LOOKSFAM

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Find the swell of a soil that weighs 1661 kg/m3 in its natural state and 1186 kg/m3 after excavation.

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1

Find the swell of a soil that weighs 1661 kg/m3 in its natural state and 1186 kg/m3 after excavation.

40%

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2

Find the shrinkage of a soil that weighs 1661 kg/m3 in its natural state and 2077 kg/m3 after compaction.

20%

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3

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

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4

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

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5

Find the base width and height of a triangular spoil bank containing 76.5 BCM if the pile length is 9.14 m, the soilโ€™s angle of repose is 37ยฐ, and its swell is 25%.

B = 7.45 m; H = 2.80 m

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6

Find the base diameter and height of a conical spoil pile that will contain 76.5 BCM of excavation if the soilโ€™s angle of repose is 32ยฐ and its swell is 12%.

D = 10.16 m; H = 3.17 m

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7

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.

๐Ÿ๐Ÿ“๐Ÿ“ ๐๐‚๐Œ

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8

Estimate the actual bucket load in bank cubic meters for a loader bucket whose heaped capacity is 3.82 m3. The soilโ€™s bucket fill factor is 0.90 and its load factor is 0.80.

๐Ÿ.๐Ÿ•๐Ÿ“ ๐๐‚๐Œ

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9

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

๐Ÿ๐Ÿ๐Ÿ‘ ๐‹๐‚๐Œ/๐ก๐ซ

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10

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

๐Ÿ๐Ÿ—๐ŸŽ ๐‹๐‚๐Œ

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11

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

๐Ÿ๐Ÿ”๐Ÿ“ ๐‹๐‚๐Œ/๐ก๐ซ

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12

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.

๐Ÿ“๐Ÿ‘ ๐‹๐‚๐Œ/๐ก๐ซ

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13

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

๐“๐จ๐ญ๐š๐ฅ ๐‘๐ž๐ฌ๐ข๐ฌ๐ญ๐š๐ง๐œ๐ž = ๐Ÿ—๐Ÿ๐ŸŽ๐ŸŽ ๐ค๐ ; ๐„๐Ÿ๐Ÿ๐ž๐œ๐ญ๐ข๐ฏ๐ž ๐†๐ซ๐š๐๐ž = ๐Ÿ๐ŸŽ%

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14

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

๐“๐จ๐ญ๐š๐ฅ ๐‘๐ž๐ฌ๐ข๐ฌ๐ญ๐š๐ง๐œ๐ž = ๐ŸŽ ๐ค๐ ; ๐„๐Ÿ๐Ÿ๐ž๐œ๐ญ๐ข๐ฏ๐ž ๐†๐ซ๐š๐๐ž = ๐ŸŽ%

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15

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

๐๐ž๐œ๐š๐ฎ๐ฌ๐ž ๐ญ๐ก๐ž ๐ฆ๐š๐ฑ๐ข๐ฆ๐ฎ๐ฆ ๐ฉ๐ฎ๐ฅ๐ฅ ๐š๐ฌ ๐ฅ๐ข๐ฆ๐ข๐ญ๐ž๐ ๐›๐ฒ ๐ญ๐ซ๐š๐œ๐ญ๐ข๐จ๐ง ๐ข๐ฌ ๐ฅ๐ž๐ฌ๐ฌ ๐ญ๐ก๐š๐ง ๐ญ๐ก๐ž ๐ซ๐ž๐ช๐ฎ๐ข๐ซ๐ž๐ ๐ฉ๐ฎ๐ฅ๐ฅ, ๐ญ๐ก๐ž ๐ญ๐ซ๐š๐œ๐ญ๐จ๐ซ ๐œ๐š๐ง๐ง๐จ๐ญ ๐ฉ๐ž๐ซ๐Ÿ๐จ๐ซ๐ฆ ๐ฎ๐ง๐๐ž๐ซ ๐ญ๐ก๐ž๐ฌ๐ž ๐œ๐จ๐ง๐๐ข๐ญ๐ข๐จ๐ง๐ฌ.

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16

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?

๐Ÿ“๐Ÿ“๐Ÿ‘๐Ÿ“ ๐ค๐ 

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17

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

๐Ÿ๐Ÿ•๐Ÿ ๐‹๐‚๐Œ/๐ก๐ซ

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18

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

๐Ÿ๐Ÿ”๐Ÿ– ๐‹๐‚๐Œ/๐ก๐ซ

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19

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

๐Ÿ๐Ÿ—๐Ÿ ๐๐‚๐Œ/๐ก๐ซ

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20

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

๐›๐š๐œ๐ค-๐ญ๐ซ๐š๐œ๐ค = ๐Ÿ‘; ๐œ๐ก๐š๐ข๐ง = ๐Ÿ

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21

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

๐Ÿ๐Ÿ๐Ÿ ๐๐‚๐Œ/๐ก๐ซ

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22

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

๐๐จ. ๐จ๐Ÿ ๐ญ๐ซ๐ฎ๐œ๐ค๐ฌ ๐ซ๐ž๐ช๐ฎ๐ข๐ซ๐ž๐ = ๐Ÿ๐Ÿ; ๐„๐ฑ๐ฉ๐ž๐œ๐ญ๐ž๐ ๐ฉ๐ซ๐จ๐๐ฎ๐œ๐ญ๐ข๐จ๐ง = ๐Ÿ๐Ÿ๐Ÿ ๐๐‚๐Œ/๐ก

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23

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

๐Ÿ๐Ÿ–๐Ÿ” ๐๐‚๐Œ/๐ก

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24

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?

๐Ÿ๐Ÿ‘.๐Ÿ ๐ก

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25

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Ÿ”.๐Ÿ– ๐ฆยณ/๐ฆ

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26

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Ÿ.๐Ÿ— ๐ฆยฒ

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27

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%)]

๐Ÿ๐Ÿ๐Ÿ‘ ๐ญ๐จ๐ง/๐ก

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28

[NOV 2022] Calculate the volume of plastic concrete that will be produced by the mix design given in the table.

๐ŸŽ.๐Ÿ“๐Ÿ ๐ฆยณ

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29

Determine the actual weight of each component to be added if the sand contains 5% excess moisture and the gravel contains 2% excess moisture.

๐–๐š๐ญ๐ž๐ซ = ๐Ÿ”๐Ÿ‘ ๐ค๐ ; ๐’๐š๐ง๐ = ๐Ÿ’๐Ÿ’๐Ÿ• ๐ค๐ ; ๐†๐ซ๐š๐ฏ๐ž๐ฅ = ๐Ÿ“๐Ÿ”๐ŸŽ ๐ค๐ 

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30

Refer to the previous problem. Determine the weight of each component required to make a three-bag mix and the mix volume.

๐‚๐ž๐ฆ๐ž๐ง๐ญ = ๐Ÿ๐Ÿ๐Ÿ•.๐Ÿ– ๐ค๐ ; ๐’๐š๐ง๐ = ๐Ÿ‘๐Ÿ•๐ŸŽ ๐ค๐ ; ๐†๐ซ๐š๐ฏ๐ž๐ฅ = ๐Ÿ’๐Ÿ”๐Ÿ’ ๐ค๐ ; ๐–๐š๐ญ๐ž๐ซ = ๐Ÿ“๐Ÿ ๐ค๐ ; ๐Œ๐ข๐ฑ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ž = ๐ŸŽ.๐Ÿ’๐Ÿ ๐ฆ๐Ÿ‘

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