Find the swell of a soil that weighs 1661 kg/m3 in its natural state and 1186 kg/m3 after excavation.
40%
Find the shrinkage of a soil that weighs 1661 kg/m3 in its natural state and 2077 kg/m3 after compaction.
20%
A soil weighs 1163 kg/LCM, 1661 kg/BCM, and 2077 kg/CCM. Find the load factor and shrinkage factor for the soil.
Load factor = 0.70 Shrinkage factor = 0.8
A soil weighs 1163 kg/LCM, 1661 kg/BCM, and 2077 kg/CCM. How many bank cubic meters (BCM) and compacted cubic meters (CCM) are contained in 593,300 LCM of this soil?
bank vol. = 415,310 BCM; compacted vol. = 332,248 CCM
Find the base width and height of a triangular spoil bank containing 76.5 BCM if the pile length is 9.14 m, the soilโs angle of repose is 37ยฐ, and its swell is 25%.
B = 7.45 m; H = 2.80 m
Find the base diameter and height of a conical spoil pile that will contain 76.5 BCM of excavation if the soilโs angle of repose is 32ยฐ and its swell is 12%.
D = 10.16 m; H = 3.17 m
Find the volume (bank measure) of excavation required for a trench 0.92 m wide, 1.83 m deep, and 152 m long. Assume that the trench sides will be approximately vertical.
๐๐๐ ๐๐๐
Estimate the actual bucket load in bank cubic meters for a loader bucket whose heaped capacity is 3.82 m3. The soilโs bucket fill factor is 0.90 and its load factor is 0.80.
๐.๐๐ ๐๐๐
Find the expected production in loose cubic meters (LCM) per hour of a small hydraulic excavator. Heaped bucket capacity is 0.57 m3. The material is sand and gravel with a bucket fill factor of 0.95. Job efficiency is 50 min/h. Average depth of cut is 4.3 m. Maximum depth of cut is 6.1 m and average swing is 90. Hint: Cycle output = 250 cycles/60 min Swing-depth factor= 1.00
๐๐๐ ๐๐๐/๐ก๐ซ
Find the expected production in loose cubic meters (LCM) per hour of a 2.3-m3 hydraulic shovel equipped with a front-dump bucket. The material is common earth with a bucket fill factor of 1.0. The average angle of swing is 75ยฐ and job efficiency is 0.80. Hint: Standard cycles = 150/60 min Swing factor = 1.05
๐๐๐ ๐๐๐
Determine the expected dragline production in loose cubic meters (LCM) per hour based on the following information: Dragline size = 1.53 m3 Swing angle = 120ยฐ Average depth of cut = 2.4 m Material = common earth Job efficiency = 50 min/h Soil swell op = 25% Hint: Ideal output = 176 BCM/h Swing-depth factor = 0.90
๐๐๐ ๐๐๐/๐ก๐ซ
Estimate the production in loose cubic meters per hour for a medium-weight clamshell excavating loose earth. Heaped bucket capacity is 0.75 m3. The soil is common earth with a bucket fill factor of 0.95. Estimated cycle time is 40 s. Job efficiency is estimated at 50 min/h.
๐๐ ๐๐๐/๐ก๐ซ
A wheel tractor-scraper weighing 91 t is being operated on a haul road with a tire penetration of 5 cm. What is the total resistance (kg) and effective grade when the scraper is ascending a slope of 5%? Hint: RRF = 20 + (6 x cm penetration) GRF = 10 x grade (%) Effective grade = Grade (%) + RRF/10
๐๐จ๐ญ๐๐ฅ ๐๐๐ฌ๐ข๐ฌ๐ญ๐๐ง๐๐ = ๐๐๐๐ ๐ค๐ ; ๐๐๐๐๐๐ญ๐ข๐ฏ๐ ๐๐ซ๐๐๐ = ๐๐%
A wheel tractor-scraper weighing 91 t is being operated on a haul road with a tire penetration of 5 cm. What is the total resistance (kg) and effective grade when the scraper is descending a slope of 5%? Hint: RRF = 20 + (6 x cm penetration) GRF = 10 x grade (%) Effective grade = Grade (%) + RRF/10
๐๐จ๐ญ๐๐ฅ ๐๐๐ฌ๐ข๐ฌ๐ญ๐๐ง๐๐ = ๐ ๐ค๐ ; ๐๐๐๐๐๐ญ๐ข๐ฏ๐ ๐๐ซ๐๐๐ = ๐%
A four-wheel-drive tractor weighs 20,000 kg and produces a maximum rimpull of 18,160 kg at sea level. The tractor is being operated at an altitude of 3050 m on wet earth. A pull of 10,000 kg is required to move the tractor and its load. Can the tractor perform under these conditions? Hint: Derating factor = [Altitude (m) - 915]/102 Coefficient of traction = 0.45
๐๐๐๐๐ฎ๐ฌ๐ ๐ญ๐ก๐ ๐ฆ๐๐ฑ๐ข๐ฆ๐ฎ๐ฆ ๐ฉ๐ฎ๐ฅ๐ฅ ๐๐ฌ ๐ฅ๐ข๐ฆ๐ข๐ญ๐๐ ๐๐ฒ ๐ญ๐ซ๐๐๐ญ๐ข๐จ๐ง ๐ข๐ฌ ๐ฅ๐๐ฌ๐ฌ ๐ญ๐ก๐๐ง ๐ญ๐ก๐ ๐ซ๐๐ช๐ฎ๐ข๐ซ๐๐ ๐ฉ๐ฎ๐ฅ๐ฅ, ๐ญ๐ก๐ ๐ญ๐ซ๐๐๐ญ๐จ๐ซ ๐๐๐ง๐ง๐จ๐ญ ๐ฉ๐๐ซ๐๐จ๐ซ๐ฆ ๐ฎ๐ง๐๐๐ซ ๐ญ๐ก๐๐ฌ๐ ๐๐จ๐ง๐๐ข๐ญ๐ข๐จ๐ง๐ฌ.
A crawler tractor weighing 36 t is towing a rubber-tired scraper weighing 45.5 t up a grade of 4%. What is the total resistance (kg) of the combination if the rolling resistance factor is 50 kg/t?
๐๐๐๐ ๐ค๐
A power-shift crawler tractor has a rated blade capacity of 7.65 LCM. The dozer is excavating loose common earth and pushing it a distance of 61 m. Maximum reverse speed in third range is 8 km/h. Estimate the production of the dozer if job efficiency is 50 min/h. Hint: Fixed time = 0.05 min Dozing speed = 4.0 km/h
๐๐๐ ๐๐๐/๐ก๐ซ
Estimate the hourly production in loose volume (LCM) of a 2.68-m3 wheel loader excavating sand and gravel (average material) from a pit and moving it to a stockpile. The average haul distance is 61 m, the effective grade is 6%, the bucket fill factor is 1.00, and job efficiency is 50 min/h. Hint: Basic cycle time = 0.50 min Travel time = 0.30 min
๐๐๐ ๐๐๐/๐ก๐ซ
Estimate the production of a single-engine two-axle tractor scraper based on the following information: Maximum heaped volume = 24 LCM Maximum payload = 34,020 kg Material: Sandy clay, 1898 kg/BCM, 1571 kg/LCM, rolling resistance 50 kg/t Job efficiency = 50 min/h Operating conditions = average Single pusher Haul route: Section 1. Level loading area Section 2. Down a 4% grade, 610 m Section 3. Level dumping area Section 4. Up a 4% grade, 610 m Section 5. Level turnaround, 183 m Hint: Travel time: Section 2 = 1.02 min Section 4 = 1.60 min Section 5 = 0.45 min Fixed cycle: Load spot = 0.3 min Load = 0.6 min Maneuver and dump = 0.7 min
๐๐๐ ๐๐๐/๐ก๐ซ
The estimated cycle time for a wheel scraper is 6.5 min. Calculate the number of pushers required to serve a fleet of nine scrapers using single pushers. Determine the result for both backtrack and chain-loading methods. Hint: Typical pusher cycle time (min) Back-track (single pusher) = 1.5 Chain or shuttle (single pusher) = 1.0
๐๐๐๐ค-๐ญ๐ซ๐๐๐ค = ๐; ๐๐ก๐๐ข๐ง = ๐
Find the expected production of the scraper fleet of the previous problem if only one pusher is available and the chain-loading method is used. Expected production of a single scraper assuming adequate pusher support is 173 BCM/h. Hint: Number of pushers required to fully serve fleet = 1.4
๐๐๐ ๐๐๐/๐ก๐ซ
Given the following information on a shovel/truck operation, calculate the number of trucks theoretically required and the production of this combination. Shovel production at 100% efficiency = 283 BCM/h Job efficiency = 0.75 Truck capacity = 15.3 BCM Truck cycle time, excluding loading = 0.5 h
๐๐จ. ๐จ๐ ๐ญ๐ซ๐ฎ๐๐ค๐ฌ ๐ซ๐๐ช๐ฎ๐ข๐ซ๐๐ = ๐๐; ๐๐ฑ๐ฉ๐๐๐ญ๐๐ ๐ฉ๐ซ๐จ๐๐ฎ๐๐ญ๐ข๐จ๐ง = ๐๐๐ ๐๐๐/๐ก
Given the following information on a shovel/truck operation, calculate the expected production if two trucks are removed from the fleet. Shovel production at 100% efficiency = 283 BCM/h Job efficiency = 0.75 Truck capacity = 15.3 BCM Truck cycle time, excluding loading = 0.5 h
๐๐๐ ๐๐๐/๐ก
24.1 km of gravel road require reshaping and leveling. You estimate that six passes of a motor grader will be required. Based on operator skill, machine characteristics, and job conditions, you estimate two passes at 6.4 km/h, two passes at 8.0 km/h, and two passes at 9.7 km/h. If job efficiency is 0.80, how many grader hours will be required for this job?
๐๐.๐ ๐ก
Trial blasting indicates that a rectangular pattern of drilling using 7.6-cm holes spaced on 2.75-m centers and 6.1 m deep will produce a satisfactory rock break with a particular explosive loading. The effective hole depth resulting from the blast is 5.5 m. Determine the rock volume produced per meter of drilling.
๐.๐ ๐ฆยณ/๐ฆ
A jaw crusher is producing 227 t/h of crushed gravel and discharging it onto a three-screen deck. The top screen in the deck is a 38-mm screen. The gradation of crusher output shows 100% passing 76 mm, 92% passing 38 mm, and 80% passing 19 mm. Material weight is 1842 kg/m3. Find the minimum size of the 38-mm screen to be used. Check both total screen load and screen passing capacity. Hint: Basic capacity: Total feed = 62/t/h/m2 Passing screen = 34 t/h/m2 Deck position factor (top) = 1.0 Halfsize factor (80%) = 1.8 Oversize factor (8%): Total feed = 0.96 Passing screen = 1.04 Weight factor = 1.15
๐.๐ ๐ฆยฒ
Calculate the maximum hourly production of an asphalt plant based on the data in the following list. Mix composition: Asphalt = 6% Aggregate composition: Coarse A = 42% Coarse B = 35% Sand = 18% Mineral filler = 5% Aggregate moisture = 8% Dryer capacity at 8% moisture removal = 110 ton/h Hint: Plant capacity = [Dry capacity ร 10^4]/[(100 - asphalt%)(100 - fines%)]
๐๐๐ ๐ญ๐จ๐ง/๐ก
[NOV 2022] Calculate the volume of plastic concrete that will be produced by the mix design given in the table.
๐.๐๐ ๐ฆยณ
Determine the actual weight of each component to be added if the sand contains 5% excess moisture and the gravel contains 2% excess moisture.
๐๐๐ญ๐๐ซ = ๐๐ ๐ค๐ ; ๐๐๐ง๐ = ๐๐๐ ๐ค๐ ; ๐๐ซ๐๐ฏ๐๐ฅ = ๐๐๐ ๐ค๐
Refer to the previous problem. Determine the weight of each component required to make a three-bag mix and the mix volume.
๐๐๐ฆ๐๐ง๐ญ = ๐๐๐.๐ ๐ค๐ ; ๐๐๐ง๐ = ๐๐๐ ๐ค๐ ; ๐๐ซ๐๐ฏ๐๐ฅ = ๐๐๐ ๐ค๐ ; ๐๐๐ญ๐๐ซ = ๐๐ ๐ค๐ ; ๐๐ข๐ฑ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ = ๐.๐๐ ๐ฆ๐