Homework 3 - Excavation Shrink swell factor PDF

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Excavation Shrink swell factor...

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Earthmoving I + II Due by: Oct. 6, by midnight (Submit it to Canvas) Show your work in details; answers only will not get full credits. 1. In a school project, 10,000 compacted cubic yards are required for the fill. Tests on the fill material indicate a bank wet unit weight of 102 pounds per cubic foot and a bank dry unit weight of 97 pounds per cubic foot. In addition, its loose wet unit density is 88 pounds per cubic foot. When compacted to the desired density, the wet soil weighs 115 pounds per cubic foot. (1) What is the percent swell for the fill material? (2) What is the load factor for the fill material? (3) What is the percent shrinkage for the fill material? (4) What is the shrinkage factor for the fill material? (5) How many bank cubic yards must be excavated and loaded on trucks to construct the fill? (6) How many loose cubic yards must be hauled in trucks from the borrow site to construct the fill? (7) What is the moisture content of the fill material (in the bank state)? (8) How many gallons of water should be added to the fill material delivered to the project site to increase the soil moisture to 8%? (1 gallon water= 8.34 lb)

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Units: CF: Cubic Foot PCF: Pounds per cubic foot CY: Cubic yard gal: Gallon

1. Required: 10,000CCY bank dry unit weight: 97 PCF bank wet unit weight: 102 PCF 𝐷𝐵 loose wet unit density: 88 PCF 𝐷𝐿 compacted wet soil density: 115 PCF 𝐷𝐶 Assumption: The moisture percentage for loose, bank and compacted soil is the same. Further compacted soil shows no volume change between wet and dry state. (1) Swell(%)=15.9% 𝐷𝐵 − 𝐷𝐿 VL − 𝑉𝐵 100% = 100% Swell(%) = 𝑉𝐵 𝐷𝐿 102PCF − 88PCF 100% = 15.9% Swell(%) = 88PCF (2) Load factor=0.86

𝐷𝐿 1 = 𝐷𝐵 1 + Swell 88PCF Load factor = = 0.86 102PCF

Load factor = (3) Shrinkage(%)=11.3%

𝑉𝐵 − 𝑉𝐶 𝐷𝐶 − 𝐷𝐵 100% 100% = 𝐷𝐶 𝑉𝐵 115PCF − 102PCF Shrinkage(%) = = 11.3% 115PCF (4) Shrinkage factor=0.89 𝐷𝐵 = 1 − shrinkage Shrinkage factor = 𝐷𝐶 102PCF = 0.89 Shrinkage factor = 115PCF (5) 11274.5 BCY Shrinkage(%) =

10,000CY 115PCF = 11274.5CY 102PCF

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(6) 13068.2 LCY

10,000CY 115PCF = 13068.2CY 88PCF (7) 5.15%

Moisture Content (%) =

Moist unit weight − Dry unit weight Dry unit weight

102PCF − 97PCF = 5.15% 97PCF (8) 0.33 gal/cf ≡ 94 gal/CY (8% − 5.15%)97PCF = 2.76PCF → 0.33 gal/cf ≡ 94 gal/CY

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2. A contractor has obtained a set of plans and specification for the construction of a small building. ✓ Topsoil depth = 1 ft ✓ Depth of building foundation excavation = 18 ft ✓ Required work area = 2 feet from all sides of the building foundation walls. ✓ Required sloping angle = 45° ✓ Percent Shrinkage = 15% ✓ Percent Swell = 20% Please determine the quantities for each the following work packages: (1) Top soil removal (including parking lot area): 1,124.1 BCY (2) Building Foundation Excavation a. Excavation for stockpile: 3,586.9 BCY b. Excavation for disposal: 7,895.6 LCY (3) Building Foundation Backfill: 3,048.9 CCY

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of

The volume of a rectangular truncated pyramid is:

(1) Topsoil:

𝑆 = slope ratio 4 𝑉 = 𝐴𝐵𝐻 + (𝐴 + 𝐵)𝐻𝑆 + 𝐻𝑆 2 where ABH = work excavation 3 4 𝑉 = 𝐻 (𝐴𝐵 + 𝑆 (𝐴 + 𝐵 + 𝑆)) 3

𝐴 = 150′ , 𝐵 = 200′ , 𝐻 = 1′ Volume including sloping of topsoil: 𝑉𝑡𝑜𝑝 = 30,351 ft 3 = 1,124.1 BCY Volume excluding sloping of topsoil: 𝑉𝑡𝑜𝑝 = 30,000 ft 3 = 1111.1 BCY

Assumption: Total excavation depth is 18’, i.e. after 1’ topsoil removal, 17’ soil will be removed (see drawing) Building Excavation [BCY] 𝑉𝑏𝑢𝑖𝑙𝑑 = 110′ (95′ )17′ = 177,650 ft 3 = 6579.6 CY Work Area [BCY] 𝑉𝑤𝑜𝑟𝑘 = (110′ + 2(2′ ))(95′ + 2(2′ ))17′ − 𝑉𝑏𝑢𝑖𝑙𝑑 = 14,212 ft3 = 526 .4 CY Total excavation volume: [BCY] Volume of pyramid based on work area: 𝑉𝑡𝑜𝑡 = 17′ (114′ (99′ ) + 17′ (114′ + 99′ +

4 17′ )) = 259,970ft3 = 9,628.5 CY 3

Sloping [BCY]

𝑉𝑠𝑙𝑜𝑝𝑒 = 𝑉𝑡𝑜𝑡 − 𝑉𝑏𝑢𝑖𝑙𝑑 − 𝑉𝑤𝑜𝑟𝑘 = 64266.7 ft3 = 2522.5 CY (3) Building Foundation Backfill: 3,048.9 CCY Backfill Volume = Total excavation volume - Building Excavation volume 𝑉𝐵𝑎𝑐𝑘 = 𝑉𝑡𝑜𝑡 − 𝑉𝑏𝑢𝑖𝑙𝑑 𝑉𝐵𝑎𝑐𝑘 = 9,628 .5 CY − 6579.6 CY = 3,048.9 CY (CCY) Page 5 of 10

(2) Building Foundation Excavation a. Excavation for stockpile: 3,586.9 BCY

𝑉𝐶 100% 100% − Shrinkage(%) 𝑉𝐶 = 𝑉𝐵𝑎𝑐𝑘 3,048.9 CY 100% = 3,586.9 CY (BCY) = 𝑉𝐵 = 100% − 15% 𝑉𝐵 =

𝑉𝑆𝑡𝑜𝑐𝑘

b. Excavation for disposal: 7,895.6 LCY 𝑉𝐷𝑖𝑠𝑝 = 𝑉𝑏𝑢𝑖𝑙𝑑 (1 + Swell(%)) 𝑉𝐷𝑖𝑠𝑝 = 6579.6 CY(1 + 20%) = 7,895.6 CY (LCY)

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3. A wheeled tractor with a 300-hp engine is operated over a haul road having a slope of +2.0% and a rolling resistance factor of 84 lb/ton. The coefficient of traction is estimated to be 0.45. The construction site is located at an elevation of 4,000 ft. In the fully loaded condition, 52% of the total vehicle weight is on the drive wheels. Gear First Second Third Fourth

Speed (mph) 3.6 6.5 11.5 20.5

Engine: four-cycle, naturally respirated Flywheel horsepower: 300 HP Loaded weight = 82,000 lb Gear train efficiency is 90%

a. What is the required rimpull to overcome the total resisting force? b. What is the maximum tractive force prior to slippage of the wheels? Can this tractor perform the hauling task? c. Will the power of this machine be affected by the altitude? If so, what is the derating factor? d. What is the maximum available rimpull in each gear, if the gear train efficiency is 90%? e. What is the maximum speed the scraper can operate at on the haul road?

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Rolling Resistance Factor (lb/ton) = 40 lb/ton + 30 lb/ton × inches of tire penetration Rolling Resistance Factor (lb/ton) = 84 lb/ton Loaded weight = 82,000 lb = 41 ton Rolling Resistance Force (lb) = Rolling Resistance Factor (lb/ton) × Total Machine Weight (ton) Rolling Resistance Force (lb) = 84 lb/ton × 41 ton = 3444 lb lb/ton ) × Grade(%) [% cancel] Grade Resistance Factor (lb/ton) = 20 ( % slope Uphill (+), Downhill (−) Grade Resistance (lb) = Vehicle Weight (ton) × Grade Resistance Factor (lb/ton) Grade Resistance (lb) = Vehicle Weight (lb) × Grade (%) lb/ton Grade Resistance Factor (lb/ton) = 20 ( ) × 2% = 40 lb/ton % slope Grade Resistance (lb) = 41 ton × 40 lb/ton = 1640 lb Grade Resistance (lb) = 82,000 lb × 2% = 1640 lb Total Resistance (lb) = Rolling Resistance Force (lb) + Grade Resistance Force (lb) Rolling Resistance Factor (lb/ton) Effective Grade = Grade (%) + 20 Total Resistance (lb) = 3444 lb + 1640 lb = 5084 lb 84 lb/ton Effective Grade = 2% + = 6.4% 20 lb/ton

Altitude

Altitude (ft) − 1000 ft 1000 ft No derating for turbocharged engines up to 10,000 ft. 4000 ft − 1000 ft = 0.09 Derating Factor = 0.03 1000 ft HP × Gear Train Efficiency Available Rimpull (lb) = 375 Speed (mph) 300hp × 0.91 × 90% Available Rimpull (lb) = 375 = 4495 lb 20.5 mph Coefficient of Traction maximum force (drawbar or rimpull)(lb) = CT × Weight on drive axles Derating Factor = 0.03

CT = 0.45 Weight on drive axles = 52% × 82,000 lb = 42 ,640 lb maximum force (drawbar or rimpull)(lb) = 0.45 × 42,640 lb = 19 ,188 lb

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