Question

3.6 Complete the earthwork calculation sheet here and plot the resulting mass diagram. Divide ccy by 0.88 to convert to bcy. Calculate the area under the mass diagram curve. What quantity of material must be hauled along the length of the project? What is the average haul distance, in stations, for this material?

Answer 1

ANSWER :
method . ( xd cut of fill 2 volume of the cut An Area of first station looft ; Ao=oft2 Now we have to write the expression to determine the volume of wt by average and area Ant Ani V cutoffill : Veut of fill An-1 V-( quto ) (100) 4700 bey. area of previous station da distance between the two stations Now substitute oft? - Ay, Az=oft2, adu-g=loofte eioo) = 300 cubic ft 6to = 300 21 = l! - il beyi calculate for fill volume 2 1 - V= 2 = 4700 21 (v=114.07.cey)

(2 Now tabulate the earth work sheet as shown. Endarea End area volume cut (sf) bill (512) of uit bay volume of fill Stripping cut (boy) stropping fou cacy) station, (cey) 2 3 4 5 6 7 0 0 0 +00 0 D 0 Hoo 94 0 174.67 2 too 0 284 o 100.0 o o 0 3+00 O 270 1025-92 u too G 220 33-33 740-74 5700 12 77.77 180 114.815 O o 6+60 120 685-185 o 250 0 0 7+00 0 990.74 0 0 8+00 285 0 108703 o 302 atoo 0 903-10 0 O 0 186 10 +00 0 0 344.44 o 0 11 +00 0 0 0 0 0 12+00 0 0 Now Area = 1 (b, hitbehr.) 2. sulestitute hi- 3422.511 bi 400 hra 4352..35 br = 600

Area (400 x 3422:511+ 600x4352.35) Ś (136900 4,4+2611410). 3980 414.4 2 1990207 2 Convert area into yourds. ご area 1990 2072 27 area 73711.38 yd ? area Maximum vertical distance = 3422.511 ft. enclosed by balance line a daga Average haul distance = mase-vertical distance blw balance and diagra mcurve. 73711:38 = 21.5 34220511 Average haul distance = 21.5 stations