---
title: "Filling volume, 50 m, formation 1081.50"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:volume-filling-50m-gl-1080-50"
source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-filling-50m-gl-1080-50
site: BCT Engineering
---

# Filling volume, 50 m, formation 1081.50

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2075 Asoj, 8 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-50m-gl-1080-50
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9

## Question

Find the volume of filling in a length of 50 m with the following data for a two level section, using the trapezoidal and prismoidal formula, where formation width = 10 m, side slope 2:1, transverse slope = 8:1. The ground levels at 25 m interval are given below.

| Chainage (m) | 0 | 25 | 50 |
|---|---|---|---|
| RL of GL (m) | 1080.50 | 1079.80 | 1078.40 |

The formation has a downward slope of 1 in 50 with the formation level at 0 chainage being 1081.50 m.

## Short answer

Filling heights from formation falling at 1 in 50. Two-level embankment areas computed with n=8, s=2, b=10. Trapezoidal volume 996.05 m³, prismoidal 942.45 m³, correction 53.60 m³.

## Answer

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 10$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse ground slope $n:1 = 8:1$, section spacing $L = 25$ m.

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

The formation falls at 1 in 50, so it loses $25/50 = 0.50$ m per 25 m: $FL = 1081.50 - 0.50 \times (\text{number of intervals})$. The centre height of fill is $h = FL - GL$.

| Chainage (m) | GL (m) | FL (m) | Centre height of fill h (m) |
|---|---|---|---|
| 0 | 1080.50 | 1081.50 | 1.00 |
| 25 | 1079.80 | 1081.00 | 1.20 |
| 50 | 1078.40 | 1080.50 | 2.10 |

#### 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, $h = 1.00$ m:

$$
\begin{aligned} h_1 &= \frac{8 \times 1.00 - 5}{8 + 2} = 0.300\ \text{m}, & h_2 &= \frac{8 \times 1.00 + 5}{8 - 2} = 2.167\ \text{m} \\ d_1 &= 5 + 2 \times 0.300 = 5.600\ \text{m}, & d_2 &= 5 + 2 \times 2.167 = 9.333\ \text{m} \\ A &= \tfrac12(5.600 + 9.333)\left(1.00 + \tfrac{10}{4}\right) - \tfrac{100}{8} = 13.633\ \text{m}^2 \end{aligned}
$$

| Chainage (m) | h (m) | h₁ (m) | h₂ (m) | d₁ (m) | d₂ (m) | Area A (m²) |
|---|---|---|---|---|---|---|
| 0 | 1.00 | 0.300 | 2.167 | 5.600 | 9.333 | 13.633 |
| 25 | 1.20 | 0.460 | 2.433 | 5.920 | 9.867 | 16.705 |
| 50 | 2.10 | 1.180 | 3.633 | 7.360 | 12.267 | 32.641 |

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

$$
V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{25}{2}\,(13.633 + 2 \times 16.705 + 32.641) = 996.05\ \text{m}^3
$$

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

$$
V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{25}{3}\,(13.633 + 4 \times 16.705 + 32.641) = 942.45\ \text{m}^3
$$

#### Step 5: Prismoidal correction

$$
C_p = V_T - V_P = 996.05 - 942.45 = 53.60\ \text{m}^3
$$

The trapezoidal formula overestimates the volume by 53.60 m³, so the prismoidal volume is the more accurate one.

**Answer: Volume of filling by the trapezoidal formula = 996.05 m³; by the prismoidal formula = 942.45 m³; prismoidal correction = 53.60 m³.**

## Check yourself

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

1. The formation falls 1 in 50 from 1081.50 m with 25 m spacing. What is the centre height of fill at chainage 25 (GL = 1079.80 m)?
   A. 1.00 m
   B. 1.20 m
   C. 1.70 m
   D. 0.70 m
   Answer: B. FL = 1081.50 - 0.50 = 1081.00 m, so h = 1081.00 - 1079.80 = 1.20 m. (Key point: Centre height of fill)

2. For an embankment on cross-sloping ground, which pair of expressions gives the heights at the high-ground and low-ground side ends?
   A. h1 = (nh + b/2)/(n - s); h2 = (nh - b/2)/(n + s)
   B. h1 = (nh - b/2)/(n + s); h2 = (nh + b/2)/(n - s)
   C. h1 = (nh - b/2)/(n - s); h2 = (nh + b/2)/(n + s)
   D. h1 = nh/(n + s); h2 = nh/(n - s)
   Answer: B. For fill, the high-ground side is h1 = (nh - b/2)/(n + s) and the low side h2 = (nh + b/2)/(n - s), the reverse of the cutting case. (Key point: Fill side heights)

3. Which formula gives the area of the two-level section?
   A. A = 0.5 (d1 + d2)(h + b/2s) - b^2/4s
   B. A = 0.5 (d1 + d2) h
   C. A = (d1 + d2)(h + b/2s) + b^2/4s
   D. A = b h + s h^2
   Answer: A. The area is the big triangle between extended side slopes and ground line, 0.5(d1+d2)(h+b/2s), minus the triangle b^2/4s at the formation. (Key point: Area of two-level section)

4. VT = 996.05 m^3 and VP = 942.45 m^3. What is the prismoidal correction and what does it mean?
   A. 53.60 m^3; trapezoidal overestimates
   B. 53.60 m^3; trapezoidal underestimates
   C. 1938.50 m^3; prismoidal is wrong
   D. 0; the formulas agree
   Answer: A. Cp = VT - VP = 53.60 m^3, positive, so the trapezoidal formula overestimates and the prismoidal volume is better. (Key point: Interpreting the correction)

## Close variants set in other exams

- [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×
- [Cutting volume, 120 m, upward slope 1 in 40](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-120m-formation-380) — Surveying I, chapter 9, asked 8×

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