--- title: "Cutting volume, depths 2 m, 3 m, 4 m" subject: "Surveying I" chapter: 9 key: "semester_3/surveying_i:volume-cutting-8m-depths-2-3-4" source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-cutting-8m-depths-2-3-4 site: BCT Engineering --- # Cutting volume, depths 2 m, 3 m, 4 m Surveying I · Chapter 9: Computation of Area and Volume - Asked in: 2078 Kartik, 5+1 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/volume-cutting-8m-depths-2-3-4 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9 ## Question The width of formation level of a certain cutting is 8 m and the side slopes are 1:1. The surface of the ground has a transverse slope of 1 in 6. If the depths of cutting at the centre lines of three sections 30 m apart are 2 m, 3 m and 4 m respectively, determine the volume of earth work involved in this length of cutting by trapezoidal approach and prismoidal approach. Also find the prismoidal correction. ## Short answer Compute cross-sectional areas for cutting with side slopes 1:1, transverse slope 6:1, formation width 8 m. Trapezoidal volume = 2094.87 m³, prismoidal volume = 2084.58 m³, prismoidal correction = 10.29 m³. ## Answer A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 8$ m, side slope $s:1 = 1:1$ (horizontal : vertical), transverse ground slope $n:1 = 6:1$, section spacing $L = 30$ m. #### Step 1: Centre heights The centre depths of cut are given: $h = 2$, $3$ and $4$ m at chainages 0, 30 and 60 m. The transverse slope 1 in 6 means $n = 6$ and the side slope 1:1 means $s = 1$. #### Step 2: Cross-sectional areas Along the cross-section the ground rises on one side and falls on the other. The depths of cut at the two side slope ends (where the side slope meets the ground) 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 below the formation width). Worked section at chainage 0, $h = 2.00$ m: $$ \begin{aligned} h_1 &= \frac{6 \times 2.00 + 4}{6 - 1} = 3.200\ \text{m}, & h_2 &= \frac{6 \times 2.00 - 4}{6 + 1} = 1.143\ \text{m} \\ d_1 &= 4 + 1 \times 3.200 = 7.200\ \text{m}, & d_2 &= 4 + 1 \times 1.143 = 5.143\ \text{m} \\ A &= \tfrac12(7.200 + 5.143)\left(2.00 + \tfrac{8}{2}\right) - \tfrac{64}{4} = 21.029\ \text{m}^2 \end{aligned} $$ | Chainage (m) | h (m) | h₁ (m) | h₂ (m) | d₁ (m) | d₂ (m) | Area A (m²) | |---|---|---|---|---|---|---| | 0 | 2.00 | 3.200 | 1.143 | 7.200 | 5.143 | 21.029 | | 30 | 3.00 | 4.400 | 2.000 | 8.400 | 6.000 | 34.400 | | 60 | 4.00 | 5.600 | 2.857 | 9.600 | 6.857 | 49.829 | #### Step 3: Volume by the trapezoidal formula $$ V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(21.029 + 2 \times 34.400 + 49.829) = 2094.87\ \text{m}^3 $$ #### Step 4: Volume by the prismoidal formula $$ V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(21.029 + 4 \times 34.400 + 49.829) = 2084.58\ \text{m}^3 $$ #### Step 5: Prismoidal correction $$ C_p = V_T - V_P = 2094.87 - 2084.58 = 10.29\ \text{m}^3 $$ The trapezoidal formula overestimates the volume by 10.29 m³, so the prismoidal volume is the more accurate one. **Answer: Volume of cutting by the trapezoidal formula = 2094.87 m³; by the prismoidal formula = 2084.58 m³; prismoidal correction = 10.29 m³.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Transverse slope 1 in 6 and side slope 1:1 give A. n = 1 and s = 6 B. n = 6 and s = 1 C. n = 6 and s = 6 D. n = 1 and s = 1 Answer: B. Transverse slope 1 in 6 is n = 6; side slope 1:1 means s = 1. (Key point: Reading the slopes) 2. Why is a two-level section used? A. Ground rises on one side and falls on the other across the road B. Ground is level C. Only fill occurs D. Depth is constant Answer: A. Cross-fall gives different depths at the two sides. (Key point: Section type) 3. Why is the prismoidal volume (2084.58) less than the trapezoidal (2094.87)? A. Prismoidal ignores a section B. Trapezoidal ignores the middle area C. Calculation error D. Trapezoidal overestimates when middle area is less than the mean of the end areas Answer: D. Areas 21.03, 34.40, 49.83: middle is below the mean (35.43), so V_T > V_P. (Key point: Trapezoidal vs prismoidal) 4. Prismoidal correction is A. 2094.87 m^3 B. 20.58 m^3 C. 10.29 m^3 D. 0 Answer: C. 2094.87 - 2084.58 = 10.29 m^3. 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