3/17/2015

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Emma Akmalah, Ph.D.

Grade resistance (lb) = vehicle weight (tons) x Grade

resistance factor (lb/ton)

Grade resistance (lb) = vehicle weight (lb) x Grade

Effective Grade (%) = Grade (%) +

Effective Grade (%) = Grade (%) +

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Soal 1:

The haul road from the borrow pit to the fill has an

adverse grade of 4%. Wheel-type hauling units will be

used on the job, and it is expected that the haul-road

rolling resistance will be 100 lb per ton.

a) What will be the effective grade for the haul?

b) Will the units experience the same effective grade

for the return trip?

c) If the haul unit has a gross weight of 47 tons and

an empty vehicle weight of 22 tons, what is the

total resistance experienced during the haul and

during the return?

Solution:

a) Equivalent grade (RR) = = 5%

b) Effective grade (TRhaul) = 5% RR + 4% GR = 9%

Effective grade (TRreturn) = 5% RR - 4% GR = 1%

c) Total resistancehaul = 47 tons x 9% x 20 lb/ton

= 8460 lb

Total resistancereturn = 22 tons x 1% x 20 lb/ton

= 440 lb

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Solution:

RR = ((72,000 + 100,000)/2,000) x 80 = 6,880 lb

GR = (72,000 + 100,000) x 0.05 = 8,600 lb

TR = 6,880 + 8,600 = 15,480 lb

Soal 3:

An off-highway truck weighs 60,000 lb empty and

can carry a payload of 100,000 lb. The

haul route requires the truck to travel down a 3%

grade and return empty on the same route.

The haul road has a rolling resistance factor of 100

lb/ton. What are the total resistance and

effective grade for each portion of the haul route?

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Haul:

EG = -3 + (100/20) = +2%

TR = (60,000 + 100,000) x 0.02 = 3,200 lb

Return:

EG = 3 + (100/20) = +8%

TR = (60,000) x 0.08 = 4,800 lb

Solution:

Maximum Rimpull = 0.30 x 18,000 lb = 5,400 lb

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Solution:

Maximum Rimpull = 108,000 lb x 0.5 x 0.6 = 32,400 lb.

Soal 6:

A wheel 140 HP tractor weighs 12.4 tons and

has a maximum speed of 3.3 mph in first gear.

If it is operated on a haul road with a positive

slope of 2% and a rolling resistance of 100

lb/ton, what is the pull available for towing a

load? The efficiency of the tractor is 0.85.

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Maximum rimpull =

Maximum rimpull = = 13,523 lb

Forces requires to overcome rolling resistance

= 12.4 tons x 100 lb/ton = 1,240 lb.

Forces requires to overcome grade resistance

= 12.4 tons x (20 lb/ton) x 2% = 496 lb.

Total resistance = 1,240 lb + 496 lb = 1,736 lb

Power available for towing a load

= 13,523 lb 1,736 lb = 11, 787 lb.

Solution:

15 tons x (180 lb/ton 110 lb/ton) = 1,050 lb

The effective drawbar pull = 5,685 1,050

= 4,635 lb.

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Soal 7

A wheel tractor scraper is used on a road project. When

the project begins, the scraper will experience high

rolling and grade resistance at one work area. The rimpull

required to maneuver in this work area is 42,000 lb. In

the fully loaded condition, 52% of the total vehicle weight

is on the drive wheels. The fully loaded vehicle weight is

230,880 lb. What minimum value coefficient of traction

between the scraper wheels and the traveling surface is

needed to maintain maximum possible travel speed?

Solution:

Weight on the drive wheels

= 0.52 x 230,880 lb = 120,058 lb

Minimum required coefficient of traction

=

= 0.35

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Soal 9

A wheel tractor-scraper weighing 100 tons is

being operated on a haul road with a tire

penetration of 2 in. What is the total resistance

and effective grade when the scraper is a)

ascending a slope of 5%; and b) descending a

slope of 5%?

RR = [40 + (30 x TP)] x GVW

RR = rolling resistance in lb/ton

TP = tire penetration in inches

GVW = gross vehicle weight in tons

a) RR factor = 40 + (30 x 2) = 100 lb/ton

RR = 100 lb/ton x 100 tons = 10,000 lb.

GR = 100 tons x 2000 lb/ton x 0.05 = 10,000 lb

TR = RR + GR = 10,000 + 10,000 = 20,000 lb.

Effective grade = 5 + (100/20) = 10%

b) GR = 100 tons x 2000 lb/ton x (-0.05) = -10,000 lb

TR = 10,000 lb 10,000 lb = 0 lb

Effective Grade = 5 + (-100/20) = 0%

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Soal 10

A four-wheel drive tractor weighs 44,000 lb (20,000

kg) and produces a maximum rimpull of 40,000 lb

(18,160 kg) at sea level. The tractor is being operated

at an altitude of 10,000 ft (3050 m) on wet earth. A

pull of 22,000 lb (10,000 kg) is required to move the

tractor and its load. Can the tractor perform under

these conditions?

Solution:

Derating factor = 3 x = 21%

or

Derating factor = 3 x = 21%

Percent rated power available = 100 21 = 79%

Maximum available power = 40,000 x 0.79 = 31,600 lb

or

Maximum available power = 18160 x 0.79 = 14346 kg

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If the coefficient of traction is 0.45:

Maximum usable pull = 0.45 x 31,600 = 14,220 lb

or

Maximum usable pull = 0.45 x 14,346 = 6456 kg

Because the maximum pull is limited by traction is

less than the required pull, the tractor cannot

perform under these conditions. For the tractor to

operate, it would be necessary to reduce the required

pull (total resistance), increase the coefficient of

traction or increase the tractors weight on the

drivers.

Soal 11

A power-shift crawler tractor has a rated blade

capacity of 10 LCY (7.65 Lm3). The dozer is

excavating loose common earth and pushing it a

distance of 200 ft (61 m). Maximum reverse speed in

third range is 5 mi/h (8 km/h). Estimate the

production of the dozer if job efficiency is 50 min/hr.

Fixed time for power-shift transmission is 0.05 min.

Note: 1 mi/h = 88 ft/min; 1 km/h = 16.7 m/min.

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Typical Dozer Operating Speed

Operating Conditions Speeds

Dozing

Hard materials, haul 100 ft (30m) or less

Hard materials, haul over 100 ft (30 m)

Loose materials, haul 100 ft (30m) or less

Loose materials, haul over 100 ft (30 m)

1.5 mi/h (2.4 km/h)

2.0 mi/h (3.2 km/h)

2.0 mi/h (3.2 km/h)

2.5 mi/h (4.0 km/h)

Return

100 ft (30m) or less

over 100 ft (30 m)

Maximum reverse speed in second range (power shift) or reverse speed in gear used for dozing (direct drive)

Maximum reverse speed in third range (power shift) or highest reverse speed (direct drive)

Solution:

Dozing speed = 2.5 mi/h

Dozing time = = 0.91 min

or:

Dozing time = = 0.91 min

Return time = = 0.45min

or:

Return time = = 0.45 min

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Cycle time = 0.05 + 0.91 + 0.45 = 1.41 min

Production = 10 x = 355 LCY/h

or

Production = 7.65 x = 271 Lm3/h

A crawler tractor with 28-heaped-cu-yd scraper (loose

cubic yard) travels to and from the job in second gear (2.2

mph). Travel distance is 900 ft. Material is clay gravel. The

job efficiency is 50 min/hr. What is the daylight hourly

production rate, if the fixed time is 2 minutes?

Solution:

Production = capacity per trip x trips per hour

Time per one-way trip =

= 4.56 min per trip

Time round trips hauling full and returning empty

= 2 x 4.56 = 9.12 min.

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Cycle time = fixed time + travel time

Cycle time = 2 + 9.12 = 11.12 min per round trip.

Trips per hour =

= 4.5 trips/hour

Production loose cubic yard = 28 x 4.5 = 126 LCY/hr

A wheeled tractor pulling a 35-loose-cu-yd scraper is

operating at an elevation of 9000 ft. Material worked

is a gravel (unit weight = 2900 lb per cu yd) to be

used in a highway fill 1.2 miles from the excavation.

Percent swell = 14%. Based on the use of a four-

wheel tractor, weighing 69,000 lb, with a 35-cu-yd

heaped scraper, empty weight 40,000 lb, and when

loaded, 40% of overall weight goes on drive wheels

what is the bank-cu-yd production for a 10-hr work-

shift? The coefficient of traction is 0.3. Push tractors

are being used. Haul road is loose gravel on a 3%

uphill grade. Gravel surface has rolling resistance of

10%. Fixed time is 1.8 minutes and the job efficiency

is 45 min/hr.

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Tractor speed and rimpulls are as shown below:

The solution of the above problem depends on

determination of several factors: (1) gear and speed

to be used; (2) cycle time; and (3) production.

Gear Speed (mph) Rimpull (lb)

First 7 38,000

Second 17 17,500

Third 36 7,000

(1) Speeds

Loaded weight = 35-cu-yd capacity x 2900 lb per cu

yd = 101,500 lb.

Total weight tractor and loaded scraper = 69,000 +

40,000 + 101,500 = 210,500 lb.

Weight on drive wheels = 40% x 210,500 = 84,200 lb.

Pull required for job = coefficient of traction x weight

on drivers = 0.3 x 84,200 = 25,260 lb.

25,260 lb rimpull required means first gear at 7 mph

when scraper when scraper is loaded.

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(b) Total weight tractor and empty scraper

= 109,000 lb.

Weight on drive wheels = 40% x 109,000 = 43,600 lb.

Pull required for job = 0.3 x 43,600 = 13,080 lb.

13,080 lb rimpull required means second gear at

17 mph, empty.

(c) Altitude correction

Derate 3% each 1000 ft above 3000 ft.

Derating factor = 3 x = 18%

Pull available in first gear (loaded scraper)

= 38,000 lb.

Pull at 9000 ft = (38,000) (100 18) = 31,160 lb.

First gear will work, loaded as 31,160 > 25,260.

1000

30009000

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Pull available in second gear (empty scraper)

= 17,500 lb.

Pull at 9000 ft = (17,500) (100 18) =

14,350 lb.

Second gear will work, empty as 14,350 >

13,080.

(d) Rolling resistance

Required rimpull = gross weight x rolling resistance

Loaded scraper = 210,500 x 0.1 = 21, 050 lb

31,160 lb available in first gear; therefore will work.

Empty scraper = 109,000 x 0.1 = 10,900 lb

14,350 lb available in second gear; therefore will

work.

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(e) Grade

Required rimpull = gross weight x grade resistance

= 210,500 x 0.03 = 6315 lb loaded

= 109,000 x 0.03 = 3270 lb empty

These values combined with rolling resistance of (d)

above still do not exceed power available in first and

second gear at 9000-ft elevation.

(f) Based on the preceding, haul gear is first at 7 mph

and empty return gear is second at 17 mph.

(2) Cycle time

Cycle time = fixed time + travel time

Fixed time = 1.8 minutes

Travel time = haul time + return time

= (distance/ speed) + (distance/speed)

= (1.2/7) + (1.2/17) = 0.2420 hr

= 14.5 min

Cycle time = 1.8 + 14.5 = 16.3 min.

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(3) Production

Production = capacity per trip x trips per hour x no. hr.

Trips per hr = (45 min/hr)/(16.3 min) = 2.7 trips per

hour

Production = 35 x 2.76 x 10 = 966 cu yd loose per 10

hr.

Percent swell = 14%

Loose volume must be reduced by 14% to get bank

volume.

966 cu yd loose x 0.14 = 135.2 cu yd

966 - 135.2 = 830.8 cu yd bank per 10 hours.

Capacity per trip Trips (cycles) per hour

Cycle Time Efficiency

Fixed Time Variable/Travel Time

Distance Speed

Available Power

Usable Power

Required Power

Rimpull, DBP

Traction RR, GR

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Production = capacity per hour x trips (cycles) per hour

Cycle time = fixed time + variable time

Trips per hour = job efficiency/cycle time

Travel time depends on the distance and speed

Speed depends on the gear selected

The gear selected depends on the available, usable, and required power

The available power is limited by altitude, temperature, and traction.

Traction = coefficient of traction x weight on drive wheels.

The total resistance (rolling resistance + grade resistance) must be overcome by the usable power.