- 2075 Asoj · 12 marks
Hill road earthwork, chainage 0-600 (12 marks)
Short answer
Level-ground sections with A=(B+sh)h, B=10m, cut slope 1.5:1, fill slope 2:1. Mean-area volumes computed from centre heights. Earthwork: cutting 1460.25 m³, filling 2615.50 m³.
Find the quantity of earth work of a hill road from the following data. Formation width is 10 m, side slope in filling and cutting 2:1 and 1½:1 respectively.
| Chainage (m) | 0 | 100 | 200 | 300 | 400 | 500 | 600 |
|---|---|---|---|---|---|---|---|
| RL of Ground (m) | 1115.20 | 1116.10 | 1116.85 | 1118.00 | 1118.25 | 1118.10 | 1117.75 |
Formation: RL at chainage 0 is 1116.5 m, upward gradient 1 in 200 up to chainage 300 m. Downward gradient 1 in 400 from chainage 300 m onwards.
Given data and assumptions
- Formation width m; side slope in filling , in cutting . No cross slope is given, so the ground is level across the road.
- Formation level 1116.50 m at 0, rising 1 in 200 up to 300 m (1118.00 m) and then falling 1 in 400.
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 | 1,115.200 | 1,116.500 | filling 1.300 |
| 0+100 | 1,116.100 | 1,117.000 | filling 0.900 |
| 0+200 | 1,116.850 | 1,117.500 | filling 0.650 |
| 0+300 | 1,118.000 | 1,118.000 | 0.000 |
| 0+400 | 1,118.250 | 1,117.750 | cutting 0.500 |
| 0+500 | 1,118.100 | 1,117.500 | cutting 0.600 |
| 0+600 | 1,117.750 | 1,117.250 | cutting 0.500 |
Sectional areas
| Chainage | Centre ht. (m) | Cutting area (m²) | Filling area (m²) |
|---|---|---|---|
| 0+000 | fill 1.30 | 0.000 | 16.380 |
| 0+100 | fill 0.90 | 0.000 | 10.620 |
| 0+200 | fill 0.65 | 0.000 | 7.345 |
| 0+300 | 0.00 | 0.000 | 0.000 |
| 0+400 | cut 0.50 | 5.375 | 0.000 |
| 0+500 | cut 0.60 | 6.540 | 0.000 |
| 0+600 | cut 0.50 | 5.375 | 0.000 |
Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---|---|---|
| 0+000 – 0+100 | 100.0 | 0.00 | 1,350.00 |
| 0+100 – 0+200 | 100.0 | 0.00 | 898.25 |
| 0+200 – 0+300 | 100.0 | 0.00 | 367.25 |
| 0+300 – 0+400 | 100.0 | 268.75 | 0.00 |
| 0+400 – 0+500 | 100.0 | 595.75 | 0.00 |
| 0+500 – 0+600 | 100.0 | 595.75 | 0.00 |
| Total | 600.0 | 1,460.25 | 2,615.50 |
Answer: Earthwork in cutting = 1,460.25 m³; earthwork in filling = 2,615.50 m³.
✓ Answer checked by Claude Opus 5.5 (high effort)
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