23.63 lht4. FUEL: 850 x 0.020 gaVlv @ $!-QQ /Sal

ORPOWER:l,r @ $_/la/v-hr: $ 17.00 /1r

5. LUBRICATION: U3 gaVlv @$l7.OO lgal-hr: $ 5.67 /lr6. AIIXILIARY FUEL: $ ---

7. LABOR: I operator @$ 12,90 /lr: $ 12.00 lhroiler @lhr: lfuhelper @ lhr: lhrTotal $12.00 /hr

+ 35Vo benefits $ 4.20 lfu: $ t6.20 lhTOTAL OPERATING COSTS : $ 77.71 III

TOTAL OWNERSHIP AND OPERATING COSTS = $139.50/Tr

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TINIT COST $759.29 lt + l25O ton/hr: $0.61 /ton(Own and Operate) 5 trucks @ $139.50: $697.50 /hr(Own) I standby @8 61.79: $ 61.79 llr5759.29/hr

Cost Estimation Form

b. - Salvage value (_%):c. Freight lb @ $_/cwt =

d. Unloading and moving cost =

e. Delivered price:

f. Operating period hrlyr _

tires

g. Economic life hr: yr (n)h. Depreciation: $_(del. price less tire cost) : $_/hr

hr

2. INTEREST, TAXES, INSURA}ICE A}{D STORAGEa. Rate : interest Yo + taxesYo * other Yo: %o

r+l

b. Average annual investment rate: 2n

c. Average annual investment:$_x oZ: $

d. Annual fixed charge: $_x Yo: $

e. Fixed charge: $

hrlyr: $_/hr

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TOTAL OWNERSHIP COSTS : $ /hr

B. Operating Costs


Purchase price 1 set of tires: $

Tire life hr

Tire cost: $

hr: $ /tr2, TIRE REPAIR COST: o/o x $_tire cost: $_/hr

3. REPAIRS & MAINTENAIICE: o/o x $_x_deprec. : $_/hr

4. FUEL: xgaVlv @$_/galOR POWER: lov @ $_/kw-hr : $_/hr

5. LUBRICATION: gal/hr @$_/gal-ltr: $_/hr6. AIIXLIARY FUEL: $

7. LABOR: operator @$_/ltr: $_/hroiler @ llv: llvhelper @ ltv: llrTotal $ /h

+ 35yo benefits $ /lr: $ /hrTOTAL OPERATING COSTS : $ /hf

TOTAL OWNERSHIP AlllD OPERATING COSTS : $ /hr

UNIT COST $_/hr + tonlhr: $_/ton

(Own and Operate)_trucks @ $_: $_/hr

(Own)standby@$:$_/hr

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:$/hrE. The discussion on trucks would also apply to:

1. Shuttle cars:

Selpropelled rubber-tired haulage vehicle designed for underground mining, primarily in coal mines.

Drive mechanism is located on the sides of the vehicle and the load is carried in the center.

A chain and flight conveyor distributes the load when the car is being loaded by a loading machine orcontinuous miner and also discharges the load onto a conveyor belt or into a mine car.

As the name implies, shuttle car shuttles back and forth between the working face and the unloadingpoint and is not required to turn round.

2.LHD, Scoop Trans:

Combines certain characteristics of conventional front end loaders and dump trucks.

Design intent is to provide one vehicle with one man, with the vehicle loading itself, hauling the loadover level or inclined haulageways and dumping the load.

3. Mine Trucks:

Cost Estimation

Haulage Unit (Truck)A. OWNERSHIP COSTS

1. Depreciation

a. Purchase Price: estimate as $7650/ton of truck capacity (live load)

b. Salvage value: recoverable, estimate as l5Yo of purchase price

c. Freight: estimate truck weight (dead load) as 1400 lb/ton of truck capacity (live loadO

Use freight charge as $4.00/cwt (cwt: 1001b)d. Unloading and moving cost: estimate as l0% of freight cost

e. Tire cost: estimate as 5% Oftruck purchase price

f. Delivered price: a * c * d - b - e

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g. Operating period: assume 2000hr/yr for I shift/dayh. Economic life: should be based on IRS guidelines

i. Depreciation: delivered price / economic life

2. Fixed Charge

a. Rate: interest * t

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1. Cost of owning and operating x units-in-run: (x) (A3 + 87): (*) (c)2. Cost of owning y spare tires: (y) (43)3. Total hour cost of owning and operating x units-in-run and owning y spareunits: Dl + D2

4. Unit cost: D3 / hourly production

NOTE: Replace all estimated values with known values if possible.

EXAMPLE:

Determinethe total hourly and unit cost of owning and operating five 85-ton trucks and one spare 85-ton truck if the trucks operate on an 8-hour shift and the following information is provided.

Operating conditions: average

Truck and tire life: unfavorable

Diesel fuel costs: $1.00/gal

Operator wage: $ 12.00/hr

Total output per shift: $10,000 tons

Interest : l4%o, taxes : 2Yo, other :2Yo

Cost Estimation Form

85-ton-Truck Mining Equipment Unit

A. Ownership Costs

I. DEPRECIATION

a. Purchase price: $650,250

b. Salvage value (15%): - 97.538c. Freight 810.000 lb @ $4.00/cwt : 4.760

d. Unloading and moving cost:476

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e. Delivered price:557.948 say $525.000

f. Operating period 20OA fulyr - 32,512 tires

g. Economic live 15.000 hr:7.:L yr(n)h. Depreciation: $525.000 (del. Prices less tire cost) :$ 35.00 /hr15.000 hr

2. INTEREST, TAXES,INSURA}ICE, A}ID STORAGEa. Rate : interest l4Yo + taxes2Yo * other 2Yo: l9yo

b. Average annual investment rate: (n+l)/2n: 8.5/15 :$ 56.7 yoc. Average annual investment: $525.000 x 56.7yo: $297.5g2d. Annual fixed charge: $297.582x18Yo: $53.582e. Fixed charge : $53.582 :$ 26.79 lhr 2000 tr/yrTOTAL OWNERSHIP COSTS :$ 61.79 /lrB. Operating Costs


Purchase price I set of tires : $ 32.512

Tire life 25001r

Tire cost : $ 32.512:$13.0_O /lr2500 hr

2. TIRE REPAIR COST: n_%X $ 13.00 tire cost:$ 2.21 /tr3. REPATRS, MAINTENANCE: 45% X $35.00 x l5/10 deprec. :$23.63 lhr4. FUEL: 850 x 0.020 galltu @ $_!.OQ/gal

OR POWER: kW @ $ /k\ry-hr :$ 17.00 /hr5. LUBRICATION: l/3 gaVtv @ $ 1700 ltu:S 5.67 ltu

6. AIDLIARY FUEL: :$ - /hr

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7. LABOR: I operator @S t2.00 /hr: $ IZ.OO lItIoiler@$/hr:$/trhelper @g_ltu: $ /hr

Total $ 12.00 lly+ 35Yo benefits $ 4.20 /lr:$ 16.20 lhrTOTAL OPERATING COSTS :$ 77.71 ltrTOTAL OWNERSHIP At{D OPERATING COSTS :$ 139.50 /hUNIT COST S 759.29 ltv + 1250 tonftr : $ 0.61 /ton

(Own and Operate) 5 trucks @ $139.50 : 697.501tr(Own) I standvy @561.79:61.79S759.2glltr

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desired.

,.,,2.,,'Determin

'.'.loader,,.,CyCle, time

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load time t- maneuver time f travel time + Dump time. return time + delays

fl min

b) cycles per hour @ lxo%efficiency total clule time min'

NOTES: i. Travel time is not included in wheel loaders

ii. Load time is dependent on the type of material

iii. Correction in cycles per hour has to be made using efficiency factors for the loader

3. Determine payload per cycle in tons, pounds, and loose cubic yards.

_

hourly pruduction

a) Required production per cycle # of cycles/hr

rcquird uaylnad/cwleb) volume or loose cubic yards per cycle matrial specifrc weigt (tons' Ih/Lyd)

_

vrlume rcquired per clule

4. Determine bucket size buclct fill fator

5. Make machine selection using bucket size and payload as criteria to meet production

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requirements.

6. Compare the cycle times of the loader, used in calculations, to the cycle time of the selectedmachine. If there is a difference, rework the process beginning at step 2.

7. a) The required operating capacity ofthe machine

: (bucket size, yd3)(density or specific weight of the material)

b) When the material is loose, as in stockpile loading, the bucket load is estimated in loose yd3 by abucket factor.

Bucket Payload, Lyd3 : (Rated bucket capacity)@ucket fill factor)

c) When material is in bank state, as in excavation, productivity is measure in bank yd3

Bucket Payload, Byd3 : (Rated bucket capacity)(material swell factor)(bucket fill factor)

8. Loader production assuming lO0% efficiency and availability can be determined as

p, =

60 xClxFfxFt' tar"

Where: P, : loader production, byd3/tr

C, : loader rated capacity, yd,

Fr: loader bucket fill factor

F. : material swell factor

T,l": loader cycle time for load and carry, min

load + travel + dump * return * maneuver + delays

9. I)ozer production assuming l00yo efficiency and/or availability can be calculated as:

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Where: Po: dozer production rate, bydr/hr

Lu: blade loading, Lyd,

F.: material swell factor

t,u: dozer cycle time in minutes

Dozer production is in bank volume of material excavated, but the blade loading and volume ofmaterial spread, dumped or piled, is in loose volume.

10. Ripper production assuming l00yo efficiency and/or availability is calculated as

.' _ 60xLx'!VxP' tt,

Where: Pr: ripper production rated, byd3/hr.

L: rip distance, length of ripped areao yd\M : rip spacing, distance between ripper passes, yd

to: ripper total cycle time, minutes.

11. Scraper Production assuming 100% availability and/or efficiency is calculated as

,, = oI:r,= ry

Where: P, : scraper production rate, byd3/hr or ton/hr

L.: scraper actual payload, byd3 or tonst,, : scraper total cycle time, minutes

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Lr=CrxF,

Where: Lr: scraper actual payload, byd3

C.: scraper rated heaped capacity (l:l), yd3F.: material swell

EXCAVATING AND LfffnVG - Transitional ExcavatorsOur discuSsion on trucks would not be complete without considering front-end loaders or shovels ordraglines which are commonly used to load materials into trucks.

A.Introduction

The major types of excavators used in earthmoving operations include:

1. Members of the crane-shovel family or transitional excavators

(shovels, draglines, hoes, clamshells, etc.)

2. Mobile excavators (dozers, front-end loaders, tractor scrapers, etc.)

B. Production of Earth Moving Equipment

The basic relationship for estimating the production of all earthmoving equipment is:

Production = (Volume per cycle) x (Cycles per hour)For estimating the production of an excavator, it is necessary to know the volume of material actuallycontained in one bucket load, such as:

1. Plate line capacity - Bucket volume contained within the bucket when following the outline of thebucket sides.

2. Struck Capacity - Bucket capacity when the load is struck offflush with the bucket sides.

3. Water Line Capacity - Assumes a level of material flush with the lowest edge of the bucket (i.e. thematerial level corresponds to the water level that would result if the bucket were filled with water.

4. Heaped Volume, expressed in Lcy - maximum volume that can be placed in the bucket withoutspillage based one specified angle of repose for the material in the bucket.

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EXAMPLE

Estimate the actual bucket load in bcy for a loader bucket with heaped capacity of 5 yd3.

Soils bucket fill factor is 0.9

Load factor is 0.8

Bucket Load: 5 x 0.9 x 0.8 :3. bcy

Bucket - Capacity Rating Method

Machine Rated Bucket Capacity

Backhoe Struck volume

Clamshell Plate line capacity or water line capacity

Dragline 90% of struckvolume

Loader Heaped capacity at2:l angle of repose

Shovel Struck volume

C. Crane - Shovel Family - Consists ofthree major assemblies:

1. Carrier or Mounting Carriers

a) Crawler mounting

Can operate on surfaces which are too soft for wheel or truck mounted equipment.

Excellent for on-site mobility.

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Speeds are too low.

b) Truck mounting

capable of higher speeds of 30 mph to 50 mph.

c) Wheel mounting

Speed is intermediate between crawler and truck mounting.

Truck and wheel mountings provide greater mobility between job sites, but are less stable than crawlermountings and require better surfaces over which to operate.

2. Revolving Deck or Turntable

Revolving superstructure containing the power and control units.

3. Front-end Attchment

The name for the particular member for the crane-shovel family is determined by the front endattachment used. Thus, a crane-shovel with a shovel attachment is referred to as simply a shovel etc.

D. Shovels

(The figure shows members of the crane-shovel family according to the US Dept. of the Army).Whelher dbsigned primarily for stripping or loading, has its greatest application in handling tight orpoorly fractured overburden because of its crowding action, which allows a higher breakout force tobe applied.

Can handle loose material as well as rock and block-like material and will have less overall effect on itsability to load efficiently.

Since it is designed with a solid connector between dipper and drive mechanism, positive control isalways possible and generally a higher loading factor will result.

Cycle times are lower because most of the material loaded is confined to an area directly

in front ofthe unit and dumped within a relatively short radius of the loading point.

Ordinarily a shovel works from a position directly on top of a coal deposit, eliminating the need forpreparation of a working bench.

Relatively thin overburden, which permits low stripping ratio.

Combinations of shovel and truck move the overburden quickly and farther than any stripping capacitydragline.

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Limiting factors of a shovel include:

a. Less flexibility and maneuverabil.

b. Slow tramming rates.

c. Bank or spoil slides and water seepage into the pit will tend to impair the operation of a shovel.

1. Two Types of Shovels:

a) Stripping shovels

Strip overburden to expose minerals such as coal, phosphate, bauxite, gJsum, iron, stone, etc.

Remove overburden and waste by a dry excavation process.

Remove overburden in long nanow cuts from the pay material and deposit it by a simple overcastingprocess in an adjoining mined out pit.

Simple overcasting.

Tandem operations with other machines to remove overburden. Tandem operations can note the useof two or more machines, each removing and spoiling overburden with one machine followinganother.

Shovel-pull-back, a shovel in connection with a dragline operating "pull back".

Generally offers lower overall operating costs than draglines or bucket wheel excavators ofcomparable size. This is a consequence of a number of factors including:

a. Lower power requirement costs.

b. Lower wire rope costs/yd3 of dipper.

c. Higher availability factor.

d. Lower level of bank preparation (e.g. blasting).

e. Lower labor requirement.

f. Shorter distance for moving the overburden.

Range from 25 yd3 to 180 yd3 with boom lengths of up to 235 ft.and can handle lengths of up to 43 m(140 ft) of available stripping.

Price of a stripping shovel with 100 yd3 dipper and200ft boom including shipment and erection costs12.0 x 106.

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b) Ouarry mine shovels

Excavate materials already prepared for digging.

Used in open pit mining.

Capacity - normally struck volume.

Used to excavate earth and load into trucks or tractor pulled wagons or onto conveyors.

Capable of excavating all kinds of earth except solid rock without prior loosening.

Can be crawler, truck, or wheel mounted.

2. Shovel Size:

The size of the shovel is indicated by the size of the dipper, expressed in yd3.

The earth is struck even with the contour of the dipper. This is referred to as struck volume,distinguished from heaped volume which the dipper may pick up in loose soil.

Due to the swelling of a soil when it is loosened, the bank measure volume of a dipper will be less thanits loose volume. It is possible that a dipper may be heaped sufficiently to give a bank measure volumeequal to the rated size of the d ipper. This condition will not occur except for in easy digging soils.

A cable-operated shovel digs with a combination of crowd action and hoist action. The dipper isforced out (crowded) or pulled back (retracted) by the turning of the dipper shaft. Dumping isaccomplished by releasing the dipper door latch, a llowing the dipper door to swing open.

In the digging action, if the depth of the face is just right, considering the type of soil and the size ofthe dipper, the dipper will be filled as it reaches the top of the face.

If the depth ofthe face, referred to as the depth of cut, is too shallow, it will not be possible to fill thedipper completely without excessive crowding and hoisting.

3. Selecting the Type and Size of a Power Shovel

In selecting the size of a shovel, the two primary factors which should be considered are the cost percubic yard of material excavated and the job conditions under which the shovel will operate. Acrawler-mounted shovel usually is less expensive than the rubber tired mounted unit and can operateon ground surfaces which are not firm enough to support the latter type unit.

Thc follswing onditions should be considered in selecting the size of a shovel;

a. High lifts to deposit earth from a basement or trench into trucks or natural ground level.

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b. If blasted rock is to be excavated, the large size dipper will handle bigger rocks.

c. Ifthe material to be excavated is hard and tough, the dipper of the large shovel which exerts higherdigging pressures will handle the material more easily.

d. If the time allotted to the completion of the job requires a high hourly output, alarge shovel mustbe used.

e. The size of available hauling units must be considered in selecting the size of the shovel.

f Weight limitations imposed by most states for hauling on highways may restrict the size of a shovelif it is to be hauled over state highways.

4. Optimum Depth of Cut

The optimum depth of cut is that depth which produces the greatest output and atwhich the dippercomes up with a full load without undue crowding, The depth varies with the class of soil and the sizeof the dipper.

5. Output of Power Shovels - Depends on:

a.) Class of material

b.) Depth of Cut

If the depth of the face from which a shovel is working is too shallow, it vvill be difficult or impossibleto fill the dipper in one pass up the face.

If the depth of the face is greater than the minimum required to fill the dipper, the operator may doone of three things:

i. Reduce the depth of penetration of the dipper into the face in order to fill the dipper in one fullstroke - results in an increase in cycle time.

ii. Dig above the base and remove the lower portion for the face later. iii. Run dipper up the full heightof the face and let the excess earth spill,

c.) Angle of Swing

Angle of swing of a power shovel is the horizontal angle expressed in degrees between the position ofthe dipper when it is excavating and the position when it is discharging the load.

Total time in a cycle includes:

i. Digging.

ii. Swinging to the dumping position.

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iii. Returning to the digging position.

If angle of swing increases, time for a cycle increases and vice versa.

d.) Job Conditions

No two excavating jobs are alike. There are certain conditions at every job over which the owner ofthe shovel has no control. These conditions are considered in estimating the probable output of ashovel. Consequently job conditions may be classified as excellent, goo, faii, and poor.Ajob having excellent job conditions would include:i. A shovel operates in a large, open pit with a firm well drained floor, where trucks can be spotted oneither side ofthe shovel to eliminate lost time. ii. Terrain of the natural ground is uniformly level sothat the depth of cut is always an o ptimum.

iii. Haul road is not affected by climatic conditions such as rain or snow.

hauling units.

Ajob having poorjob conditions would include:i. Another shovel may be used to excavate material for a highway cut through a hill. Depth of cut mayvary from zero to considerably more than the optimum depth.

ii. The sides of the cut must be carefully sloped. The sides ofthe cut may be so narrow that a loadedtruck must move out before an empty truck can back onto loading positions. As the truck must bespotted behind the shovel, the angle of swing will approximate 180o.

iii. The floor of the cutmay be muddy which will delay the movement of trucks. Light rains may delayoperations for several days.

e) Management Conditions

The attitude of the owner of a shovel will also affect productivity. While the owner may not improvejob conditions, he or she can improve management conditions. The following are just u f.* goathings he or she can do:

a. Provide a competent supervisor.

b. Pay bonuses to crews.

c. Provide and maintain adequate trucks.

d. Provide and maintain adequate shovels.

f) Skill of the operator

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g) Ph)sical conditions of the shovelEXAMPLE 1

Consider a3/4 cu-yd shovel excavating earth with 90 swing and with a 2l sec. cycle time.Assume dipper & truck are operated at their heaped capacities.

Assume that the number of dippers required to fill a truck: capacity oftruck + size ofthe dipper.Assume that the travel cycle which includes traveling to the dump, dumping and returning to shovel:6 minutes.

If3 cu-yd trucks re used.

Number of dippers required: + - 4 rlpers3t{

Cycle time for the shovel = 2l secondsTime required to fill a truck : 2l x 4 : g4 seconds : 1.4 minutesMinimum round trip cycle for a truck : 6 + 1.4 : 7.4 minutes

Minimum number of trucks required to keep shovel busy: 7.4 + 1.4: 5.3Thus, 6 trucks will be necessary to keep the shovel busy or else permit the shovel to idle betweentrucks.

Time to load six trucks:6 x 1.4 : 8.4 min.

Lost time per truck : 8.4 - 7.4: I nnn.This produces an operating factor of 7.4 + g.4 : gg% for the trucks.

ff6 cu-yd trucks are used.

Number of dippers required : 6 - 8.

3t4

Cycle time for a shovel:21 seconds

Time required to fill a truck: 8 x 2r seconds : 16g seconds:2.g minutesMinimum round trip cycle for a truck: + 2.g : g.g secondsMinimum number oftrucks required to keep shovel busy: g.g + 2.g :3.15

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Time required to load 3 trucks: 3 x 2.8 : 8.4 minutes

This produces an operating factor : 8.4 + 8.8 : .95 :95yotruck time,

or loss of (8.8 - 8.4y8.8 :SYo

Time required to load 4 trucks : 4 x2.B: ll.2 minutesThis produces an operating factor : 8.8 + ll.2: 79yo trucktimeor a loss of (11.2-8.8)/11.2:2tYoAssume shovel is operating80%o efliciency.

# of cycles per minute : 60/21:2.86

# of cycles per hour: 60 x 2.86: 171.6

Ideal output per hour :3/4 x 176.6: 128 cu-ydOutput @80% efficiency : 128 x 0.80 :102 cu-yd

Travel cycle time per truck: 6 minutes

Assume that 6 cu-yd trucks are used.

The ideal number will be 3.15 as previously determined

3.15 trucks needed for 102 cu-yd / hr

3 trucks need for 3/3.15 x 102:97 cv-ydper hour.

Consequently, if three trucks are used, the output :97 cu-ydper hour.Cost per hour for a truck & driver: $14.70

Total cost per hour for trucks (3 x 14.70) : $44.10

Truck cost while loading, (2.8 x 14.70)/60 = $0.66

Truck cost per cu-yd of earth loaded $0.6616 : $0.11

Haulingcostpercu-yd: ffi = H - $.4s6

age 3t o+2

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EXAMPLE 2

Number of units required:3.3

Units in-run:4

In run utilization :3.3/4:83Yo

Assume 807o avilability

Units required :4/.8:5 unitsNumber purchased: 5

Acceptable availability : 415 : 80Yo

Capital Utilization : 3.3/5 : 66Yo

Check shovel capacity per hour and compare to truck capacity per hour.

Check body size.

Body volume: , p.lJL4

. = Jd3 rrquired in body ar 2:I heq" demsity (bcy) J'

ffi=2sesr=26y:Actual Payload : '48 tons/pass x 6 passes- x 200f1 D/tnn = 2gg Jd 32?t0 b/yd3

Body size is adequate.

EXAMPLE 3

Consider a I cu-yd power shovel for excavating hard clay with a depth of cut of 7.5 ft. An analysis ofthe project indicates an average angle of swing of 75, job conditions will be fair, management will begood. Determine the probably output in cubic yards per hour bank measure.

Solution

From Table: ideal output :145 cu-yd/hr

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optimum depth:9 ft.

average depth of cut:7.5

Percent of optimum depth of cut : 7.5/9 : 83.3Yo

Job management factor: 0.69

Swing-depth factor = 1.04

The probable output/hr : 145 x 1.04 x 0.69: 104 cu-yd

For a 50 minute hour @fficiency: 50 min), the probable output:0.83 x 104 : 86 yd3

6. Production

Output is expressed in:

Struck bucket capacity

Heaped bucket capacity

The measure employed customarily is bank (solid) measure, the volume of material in place. Incontrast, when dealing with haulage units (truck), output is measured in loose (broken) measure.a) Output is obtained by using tables compiled by the PCSA' output is based on the followingassumptions:

100% efficiency

90% (90 degree) angle of swing

optimum depth of cut.

working time of 60 minutes/hr

bucket factor of 1.0

job-management factor of l00Yo

material loaded into haul units at grade level

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Power shovels are rated by their ideal output (yd3llv or -3/hr. The output is a function of:

bucket size

type ofmaterial

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working time

difficulty in digging

type of haulage unit

working conditions (ob & management factors)Output must be modified for "real life" conditions. Corrections must be made for:

working time

depth of cut if less than or greater than optimum.

angle of swing

bucket factor (% of bucket capacity utilized)b) Calculation:

Cycle time: digging time + swinging to dumping position+ dumping f returning to digging positionshovel capacity (yd3): (bucket size, yd3)(swell factor)(fill factor)shovel capacity (tons) : (bucket size, yd3)(swell factor)(fill factor)(weight of material bank measure, tons/yd3)

Dipper fill factor: vV +lcr ir yd t Oanlc mmrurc)(Can be greater than 1.0 for materials that heap well.)Bucket fill factor : Yo of bucket capacity utilized.

EXAMPLE

Based on the following information determine the expected shovel production in bank cubic yards perhour.

Shovel size:2yd3

Swing angle:120

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Average depth of cut:8.2 ft

Material: common earth

Job efficiency:0.75

Solution

Ideal output : 300 bcy/hr

Optimum depth of cut :10.2ft

Average depth of cut:8.2 ft

A.Ettrl dqtJr/optimrmr rl+th = 8.2 x t0tt0.2 =

80o/o

Swing depth factor : 0.86

Estimated production: 300 x .75 x .86 : 193.5 bcy/hr

E. Draglines

Draglines are used to excavate earth and load it into hauling units, such as trucks or tractor-pulledwagons, or to deposit it in levees, dams, and spoil banks near the pits from which is excavated. Ingeneral, a power shovel up to a capacity of 2.5 yd3 can be converted into a dragline by replacing theboom of the shovel with a crane boom and substituting a dragline bucket for the shovel dper. -

For some projects either a pou/er shovel or a dragline may be used to excavate materials, but forothers the dragline will have a distinct advantageiompard with a shovel. A dragline usually does nothave to go into a pit or hole in order to excavate. It may operate on natural ground while excavatingmaterial from a pit with its bucket. This will be very advantageous when earth is removed from aditch, canal, or pit containing water. If the earth is hauled with trucks, they do not have to go into thepit and contend with mud. If the earth can be deposited along a canal or ditch or near a pit, itfrequently is possible to use a dragline with a boom long enough to dispose ofthe earth in oneoperation, eliminating the need for hauling units, which will reduce the cost of handling the earth.Draglines are excellent units for excavating trenches when the sides are permitted to establish theirangles of repose, without shoring.

One disadvantage in using a dragline compared with a power shovel is the reduced output of thedragline. A comparison ofthe ideal output of various sizes of draglines with the output of powershovels shows that a dragline will excavate approximately 75 to 80 percent as much earth as a shovelofthe same size.

Some other limitations of draglines are:

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a. Does not have the positive digging action or lateral control of the shovel.

b. Bucket may bounce of move sideways during hard digging.

c. More spillage expected in loading operations.

Some assets of draglines are:

a. A greater reach and dumping radius which enable a deeper overburden to be removed.

b. Ease of maneuverability.

c. Since the dragline operates on top of the overburden, certain problems such as bank slides, waterrunoffand seepage are reduced or eliminated.

d. Compensation for pitches and rolls in the mineral deposit can be made and if necessary, the draglinemay be used to chop down a small portion of the overburden to prepare it a working bench.

e. Reduced maintenance (walking dragline)

f Pressure exerted on the ground range from 19 psi for large machines to bearing pressures of 9.5 psifor smaller machines making them extremely suitable for soft or wet formations.

g. Can operate on the tops of spoil either as a primary stripping tool or for reclamation purposes.

h. Ideal for producing the initial or box cut in a stripping operation, a procedure that is almostimpossible for a shovel.

i. Can be effective in operations such as building roads and ditches in addition to being able to clean arough coal seam for direct loading into transportation vehicles.

j. Bucket loading requires a higher degree of operator skill due to the design of the dragline.k. Cycle times are high and small amounts of spillage occur.

l. A dragline will excavate approximately 75Yo to 80% as much earth as a shovel of the same size.

m. Dippers shovels

Buckets Draglines

Draglines have reached their physical limitations the largest one being built has a220 yd3 bucket.However because of its newer design and introduction of new components, it has not had thereliability and performance necessary to encourage others to be built. The largest draglines built todayhave capacities from 100 to 150 yd3.

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1. Types of Draglines

crawler mounted

truck mounted

wheel mounted (wagon mounted & walking draglines)

2. Size of Draglines

Size is indicated by the size of the bucket expressed in yd3 which in general is the same size as thedipper of the shovel into which it may be converted.

3. Operation

Excavating is started by swinging the empty bucket to the digging position, at the same time stackingoffthe drag and hoist cables.

Excavating is accomplished by pulling the bucket toward the machine while regulating the diggingdepth by means of the tension maintained in the hoist cable.

The cycle is made up of

a. Digging

b. Hoisting

c. Swinging

d. Dumping

It is more difficult to control the accuracy of dumping from a dragline as compared with a powershovel; it is desirable to use larger hauling units for dragline loading in order t reduce the spillage.

4. Optimum Depth of Cut

A dragline will produce its greatest output if the job is planned to permit the earth to be excavated atthe optimum depth where possible.

5. Output of Draglines

The output of a dragline will vary with the following factors:

class of material

Depth of cut

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angle of swing

size and type of bucket

length of boon

job conditionsmanagemeil conditions

rnethod ffud, orstillg, orloading truckssi dllanig ws

skM dqer-aor

psicd condition of the rnachine

6. Buckets for draglines are available in three types:

LiU[ duty buckets - used for excavating materials which are easily dug such as sand, or sand clay.Medium duty buckets - used for general excavating service as digging clay, soft shale or loosegravel.

Heavy duty buckets - used for mine stripping, handling blasted rock, excavating hardpan and highlyabrasive materials.

Buckets are sometimes perforated to permit excess water to drain from the loads. In selecting themost suitable size of bucket to use for use with a given dragline, it is desirable to know the w:eight ofthe loosened material to be handled, expressed in lb/ft3. Combined weight of the bucket and loadshould not exceed the safe load recommended by the dragline.

The production of a dragline can be estimated using the tables based on PCSA data or by calculation.The procedure using the PCSA data to determine the production of a dragline is similar io that for apower shovel. The table for a dragline with short booms is given on the next page. The procedure forcalculation production for a dragline is similar to that for a power shovel.

EXAMPLE

Determine the expected dragline production in bank cubic yards (gr3) per hour on the followinginformaton.

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Dragline bucket size:2 yd3

Swing angle:720

Average depth of cut :7.9 ft.

Material: common earth

Job efficiency: 50 min/hr

Solution

Ideal output : 230 bcylhr

Optimum depth of cut:9.9 ft.

Actual deptlr/optimum depth :7.919.9: 0.80

Swing depth factor:0.9

Efficiency factor: 50/60 : 0.833

Estimated production : (230 x 0.9 x 0.833) bcy: 172.4bcyF. Bucket Wheel Excavators

The largest, most complicated, and under favorable conditions, the most productive excavatingmachines used in surface mining.

Remove very effectively unconsolidated overburden such as earth glacial till, clay and soft shale thatdoes not require blasting.

Will excavate material from a highwall on one side of a mine and deposit it as spoil at the rate of 1500to 2000 yd3lfu up to 426'away on the opposite side of the pit.

Largest unit in the world produces 12,000 yd3lhr, operates in a German brown coal seam.

Long stacker allows a much greater discharge radius to be achieved while consuming less powerd3of materials removed compared to other equipment.

Advantages

a. A bucket excavator is actually smaller than a dragline or shovel for a given output.

b. It is a continuous excavator and has not cycle times.

c. Lower instantaneous porer requirements and no shock roading

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d. Larger discharge radius and can be designed to operate above or below the active mine level.

e. Can load mined material efficiently onto a variety of haulage modes, including truc haulage, orbelt conveyor.

fl Can be designed to operate efficiently tkough a wide range of mining heights.

g. Produces a smaller size consistency of material

h- Ground bearing pressures as low as I I psi and as high as 20 psi. Its wheels produce more stableslopes and wider benches

i' Close control of selective mining of interbedded coal and mall partings such as high and low sulfurhorizons is possible.

j. Has flexibility to deliver the mines material above or below the working level.k. There is little or now dead weight structure required as counterweight because the stacker and driveare used.

Disadvantages

a. Machinery is highly complicated and therefore has reduced availabil.b. Less flexibility due to lack of mobility.

c. Can't handle consolidated hard materials

d. Require large maintenance force and has high initial cost for a given production rate.

e. Unit restricted to mining thick coal seams only. Thickness of coal seam would at least be 0.7 timesthe wheel diameter. This is because bucket configuration and wheel diameter can cause problems intryrng to follow the bottom of the coal seam.

Wheels are designed with either celled or cell-less construction. Most modern wheels are cell-less. Inthe cellJess type, the buckets in the digging position can continuously empty into an annular ring.

EXAMPLE 1

A wheel loader must produce (230m3lhr) :00 yd3ltr in a truck loading application. Estimated cycle

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time is 0.6 min. and job efficiency is75% (or working 45min/hr). Bucket fill factor is95Yo andmaterial density is (1780 kglrn3) 3000 lb/yd3.

Determine bucket size and machine model:

Solution

#ofcycteslhr= fffi - o.?s= 75 cycles/hr

Let bucket size be x yd3

Volume of material : 0.95x

Production/hr : (75 cycles/hr)(0. 95 x yd3lcycle)

:7l.25xyd3/nr.

Required production : 300 yd3 /nr.

7 1.25x yd3 /hr: 300 yd3 hrx:4.2

Useg66fwith 4yd3bucket dependent on material dens and bucket capacity. Refer to pg. 12-40 inCaterpillaf.

EXAMPLE 2

A wheel loader is required to produce 496 tons/lr for 3/8" gravel in 20 ft high stockpile. The densityof material is 28001b/yd3. Cycle time is 0.5 min. with actual working time of 50 midhr. The bucket fillfactor is 0.95. Trucker are in 8-12 yd3 . Determine the required rated bucket capacity.

Solution

#ofcycles/hr= .5=.0.!s.. - looclrles/lu.- 0.5mircpb

Volume ofmaterial required: ffiffi: l.4ton/yd3

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Required production: ffi - 3s4 yd3ihr

Volumepercycle: ==l54s- = ss4yd3/cycblll0 culeeiluur

Rated bucket capacity: 354 yd3/cfle0.95 fill fator

:3.72 yd3l cycle

The bucket size required and material density lead to a machine model of 950G series IL

Required operation capacity :3.75 yd x 2800 lb/yd3: 10,500lbs.

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