Question

Part 1 Complete the following Earthwork Sheet Material: Earth and Gravel, Swell Factor: 83 Strip fill(cy Total till och Total


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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 equipment
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Answer #1

STEP 1: COMPLETING THE EARTHWORK TABLE

First we will explain all columns to understand how to calculate the earthwork using given data:

1) Station

This is the station where the measurements (i.e. Length, depth) of a road are recorded. It is the distance of that point from a base station and the distance is also known as chainage of that station. Chainages are represented as “00+00.00” form where, digits before “+” are 100s of feet and after “+” foot.

Stations are generally taken at 100 feet interval but may be taken intermediate as per requirement.

2) End area cut

This is the cross sectional area of road at that station which need to be excavated to reach the desired sub–grade level. This is generally measured in SF (Square feet).

3) End area fill

This is the cross sectional area of road at that station which need to be filled to reach the desired sub–grade level. This is generally measured in SF (Square feet).

4) Volume of cut/ fill

Volume of cut/fill calculated by below formula:

Volume = Average End area of two continuous stations (SF) x Distance between stations (ft) / 27

5) Stripping cut/ fill

This is for additional cutting/ filling of road layer.

6) Total cut/ fill

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

7) Adjusted fill

The soil tends to shrink when filled thus its volume get decreased from the original volume. Hence a swell factor need to be applied to the fill quantity to calculate the bank volume of soil required. It’s value varies by type of soil. Swell factor for our problem = 0.83

8) Algebraic Sum

Considering cut as Positive (+) and fill as Negative (-), this is the sum of both cut and fill volumes at that stations.

9) Mass Ordinates

This is cumulative value of earthwork at that station by adding the algebraic sums up to that station.

Part 1: Earth work quantity:

.

Earth work is computed below:

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

Since the bucket does not load short quantity, number of cycles would be 10.

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

Since the bucket does not load short quantity, number of cycles would be 7.

Part 2 (a) (ii) Number of trucks

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 below:

Accordingly graph is plotted below:

Part 2 (b) Time for hauling the earth

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

Quantity to be imported from part 1 = 1955.38 cy

Required time to complete import operation = 1955.38/972 = 2.01 hours

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