---
title: "Cutting volume, 60 m, GL 540.70, formation 538.20"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:volume-cutting-60m-gl-540-70"
source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-cutting-60m-gl-540-70
site: BCT Engineering
---

# Cutting volume, 60 m, GL 540.70, formation 538.20

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2072 Chaitra, 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-cutting-60m-gl-540-70
- 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 (average end area). Also calculate the prismoidal correction. Formation width = 10 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) | 540.70 | 541.00 | 541.60 |

The formation has a downward slope of 1 in 40 with the formation level at 0 chainage being 538.20 m.

## Short answer

Cutting depths from formation falling 1 in 40. Two-level cut areas with n=6, s=2, b=10. Trapezoidal volume 4412.58 m³, prismoidal 4338.84 m³, correction 73.74 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 $n:1 = 6:1$, section spacing $L = 30$ m.

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

The formation falls at 1 in 40, so it loses $30/40 = 0.75$ m per 30 m: $FL = 538.20 - 0.75 \times (\text{number of intervals})$. The centre depth of cut is $h = GL - FL$.

| Chainage (m) | GL (m) | FL (m) | Centre depth of cut h (m) |
|---|---|---|---|
| 0 | 540.70 | 538.20 | 2.50 |
| 30 | 541.00 | 537.45 | 3.55 |
| 60 | 541.60 | 536.70 | 4.90 |

#### 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, $h = 2.50$ m:

$$
\begin{aligned} h_1 &= \frac{6 \times 2.50 + 5}{6 - 2} = 5.000\ \text{m}, & h_2 &= \frac{6 \times 2.50 - 5}{6 + 2} = 1.250\ \text{m} \\ d_1 &= 5 + 2 \times 5.000 = 15.000\ \text{m}, & d_2 &= 5 + 2 \times 1.250 = 7.500\ \text{m} \\ A &= \tfrac12(15.000 + 7.500)\left(2.50 + \tfrac{10}{4}\right) - \tfrac{100}{8} = 43.750\ \text{m}^2 \end{aligned}
$$

| Chainage (m) | h (m) | h₁ (m) | h₂ (m) | d₁ (m) | d₂ (m) | Area A (m²) |
|---|---|---|---|---|---|---|
| 0 | 2.50 | 5.000 | 1.250 | 15.000 | 7.500 | 43.750 |
| 30 | 3.55 | 6.575 | 2.037 | 18.150 | 9.075 | 69.856 |
| 60 | 4.90 | 8.600 | 3.050 | 22.200 | 11.100 | 110.710 |

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

$$
V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(43.750 + 2 \times 69.856 + 110.710) = 4412.58\ \text{m}^3
$$

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

$$
V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(43.750 + 4 \times 69.856 + 110.710) = 4338.84\ \text{m}^3
$$

#### Step 5: Prismoidal correction

$$
C_p = V_T - V_P = 4412.58 - 4338.84 = 73.74\ \text{m}^3
$$

The trapezoidal formula overestimates the volume by 73.74 m³, so the prismoidal volume is the more accurate one.

**Answer: Volume of cutting by the trapezoidal formula = 4412.58 m³; by the prismoidal formula = 4338.84 m³; prismoidal correction = 73.74 m³.**

## Check yourself

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

1. Formation falls 1 in 40 from 538.20 m, spacing 30 m. What is the centre depth of cut at chainage 30 (GL = 541.00 m)?
   A. 2.50 m
   B. 3.55 m
   C. 4.30 m
   D. 2.80 m
   Answer: B. FL = 538.20 - 0.75 = 537.45 m, so h = 541.00 - 537.45 = 3.55 m. (Key point: Formation level and depth of cut)

2. At chainage 0 (h = 2.50, n = 6, s = 2, b = 10), what is the depth h1 on the high-ground side?
   A. 5.000 m
   B. 1.250 m
   C. 3.750 m
   D. 2.500 m
   Answer: A. h1 = (nh + b/2)/(n - s) = (15 + 5)/4 = 5.000 m. (Key point: Side depths in cutting)

3. How are the horizontal distances d1 and d2 from the centre line found?
   A. d = s h
   B. d = b/2 + s h_side
   C. d = b + s h_side
   D. d = b/2 - s h_side
   Answer: B. The distance is half the formation width plus the side slope run: d = b/2 + s x (side depth). (Key point: Horizontal distance to side slope end)

4. VT = 4412.58 m^3 and VP = 4338.84 m^3. Which statement is correct?
   A. Prismoidal correction is 73.74 m^3 and trapezoidal overestimates
   B. Prismoidal correction is 73.74 m^3 and trapezoidal underestimates
   C. The correction is negative
   D. The trapezoidal volume is more accurate
   Answer: A. Cp = VT - VP = 73.74 m^3, so the trapezoidal formula overestimates and the prismoidal value is more accurate. (Key point: Trapezoidal vs prismoidal result)

## 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 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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