- Asked 2/22
- 2068 Baisakh (old course) · 8 marks
Road earthwork, grades 1 in 120 and 1 in 90 (0-120)
Short answer
Calculate formation levels from given grades and reference, compute cut/fill depths. Using level-section area A=(B+s h)h, determine volumes by mean area method: 1,745.94 m³ cutting and 187.07 m³ filling.
Estimate the quantity of earthwork of a portion of road from the following data: Formation width of the road = 10 m. Side slope in banking = 2:1 (H:V), side slope in cutting = 1:1. Downward grade 1 in 120 from distance 0 m to 30 m while it remains in level from distance 30 m to 90 m and has again upward grade 1 in 90 from distance 90 to 120 m. The formation level at distance 60 m = 1197.50 m. The ground levels of the centre line of the road are as under:
| R.L. of ground | 1198.65 | 1196.40 | 1199.30 | 1200.40 | 1198.10 |
|---|---|---|---|---|---|
| Distance in m | 0 | 30 | 60 | 90 | 120 |
Given data and assumptions
- Formation width m; banking slope , cutting slope . Ground is level across the road.
- Formation levels: 1197.50 m at 60 m; level from 30 m to 90 m; falling 1 in 120 from 0 to 30 m, so the level at 0 m is 1197.50 + 30/120 = 1197.75 m; rising 1 in 90 from 90 to 120 m, so the level at 120 m is 1197.50 + 30/90 = 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+000 | 1,198.650 | 1,197.750 | cutting 0.900 |
| 0+030 | 1,196.400 | 1,197.500 | filling 1.100 |
| 0+060 | 1,199.300 | 1,197.500 | cutting 1.800 |
| 0+090 | 1,200.400 | 1,197.500 | cutting 2.900 |
| 0+120 | 1,198.100 | 1,197.833 | cutting 0.267 |
Sectional areas
| Chainage | Centre ht. (m) | Cutting area (m²) | Filling area (m²) |
|---|---|---|---|
| 0+000 | cut 0.90 | 9.810 | 0.000 |
| 0+030 | fill 1.10 | 0.000 | 13.420 |
| 0+060 | cut 1.80 | 21.240 | 0.000 |
| 0+090 | cut 2.90 | 37.410 | 0.000 |
| 0+120 | cut 0.27 | 2.738 | 0.000 |
Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---|---|---|
| 0+000 – 0+030 | 30.0 | 66.22 | 110.71 |
| 0+030 – 0+060 | 30.0 | 197.75 | 76.36 |
| 0+060 – 0+090 | 30.0 | 879.75 | 0.00 |
| 0+090 – 0+120 | 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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