--- title: "Road earthwork, four lane NRS-1070" subject: "Estimating and Costing" chapter: 6 key: "semester_7/estimating_and_costing:road-earthwork-g-four-lane" source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/road-earthwork-g-four-lane site: BCT Engineering --- # Road earthwork, four lane NRS-1070 Estimating and Costing · Chapter 6: Detailed Estimate - Asked in: 2079 Bhadra, 10 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/road-earthwork-g-four-lane - Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6 ## Question Workout the quantities of earthwork in embankment and cutting based on the data provided below. Formation width of road: Dedicated four lane width based on NRS-1070. Side slope in embankment and excavation is 1:2 and 1:1.5 respectively. | Chainage | 0+000 | 0+030 | 0+060 | 0+090 | 0+120 | 0+150 | 0+180 | |---|---|---|---|---|---|---|---| | RL of GL (m) | 152.00 | 152.35 | 152.60 | 152.80 | 153.00 | 152.65 | 152.20 | | RL of FL (m) | | | | 151.80 / 152.45 (as printed) | | | 153 | | Cross slope | Plain | Plain | 1:12 | 1:10 | 1:11 | 1:11 | 1:13 | | Gradient of road | 1:200 rising (up to 0+090) | | | | 1:300 rising (0+090 onwards) | | | ## Short answer From cross sections with given cross slopes, compute cut and fill areas at each chainage. Apply mean area method over 30 m intervals. Total earthwork: cutting 1315.99 m³, filling 610.78 m³. ## Answer #### Given data and assumptions - Formation width of a dedicated four-lane road (NRS 2070): four lanes of 3.5 m, a 1.5 m median and two shoulders of 2.5 m give about 21.0 m; $B = 21$ m is used (check with the standard in use). - Side slope in embankment $2:1$ and in excavation $1.5:1$ (the question writes 1:2 and 1:1.5, i.e. 1 vertical to 2 and 1.5 horizontal). Cross slopes: "plain" is taken as level ground. - The printed formation levels are unclear. Formation level at 0+000 is taken as 152.00 m, rising 1 in 200 up to 0+090 (152.45 m, as printed) and then 1 in 300 to 0+180 (152.75 m). #### Method Ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$: $$ \begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned} $$ Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line and $d_u, d_d$ the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling. | Chainage | GL (m) | FL (m) | Depth of cutting / height of filling (m) | |---|---:|---:|---:| | 0+000 | 152.000 | 152.000 | 0.000 | | 0+030 | 152.350 | 152.150 | cutting 0.200 | | 0+060 | 152.600 | 152.300 | cutting 0.300 | | 0+090 | 152.800 | 152.450 | cutting 0.350 | | 0+120 | 153.000 | 152.550 | cutting 0.450 | | 0+150 | 152.650 | 152.650 | 0.000 | | 0+180 | 152.200 | 152.750 | filling 0.550 | #### Sectional areas | Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) | |---|---:|---:|---:| | 0+000 | 0.00 | 0.000 | 0.000 | | 0+030 | cut 0.20 | 4.260 | 0.000 | | 0+060 | cut 0.30 | 9.467 | 2.381 | | 0+090 | cut 0.35 | 11.529 | 3.062 | | 0+120 | cut 0.45 | 12.563 | 1.711 | | 0+150 | 0.00 | 5.803 | 6.125 | | 0+180 | fill 0.55 | 0.488 | 14.160 | Working for one section: Take chainage 0+060: centre cutting depth $h = 0.30$ m, $n = 12$, $B = 21.0$ m. - Ground height above formation at the uphill edge $= 1.175$ m, at the downhill edge $= -0.575$ m. - Toe of the side slope on the uphill side at $x = 12.514$ m from the centre line (ground above formation by 1.343 m); downhill toe at $x = 11.880$ m (ground below formation by 0.690 m). - Area from the co-ordinates of the polygon: cutting $= 9.467$ m², filling $= 2.381$ m². #### Volumes | Between | Distance (m) | Cutting (m³) | Filling (m³) | |---|---:|---:|---:| | 0+000 – 0+030 | 30.0 | 63.90 | 0.00 | | 0+030 – 0+060 | 30.0 | 205.91 | 35.71 | | 0+060 – 0+090 | 30.0 | 314.95 | 81.64 | | 0+090 – 0+120 | 30.0 | 361.39 | 71.61 | | 0+120 – 0+150 | 30.0 | 275.49 | 117.54 | | 0+150 – 0+180 | 30.0 | 94.36 | 304.28 | | **Total** | 180.0 | **1,315.99** | **610.78** | **Answer: Earthwork in cutting = 1,315.99 m³; earthwork in filling = 610.78 m³.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the 21 m formation width of the four-lane road obtained? A. Four lanes of 3.5 m + 1.5 m median + two 2.5 m shoulders B. Four lanes of 3.0 m only C. Two lanes of 7 m + 2 m median D. Four lanes of 3.5 m + two 1.5 m shoulders Answer: A. 14 + 1.5 + 5 = 20.5, approximately 21 m, which is used as B (to be checked against the standard in use). (Key point: Formation width) 2. Formation level at 0+000 is 152.00 m rising 1 in 200. What is it at 0+060? A. 152.60 m B. 152.30 m C. 152.06 m D. 152.20 m Answer: B. 60/200 = 0.30 m rise, so FL = 152.30 m. (Key point: Formation level from gradients) 3. At 0+060 the centre is in cutting 0.30 m yet there is also some filling area. Why? A. With a 1:12 cross slope the downhill edge of the road is below formation level, so that side needs filling B. The side slopes are reversed C. The area formula double counts D. Filling is added for shoulders Answer: A. The ground falls below formation at the downhill edge (-0.575 m), giving a filling area alongside the cutting. (Key point: Part cutting, part filling) 4. The question gives side slopes 1:2 for embankment and 1:1.5 for excavation. The answer interprets them as: A. 2 horizontal to 1 vertical (filling) and 1.5 horizontal to 1 vertical (cutting) B. 1 horizontal to 2 vertical C. 1.5 vertical to 1 horizontal D. Slopes of 2% and 1.5% Answer: A. They are read as V:H, i.e. the usual 2:1 and 1.5:1 (H:V). 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