Skip to main content

Side-long ground hill road, 0-400 m

Short answer

Formation levels are computed from downward gradient 1 in 200, centre heights determined. With cross-slope 1 in 5, side-hill areas calculated and mean area method yields volumes. Earthwork: cutting 225.78 m³, filling 8262.11 m³.

Calculate the quantities of earthwork of a hill road in side long ground from 0 m to 400 m partly in cutting and partly in filling with the following data: width of road = 10 m, side slope in cutting and filling = (1:1) and (2:1). The road has a downward gradient of 1 in 200. The cross slope of ground = 1 in 5. Formation level at 0 m = 1203.50 m.

Ground level1202.501201.971202.351199.661200.50
Distance0100200300400

Given data and assumptions

  • Formation width B=10B = 10 m; side slope in cutting 1:11:1, in filling 2:12:1; cross slope of ground 11 in 55 (n=5n = 5) throughout.
  • Formation level at 0 m is 1203.50 m and falls 1 in 200, so it is 1203.50, 1203.00, 1202.50, 1202.00, 1201.50 m at 0, 100, 200, 300, 400 m.
  • Ground levels are at the centre line.

Method

Ground has a transverse (cross) slope n:1n:1 (H:V), so each section is a side-hill section. With centre height hh (cutting +, filling −) above/below formation level, formation width BB and side slope s:1s:1:

xu=n (B/2+sh)n−s,du=h+xunxd=n (B/2+sh)n+s,dd=h−xdnA=12Bh+12[B2(du+dd)+h (xu+xd−B)]\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 xu,xdx_u, x_d are the horizontal distances of the up-hill and down-hill toes from the centre line and du,ddd_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, xd=n(B/2+sh)/(n−s)x_d = n(B/2 + s h)/(n - s) and xu=n(B/2+sh)/(n+s)x_u = n(B/2 + s h)/(n + s), with hh the filling height and ss 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.

ChainageGL (m)FL (m)Depth of cutting / height of filling (m)
0+0001,202.5001,203.500filling 1.000
0+1001,201.9701,203.000filling 1.030
0+2001,202.3501,202.500filling 0.150
0+3001,199.6601,202.000filling 2.340
0+4001,200.5001,201.500filling 1.000

Sectional areas

ChainageCentre ht. hh (m)Cutting area (m²)Filling area (m²)
0+000fill 1.000.00016.667
0+100fill 1.030.00017.169
0+200fill 0.152.2585.510
0+300fill 2.340.00043.275
0+400fill 1.000.00016.667

Working for one section:

Take chainage 0+000: centre filling height h=1.00h = 1.00 m, n=5n = 5, B=10.0B = 10.0 m.

  • Ground height above formation at the uphill edge =0.000= 0.000 m, at the downhill edge =−2.000= -2.000 m.
  • Toe of the side slope on the uphill side at x=5.000x = 5.000 m from the centre line (ground above formation by 0.000 m); downhill toe at x=11.667x = 11.667 m (ground below formation by 3.333 m).
  • Area from the co-ordinates of the polygon: cutting =0.000= 0.000 m², filling =16.667= 16.667 m².

Volumes

BetweenDistance (m)Cutting (m³)Filling (m³)
0+000 – 0+100100.00.001,691.77
0+100 – 0+200100.0112.891,133.96
0+200 – 0+300100.0112.892,439.28
0+300 – 0+400100.00.002,997.10
Total400.0225.788,262.11

Answer: Earthwork in cutting = 225.78 m³; earthwork in filling = 8,262.11 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Estimating and Costing chapter 6 questions →