- Asked 3/22
- 2067 Asar (old course) · 7 marks
Road earthwork, grades from 90 m to 210 m (s)
Short answer
Determine formation levels from given grades and reference, compute depths. Using level-section area formula A=(B+s h)h, compute volumes by mean area method: 1,745.94 m³ cutting and 187.07 m³ filling.
Find out the quantity of earthwork of a portion of road to be constructed with the following data: Formation width of the road throughout = 10 m. Side slope in banking (2:1) and side slope in cutting (1:1). Downward grade 1 in 120 from distance 90 m to 120 m. While it remains in level from distance 120 m to 180 m and has again upward grade 1 in 90 from distance 180 m to 210 m. The formation level at distance 180 m = 1197.50 m. The ground levels are as under:
| R.L. of ground | 1198.65 | 1196.40 | 1199.30 | 1200.40 | 1198.10 |
|---|---|---|---|---|---|
| Distance (m) | 90 | 120 | 150 | 180 | 210 |
Given data and assumptions
- Formation width m throughout; banking slope , cutting slope ; ground level across the road.
- Formation level at 180 m is 1197.50 m; level from 120 m to 180 m; falling 1 in 120 from 90 to 120 m, so level at 90 m = 1197.50 + 30/120 = 1197.75 m; rising 1 in 90 from 180 to 210 m, so level at 210 m = 1197.833 m.
Method
Level (plain-ground) sections are used. With centre height , formation width and side slope :
Volume between two sections = mean area × distance; where a section changes from cutting to filling, the length of each part is found by proportion of the depths and the volume of each part = ½ × end area × length.
| Chainage | GL (m) | FL (m) | Depth of cutting / height of filling (m) |
|---|---|---|---|
| 0+090 | 1,198.650 | 1,197.750 | cutting 0.900 |
| 0+120 | 1,196.400 | 1,197.500 | filling 1.100 |
| 0+150 | 1,199.300 | 1,197.500 | cutting 1.800 |
| 0+180 | 1,200.400 | 1,197.500 | cutting 2.900 |
| 0+210 | 1,198.100 | 1,197.833 | cutting 0.267 |
Sectional areas
| Chainage | Centre ht. (m) | Cutting area (m²) | Filling area (m²) |
|---|---|---|---|
| 0+090 | cut 0.90 | 9.810 | 0.000 |
| 0+120 | fill 1.10 | 0.000 | 13.420 |
| 0+150 | cut 1.80 | 21.240 | 0.000 |
| 0+180 | cut 2.90 | 37.410 | 0.000 |
| 0+210 | cut 0.27 | 2.738 | 0.000 |
Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---|---|---|
| 0+090 – 0+120 | 30.0 | 66.22 | 110.71 |
| 0+120 – 0+150 | 30.0 | 197.75 | 76.36 |
| 0+150 – 0+180 | 30.0 | 879.75 | 0.00 |
| 0+180 – 0+210 | 30.0 | 602.22 | 0.00 |
| Total | 120.0 | 1,745.94 | 187.07 |
Answer: Earthwork in cutting = 1,745.94 m³; earthwork in filling = 187.07 m³.
✓ Answer checked by Claude Opus 5.5 (high effort)
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