Volume from central depths 1.85, 2.15, 2.45

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 12$ m, side slope $s:1 = 2:1$ (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 directly: $h = 1.85$, $2.15$ and $2.45$ m at chainages 0, 30 and 60 m. A cutting is assumed because a depth of cut is given.

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, $h = 1.85$ m:

$$\begin{aligned} h_1 &= \frac{6 \times 1.85 + 6}{6 - 2} = 4.275\ \text{m}, & h_2 &= \frac{6 \times 1.85 - 6}{6 + 2} = 0.638\ \text{m} \\ d_1 &= 6 + 2 \times 4.275 = 14.550\ \text{m}, & d_2 &= 6 + 2 \times 0.638 = 7.275\ \text{m} \\ A &= \tfrac12(14.550 + 7.275)\left(1.85 + \tfrac{12}{4}\right) - \tfrac{144}{8} = 34.926\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(34.926 + 2 \times 41.676 + 48.831) = 2506.64\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(34.926 + 4 \times 41.676 + 48.831) = 2504.61\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 2506.64 - 2504.61 = 2.03\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 2.03 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of cutting by the trapezoidal formula = 2506.64 m³; by the prismoidal formula = 2504.61 m³; prismoidal correction = 2.03 m³.