Cutting volume, 120 m, upward slope 1 in 40

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 8$ m, side slope $s:1 = 1.5:1$ (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 rises at 1 in 40, so it gains $30/40 = 0.75$ m in every 30 m: $FL = 380.00 + 0.75 \times (\text{number of intervals})$. 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, $h = 0.80$ m:

$$\begin{aligned} h_1 &= \frac{6 \times 0.80 + 4}{6 - 1.5} = 1.956\ \text{m}, & h_2 &= \frac{6 \times 0.80 - 4}{6 + 1.5} = 0.107\ \text{m} \\ d_1 &= 4 + 1.5 \times 1.956 = 6.933\ \text{m}, & d_2 &= 4 + 1.5 \times 0.107 = 4.160\ \text{m} \\ A &= \tfrac12(6.933 + 4.160)\left(0.80 + \tfrac{8}{3}\right) - \tfrac{64}{6} = 8.562\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = L\left[\frac{A_1 + A_{5}}{2} + A_2 + \dots + A_{4}\right] = 30\left[\frac{8.562 + 33.714}{2} + 17.782 + 25.687 + 21.974\right] = 2597.43\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\left[A_1 + A_{5} + 4(A_2 + A_4) + 2A_3\right] = \frac{30}{3}\left[8.562 + 33.714 + 4(17.782 + 21.974) + 2(25.687)\right] = 2526.74\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 2597.43 - 2526.74 = 70.69\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 70.69 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of cutting by the trapezoidal formula = 2597.43 m³; by the prismoidal formula = 2526.74 m³; prismoidal correction = 70.69 m³.