How is the 21 m formation width of the four-lane road obtained?

Four lanes of 3.5 m + 1.5 m median + two 2.5 m shoulders

Four lanes of 3.5 m + two 1.5 m shoulders

At 0+060 the centre is in cutting 0.30 m yet there is also some filling area. Why?

With a 1:12 cross slope the downhill edge of the road is below formation level, so that side needs filling

The question gives side slopes 1:2 for embankment and 1:1.5 for excavation. The answer interprets them as:

2 horizontal to 1 vertical (filling) and 1.5 horizontal to 1 vertical (cutting)

2079 Bhadra · 10 marks

Road earthwork, four lane NRS-1070

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.

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)

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Road earthwork, four lane NRS-1070

Given data and assumptions

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