- 2076 Chaitra · 10 marks
Road earthwork, rising 1 in 200, falling 1 in 120
Short answer
Using level section formula A=(B+sh)h with B=10m and side slopes 1:1 cutting, 2:1 filling, compute areas from centre heights and then mean-area volumes. Earthwork: cutting 1623.40 m³, filling 137.02 m³.
Estimate quantities of earthwork of a portion of road from the following data: (i) Formation width of road is 10 m. (ii) Side slope in cutting and filling (1:1) and (2:1) (H:V) respectively.
| Distance | 0 | 30 | 60 | 90 | 120 | 150 | 180 |
|---|---|---|---|---|---|---|---|
| R.L. of ground | 102.60 | 103.00 | 102.65 | 102.20 | 101.50 | 101.20 | 100.65 |
| R.L. of formation | 101 | 102.15 | |||||
| Gradient | Rising gradient 1 in 200 | Falling gradient 1 in 120 |
Given data and assumptions
- Formation width m; side slope in cutting , in filling . Ground is level across the road.
- Formation level at 0 m is 101.00 m, rising 1 in 200 up to 120 m (101.60 m) and falling 1 in 120 from 120 m to 180 m (101.10 m). The printed 102.15 at 120 m does not fit a 1 in 200 gradient, so the stated gradients are used.
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 | 102.600 | 101.000 | cutting 1.600 |
| 0+030 | 103.000 | 101.150 | cutting 1.850 |
| 0+060 | 102.650 | 101.300 | cutting 1.350 |
| 0+090 | 102.200 | 101.450 | cutting 0.750 |
| 0+120 | 101.500 | 101.600 | filling 0.100 |
| 0+150 | 101.200 | 101.350 | filling 0.150 |
| 0+180 | 100.650 | 101.100 | filling 0.450 |
Sectional areas
| Chainage | Centre ht. (m) | Cutting area (m²) | Filling area (m²) |
|---|---|---|---|
| 0+000 | cut 1.60 | 18.560 | 0.000 |
| 0+030 | cut 1.85 | 21.922 | 0.000 |
| 0+060 | cut 1.35 | 15.323 | 0.000 |
| 0+090 | cut 0.75 | 8.062 | 0.000 |
| 0+120 | fill 0.10 | 0.000 | 1.020 |
| 0+150 | fill 0.15 | 0.000 | 1.545 |
| 0+180 | fill 0.45 | 0.000 | 4.905 |
Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---|---|---|
| 0+000 – 0+030 | 30.0 | 607.24 | 0.00 |
| 0+030 – 0+060 | 30.0 | 558.68 | 0.00 |
| 0+060 – 0+090 | 30.0 | 350.78 | 0.00 |
| 0+090 – 0+120 | 30.0 | 106.71 | 1.80 |
| 0+120 – 0+150 | 30.0 | 0.00 | 38.47 |
| 0+150 – 0+180 | 30.0 | 0.00 | 96.75 |
| Total | 180.0 | 1,623.40 | 137.02 |
Answer: Earthwork in cutting = 1,623.40 m³; earthwork in filling = 137.02 m³.
✓ Answer checked by Claude Opus 5.5 (high effort)
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