Cutting volume, 60 m, GL 1181.50, formation 1179.00

Given: formation width $b = 9$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse (cross) slope of ground $n:1 = 6:1$, length 60 m with sections at 0, 30 and 60 m. The formation falls 1 in 40, so the fall in 30 m = 30/40 = 0.75 m.

Step 1: Depth of cutting at the centre line

Formation level at 30 m = 1178.25 m, at 60 m = 1177.50 m.

$$h = \text{GL} - \text{formation level}:\quad h_0 = 2.50,\ h_{30} = 3.55,\ h_{60} = 4.90\ \text{m}$$

Step 2: Area of a two-level section

For ground rising at $1$ in $n$ on one side, the horizontal distances of the two cut edges from the centre line are

$$x_1 = \frac{n\left(\frac{b}{2} + s h\right)}{n - s},\qquad x_2 = \frac{n\left(\frac{b}{2} + s h\right)}{n + s}$$

and the depths of the cut at these edges are $h_1 = h + x_1/n$ and $h_2 = h - x_2/n$. The area is

$$A = \frac{b\,h}{2} + \frac{\left(\frac{b}{2} + s h\right)(h_1 + h_2)}{2}$$

Example for the section at chainage 0: $h = 2.50$, $\frac{b}{2} + sh = 4.5 + 2(2.50) = 9.50$, $x_1 = 6(9.50)/4 = 14.250$, $x_2 = 6(9.50)/8 = 7.125$, $h_1 = 4.875$, $h_2 = 1.3125$, $A_0 = \frac{9(2.50)}{2} + \frac{9.50(6.1875)}{2} = 40.641$ m².

Step 3: Volumes

(a) Trapezoidal (average end area) formula with 30 m between sections:

$$V_t = \frac{d}{2}\left[A_0 + 2A_1 + A_2\right] = \frac{30}{2}\left[40.641 + 2(65.565) + 104.901\right] = 4150.07\ \text{m}^3$$

(b) Prismoidal formula (three sections, i.e. two equal intervals):

$$V_p = \frac{d}{3}\left[A_0 + 4A_1 + A_2\right] = \frac{30}{3}\left[40.641 + 4(65.565) + 104.901\right] = 4078.01\ \text{m}^3$$

(c) Prismoidal correction

$$C_p = V_t - V_p = 4150.07 - 4078.01 = 72.06\ \text{m}^3$$

The trapezoidal formula overestimates the volume by this amount, so the prismoidal volume is the more accurate value.

Answer: volume of cutting by trapezoidal formula = 4150.1 m³; by prismoidal formula = 4078.0 m³; prismoidal correction = 72.1 m³ (to be subtracted from the trapezoidal volume).