Cutting volume, 60 m, GL 540.70, formation 538.20

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 10$ 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 formation falls at 1 in 40, so it loses $30/40 = 0.75$ m per 30 m: $FL = 538.20 - 0.75 \times (\text{number of intervals})$. The centre depth of cut is $h = GL - FL$.

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, $h = 2.50$ m:

$$\begin{aligned} h_1 &= \frac{6 \times 2.50 + 5}{6 - 2} = 5.000\ \text{m}, & h_2 &= \frac{6 \times 2.50 - 5}{6 + 2} = 1.250\ \text{m} \\ d_1 &= 5 + 2 \times 5.000 = 15.000\ \text{m}, & d_2 &= 5 + 2 \times 1.250 = 7.500\ \text{m} \\ A &= \tfrac12(15.000 + 7.500)\left(2.50 + \tfrac{10}{4}\right) - \tfrac{100}{8} = 43.750\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(43.750 + 2 \times 69.856 + 110.710) = 4412.58\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(43.750 + 4 \times 69.856 + 110.710) = 4338.84\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 4412.58 - 4338.84 = 73.74\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 73.74 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of cutting by the trapezoidal formula = 4412.58 m³; by the prismoidal formula = 4338.84 m³; prismoidal correction = 73.74 m³.