Given the end area of cut and fill of soil as in Table Q4(c). If the material bulks after excavation about 10% and shrink 12%. Given, distance between chainage is 100 m. Draw the mass haul diagram and estimate how much excess cut or fill is required.
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- WA road section is 1200 m long. The cut and fill areas are to be computed between the station as shown in table below, determine the net volume of cut and fill between the stations, if you know that the material in fill will be consolidate by 12 percent after compaction. Given 100 m between each station. Station 0+00 2+00 4+00 6+00 8+00 10+00 12+00 1.0 5.0 Cut Fill 4.0 2.6 7.0 2.0 9.5 0.5 2.0 End Area (m2) 3.0 8.0 6.0The answer is 3830.43 LCM 6. As part of a road construction project, an excavation that is 2km long by 3m wide and 0.5m deep must be backfilled and compacted. The excavation will be backfilled with sand and gravel. What volume of (in LCM) of sand and gravel must be brought to site to complete this job?A road section is 600 m long (20 Stations). The cut and fill volumes are to be computed between each station. The table below lists the station numbers, the end area values (m:) between each station that are in cut (column 2) and that are in fill (column 3). Material in a fill section will consolidate (known shrinkage), and for this road section, is 10 as percent. Determine the accumulated volumes? End Area (m?) 1 Station 3 Cut Fill 3 18 2 50 2 97 4 130 8 51 40 45 45 20 80 122 2 6. 130 10 140 11 100 12 80 30 13 75 20 14 50 50 15 20 80 16 10 100 17 120 18 120 19 40 50 20 30 30 012345678
- An approach road for a bridge needs 60000 m3 of soil compacted to a void ratio of 0.7. There are two borrow pits A and B from where the required soil can be taken and transported to the site. The borrow pit A has void ratio 0.85 and borrow pit B has void ratio 0.95. If the cost of excavation of soiland transportation is Rs. 20/10 m3for area A and Rs. 22/10 m3for area B, which of the borrow area is more economical? G=2.676.5 Complete the earthwork calculation sheet here and plot the resulting mass diagram. Divide ccy by 0.9 to convert to bcy (column 10). Calculate th average haul (trapezoidal formula) for the balances on this project. Is this a waste or borrow project? Alge- braic Strip- ping fill Strip- ping cut Vol. of End- Vol. of End- Total fill Adj. fill Total Mass ordi- area area cut sum fill fill cut cut (ccy) (bcy) (ccy) (8) (bcy) (10) (bcy) (11) (bcy) (ссу) (Ьсy) (5) nate (sf) (2) (sf) (3) Station (9) (7) (12) (6) (4) (1) 0.90 10 + 00 80 580 11 + 00 90 2100 0. 12 + 00 100 13 + 00 4650 100 6000 14 + 00 60 80 560 3250 15 + 00 80 1620 16 + 00 1300 85 80 2450 700 17 + 00 100 7800 18 + 00 90 3620 1980 19 + 00 80 20 + 00 80 1310 21 + 00 10 80 860 580 22 + 00 100 10 23 + 00 1620 250 100 3850 24 + 00 100 25 + 00 2600In a certain portion of a road construction, the following data were taken; Cost of haul = P30 per cu. m per station Cost of borrow = P360 per cu. m. Free haul distance = 60m Volume of burrow = 7700 cu. m. Volume of waste = 7000 cu. m. Unit cost of excavation = P300 per cu. m. A) Compute the length of the Economical Haul. B) Calculate the Cost of Excavation.
- A road section is 600 m long (20 Stations). The cut and fill volumes are to be computed between each station. The table below lists the station numbers, the end area values (m?) between each station that are in cut (column 2) and that are in fill (column 3). Material in a consolidate (known shrinkage), and for this road section, is 10 fill section will as percent. Determine the accumulated volumes? End Area (m2) 1 3 Station Cut Fill 3 18 1 2 50 2 97 4 130 4 8 51 40 45 6. 45 20 7 80 8 122 2 130 10 140 11 100 12 80 30 13 75 20 14 50 50 15 20 80 16 10 100 17 120 18 120 19 40 50 20 30 304. Given the approximate flat area and depth of excavation of the following borrow pits: a. 3750.0 sq. m. and 758.0 cm b. 0.035 sq. km and 180.0 m c. 15.6 ares and 495.0 m d. 9.250 ha and 250.0 m e. 46750 sq. m. and 195.0 m Determine the volume of earth removed from each pit in cubic meters.And graph it using compaction curve. Show solution.
- 02: For the soil profile and loading condition shown in Figure (2). Find: a) Final consolidation settlement. b) Time required for the settlement to be (42) mm. c) The settlement after 2 years. Surcharge Aq= 100 kPa 2 m Sand y= 19.5 kN/m 3 m Sand Y=21 kN/ m Clay ö, = 200 kN/m C =0.2 C, - 3.4 m/yr 8 m Ysa = 19 kN/m e,-0.85 C, = 0.04 T, = Impervious layer d2 Fig. 2 T, = Ua 2 for Uav S 60% T, = 1.781 - 0.933 log( 100 – Uap% ) for Uav > 60% O'po + Ao', d'vo Ser H, log Ce Sef H, log %3D 1+ ee 1+ eo Ce o'vo + Ao', H, log Ser H, log 1 + eo 1 + eo o'pFrom the given figure. From the e-log p plot, two points were given. p1 = 80 kPa e1 = 0.60 p2 = 150 kPa e2 = 0.40 Determine the consolidation settlement in mm.Estimate the time required (in hours) for a dragline to excavate material to be placed and compacted on a parking lot construction site, where 7200 CCM of material is needed. Use the following information: Dragline size = 1.53 m³ Swing angle = 90° 4 Average depth of cut - 3.36 m Material sand and gravel Job efficiency = 45 min/h Soil shrinkage 12% Your Answer: Answer