--- title: "Earthwork volume, formation 1201.85, 1.65 m depth" subject: "Surveying I" chapter: 9 key: "semester_3/surveying_i:volume-earthwork-formation-1201-85" source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-earthwork-formation-1201-85 site: BCT Engineering --- # Earthwork volume, formation 1201.85, 1.65 m depth Surveying I · Chapter 9: Computation of Area and Volume - Asked in: 2076 Chaitra, 6 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-earthwork-formation-1201-85 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9 ## Question Find the volume of earthwork by trapezoidal and prismoidal formula in three consecutive sections at 30 m interval. Formation level of starting chainage = 1201.85 m. Formation width = 6 m. Downward slope of formation = 100:1, side slope = 2:1 and transverse slope = 6:1. The ground has an upward gradient of 50:1. The depth of cutting at 0 chainage is 1.65 m. Compute the prismoidal correction also. ## Short answer Determine ground levels and formation levels from given gradients, then compute cross-sectional areas for cutting with side slopes 2:1, transverse slope 6:1, formation width 6 m. Trapezoidal volume = 1999.04 m³, prismoidal volume = 1980.81 m³, prismoidal correction = 18.23 m³. ## Answer A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 6$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse ground slope $n:1 = 6:1$, section spacing $L = 30$ m. #### Step 1: Formation levels and depths at the centre line The ground level at chainage 0 is $1201.85 + 1.65 = 1203.50$ m. The ground rises 1 in 50, i.e. $30/50 = 0.60$ m per 30 m, and the formation falls 1 in 100, i.e. $30/100 = 0.30$ m per 30 m. The depth of cut therefore increases by $0.60 + 0.30 = 0.90$ m per 30 m. | Chainage (m) | GL (m) | FL (m) | Centre depth of cut h (m) | |---|---|---|---| | 0 | 1203.50 | 1201.85 | 1.65 | | 30 | 1204.10 | 1201.55 | 2.55 | | 60 | 1204.70 | 1201.25 | 3.45 | #### 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, $h = 1.65$ m: $$ \begin{aligned} h_1 &= \frac{6 \times 1.65 + 3}{6 - 2} = 3.225\ \text{m}, & h_2 &= \frac{6 \times 1.65 - 3}{6 + 2} = 0.862\ \text{m} \\ d_1 &= 3 + 2 \times 3.225 = 9.450\ \text{m}, & d_2 &= 3 + 2 \times 0.862 = 4.725\ \text{m} \\ A &= \tfrac12(9.450 + 4.725)\left(1.65 + \tfrac{6}{4}\right) - \tfrac{36}{8} = 17.826\ \text{m}^2 \end{aligned} $$ | Chainage (m) | h (m) | h₁ (m) | h₂ (m) | d₁ (m) | d₂ (m) | Area A (m²) | |---|---|---|---|---|---|---| | 0 | 1.65 | 3.225 | 0.862 | 9.450 | 4.725 | 17.826 | | 30 | 2.55 | 4.575 | 1.537 | 12.150 | 6.075 | 32.406 | | 60 | 3.45 | 5.925 | 2.213 | 14.850 | 7.425 | 50.631 | #### Step 3: Volume by the trapezoidal formula $$ V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(17.826 + 2 \times 32.406 + 50.631) = 1999.04\ \text{m}^3 $$ #### Step 4: Volume by the prismoidal formula $$ V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(17.826 + 4 \times 32.406 + 50.631) = 1980.81\ \text{m}^3 $$ #### Step 5: Prismoidal correction $$ C_p = V_T - V_P = 1999.04 - 1980.81 = 18.23\ \text{m}^3 $$ The trapezoidal formula overestimates the volume by 18.23 m³, so the prismoidal volume is the more accurate one. **Answer: Volume of cutting by the trapezoidal formula = 1999.04 m³; by the prismoidal formula = 1980.81 m³; prismoidal correction = 18.23 m³.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Ground rises 1 in 50 and formation falls 1 in 100. By how much does the centre depth of cut change per 30 m section? A. 0.30 m B. 0.60 m C. 0.90 m D. 0.30 m decrease Answer: C. Ground rises 30/50 = 0.60 m and formation falls 30/100 = 0.30 m, both increasing the cut, so the depth grows by 0.90 m. (Key point: Centre depth change per section) 2. In a cutting with transverse slope n:1 and side slope s:1, which expression gives the depth of cut at the high-ground side end? A. (nh - b/2)/(n + s) B. (nh + b/2)/(n - s) C. (nh + b/2)/(n + s) D. (nh - b/2)/(n - s) Answer: B. On the high-ground side the formula is h1 = (nh + b/2)/(n - s); the low side uses (nh - b/2)/(n + s). (Key point: Side height formulas for cut) 3. For three sections at spacing L, which pair of formulas is used? A. Trapezoidal: (L/3)(A1+4A2+A3); prismoidal: (L/2)(A1+2A2+A3) B. Trapezoidal: (L/2)(A1+2A2+A3); prismoidal: (L/3)(A1+4A2+A3) C. Both use (L/2)(A1+2A2+A3) D. Trapezoidal: L(A1+A2+A3); prismoidal: (L/3)(A1+A2+A3) Answer: B. The trapezoidal rule is (L/2)(A1+2A2+A3) and the prismoidal (Simpson) rule is (L/3)(A1+4A2+A3). (Key point: Trapezoidal vs prismoidal formula) 4. The trapezoidal volume is 1999.04 m^3 and the prismoidal volume 1980.81 m^3. What does the correction of 18.23 m^3 indicate? A. Trapezoidal underestimates the volume B. Trapezoidal overestimates, so the prismoidal value is more accurate C. Both are equally accurate D. Prismoidal overestimates the volume Answer: B. Cp = VT - VP is positive, so the trapezoidal formula overestimates; the prismoidal volume is the more accurate one. 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