Earthwork volume, formation 1201.85, 1.65 m depth

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 6$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse ground slope $n:1 = 6:1$, section spacing $L = 30$ m.

Step 1: Formation levels and depths at the centre line

The ground level at chainage 0 is $1201.85 + 1.65 = 1203.50$ m. The ground rises 1 in 50, i.e. $30/50 = 0.60$ m per 30 m, and the formation falls 1 in 100, i.e. $30/100 = 0.30$ m per 30 m. The depth of cut therefore increases by $0.60 + 0.30 = 0.90$ m per 30 m.

Step 2: Cross-sectional areas

Along the cross-section the ground rises on one side and falls on the other. The depths of cut at the two side slope ends (where the side slope meets the ground) are

$$h_1 = \frac{nh + b/2}{n - s} \quad (\text{high-ground side}), \qquad h_2 = \frac{nh - b/2}{n + s} \quad (\text{low-ground side})$$

The horizontal distances from the centre line to these points are $d_1 = b/2 + s h_1$ and $d_2 = b/2 + s h_2$, and the area of the section is

$$A = \frac{1}{2}(d_1 + d_2)\left(h + \frac{b}{2s}\right) - \frac{b^2}{4s}$$

(the triangle between the two side slopes extended and the ground line, minus the triangle below the formation width).

Worked section at chainage 0, $h = 1.65$ m:

$$\begin{aligned} h_1 &= \frac{6 \times 1.65 + 3}{6 - 2} = 3.225\ \text{m}, & h_2 &= \frac{6 \times 1.65 - 3}{6 + 2} = 0.862\ \text{m} \\ d_1 &= 3 + 2 \times 3.225 = 9.450\ \text{m}, & d_2 &= 3 + 2 \times 0.862 = 4.725\ \text{m} \\ A &= \tfrac12(9.450 + 4.725)\left(1.65 + \tfrac{6}{4}\right) - \tfrac{36}{8} = 17.826\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(17.826 + 2 \times 32.406 + 50.631) = 1999.04\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(17.826 + 4 \times 32.406 + 50.631) = 1980.81\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 1999.04 - 1980.81 = 18.23\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 18.23 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of cutting by the trapezoidal formula = 1999.04 m³; by the prismoidal formula = 1980.81 m³; prismoidal correction = 18.23 m³.