Filling volume, 50 m, formation 1081.50

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 $n:1 = 8:1$, section spacing $L = 25$ m.

Step 1: Formation levels and depths at the centre line

The formation falls at 1 in 50, so it loses $25/50 = 0.50$ m per 25 m: $FL = 1081.50 - 0.50 \times (\text{number of intervals})$. The centre height of fill is $h = FL - GL$.

Step 2: Cross-sectional areas

For an embankment on cross-sloping ground, the heights of fill at the two side slope ends 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 above the formation width).

Worked section at chainage 0, $h = 1.00$ m:

$$\begin{aligned} h_1 &= \frac{8 \times 1.00 - 5}{8 + 2} = 0.300\ \text{m}, & h_2 &= \frac{8 \times 1.00 + 5}{8 - 2} = 2.167\ \text{m} \\ d_1 &= 5 + 2 \times 0.300 = 5.600\ \text{m}, & d_2 &= 5 + 2 \times 2.167 = 9.333\ \text{m} \\ A &= \tfrac12(5.600 + 9.333)\left(1.00 + \tfrac{10}{4}\right) - \tfrac{100}{8} = 13.633\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{25}{2}\,(13.633 + 2 \times 16.705 + 32.641) = 996.05\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{25}{3}\,(13.633 + 4 \times 16.705 + 32.641) = 942.45\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 996.05 - 942.45 = 53.60\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 53.60 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of filling by the trapezoidal formula = 996.05 m³; by the prismoidal formula = 942.45 m³; prismoidal correction = 53.60 m³.