--- title: "Hill road earthwork, chainage 0-600 (12 marks)" subject: "Estimating and Costing" chapter: 6 key: "semester_7/estimating_and_costing:hill-road-l-0-600" source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-l-0-600 site: BCT Engineering --- # Hill road earthwork, chainage 0-600 (12 marks) Estimating and Costing · Chapter 6: Detailed Estimate - Asked in: 2075 Asoj, 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-l-0-600 - Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6 ## Question 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. ## 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. ## Answer #### Given data and assumptions - Formation width $B = 10$ m; side slope in filling $2:1$, in cutting $1.5:1$. 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 $h$, formation width $B$ and side slope $s:1$: $$ A = (B + s\,h)\,h $$ 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. $h$ (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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. FL is 1116.50 m at 0, rising 1 in 200. FL at 200 m is: A. 1117.50 m B. 1117.00 m C. 1116.70 m D. 1118.50 m Answer: A. 200/200 = 1 m rise, so 1117.50 m. (Key point: Formation level) 2. At 0+100, GL = 1116.10 and FL = 1117.00. The section is: A. Filling 0.9 m B. Cutting 0.9 m C. Filling 0.1 m D. Cutting 0.1 m Answer: A. Ground is below formation by 0.9 m, so filling. (Key point: Cutting or filling) 3. With B = 10 m and filling slope 2:1, area at h = 0.9 m is: A. (10 + 1.8) x 0.9 = 10.62 m2 B. (10 + 0.9) x 0.9 = 9.81 m2 C. 10 x 0.9 = 9 m2 D. (10 + 2.7) x 0.9 Answer: A. A = (B + s h)h = 11.8 x 0.9 = 10.62 m2. (Key point: Filling area) 4. At 300 m GL = FL = 1118.00. How does this affect the volumes either side? A. The 200-300 filling and 300-400 cutting each use mean of the end area and zero B. The volumes at 300 are doubled C. No volume is calculated for these segments D. Cutting and filling are subtracted Answer: A. Zero area at 300 m gives mean areas such as (7.345 + 0)/2 x 100 = 367.25 m3. 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