---
title: "Filling volume, 60 m, formation 1281.49"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:volume-filling-60m-gl-1280-50"
source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-filling-60m-gl-1280-50
site: BCT Engineering
---

# Filling volume, 60 m, formation 1281.49

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2080 Baisakh, 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-filling-60m-gl-1280-50
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9

## Question

Find the volume of filling in a length of 60 m with the following data for a two level section, using the trapezoidal and prismoidal formulae.

| Chainage (km) | 0+000 | 0+030 | 0+060 |
|---|---|---|---|
| RL of GL (m) | 1280.50 | 1280.80 | 1281.40 |

Formation width = 12 m, side slope = 2:1, transverse slope = 8:1. The formation has an upward gradient of 1 in 60 with the formation level at 0+000 chainage being 1281.49 m. Also compute the prismoidal correction.

## Short answer

Compute formation levels and centre heights, then cross-sectional areas using two-level section formulas for embankment with side slopes 2:1, transverse slope 8:1. Trapezoidal volume = 1088.34 m³, prismoidal volume = 1114.62 m³, prismoidal correction = -26.28 m³.

## Answer

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 12$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse ground slope $n:1 = 8:1$, section spacing $L = 30$ m.

#### Step 1: Formation levels and depths at the centre line

The formation rises at 1 in 60, so it gains $30/60 = 0.50$ m per 30 m: $FL = 1281.49 + 0.50 \times (\text{number of intervals})$. The centre height of fill is $h = FL - GL$.

| Chainage (m) | GL (m) | FL (m) | Centre height of fill h (m) |
|---|---|---|---|
| 0 | 1280.50 | 1281.49 | 0.99 |
| 30 | 1280.80 | 1281.99 | 1.19 |
| 60 | 1281.40 | 1282.49 | 1.09 |

#### Step 2: Cross-sectional areas

For an embankment on cross-sloping ground, the heights of fill at the two side slope ends 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 above the formation width).

Worked section at chainage 0, $h = 0.99$ m:

$$
\begin{aligned} h_1 &= \frac{8 \times 0.99 - 6}{8 + 2} = 0.192\ \text{m}, & h_2 &= \frac{8 \times 0.99 + 6}{8 - 2} = 2.320\ \text{m} \\ d_1 &= 6 + 2 \times 0.192 = 6.384\ \text{m}, & d_2 &= 6 + 2 \times 2.320 = 10.640\ \text{m} \\ A &= \tfrac12(6.384 + 10.640)\left(0.99 + \tfrac{12}{4}\right) - \tfrac{144}{8} = 15.963\ \text{m}^2 \end{aligned}
$$

| Chainage (m) | h (m) | h₁ (m) | h₂ (m) | d₁ (m) | d₂ (m) | Area A (m²) |
|---|---|---|---|---|---|---|
| 0 | 0.99 | 0.192 | 2.320 | 6.384 | 10.640 | 15.963 |
| 30 | 1.19 | 0.352 | 2.587 | 6.704 | 11.173 | 19.453 |
| 60 | 1.09 | 0.272 | 2.453 | 6.544 | 10.907 | 17.687 |

#### Step 3: Volume by the trapezoidal formula

$$
V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(15.963 + 2 \times 19.453 + 17.687) = 1088.34\ \text{m}^3
$$

#### Step 4: Volume by the prismoidal formula

$$
V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(15.963 + 4 \times 19.453 + 17.687) = 1114.62\ \text{m}^3
$$

#### Step 5: Prismoidal correction

$$
C_p = V_T - V_P = 1088.34 - 1114.62 = -26.28\ \text{m}^3
$$

The correction is negative here because the middle area is larger than the mean of the two end areas. The trapezoidal formula therefore underestimates the volume, and 26.28 m³ has to be added to it to get the prismoidal volume.

**Answer: Volume of filling by the trapezoidal formula = 1088.34 m³; by the prismoidal formula = 1114.62 m³; prismoidal correction = -26.28 m³.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is h found in an embankment?
   A. h = GL - FL
   B. h = FL - GL
   C. h = average of GL
   D. h = FL + GL
   Answer: B. Fill occurs where formation is above ground. (Key point: Centre height of fill)

2. For fill on cross-sloping ground, which gives the high-ground side height h1?
   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: A. The high-ground side uses (nh - b/2)/(n + s); the low side uses (nh + b/2)/(n - s). (Key point: Side heights)

3. With 3 sections, the prismoidal volume is
   A. L/2 (A1 + 2A2 + A3)
   B. L (A1 + A2 + A3)
   C. L/3 (A1 + 2A2 + A3)
   D. L/3 (A1 + 4A2 + A3)
   Answer: D. Simpson's rule on three sections. (Key point: Prismoidal formula)

4. Cp = -26.28 m^3 here. What does this mean?
   A. Trapezoidal overestimates
   B. Calculation error
   C. Trapezoidal underestimates; correction is added
   D. Volume is negative
   Answer: C. The middle area exceeds the mean of the end areas, so V_T < V_P. (Key point: Negative correction)

## Related in Surveying I

- [Cutting volume, 60 m, GL 1181.50, formation 1179.00](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-60m-gl-1181-50-formation-1179) — Surveying I, chapter 9, asked 8×
- [Filling volume, 50 m, formation 1081.50](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-filling-50m-gl-1080-50) — Surveying I, chapter 9, asked 4×
- [Cutting volume, 60 m, GL 540.70, formation 538.20](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-60m-gl-540-70) — Surveying I, chapter 9, asked 7×
- [Cutting volume, 60 m, formation level 279.00](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-60m-gl-281-50-formation-279) — Surveying I, chapter 9, asked 7×
- [Cutting volume, 60 m, formation level 278.00](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-60m-formation-278-upward) — Surveying I, chapter 9, asked 7×
- [Level book and longitudinal section, RL 525.50](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-chainage-200m-rl-525-50) — Surveying I, chapter 5

## Related in other subjects

- [RL of hill station P (A-B 125 m, RL B 1280.50)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-p-rod-2m-at-b) — Surveying II, chapter 3, asked 2×
- [Road earthwork with RL of GL and FL (0-600)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-b-gl-fl) — Estimating and Costing, chapter 6
- [Road earthwork, chainage 0-80 (s)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-s-table-0-80) — Estimating and Costing, chapter 6, asked 2×
- [Road earthwork, 80 m length (v)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-v-0-80) — Estimating and Costing, chapter 6, asked 2×