latihan soal peralatan konstruksi
DESCRIPTION
Membahas cara perhitungan produktifitas beberapa alat berat pada dunia konstruksi bangunan.TRANSCRIPT
-
3/17/2015
1
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 (%) +
-
3/17/2015
2
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
-
3/17/2015
3
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?
-
3/17/2015
4
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
-
3/17/2015
5
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.
-
3/17/2015
6
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.
-
3/17/2015
7
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
-
3/17/2015
8
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%
-
3/17/2015
9
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
-
3/17/2015
10
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.
-
3/17/2015
11
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
-
3/17/2015
12
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.
-
3/17/2015
13
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.
-
3/17/2015
14
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.
-
3/17/2015
15
(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
-
3/17/2015
16
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.
-
3/17/2015
17
(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.
-
3/17/2015
18
(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
-
3/17/2015
19
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.