Filling volume, 60 m, formation 1281.49

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 $n:1 = 8:1$, section spacing $L = 30$ m.

Step 1: Formation levels and depths at the centre line

The formation rises at 1 in 60, so it gains $30/60 = 0.50$ m per 30 m: $FL = 1281.49 + 0.50 \times (\text{number of intervals})$. 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, $h = 0.99$ m:

$$\begin{aligned} h_1 &= \frac{8 \times 0.99 - 6}{8 + 2} = 0.192\ \text{m}, & h_2 &= \frac{8 \times 0.99 + 6}{8 - 2} = 2.320\ \text{m} \\ d_1 &= 6 + 2 \times 0.192 = 6.384\ \text{m}, & d_2 &= 6 + 2 \times 2.320 = 10.640\ \text{m} \\ A &= \tfrac12(6.384 + 10.640)\left(0.99 + \tfrac{12}{4}\right) - \tfrac{144}{8} = 15.963\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(15.963 + 2 \times 19.453 + 17.687) = 1088.34\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(15.963 + 4 \times 19.453 + 17.687) = 1114.62\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 1088.34 - 1114.62 = -26.28\ \text{m}^3$$

The correction is negative here because the middle area is larger than the mean of the two end areas. The trapezoidal formula therefore underestimates the volume, and 26.28 m³ has to be added to it to get the prismoidal volume.

Answer: Volume of filling by the trapezoidal formula = 1088.34 m³; by the prismoidal formula = 1114.62 m³; prismoidal correction = -26.28 m³.