1. Refisa Jiru
HAwassa University IOT
Example 2 – Practical Exercise
(Truck Production)
Ten 25 ton/18 CY rear dump trucks with low pressure tires are available to haul a rock
dirt/gravel mixture. The wheel loader being used has a 4.25 CY bucket.
The haul and return, over poorly maintained earth, is 3 miles. It is a downhill grade of
1.25% going to the dump area. Net truck weight is 36,860 lb. Work hour efficiency is
55 minutes
Solution
STEP 1 BUCKET LOADS
Number of bucket loads required to fill the truck
Truck capacity = 18 CY
Dirt/gravel mixture
From Table 4.3, loose unit weight is 2,600 lb/LCY
Fill Factor wheel loader (Table 8.6) 100- 120%, use average 110%
Loader bucket capacity = loader bucket x Fill Factor =4.25 x 1.1= 4.675 CY
Bucket loads =
Truck capacity
Loader bucket capacity
=
18
4.675
= 3.85
Bucket loads must be an integer number, Therefore use either 3 or 4 bucket loads
Page 1
2. Refisa Jiru
HAwassa University IOT
STEP 2 LOAD TIME
For 4.25 cy bucket Table 8.9. Bucket cycle time 30 - 33 sec, use average 31.5sec=0.524
min
3 bucket loads:
Truck volume =No. bucket x Loader bucket capacity =3 x 4.675= 14 LCY
Check load weight against gravimetric capacity of truck
Load weight=Truck Volume×Material unit weight= 14 LCY×
2,600
2,000
= 18.2 tn < 25 tn ok !
Load time=Number of bucket×Cycle time=3x 0.525 = 1.58 min.
4 bucket loads:
Truck volume = 18 LCY and the excess spills off
Check load weight against gravimetric capacity of truck
Load weight=Truck Volume×Material unit weight= 18 LCY×
2,600
2,000
= 23.4 tn < 25 tn ok !
Load time=Number of bucket×Cycle time=4x 0.525 = 2.10 min.
STEP 3 HAUL TIME
If tire penetration is unknown Rolling Resistance (lb/ton) can be estimated from the
information in Table 5.1
For Poorly maintained earth
Page 2
3. Refisa Jiru
HAwassa University IOT
Resistance factor =70 - 100 lb/ton,
Use 85 lb/ton
Roll resistance=
Resistance factor (lb/tn)
20 lb/tn
=
85
20
=4. 25%
When traveling downhill, grade assistance is negative.
Total Resistance(@3%) = RR+GA=4.25 %+ (-1.25%) =
3.0%
4.25%
×85=60 lb/ton
Net truck weight 36,860 lb or 18.43 ton(1 ton=2000 lb)
3 BUCKET LOAD
Gross wt. =Material weight + Empty vehicles weight=18.43 + 18.20 =36.63 tons
Rimpull =Total Resistance ×payload= 60 lb/ton × 36.63 ton= 2,200 lb
From table 3.1 correspondent to Rimpull of 2,200lb, haul speed is 35 mph
Haul time =
Haul distance( miles)× 5,280 ft/mile
88 ×Haul speed
=
3×5280
88×35
=5.143min
4 BUCKET LOAD
Payload =Empty vehicle weight + materials weight = 18.43 + 23.40 =41.83 tons
Rimpull = 60 lb/ton x 41.83 ton= 2,510 lb
From table 3.1 correspondent to Rimpull of 2,510 lb, haul speed is 22 mph
Page 3
4. Refisa Jiru
HAwassa University IOT
Haul time =
Haul distance( miles)× 5,280 ft/mile
88 ×Haul speed
=
3×5280
88×22
=8.182min
The rimpull is the equivalent of all resistance forces which are in opposition to the
truck movement. The rimpull is the result of three factors: Road grade, Rolling
resistance of the ground (soil friction) and Vehicle weights
STEP 4 RETURN TIME
Net truck weight 36,860 lb or 18.43 tons (1ton=2000pounds)
When the vehicle is traveling up a grade, grade resistance is positive.
Rolling resistance, 4.25%
Grade resistance load to dump 1.25%
Page 4
5. Refisa Jiru
HAwassa University IOT
Total Resistance = 4.25 + 1.25%=
5.5%
4.25%
×85 = 110 lb/ton
Rimpull = Total Resistance×Empty truck weight = 110 lb/ton×18.43 ton= 2,030 lb
From table 3.1 correspondents to Rimpull of 2030lb
Return speed is 35 mph
RETURN TIME =
RETURN DISTANCE (FT)
88×HAUL SPEED
=
3 miles ×5,280 ft / mile
88×35
=5.143min
STEP 5 DUMP TIME
Rear Dump Trucks must be spotted before dumping
Total dump time averages about 2 minutes. Use 2.0 minutes
STEP 6 TRUCK CYCLE TIME
3 bucket loads
Load time. 1.580 min
+ Haul time 5.143 min
+ Dump time 2.000 min
+ Return time 5.143 min
Cycle time 13.866 min
4 bucket loads
Load time 2.100 min
Page 5
6. Refisa Jiru
HAwassa University IOT
+ Haul time. 8.183 min
+ Dump time 2.000 min
+ Return time 5.143 min
Cycle time = 17.426 min
STEP 7 TRUCKS REQUIRED
3 bucket loads
Trucks required = 1+
Truck Cycle time
Load Cycle time
=
13.866
1.58
+ 1
Trucks required = 9.8
Must be an integer number therefore 10 trucks
4 bucket loads
Trucks required = 1+
Truck Cycle time
Load Cycle time
=
17.426
2.10
+ 1
Trucks required = 9.3
Must be an integer number therefore 10 trucks
STEP 8 EFFICIENCY
Working hour efficiency ( E)=
55 min
60min
=0.917
STEP 9 PRODUCTIONS
3 bucket loads
Page 6
7. Refisa Jiru
HAwassa University IOT
The loader will control production if at least one extra truck is used.
Production (LCY/hr )=
60×Truck Load ( LCY)
Load time ( min)
×Efficiency
PRODUCTION ( LOADER)=
14.025 LCY
(1.58 /60)hr
×0.917=488 LCY/hr
4 bucket loads
The loader will control production if at least one extra truck is used
PRODUCTION ( LOADER)=
60×Truck Load (LCY)
Load time (min)
×Efficiency=
18 LCY
(2.10/60) hr
×0.917=471 LCY/hr
For 3 bucket loads 10 trucks production is488 LCY/hr
For 4 bucket loads 10 trucks prduction is471 LCY/hr
If only 8 trucks are AVAILABLE the trucks control production
3 bucket loads
PRODUCTION (TRUCK )=60×
Truck Load (LCY)
Truck cycle time (min)
×Efficiency×Number of trucs
PRODUCTION(TRUCK )=60×
14.025 LCY
13.866 min
×0.917×8=445 LCY/hr
4 bucket loads
PRODUCTION (TRUCK )=60×
Truck Load (LCY)
Truck cycle time (min)
×Efficiency×Number of trucs
PRODUCTION(TRUCK )=60×
18 LCY
17.426 min
×0.917×8=454 LCY/ hr
Page 7
8. Refisa Jiru
HAwassa University IOT
Truck Available 3 Buckets (LCY) 4 Buckets (LCY)
10 488 471
9 488 471
8 445 454
Example 3
Given: its Caterpillar 725 Articulated Truck:
Truck Net empty weight = 22,260 kg
Truck Payload = 23,590 kg
Truck heaped capacity = 14.4 m3
Excavator heaped capacity = 1.9 m3
Quantity of excavation material = 20000 m3
Haul road type = smooth roadway (rolling resistance = 1.5%) (Table 4.5)
Haul material type is dry clay (loose material weight = 1480 kg/m3
, bucket fill factor is
90%, excavator cycle time is 23 seconds) (Table 4.4). The haul road from the borrow
site to the dump is 4 km uphill grade of 2%, Job efficiency is 50 minutes per hour with
road legal speed is 90 km/hr . Operators are good = 0.95
Estimate the earthmoving productivity
Solution
Page 8
9. Refisa Jiru
HAwassa University IOT
Step1: BUCKET LOADS
Number of bucket loads required to fill the truck
Truck capacity = 14.4 LCM
Dirt/gravel mixture
Fill Factor wheel loader (Table 4.3) is 90%
Excavator bucket capacity = Excavator bucket x Fill Factor =1.90 x 0.9= 1.71LCM
Balanced number of buckets( Bucket Loads) =
14.4
1.71
= 8.42
The actual number of buckets must be an integer numbers we have 8 or 9 buckets
Step 2: Load time
Load time =Bucket cycle time×No. of bucket loads
8 buckets
Load time =Number of bucket x Cycle time =8×
[23
60 ]= 3.067 min
Load volume =No of bucket×Excavator bucket x Fill Factor = 8 x 1.9 x 0.9 = 13.68 m3
< 14.4 m3
OK !
Load weight =Load volume×Loose unit weight= 13.68 x 1480 = 20,246.4 kg< Payload OK !
9 buckets
Load time =Number of bucket x Cycle time =9×
[23
60 ]= 3.45 min
Page 9
10. Refisa Jiru
HAwassa University IOT
Load volume =No of bucket×Excavator bucket x Fill Factor = 9 x 1.9 x 0.9 = 15.39 m3
> 14.4 m3
Use 14.4 m3
Load weight =Load volume×Loose unit weight= 14.4 x 1480 = 21,312 kg< Payload OK !
Step 3: Haul time
Roll resistance=
Resistance factor (lb/tn)
20 lb/tn
Roll Resistance factor (1.5%) = 1.5 x 20 lb/ton = 30 lb/ton,
Grade Resistance factor(+) = 20 x percent grade
Grade Resistance (+2%) = 2 x 20 lb/ton = +40lb/ton
Total Resistance = 70 lb/ton
Engine horsepower = 301 hp,
1hp=550
ft−lb
sec
=375
mile−lb
hr
=1,980,000
ft−lb
hr
=603.504
km−lb
hr
1kg=909.09 ton, 1miles=5280ft, 1 ton=2,000 lb, 1miles=1.60934km
For 8 buckets
Weight fully loaded = 22,260 + 20,246. 4 =
42,506. 4
909.09
=46.757 lton (for haul speed )
Haul Speed=
Engine horse power
Total Resisance×Weight of fully loaded
Haul Speed =
375 × 301
46.757×70
× 1.61= 55.52 km/hr
Page 10
11. Refisa Jiru
HAwassa University IOT
Haul time =
Haul distance
Haul speed
=
4
55.52
×60= 4 .3227 min
For 9 buckets
Weight fully loaded = 22,260 + 21,312 =
43,572
909.09
=47.929 lton ( for haul speed)
Haul Speed(km/hr)=
375×Engine horse power(hp)
Total Resisance×F ully loaded weight
×1.6093
Haul Speed =
301× 375
47.929 ×70
= 33.643miles/hr ⇒54.166 km/hr
Haul time =
Haul distance
Haul Speed
=
4
54 .166
x 60 = 4.4308 min
Step 4: Return time
The effect of gravitational force in aiding movement of a vehicle down a slope
Roll Resistance = 1.5 x 20 lb/ton = 30 lb/ton,
Grade Assistance = -2 x 20 lb/ton = -40 lb/ton
Total Resistance = -10 lb/ton
Truck empty weight = 22,260 kg =
22,260
909.09
=24.486ton ( for return speed)
Return Speed=
Engine horse power
Total Resisance×Empty vehicle weight
=
375×301
−10×24.486
×1.61=−741.87km/hr
The speed will be in negative so use maximum road legal speed is 90km/hr
Return time =
Return Distance
Return Speed
=
4
90
x 60= 2.667 min
Page 11
12. Refisa Jiru
HAwassa University IOT
Step 5: Dump time
Rear Dump Trucks must be spotted before dumping
Total dump time averages about 2 minutes. Use 2.0 minutes
Step 6: Truck cycle time
8 buckets 9 buckets
Load time (min.) 3.067 3.450
Haul time (min.) 4.3227 4.438
Dump time (min.) 2.000 2.000
Return time (min.) 2.667 2.667
Truck cycle time(min.) 12.056 12.548
Step 7: Number of trucks:
8 buckets
Number of trucks =
Truck cycle time
Load Cycle time
=
12.0567
3.06
= 3.931
9 buckets
Number of trucks =
Truck cycle time
Load Cycle time
=
12.5478
3.45
= 3.63704
Must be an integer number therefore 4
Step 8: Production:
Page 12
13. Refisa Jiru
HAwassa University IOT
For 8 buckets
3 trucks
Production =
Load volume x no. of trucks x 60
truck cycle time
=
13.68×3×60
12.056
=204.235 LCM/hr
4 trucks
Production =
Load volume x 60
Load time
=
13.68×60
3.067
=267.62LCM /hr
5trucks
Production =
Load volume x 60
Load time
=
13.68×60
3.067
=267.62LCM /hr
For 9 buckets
3 trucks
Production =
Load volume x no. of trucks x 60
truck cycle time
=
14.40×3×60
12.548
=206.57LCM /hr
4 trucks
Production =
Load volume x 60
Load time
=
14.40×60
3.45
=250.43LCM /hr
5 trucks
Production =
Load volume x 60
Load time
=
14.40×60
3.45
=250.43LCM /hr
Truck available
8 buckets
(LCM/hr)
9 buckets (LCM/hr)
3 204.235 206.570
Page 13
14. Refisa Jiru
HAwassa University IOT
4 267.620 250.430
5 267.620 250.430
Choose maximum production = 267.62 LCM/hr
Actual production = 267.62 x 0.83 x 0.95 = 211.02 LCM/hr
At Number of trucks are 4 trucks
Number of buckets per truck = 8 buckets
Haul speed (1st
road segment) = 55.52 km/hr
Return speed (last road segment) = 90 km/hr
Example 4
Given: its Caterpillar 772 Articulated Truck
Truck Net power = 535hp
Truck Net empty weight = 35,454 kg
Truck Payload = 45,000 kg
Truck heaped capacity = 31.3 m3
Excavator heaped capacity = 2.8 m3
Haul road type: 1 Km smooth roadway 1% grade (rolling resistance (RR) = 1.5%) + 2 Km
dirt roadway 1% grade (RR = 4%) + 2 Km sand -4% grade (RR = 10%) (Table 4.5), with 90
Km/hr legal speed. Haul material type = dry gravel (loose material weight = 1690 kg/m3
,
bucket fill factor = 95%, excavator cycle time = 23 seconds and load factor = 0.89)
(Table 4.4). Job efficiency = 50 minutes per hour = 0.83. Operators are good = 0.95
Estimate the earthmoving productivity
Page 14
15. Refisa Jiru
HAwassa University IOT
Solution
Step1: BUCKET LOADS
Number of bucket loads required to fill the truck
Balanced number of buckets =
Truck heaped capacity
Excavator heaped capacity×bucket fill factor
Number of buckets=
31.3
2.8 x 0.95
= 11.77
The actual number of buckets must be an integer numbers we have 11 or 12
buckets
Step 2: Load time
11 buckets
Load time =
Number of buckets×excavator cycle time
60
=
11 x 23
60
= 4.217 min
Load volume =Number of buckets×Excavator heaped capacity×Bucket fill factor
= 11 x 2.8 x 0.95 = 29.26 m
3
<(Truck heaped capacity =31.3 m
3
) OK !
Load weight =Load volume×loose material weight =29.26 x 1690
= 49,449.4 kg>payload NOT OK !
Use 10 buckets
Load time =
Number of buckets×excavator cycle time
60
=
10 x 23
60
= 3.83 min
Load volume =Number of buckets×Excavator heaped capacity×Bucket fill factor
Page 15
16. Refisa Jiru
HAwassa University IOT
= 10 x 2.8 x 0.95 = 26.6 m3
<31.3 m3
Ok
Load weight =Load volume×loose material weight =26.6 x 1690
= 44,954 kg<payload OK !
Step 3: Haul time
10 buckets
Road 1:
RR (1.5%) = 1.5 x 20 lb/ton = 30 lb/ton,
GR (+1%) = 1 x 20 lb/ton = +20 lb/ton
TR = 50 lb/ton
Engine horse power = 535hp
Weight fully loaded = (35454 + 44954) / 909.09 = 88.45lton
Speed = ((375 x 535)/(88.45 (50))) x 1.61= 73.037 km/hr
Haul time = (1 x 60) / 73.037 = 0.82 min
Road 2:
RR = 4% = 4 x 20 lb/ton = 80 lb/ton,
GR = +1% = 1 x 20 lb/ton = +20 lb/ton
TR = 100 lb/ton
Engine horsepower = 535hp,
Weight fully loaded = (35454 + 44954) / 909.09 = 88.45 lton
Page 16
17. Refisa Jiru
HAwassa University IOT
Speed = ((375 x 535) / (88.45 (100))) x 1.61 = 36.52 km/hr
Haul time = (2 x 60) / 36.52 = 3.286 min
Road 3:
RR = 10% = 10 x 20 lb/ton = 200 lb/ton,
GR = -4% = -4 x 20 lb/ton = -80 lb/ton
TR = 120 lb/ton
Engine horsepower = 535hp,
Weight fully loaded = (35454 + 44954) / 909.09 = 88.45 lton
Speed = ((375 x 535) / (88.45 (120))) x 1.61 = 30.43 km/hr
Haul time = (2 x 60) / 30.43 = 3.943 min
Total haul time = 0.82 + 3.286 + 3.943 = 8.049 min.
Step 4: Return time:
Road 3:
RR ( 1.5% = 1.5 x 20 lb/ton = 30 lb/ton,
GR (-1%) = -1 x 20 lb/ton = -20 lb/ton
TR = 10 lb/ton
Engine horsepower = 535hp,
Weight empty = (35454) / 909.09 = 39 lton
Page 17
18. Refisa Jiru
HAwassa University IOT
Speed = ((375 x 535) / (39 (10))) x 1.61 -ve speed use 90 km/hr
Return time = (1 x 60) / 90 = 0.67 min
Road 2:
RR (4%) = 4 x 20 lb/ton = 80 lb/ton,
GR ( -1% )= -1 x 20 lb/ton = -20 lb/ton
TR = 60 lb/ton
Engine horsepower = 535hp,
Weight empty = (35454) / 909.09 = 39 lton
Speed = ((375 x 535) / (39 (60))) x 1.61 > legal speed use 90 km/hr
Return time = (2 x 60) / 90 = 1.33 min
Road 1:
RR ( 10% )= 10 x 20 lb/ton = 200 lb/ton,
GR ( +4% )= 4 x 20 lb/ton = 80 lb/ton
TR = 280lb/ton
Engine horsepower = 535hp,
Weight empty = (35454) / 909.09 = 39 lton
Speed = ((375 x 535) / (39 (280))) x 1.61 = 29.58 km/hr
Return time = (2 x 60) / 29.58 = 4.057 min
Page 18
19. Refisa Jiru
HAwassa University IOT
Total return time = 0.67 + 1.33 + 4.057 = 6.057 min
Step 5: Dump time = 2 min
Step 6: Truck cycle time:
10 buckets
Load time (min.) 3.83
Haul time (min.) 8.049
Dump time (min.) 2
Return time (min.) 6.057
Truck cycle time (min.) 19.936
Step 7: Number of trucks:
10 buckets
No. of trucks = 19.936 / 3.83 = 5.2
Step 8: Production:
10 buckets
5 trucks 6 trucks
Load volume x no. of trucks x 60 / truck cycle
time
Load volume x 60 / load time
400.28 LCM/hr 416.71 LCM/hr
Choose maximum production = 416.71 LCM/hr
Actual production = 416.71 x 0.83 x 0.95 = 328.57 LCM/hr At:
Number of trucks = 6 trucks
Page 19
20. Refisa Jiru
HAwassa University IOT
Number of buckets per truck = 10 buckets
Haul speed (1st
road segment) = 73.037 km/hr
Return speed (last road segment) = 90 km/hr
Page 20
