Cut and fill volume, 20 m formation width

Reading of the given data: the scanned slope reads "1 in 0", which is incomplete. Because the question asks for both cutting and filling, a cross-fall is needed, so it is read as 1 in 10 (n = 10). Formation width $b = 20$ m (half width 10 m), cutting side slope 1:1, filling side slope 1:2 (taken as 2 horizontal : 1 vertical, so $s = 2$), centre-line depths of excavation 0.4, 0.8 and 1.2 m at 50 m spacing.

Step 1: Type of section at each chainage

With a cross-fall of 1 in 10, the ground is 1.0 m higher than the centre at the formation edge on the high side and 1.0 m lower on the low side. If the centre depth $h < 1.0$ m, the low side is in filling and the high side in cutting (a side-long section). The ground cuts the formation level at a distance $nh = 10h$ from the centre on the low side.

Step 2: Areas of cutting and filling

For a side-long section, the cut is a triangle on the high side and the fill is a triangle on the low side.

• Cutting depth at the high-side slope end: $H = \dfrac{h + 10/10}{1 - 1/10} = \dfrac{h + 1}{0.9}$ and $A_{cut} = \tfrac12 (10 + 10h)\,H$
• Filling height at the low-side slope end: $F = \dfrac{1 - h}{1 - 2/10} = \dfrac{1 - h}{0.8}$ and $A_{fill} = \tfrac12 (10 - 10h)\,F$

At 100 m the whole width is in cutting (no filling), so the two-level formula is used with $s = 1$, $n = 10$: $h_1 = \dfrac{nh + b/2}{n - s} = 2.444$ m, $h_2 = \dfrac{nh - b/2}{n + s} = 0.182$ m (depth of cut on the low side), and $A = \tfrac12(d_1 + d_2)\left(h + \dfrac{b}{2s}\right) - \dfrac{b^2}{4s} = 26.707$ m².

Step 3: Trapezoidal volumes

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3)$$ $$\begin{aligned} V_{cut} &= 25\,(10.889 + 2 \times 18.000 + 26.707) = 1839.90\ \text{m}^3 \\ V_{fill} &= 25\,(2.250 + 2 \times 0.250 + 0.000) = 68.75\ \text{m}^3 \end{aligned}$$

Step 4: Prismoidal volumes and correction

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3)$$ $$\begin{aligned} V_{cut} &= \frac{50}{3}\,(10.889 + 4 \times 18.000 + 26.707) = 1826.60\ \text{m}^3 \\ V_{fill} &= \frac{50}{3}\,(2.250 + 4 \times 0.250 + 0.000) = 54.17\ \text{m}^3 \end{aligned}$$

Prismoidal correction $C_p = V_T - V_P$:

Answer: Cutting = 1839.90 m³ and filling = 68.75 m³ by the trapezoidal method. Prismoidal correction = 13.30 m³ for cutting and 14.58 m³ for filling (prismoidal volumes 1826.60 m³ and 54.17 m³).