---
title: "Cutting volume, 60 m, GL 1181.50, formation 1179.00"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:volume-cutting-60m-gl-1181-50-formation-1179"
source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-cutting-60m-gl-1181-50-formation-1179
site: BCT Engineering
---

# Cutting volume, 60 m, GL 1181.50, formation 1179.00

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2075 Chaitra, 6 marks; 2074 Chaitra, 6 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-1181-50-formation-1179
- 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 the prismoidal and trapezoidal 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 |
|---|---|---|---|
| Ground level (m) | 1181.50 | 1181.80 | 1182.40 |

The formation has a downward slope of 1 in 40 with the formation level at 0+000 chainage being 1179.000 m.

## Short answer

Using two-level section areas (9 m width, 2:1 side slope, 6:1 cross slope), trapezoidal 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 = 1178.25 m, at 60 m = 1177.50 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 | 1181.50 | 1179.00 | 2.50 | 14.250 | 7.125 | 4.875 | 1.3125 | 40.641 |
| 30 | 1181.80 | 1178.25 | 3.55 | 17.400 | 8.700 | 6.450 | 2.1000 | 65.565 |
| 60 | 1182.40 | 1177.50 | 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. What is the formation level at chainage 30 m?
   A. 1179.75 m
   B. 1177.50 m
   C. 1178.75 m
   D. 1178.25 m
   Answer: D. Fall in 30 m = 30/40 = 0.75 m, so 1179.00 - 0.75 = 1178.25 m. (Key point: Formation level at 30 m)

2. Which set gives centre cuts at chainages 0, 30, 60?
   A. 2.50, 3.55, 4.90 m
   B. 2.50, 3.05, 3.40 m
   C. 3.55, 2.50, 4.90 m
   D. 2.50, 4.30, 4.90 m
   Answer: A. GL minus formation: 1181.50-1179.00, 1181.80-1178.25, 1182.40-1177.50. (Key point: Centre cuts)

3. Which formula gives the prismoidal volume for three sections 30 m apart?
   A. (30/2)[A0 + 2A1 + A2]
   B. (30/3)[A0 + 4A1 + A2]
   C. 30[A0 + A1 + A2]
   D. (30/3)[A0 + 2A1 + A2]
   Answer: B. Simpson's form with the middle section coefficient 4 gives 4078.0 m3. (Key point: Prismoidal formula use)

4. Which volume is more accurate, and what is the correction?
   A. Trapezoidal; correction 72.1 m3 added
   B. Both equal
   C. Prismoidal; trapezoidal minus 72.1 m3
   D. Prismoidal; correction 4078 m3
   Answer: C. Trapezoidal overestimates, so prismoidal = trapezoidal - 72.1 m3. (Key point: Correction sign)

## Close variants set in other exams

- [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, 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, 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×
- [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×
- [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×

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