---
title: "RCC slab culvert: excavation, PCC foundation, slab"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:slab-culvert-j"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/slab-culvert-j
site: BCT Engineering
---

# RCC slab culvert: excavation, PCC foundation, slab

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2076 Asoj, 10 marks
- Video: [ESTIMATION OF 2.0M *1.5M R.C.C. SLAB CULVERT.#rccslabculvert #culverts](https://www.youtube.com/watch?v=LVxIDj1TRwA) — MR. CIVIL ENGINEER
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/slab-culvert-j
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Estimate the quantities of the following items of work from the accompanying RCC Slab Culvert drawings:
a) Earthwork in excavation in foundation
b) PCC (1:3:6) in foundation
c) PCC (1:2:4) for RCC slab
[Drawing not included in the scanned paper]

## Short answer

Based on assumed dimensions: excavation20.16 m³, PCC (1:3:6)4.032 m³, PCC (1:2:4) slab5.472 m³. Substitute actual drawing dimensions in same expressions.

## Answer

The drawing is not available, so the method is shown on a stated set of dimensions; the same steps are used for any slab culvert.

#### Assumed dimensions

- Clear span $3.00$ m, bearing of slab $0.40$ m on each abutment; slab thickness $0.30$ m.
- Width of culvert (measured across the road, i.e. along the stream) $4.80$ m (4.00 m roadway + 2 kerbs of 0.40 m); this is also the length of each abutment.
- Each abutment foundation: PCC (1:3:6) $1.40$ m wide $\times$ $0.30$ m thick $\times$ $4.80$ m long, with the bottom of excavation $1.50$ m below the ground.

#### a) Earthwork in excavation

$V = 2 \times 1.40 \times 4.80 \times 1.50 = 20.16$ m³

#### b) PCC (1:3:6) in foundation

$V = 2 \times 1.40 \times 4.80 \times 0.30 = 4.032$ m³

#### c) PCC (1:2:4) for RCC slab

Slab length $= 3.00 + 2 \times 0.40 = 3.80$ m; $V = 3.80 \times 4.80 \times 0.30 = 5.472$ m³

**Answer (for the assumed drawing): excavation = 20.16 m³; PCC 1:3:6 = 4.032 m³; concrete 1:2:4 in slab = 5.472 m³.** For the actual drawing, substitute its dimensions in the same expressions.

## Check yourself

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

1. Excavation is computed as
   A. Slab volume x 1.5
   B. Clear span x width x depth
   C. Only one abutment x depth
   D. 2 abutments x width x length x depth to the bottom of excavation
   Answer: D. 2 x 1.40 x 4.80 x 1.50 = 20.16 m3. (Key point: Excavation volume)

2. The slab length used for the slab concrete is
   A. Clear span + bearing on both ends = 3.00 + 2 x 0.40 = 3.80 m
   B. Clear span only, 3.00 m
   C. Clear span + one bearing
   D. Foundation width 1.40 m
   Answer: A. The slab rests on both abutments, so the bearings add to clear span. (Key point: Slab length)

3. Why are PCC 1:3:6 and PCC 1:2:4 measured separately?
   A. They are the same item
   B. They are different mixes in different members, foundation and slab, so rates differ
   C. One is deducted from the other
   D. The slab is not concrete
   Answer: B. Items are measured by mix and location because each has its own rate. (Key point: Different concrete items)

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