Cutting volume, depths 2 m, 3 m, 4 m

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 8$ m, side slope $s:1 = 1:1$ (horizontal : vertical), transverse ground slope $n:1 = 6:1$, section spacing $L = 30$ m.

Step 1: Centre heights

The centre depths of cut are given: $h = 2$, $3$ and $4$ m at chainages 0, 30 and 60 m. The transverse slope 1 in 6 means $n = 6$ and the side slope 1:1 means $s = 1$.

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, $h = 2.00$ m:

$$\begin{aligned} h_1 &= \frac{6 \times 2.00 + 4}{6 - 1} = 3.200\ \text{m}, & h_2 &= \frac{6 \times 2.00 - 4}{6 + 1} = 1.143\ \text{m} \\ d_1 &= 4 + 1 \times 3.200 = 7.200\ \text{m}, & d_2 &= 4 + 1 \times 1.143 = 5.143\ \text{m} \\ A &= \tfrac12(7.200 + 5.143)\left(2.00 + \tfrac{8}{2}\right) - \tfrac{64}{4} = 21.029\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(21.029 + 2 \times 34.400 + 49.829) = 2094.87\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(21.029 + 4 \times 34.400 + 49.829) = 2084.58\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 2094.87 - 2084.58 = 10.29\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 10.29 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of cutting by the trapezoidal formula = 2094.87 m³; by the prismoidal formula = 2084.58 m³; prismoidal correction = 10.29 m³.