---
title: "Materials for PCC roof 1:1.5:3 with rebar"
subject: "Estimating and Costing"
chapter: 4
key: "semester_7/estimating_and_costing:materials-pcc-rcc-roof-g"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/materials-pcc-rcc-roof-g
site: BCT Engineering
---

# Materials for PCC roof 1:1.5:3 with rebar

Estimating and Costing · Chapter 4: Analysis of Rates

- Asked in: 2079 Bhadra, 4 marks
- Video: [Quantity Estimation of PCC (Plain Cement Concrete) | Quantity of Cement, Sand and Crush stone in PCC](https://www.youtube.com/watch?v=pCw-mfBv-4Q) — 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/materials-pcc-rcc-roof-g
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-4

## Question

Calculate the quantities of materials required for PCC (1:1.5:3) for RCC roof 0.10 m thick, 20 m wide and 25 m long [assuming rebar 0.8% of volume of PCC].

## Short answer

Net concrete volume 49.6 m³ (after deducting 0.8% steel), dry volume 76.384 m³. For 1:1.5:3 mix: cement 400 bags (50 kg), sand 20.83 m³, coarse aggregate 41.66 m³, steel 3,140 kg.

## Answer

Assumptions: dry volume = 1.54 × wet volume for concrete; 1 m³ of cement = 28.8 bags of 50 kg; no wastage is added.

#### Volumes

- Gross volume of roof slab $= 0.10 \times 20 \times 25 = 50.00$ m³
- Steel (0.8%) $= 0.008 \times 50.00 = 0.400$ m³, which is deducted because the steel occupies this space.
- Net volume of concrete $= 50.00 - 0.400 = 49.600$ m³

#### Materials for concrete 1:1.5:3

- Wet volume $= 49.600$ m³; dry volume $= 1.54 \times 49.600 = 76.384$ m³ (sum of ratio $= 5.5$)
- Cement $= 1/5.5 \times 76.384 = 13.888$ m³ $= 13.888 \times 28.8 = 400.0$ bags (50 kg)
- Sand $= 1.5/5.5 \times 76.384 = 20.832$ m³
- Coarse aggregate $= 3/5.5 \times 76.384 = 41.664$ m³

#### Steel

Mass of steel $= 0.400 \times 7850 = 3,140$ kg $= 3.140$ tonne.

**Answer: cement = 400.0 bags (13.89 m³); sand = 20.83 m³; coarse aggregate = 41.66 m³; steel = 3,140 kg.**

## Check yourself

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

1. Why is 0.4 m^3 deducted from the slab volume of 50 m^3?
   A. It is the wastage of concrete
   B. It is the volume of formwork
   C. It is the shrinkage loss on drying
   D. Steel occupies that space, so concrete volume is 49.6 m^3
   Answer: D. Steel at 0.8% of 50 m^3 = 0.4 m^3 takes the place of concrete. (Key point: Net concrete volume)

2. What dry volume of materials is needed for 49.6 m^3 of wet concrete?
   A. 49.6 m^3
   B. 63.0 m^3 (1.27 x 49.6)
   C. 76.384 m^3 (1.54 x 49.6)
   D. 50 x 1.54 = 77.0 m^3 on gross volume
   Answer: C. Dry volume = 1.54 x 49.6 = 76.384 m^3. (Key point: Dry volume)

3. Cement for 1:1.5:3 from dry volume 76.384 m^3?
   A. 25.46 m^3 = 733 bags (1/3 of dry)
   B. 13.888 m^3 = 400 bags (1/5.5 of dry, x 28.8)
   C. 13.888 m^3 = 13.9 bags
   D. 20.83 m^3 = 600 bags
   Answer: B. Cement = 76.384/5.5 = 13.888 m^3; times 28.8 bags per m^3 = 400 bags. (Key point: Cement bags)

4. How is the steel mass found?
   A. 0.400 m^3 x 7850 kg/m^3 = 3,140 kg
   B. 0.400 x 1.54 = 0.616 t
   C. 0.400 x 28.8 = 11.5 kg
   D. 50 x 7850 = 392,500 kg
   Answer: A. Mass = volume of steel x density 7850 kg/m^3 = 3,140 kg. (Key point: Steel mass)

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