Embankment volume, 160 m, rising gradient 1 in 100

Reading of the given data: the scanned formation level "114.95 m" cannot be right for ground levels near 1115 m (it would put the formation about 1000 m below the ground). Read as 1114.95 m, the formation would switch between cutting and filling at every section, which does not fit an embankment. It is therefore taken here as 1116.95 m at 0+000, which keeps the formation above the ground at every chainage. "Ground level across the centre line" is taken to mean level cross-sections (no transverse slope). Formation width $b = 10$ m, side slope $s = 2$ (2:1), spacing $L = 40$ m.

Step 1: Formation levels and heights of fill

The formation rises at 1 in 100, i.e. $40/100 = 0.40$ m per 40 m: $FL = 1116.95 + 0.40 \times (\text{number of intervals})$. Height of fill $h = FL - GL$.

Step 2: Cross-sectional areas

For a level section of an embankment, the area is a trapezoid:

$$A = (b + s h)\,h$$

Step 3: Trapezoidal formula

$$V_T = L\left[\frac{A_1 + A_{5}}{2} + A_2 + \dots + A_{4}\right] = 40\left[\frac{15.625 + 1.020}{2} + 49.105 + 12.000 + 48.000\right] = 4697.10\ \text{m}^3$$

Step 4: Prismoidal formula

There are four equal intervals (an even number), so Simpson's rule applies.

$$V_P = \frac{L}{3}\left[A_1 + A_{5} + 4(A_2 + A_4) + 2A_3\right] = \frac{40}{3}\left[15.625 + 1.020 + 4(49.105 + 48.000) + 2(12.000)\right] = 5720.87\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 4697.10 - 5720.87 = -1023.77\ \text{m}^3$$

Answer: Volume of earthwork (filling) = 4697.10 m³ by the trapezoidal formula and 5720.87 m³ by the prismoidal formula (prismoidal correction = -1023.77 m³).