--- title: "Cut and fill volume, 20 m formation width" subject: "Surveying I" chapter: 9 key: "semester_3/surveying_i:volume-road-20m-width-cut-fill-1-1-1-2" source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-road-20m-width-cut-fill-1-1-1-2 site: BCT Engineering --- # Cut and fill volume, 20 m formation width Surveying I · Chapter 9: Computation of Area and Volume - Asked in: 2080 Bhadra, 6 marks - Video: [Calculation of Volume of Earthwork from Cross Sections | Road Estimate | Earthwork Calculation](https://www.youtube.com/watch?v=zMl6OVLeKWM) — Civil Engineering Edu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/volume-road-20m-width-cut-fill-1-1-1-2 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9 ## Question The width of a certain road at the formation level is 20 m, side slope 1:1 for cutting and 1:2 for filling. The ground transverse slope is 1 in 0 [?]. If the depth of excavation at the centre line of three successive sections 50 m apart are 0.4, 0.8 and 1.20 m, calculate the volume of cutting and filling by trapezoidal method and prismoidal correction also. ## Short answer For depths 0.4, 0.8, 1.2 m at 50 m spacing, trapezoidal volumes: cut 1839.90 m³, fill 68.75 m³. Prismoidal: cut 1826.60 m³, fill 54.17 m³. Prismoidal correction: cut 13.30 m³, fill 14.58 m³. ## Answer **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. | Chainage | h (m) | Section | |---|---|---| | 0 | 0.4 | cut + fill (10h = 4 m < 10 m) | | 50 | 0.8 | cut + fill (10h = 8 m < 10 m) | | 100 | 1.2 | full cutting (two-level section) | #### 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 $A_{fill} = \tfrac12 (10 - 10h)\,F$ | Chainage (m) | h (m) | H (m) | F (m) | A cut (m²) | A fill (m²) | |---|---|---|---|---|---| | 0 | 0.4 | 1.556 | 0.750 | 10.889 | 2.250 | | 50 | 0.8 | 2.000 | 0.250 | 18.000 | 0.250 | | 100 | 1.2 | 2.444 ($h_1$) | – | 26.707 | 0 | 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 $C_p = V_T - V_P$: | | Trapezoidal (m³) | Prismoidal (m³) | $C_p$ (m³) | |---|---|---|---| | Cutting | 1839.90 | 1826.60 | 13.30 | | Filling | 68.75 | 54.17 | 14.58 | **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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At centre depth 0.4 m with cross-fall 1 in 10 and half-width 10 m, what section is formed? A. Side-long section with cutting on the high side and filling on the low side B. Full cutting C. Full filling D. Level section Answer: A. The low-side ground meets formation level at 10h = 4 m from the centre, less than the 10 m half-width, so part of the width is in fill. (Key point: Section type) 2. Why is the 1.2 m depth section a two-level cutting? A. Because side slope is 1:1 B. Because it is the last section C. Because fill is not allowed D. 10h = 12 m exceeds half-width 10 m, so ground is above formation across the full width Answer: D. The low edge ground is still above formation level, so the whole width is in cut. (Key point: Full cutting at 100 m) 3. Which relation gives correction? A. Cp = V_prismoidal x 2 B. Cp = V_trapezoidal + V_prismoidal C. Cp = V_trapezoidal - V_prismoidal, separately for cut and fill D. Cp = A1 - A3 Answer: C. Cut: 1839.90 - 1826.60 = 13.30; fill: 68.75 - 54.17 = 14.58 m^3. 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