Hill road earthwork, chainage 12-15

Given data and assumptions

• Formation width $B = 10$ m; side slope in cutting $1:1$, in banking $2:1$. The length of one chain is 30 m, so the sections are 30 m apart (chainage numbers 12, 13, 14, 15 are chain numbers).

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.

Sectional areas

Working for one section:

Take chainage 12: centre cutting depth $h = 0.40$ m, $n = 10$, $B = 10.0$ m.

• Ground height above formation at the uphill edge $= 0.900$ m, at the downhill edge $= -0.100$ m.
• Toe of the side slope on the uphill side at $x = 6.000$ m from the centre line (ground above formation by 1.000 m); downhill toe at $x = 5.250$ m (ground below formation by 0.125 m).
• Area from the co-ordinates of the polygon: cutting $= 4.500$ m², filling $= 0.062$ m².

Volumes

Answer: Earthwork in cutting = 149.91 m³; earthwork in filling = 232.33 m³.