EQUIPMENT PRODUCTIVITY

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1 EQUIPMENT PRODUCTIVITY
CHAPTER 15 EQUIPMENT PRODUCTIVITY

2 15.1 Need for Heavy Equipment
Heavy construction equipment is needed to amplify human reach, speed, and capacity. The manager’s goal is to select the equipment combination that yields the maximum production at the best or most reasonable price. The cost and the rate of production combine to yield the cost per unit of production. Cost of a particular fleet of haulers and loaders is $500 per hour Production rate is 750 cubic yards per hour Unit price is $0.66 per cubic yard (= $500 / 750yard3) Equipment can be classified by the nature of its work: It typically Transforms and Transports material. Construction equipment can be divided into two major categories: Productive equipment and Support equipment. In heavy construction, large quantities of fluid or semifluid materials such as earth, concrete, and asphalt are handled and placed, leading to the use of machines. Heavy construction operations are referred to as being equipment intensive.

3 15.2 Productivity Concepts
The rate of output (capacity) for equipment: 작업능력 = 작업량  1/작업시간 Cyclic capacity : the number of units produced per cycle: 작업량 (한번에 처리하는 양) Cyclic rate : speed of an equipment piece : 1/작업시간 (한번 처리하는데 걸리는 시간의 역수) Different weight-to-volume ratio: three types of measure (a) bank cubic yards, (b) loose cubic yards, (c) compacted cubic yards. Bank Loose Compact 같은 무게에서 부피가 30% 증가함 밀도 : 보통 (2,000 / 1 = 2,000) 밀도 : 낮음 (2,000 / 1.3 = 1,538) 밀도 : 높음 (2,000 / 0.9 = 2,222) FIGURE 15.1 Volume relationships

4 15.2 Productivity Concepts
The relationship between the bulk or loose volume and the bank volume is defined by the percent swell. (예제: percent swell 30%  같은 중량에 대해 부피가 30% 증가함) The shrinkage factor relates the volume of the compacted material to the volume of the bank material. In the example, the shrinkage factor is 10% (같은 중량에 대해 부피가 10% 감소함) Load factor = 1  Percent swell = 0% 1,538 0.77 2,000

5 15.2 Productivity Concepts

6 15.2 Productivity Concepts
A front-end loader (200 bank yd3/hour) It loads a fleet of four trucks (capacity 18 loose yd3 each), which haul the earth to a fill where it is compacted with a shrinkage factor of 10% Each truck has a total cycle time of 15 min. The earth has a percent swell of 20%. The job requires a volume of 18,000 compacted yd3. How many hours will be required to excavate and haul the material to the fill? Then the loader productivity (given 20% swell) is: The truck fleet production is: 병목장비: 생산용량을 제약하는 장비

7 15.2 Productivity Concepts
shrinkage factor : 10% percent swell of 20%

8 15.3 Cycle Time and Power Requirements
In the case of heavy equipment, the speed of the machine is governed by : (a) the power required (b) the power available (c) the usable portion of the power available The power required is related to the rolling resistance (RR) & grade (경사) resistance (GR) 노면 상태가 무를수록 더 큰 힘이 필요함

9 15.3 Cycle Time and Power Requirements
FIGURE 15.2 Factors influencing rolling resistance 오르막에서 더 큰 힘이 필요함 FIGURE 15.3 Grade resistance: (a) negative (resisting) force and (b) positive (aiding) force

10 15.3 Cycle Time and Power Requirements
2 inches The rule of thumb states that the RR is approximately 40 lbs/ton plus 30 lbs/ton for each inch of penetration of the surface under wheeled traffic. If the estimated deflection is 2 inches and the weight on the wheels of a hauler is 70 tons, we can calculate the approximate RR as: RR = [40 + 2(30)] lb/ton × 70 tons = 7,000 lb A slope rises 6 feet in 100 feet of horizontal distance, the percent grade is 6. Percent grade is used to calculate the GR using the following relationship: GR = percent grade × 20 lb/(ton × % grade) × weight on wheels(tons) GR = 6% grade × 20 lb/(ton × % grade) × 70 tons = 8,400 lb Power required = RR + GR = 7,000 lb + 8,400 lb = 15,400 lb  오르막 Power required = RR − GR = 7,000 lb − 8,400 lb = − 1,400 lb  내리막

11 15.3 Cycle Time and Power Requirements
경사 저항 회전 저항

12 15.4 Power Available The power available is controlled by the engine size of the equipment and the drive train (동력 전달축) 정상출력 최대출력 10,000 lb의 출력이 필요할 때 최대 가능 속도는?

13 15.4 Power Available +70,000 lb  2,500 lb, 41 mph  5,500 lb, 19 mph Loaded 126,000 lb  Empty 56,000 lb 35-ton, off-highway truck 경로 선택 문제 경로 #1: 4.6 miles (24,344 feet) RR = 50lb/ton 경로 #2: (14,784) 2.8 miles, RR = 90lb/ton 소요 구동력 4.5%  5,500 lb 2.5% 2,500 lb  19 mph  41 mph  FIGURE 15.5 Gear requirements chart for 35-ton off-highway truck (Caterpillar Tractor Co.)

14 15.4 Power Available To break the 24,344 into two segments: 9.31분 (a) 16,000 ft : 7.2분 (b) the 8,344 ft remaining : 2.31분 = 8,344 ft / (4.1 mi/h  5,280 ft/mi)  60 min 운송 시간 추정 경로 #1: 9.5분 경로 #2: 9.31분 4.5% 9.5분 2.5% 7.2분 14,784 feet FIGURE 15.6 Travel time: (a) empty and (b) loaded (Caterpillar Tractor Co.)

15 = (coefficient of traction) × (weight on drivers)
15.5 Usable Power Typical values for rubber-tired and tracked vehicles on an assortment of surface materials. 견인력 계수 Usable pounds pull = (coefficient of traction) × (weight on drivers) 물 뿌리기

16 15.5 Usable Power Usable pounds pull = (coefficient of traction) × (weight on drivers) Driver Driver Driver FIGURE 15.7 Determination of driver weights

17 15.5 Usable Power High RR of the sand (RR = 400 lb/ton), 26-ton loads
The weight distribution characteristics of the 30-yard tractor-scraper are as follows: The difference between the total weight empty and loaded is 72,000 pounds, or 36 tons. The loaded weight with 26-ton loads would be 127,800 pounds. 차이 = 147,800 – 75,800 = 72,000 pound = 36 ton (최대 적재량) 75,800 pound + 26 ton  2000 pound/ton = 127,800 pound 필요 출력 마른 모래 제공 출력 63.9 ton 젖은 모래 제공 출력 현장에 물을 뿌림으로써 필요 출력을 제공할 수 있음

18 FIGURE 15.8 Impact of usable power constraints
고산지대의 경우, 사소 부족으로 출력 감소 현상 발생 If a tractor is operating at 5,000 feet above sea level, its power will be decreased by 6%. Usable pound pull 사용 가능 범위 FIGURE 15.8 Impact of usable power constraints

19 FIGURE 15.9 Scraper-pusher dual-cycle model
15.6 Equipment Balance Two types of equipment work together: a 30 cubic yard scraper & a 385-horsepower pusher dozer 빈차 (1.2분) 싣고 내리고 만차 (1.4분) FIGURE 15.9 Scraper-pusher dual-cycle model

20 FIGURE 15.10 Travel time nomographs (Caterpillar Tractor Co.)
15.6 Equipment Balance 2% Assume that in this case the 30-cubic yard tractor Scraper is carrying rated capacity and operating on a 3,000-foot level haul road where the RR developed by the road surface is 40 lb/ton. Effective grade = (RR)/(20 lb/ton/% grade) = (40 lb/ton)/(20 lb/ton/% grade) = 2% Pusher Load time = 0.70 Boost time = 0.15 Transfer time = 0.10 Return time = 0.28 Total = 1.23 minutes (1.4분) 2% (1.2분) FIGURE Travel time nomographs (Caterpillar Tractor Co.)

21 FIGURE 15.11 Scraper-pusher cycle times
15.6 Equipment Balance 흙을 운반하는 사이클 = 4.05분 흙을 싣는 사이클 = 1.23분 FIGURE Scraper-pusher cycle times

22 FIGURE 15.12 Productivity plot
15.6 Equipment Balance Scraper 5대 생산성 손실 = 1,463.4 / 444.4 Scraper 3대 생산성 손실 Scraper 2대 Scraper 1대 FIGURE Productivity plot

23 FIGURE 15.13 Flow of equipment operations calculation
15.6 Equipment Balance FIGURE Flow of equipment operations calculation

24 15.7 Random Work Task Durations
The influence of random durations on the scraper fleet production: the reduction in production caused by the addition of random variation of cycle activity times. 생산성 손실 FIGURE Productivity curve to include the effect of random cycle times

25 15.7 Random Work Task Durations
3대의 Scraper가 동일 간격으로 움직이고 있어야하나, Random한 요소로 인해 간격이 달라지면 Pusher에 Idle Time이 발생함  생산성 저하 발생 Scraper 1 Scraper 2 Pusher Scraper 3 FIGURE Comparison of haul unit cycles

26 15.7 Random Work Task Durations
Coefficient Of Variation (COV) : 변동계수 또는 변이계수 = 표준편차 / 평균 (평균 대비 표준편차의 크기) Lognormal 분포 가정 COV 변동성이 0.1일 때, 10% 생산성 저하 FIGURE Plot of cycle time coefficient of variation