---
title: "Cutting volume, 60 m, formation level 278.00"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:volume-cutting-60m-formation-278-upward"
source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-cutting-60m-formation-278-upward
site: BCT Engineering
---

# Cutting volume, 60 m, formation level 278.00

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2079 Bhadra, 3+3 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-cutting-60m-formation-278-upward
- 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 rule) 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 an upward slope of 1 in 40 with the formation level at 0 chainage being 278.00 m.

## Short answer

Compute formation levels and centre depths, then cross-sectional areas for cutting with side slopes 2:1, transverse slope 6:1. Trapezoidal volume = 3299.57 m³, prismoidal volume = 3261.25 m³, prismoidal correction = 38.32 m³.

## Answer

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 9$ 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 formation rises at 1 in 40, so it gains $30/40 = 0.75$ m per 30 m: $FL = 278.00 + 0.75 \times (\text{number of intervals})$. The centre depth of cut is $h = GL - FL$.

| Chainage (m) | GL (m) | FL (m) | Centre depth of cut h (m) |
|---|---|---|---|
| 0 | 281.50 | 278.00 | 3.50 |
| 30 | 281.80 | 278.75 | 3.05 |
| 60 | 282.40 | 279.50 | 2.90 |

#### 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, $h = 3.50$ m:

$$
\begin{aligned} h_1 &= \frac{6 \times 3.50 + 4.5}{6 - 2} = 6.375\ \text{m}, & h_2 &= \frac{6 \times 3.50 - 4.5}{6 + 2} = 2.062\ \text{m} \\ d_1 &= 4.5 + 2 \times 6.375 = 17.250\ \text{m}, & d_2 &= 4.5 + 2 \times 2.062 = 8.625\ \text{m} \\ A &= \tfrac12(17.250 + 8.625)\left(3.50 + \tfrac{9}{4}\right) - \tfrac{81}{8} = 64.266\ \text{m}^2 \end{aligned}
$$

| Chainage (m) | h (m) | h₁ (m) | h₂ (m) | d₁ (m) | d₂ (m) | Area A (m²) |
|---|---|---|---|---|---|---|
| 0 | 3.50 | 6.375 | 2.062 | 17.250 | 8.625 | 64.266 |
| 30 | 3.05 | 5.700 | 1.725 | 15.900 | 7.950 | 53.077 |
| 60 | 2.90 | 5.475 | 1.612 | 15.450 | 7.725 | 49.551 |

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

$$
V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(64.266 + 2 \times 53.077 + 49.551) = 3299.57\ \text{m}^3
$$

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

$$
V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(64.266 + 4 \times 53.077 + 49.551) = 3261.25\ \text{m}^3
$$

#### Step 5: Prismoidal correction

$$
C_p = V_T - V_P = 3299.57 - 3261.25 = 38.32\ \text{m}^3
$$

The trapezoidal formula overestimates the volume by 38.32 m³, so the prismoidal volume is the more accurate one.

**Answer: Volume of cutting by the trapezoidal formula = 3299.57 m³; by the prismoidal formula = 3261.25 m³; prismoidal correction = 38.32 m³.**

## Check yourself

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

1. FL at 60 m with start 278.00 and gradient 1 in 40 upward is
   A. 278.75 m
   B. 279.50 m
   C. 280.00 m
   D. 278.60 m
   Answer: B. 60/40 = 1.5 m rise. (Key point: Formation level)

2. Which pattern holds for h?
   A. 3.50, 3.05, 2.90 (GL - FL)
   B. 3.50, 3.80, 4.10
   C. 0.5 each
   D. 2.90, 3.05, 3.50 increasing
   Answer: A. GL - FL at each chainage. (Key point: Centre depths)

3. Which is the two-level section area?
   A. b x h
   B. 1/2 b h
   C. (d1 + d2) h
   D. 1/2 (d1 + d2)(h + b/2s) - b^2/(4s)
   Answer: D. The big triangle between extended side slopes less the small triangle under the formation width. (Key point: Area formula)

4. Cp = 38.32 m^3 indicates
   A. Prismoidal overestimates
   B. Both equal
   C. Trapezoidal overestimates, prismoidal is more accurate
   D. Prismoidal is less accurate
   Answer: C. Cp = V_T - V_P positive. (Key point: Cp meaning)

## 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, 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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