Hill road earthwork, 0+060 to 0+120

Given data and assumptions

• Formation width: cutting $8$ m, banking $10$ m; side slope in cutting $1:1$, in filling $2:1$.
• Sections 30 m apart; cutting at 0+060 and 0+090, filling at 0+120.

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+060: centre cutting depth $h = 0.50$ m, $n = 10$, $B = 8.0$ 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 $x = 5.000$ m from the centre line (ground above formation by 1.000 m); downhill toe at $x = 4.091$ m (ground above formation by 0.091 m).
• Area from the co-ordinates of the polygon: cutting $= 4.455$ m², filling $= 0.000$ m².

Change-over between 0+090 (wholly in cutting, 0.60 m) and 0+120 (wholly in filling, 0.70 m): the zero point is $30 \times 0.60/1.30 = 13.85$ m from 0+090, leaving $16.15$ m in filling.

• Cutting $= \tfrac12 \times 5.254 \times 13.85 = 36.38$ m³
• Filling $= \tfrac12 \times 8.565 \times 16.15 = 69.18$ m³

Volumes

Answer: Earthwork in cutting = 182.02 m³; earthwork in filling = 69.18 m³.