Road earthwork, RD 0-120, gradient 1 in 150

Given data and assumptions

• Formation width $8$ m in cutting and $10$ m in banking; side slope in cutting $1:1$, in banking $2:1$ (H:V).
• Formation level 507.00 m at RD 0, rising 1 in 150: 507.00, 507.20, 507.40, 507.60, 507.80 m. The section at RD 0 has zero centre height and is taken with the cutting width of 8 m.

Method

The 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 $d_u, d_d$ are 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. Where the centre changes from cutting to filling between two sections and one type of area is zero at one end, the change-over point is found by proportion of the centre heights, and that area is taken over its own part of the length only: $V = \tfrac12 A \times l$.

Sectional areas

Working for one section:

Take chainage 0+000: centre cutting depth $h = 0.00$ m, $n = 10$, $B = 8.0$ m.

• Ground height above formation at the uphill edge $= 0.400$ m, at the downhill edge $= -0.400$ m.
• Toe of the side slope on the uphill side at $x = 4.444$ m from the centre line (ground above formation by 0.444 m); downhill toe at $x = 5.000$ m (ground below formation by 0.500 m).
• Area from the co-ordinates of the polygon: cutting $= 0.889$ m², filling $= 1.000$ m².

Change-over between RD 30 (wholly in cutting, 0.75 m) and RD 60 (filling 0.10 m at the centre): the zero point is $30 \times 0.75/0.85 = 26.47$ m from RD 30, leaving $3.53$ m in filling.

• Cutting is present at both ends (6.720 and 0.889 m²), so the mean area is used: $\tfrac{6.720 + 0.889}{2} \times 30 = 114.14$ m³
• Filling is zero at RD 30, so it is taken over the 3.53 m only: $\tfrac12 \times 2.250 \times 3.53 = 3.97$ m³

Volumes

Answer: Earthwork in cutting = 241.61 m³; earthwork in filling = 573.43 m³.