---
title: "Cutting volume, 60 m, formation level 279.00"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:volume-cutting-60m-gl-281-50-formation-279"
source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-cutting-60m-gl-281-50-formation-279
site: BCT Engineering
---

# Cutting volume, 60 m, formation level 279.00

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2081 Bhadra, 6 marks; 2073 Shrawan, 8 marks (asked 2 times)
- 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-60m-gl-281-50-formation-279
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9

## Question

Find the volume of cutting in a length of 60 m with the following data for a two-level section using prismoidal and trapezoidal (average end area) formula. Also calculate the prismoidal correction. Formation width = 9 m, side slope = 2:1, transverse slope = 6:1. The ground levels at 30 m intervals are given below.

| Chainage (m) | 0 | 30 | 60 |
|---|---|---|---|
| GL (m) | 281.50 | 281.80 | 282.40 |

The formation has a downward slope of 1 in 40 with the formation level at 0 chainage being 279.00 m.

## Short answer

Compute two-level section areas at 0, 30, 60 m using formation width 9 m, side slope 2:1, cross slope 6:1. Trapezoidal (average end area) volume = 4150.1 m³, prismoidal volume = 4078.0 m³, prismoidal correction = 72.1 m³.

## Answer

**Given:** formation width $b = 9$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse (cross) slope of ground $n:1 = 6:1$, length 60 m with sections at 0, 30 and 60 m. The formation falls 1 in 40, so the fall in 30 m = 30/40 = 0.75 m.

#### Step 1: Depth of cutting at the centre line
Formation level at 30 m = 278.25 m, at 60 m = 277.50 m.

$$
h = \text{GL} - \text{formation level}:\quad h_0 = 2.50,\ h_{30} = 3.55,\ h_{60} = 4.90\ \text{m}
$$

#### Step 2: Area of a two-level section
For ground rising at $1$ in $n$ on one side, the horizontal distances of the two cut edges from the centre line are

$$
x_1 = \frac{n\left(\frac{b}{2} + s h\right)}{n - s},\qquad x_2 = \frac{n\left(\frac{b}{2} + s h\right)}{n + s}
$$

and the depths of the cut at these edges are $h_1 = h + x_1/n$ and $h_2 = h - x_2/n$. The area is

$$
A = \frac{b\,h}{2} + \frac{\left(\frac{b}{2} + s h\right)(h_1 + h_2)}{2}
$$

Example for the section at chainage 0: $h = 2.50$, $\frac{b}{2} + sh = 4.5 + 2(2.50) = 9.50$,
$x_1 = 6(9.50)/4 = 14.250$, $x_2 = 6(9.50)/8 = 7.125$, $h_1 = 4.875$, $h_2 = 1.3125$,
$A_0 = \frac{9(2.50)}{2} + \frac{9.50(6.1875)}{2} = 40.641$ m².

| Chainage (m) | GL (m) | Formation level (m) | Centre cut $h$ (m) | $x_1$ (m) | $x_2$ (m) | $h_1$ (m) | $h_2$ (m) | Area (m²) |
|---|---|---|---|---|---|---|---|---|
| 0 | 281.50 | 279.00 | 2.50 | 14.250 | 7.125 | 4.875 | 1.3125 | 40.641 |
| 30 | 281.80 | 278.25 | 3.55 | 17.400 | 8.700 | 6.450 | 2.1000 | 65.565 |
| 60 | 282.40 | 277.50 | 4.90 | 21.450 | 10.725 | 8.475 | 3.1125 | 104.901 |

#### Step 3: Volumes
**(a) Trapezoidal (average end area) formula** with 30 m between sections:

$$
V_t = \frac{d}{2}\left[A_0 + 2A_1 + A_2\right] = \frac{30}{2}\left[40.641 + 2(65.565) + 104.901\right] = 4150.07\ \text{m}^3
$$

**(b) Prismoidal formula** (three sections, i.e. two equal intervals):

$$
V_p = \frac{d}{3}\left[A_0 + 4A_1 + A_2\right] = \frac{30}{3}\left[40.641 + 4(65.565) + 104.901\right] = 4078.01\ \text{m}^3
$$

**(c) Prismoidal correction**

$$
C_p = V_t - V_p = 4150.07 - 4078.01 = 72.06\ \text{m}^3
$$

The trapezoidal formula overestimates the volume by this amount, so the prismoidal volume is the more accurate value.

**Answer:** volume of cutting by trapezoidal formula = **4150.1 m³**; by prismoidal formula = **4078.0 m³**; prismoidal correction = **72.1 m³** (to be subtracted from the trapezoidal volume).

## Check yourself

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

1. With a fall of 1 in 40 and 279.00 at chainage 0, what is the formation level at chainage 60?
   A. 278.25 m
   B. 279.75 m
   C. 276.75 m
   D. 277.50 m
   Answer: D. Fall in 60 m = 60/40 = 1.5 m, so 279.00 - 1.5 = 277.50 m. (Key point: Formation levels)

2. What is the centre cut at chainage 30?
   A. 3.55 m
   B. 2.50 m
   C. 4.90 m
   D. 3.05 m
   Answer: A. GL 281.80 minus formation 278.25 = 3.55 m. (Key point: Centre depth)

3. Which formula is used for the trapezoidal volume with sections 30 m apart?
   A. (30/3)[A0 + 4A1 + A2]
   B. (30/2)[A0 + 2A1 + A2]
   C. 30 x (A0 + A2)
   D. (30/6)[A0 + A2]
   Answer: B. V = (d/2)[A0 + 2A1 + A2] gives 4150.07 m3. (Key point: Trapezoidal volume)

4. What is the prismoidal correction and how is it applied?
   A. 72.06 m3, added to the trapezoidal volume
   B. 4078 m3, subtracted
   C. 72.06 m3, subtracted from the trapezoidal volume
   D. 0, since sections are equal
   Answer: C. Cp = Vt - Vp = 4150.07 - 4078.01 = 72.06 m3; the trapezoidal formula overestimates. (Key point: Prismoidal correction)

## Close variants set in other exams

- [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, 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, 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×
- [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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