A road section is 600 m long (20 Stations). Thecut and fill volumes are to be computedbetween each station. The table below lists thestation numbers, the end area values (m:)between each station that are in cut (column2) and that are in fill (column 3). Material in afill section will consolidate (knownshrinkage), and for this road section, is 10aspercent.Determine the accumulated volumes?End Area (m?)1Station3CutFill3182502974130851404545208012226.130101401110012803013752014505015208016101001712018120194050203030012345678

Introduction - A Cut and fill is a way to construct a highway, road, or canal in which the quantity…

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 (m2) between each station that are in cut (column 2) and that are in fill (column 3). Material in a fill section will consolidate (known as shrinkage), and for this road section, is 10

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 (m2) between each station that are in cut (column 2) and that are in fill (column 3). Material in a fill section will consolidate (known as shrinkage), and for this road section, is 10

Traffic and Highway Engineering

5th Edition

ISBN:9781305156241

Author:Garber, Nicholas J.

Publisher:Garber, Nicholas J.

Chapter14: Highway Surveys And Location

Section: Chapter Questions

Problem 15P

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A roadway section is 1.5km long (15 stations), the cut and fill areas shown in the table. Complete the- carthwork calculation using shrinkage of 10%. Draw the Mass Haul Diagram (M.H.D), then determine: 1. The values of balance point stations for the mass diagram for the x-axis. station cut(m2) fill (m2) 6 - 21 7 12 1 50 90 70 40 22 20 2. The difference in mass diagram ordinates between st.4+00 and st.11+00 and calculate the cost.. 3. The net volume of the earthwork and the cost." 40 12 45 7 90 3 110 (Cost of overhaul = 50 ID/m².station., Cost of borrow = 135 ID/m², Cost of Waste= 75 ID/m² , Cost of freehaul = 70 ID/m³ ). 10 80 50 11 4 23 30 13 70 %3D 14 40 100 15 10 120 234567899 234
Det using development length use NSCP 2015 with complete solutions
Solve only Subspart 9 and 10 part ok asap ok
Title: Concentration Calculation Find CLR? () Ib/ft3 (*No unit) QLR = 60.0 ft3/s CLR = ? Ib/ft3 Lodge Run = 100.0 ft/s CRR = 0 Ib/ft3 Qout = 160.0 ft3/s Cout = 150 g/L . New River
Q2. After surveying a borrow pit, the surveyor produced the following pit grade below. What is the total volume (cubic yard)? The unit in the map is ft. -20. O 6.1 5.4 5.8 20, 0 4.7 H3. 3- 4.8 5. 1 4. 6 3.9 -26. 7– 4.6 5.0 4. 4 5.5 5. 1 4. 9 4. 2 3. 7 3. 9 4. 4 3. 7 20.0
The following are the data of a simple summit mass haul diagram STATION MASS ORDINATE (m3) 0+000 -80 0+500 -130 Mass ordinate at initial point of the limit of free haul distance = +600 Mass ordinate at initial point of the economical haul distance = +200 Limit of Economic Haul = 450m Cost of Haul = P120.00 per cubic meter per meter station a. Determine the volume of borrow and its overhaul volume. b. Determine the length of overhaul if the cost of hauling is 00
Consider the following data for a proposed highway project: At station 10+500, the cross-sectional area is 20.98 m² in fill and the finished roadway slopes upward producing a cross-sectional area of 26.80 m² in cut at station 10+950. The stationing of the balancing point is 10+750. Free-haul distance is 50 m, cost of haul is PO.25 per meter-station, and cost of borrow is P4.375 per cu. m. a. Compute the limit of economical haul. Round off your answer to whole number. b. Compute the stationing of the limits of free-haul distance. Round off your answer to two decimal places. c. Compute the free-haul volume. Round off your answer to two decimal places. d. Compute the length of overhaul. Round off your answer to two decimal places.
The following net accumulations were calculated. The free haul area begins at station 4+30 and continues to 14+30. What is the total overhaul distance (ft)? Round to the nearest hundredths. from station to station net accumulation, yd³ 2+60 3+00 3.85 3+00 4+00 5.85 4+00 4+30 8.35 Response Feedback: [3.85×150+2×80+2.5×15]/8.35
The given data of a proposed road is tabulated below. The free haul distance is 50m and the cost of borrow is 420 Php per m3, the cost of excavation is 350 Php per m3, and the cost of haul is 21 Php per meter-station. The ground surface is assumed to be uniformly sloping. Area in Area in Station m2 m2 cut fill 2+440 50 2+740 3+040 70 What is the limit of economic haul? [ Select ] What is the free haul volume? [ Select ] What is the overhaul volume? [ Select ] m3 What is the waste volume? [ Select ] m3 What is the borrow volume? [ Select ] m3
using the data given in the image uploaded. After 10 yr of consolidation, an additional areal fill load of 49 kpa is placed on the site. compute the amount of additional settlement and prepare a plot of the settlement.
Q12. After surveying a borrow pit, the surveyor produced the following pit grade below. What is the total volume (cubic yard)? The unit in the map is ft, the square is 20ft x 20ft. 5.0 4.9 20.0 4.2 3.9 5.4 13.3- -13.3 4.3 4.6 5.4 4.4 -20.0 6.1 5.4 5.8 4.9 4.7 4.8 5.1 4.6 3.9 -13.3- -26.7- 4.6 5.4 4.4 |3.9 5.5 5.1 4.9 4.2 4.2 3.9 4.4 4.7 -20.0 -20.0
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 6.0