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Part 2 Based upon the results to part 1, the project requires hauling materials to, or from, the site. The following equipmenPart 1 Complete the following Earthwork Sheet Material: Earth and Gravel, Swell Factor:.83 Station Volume of fill Volume of c

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Answer #1

Part 1: complete the earthwork sheet

To complete the earth work sheet, we need to insert additional columns:

(i) Distance between stations = Chainage of current station – Chainage of previous station

Chainages are represented as “00+00.00” form where, digits before “+” are 100s of feet and after “+” foot.

(ii) Average end area cut / fill = (Cut/ Fill area of current station + Cut/ Fill area of previous station)/2

Volume of cut/fill can be calculated by below formula:

Volume = Average End area of x Distance between stations / 27

27 is the conversion factor from cf to cy.

Total cut/fill = Volume of cut / fill calculated by end area + Stripping cut/fill

Adjusted fill = Fill Volume (bcy)/ Swell factor

Considering cut as Positive (+) and fill as Negative (-), sum of both cut and fill volumes is calculated in sum column and cumulative value of earthwork at that station by adding the algebraic sums up to that station is the mass ordinate of the station.

Earthwork sheet completed below:

Since the final mass ordinate is negative, there will be imported soil required to complete the project.

Quantity of soil to be imported from outside = 1774.52 cy

Part 2 (a) (i): Number of Cycles per loader to fill each truck:

For 5 CY Loader:

Given Data:

· Bucket fill factor F = 90%

· Truck Fill factor = 100%

· Loader Bucket Capacity C = 5 CY

· Truck Capacity = 45.9 CY

For 7 CY Loader:

Given Data:

· Bucket fill factor F = 90%

· Truck Fill factor = 100%

· Loader Bucket Capacity C = 7 CY

· Truck Capacity = 45.9 CY

Part 2 (a) (ii)& (iii) Number of trucks, cost per cy and curve

Step 1: Cycle time of truck:

Given Data:

· Truck Dump Time = 0.5 min

1 mph = 88 feet per minute

Truck time = Length / Speed

Cycle time (t) = Total Haul time + Return Time + Dump Time=3.13+1.14+2.56+0.49+0.19+0.74 = 8.25 minutes

Step 2: Maximum Production of one truck:

Given Data:

· Number of buckets for 5 CY bucket n5 = 10

· Number of buckets for 7 CY bucket n7 = 7

· Fill factor of buckets F = 90%

Step 3: Maximum Production of each loader:

Given Data:

· Bucket fill factor F = 90%

· Loader Bucket Capacity C =5/7 CY

· Working Efficiency E = 50 min per hour

· Loader Cycle Time T = 0.5/ 0.6 min

Step 4: Cost of earth for various number of trucks:

Given Data:

· Maximum production of 5 CY Loader = 450 CY per hour

· Maximum production of 7 CY Loader = 525 CY per hour

· Maximum production of 1 truck with 5 CY Loader = 327.27 CY per hour

· Maximum production of 1 truck with 7 CY Loader = 320.73 CY per hour

· Cost of 5 CY capacity loader = $105/hr

· Cost of 7 CY capacity loader = $110/hr

· 45.9 CY Truck cost = $120/hr

Maximum production of both loaders = 450 + 525 = 975 CY per hour

Maximum production of2 trucks = 327.27 + 320.73 = 648 CY per hour

Total Cost of loaders = 105 + 110 = $215 per hour

Accordingly cost per CY calculated in below table:

Hence the number of trucks would be 3 for most economic haul operation.

Accordingly graph is plotted below:

Part 2 (b) Time for hauling the earth

From above table, economic production for 3 number trucks = 972 cy/hr

Quantity to be imported from part 1 = 1774.52 cy

Required time to complete import operation = Quantity/ Production = 1774.52 /972 = 1.83 hours = 1 hr 50 min

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