HW CHAPTER 10 PART 2
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Northern Virginia Community College *
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231
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Industrial Engineering
Date
Dec 6, 2023
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21.
Using the average end area method, determine the required volume of the
cuts and fills for a new road. The cuts and fills for each station are shown in
Table 10.1
. Express your answer in cubic yards.
Table 10.1
Cuts and Fills for Problem 21
Using Average End Area Method
V = (A1+A2)/2 x L
22.
The material in
Problem 21
has a shrinkage of 97 percent and a swell of 18 percent. Determine
the net import or export required for the project and the number of 11 cy truckloads needed
to haul the material to or from the project.
Stations
Cut
Fill
Vol of Cut
(ft^3)
Vol of Cut
(cy)
Vol to fill
(ft^3)
Vol of Fill
(cy)
10 + 00
0
0
0.00
0.00
0.00
0.00
10 + 50
257
0
6425.00
237.96
0.00
0.00
11 + 00
175
50
10800.00
400.00
1250.00
46.30
11 + 50
20
285
4875.00
180.56
8375.00
310.19
12 + 00
0
233
500.00
18.52
12950.00
479.63
12 + 20
0
0
0.00
0.00
2330.00
86.30
TOTAL CUT
837.04
TOTAL FILL
922.41
Calculate the remaining volume:
Remaining volume
= Fill – Cut = 922.41 – 837.04
= 85.37 cy
Shrinkage Factor = 97%
Swell Factor = 18%
Truckloads = 11 cy
Required bcy = 85.37/0.97 = 88.01 bcy
Volume of the material to be transported.
Lcy = bcy x (1 + swell)
Lcy = 88.01 x (1 + 0.18)
Lcy = 103.85
Finally,
# of truck loads needed
Loads = lcy of haul / lcy per load
Loads = 103.85 / 11
Loads = 9.44 =
10 Loads
27.
Six inches of topsoil is to be removed and stockpiled for the building in
Figure 10.48
. The
topsoil is to be removed 8 feet back from the perimeter of the building to allow for working
around the building and sloping the sides of the excavation. How many bcy of topsoil need to
be removed and stockpiled?
Depth
= 6 inches =
0.5'
All round the perimeter of Building = 8' back
Length = 69'
Width = 58'
Area of excavation site
= (69 x 58) – (20 x 35)
= 4,071 – 700
= 3,371
Bcf = Area x Depth
Bcf = 3,371 x 0.5
Bcf = 1,685.5 ft^3
Bcy = Bcf / 27
Bcy = 62.43 bcy
62.43 Bcy of topsoil needs to be removed.
31.
Determine the amount of excavation needed for the basement in
Figures 10.50
and
10.51
. A 2-
foot working distance is required and the excavation will be sloped 1.5:1 (1.5 vertical feet for
every horizontal foot).
Volume = Avg L x Avg Width x Depth
Avg L = 2 x (22’ - 8”) + 2 x (34’ -
8”)
Avg L = 2 x (22-0.667) + 2 x (34 – 0.667)
Avg L = 42.667 + 66.667
Avg L = 109.333
Volume = 109.333 x 11.167 x 10
Volume = 12,209.216 ft^3
Volume = 452.19 cy
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43.
Determine the quantity of asphalt and granular material required for the parking lot shown in
Figure 10.60
. The asphalt is 4 inches thick and the granular material is 8 inches thick.
LOT AREA =
(99’ x 200’) + (23’ x 24’)
= 19,800 + 552
= 20,352 sqft
ASPHALT 4”
= 20,352 x (26/1000)
= 20,352 x 0.026
= 529.152 Tons
GRANULAR MATEIRAL 8”
= 20,352 x (42/1000)
= 20,352 x 0.042
= 854.784 Tons