---
title: "T-beam decking: RCC and wearing coat"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:t-beam-decking-rcc-wearing"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/t-beam-decking-rcc-wearing
site: BCT Engineering
---

# T-beam decking: RCC and wearing coat

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2075 Chaitra, 5+2 marks; 2070 Chaitra, 5+3 marks (asked 2 times)
- Video: [ESTIMATION OF RCC T BEAM DECKING](https://www.youtube.com/watch?v=HmdpoBWqEWo) — ADV. LEARNING
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/t-beam-decking-rcc-wearing
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Find out the quantities of the following items of work of a T-beam decking of a bridge with 6 m span and 45 cm bearing at ends.
i) RCC work (1:2:4) excluding steel
ii) Cement concrete (1:2:4) in wearing coat
[Figure: RCC T-beam decking cross-section. Overall width 7.7 m, 7 m roadway, 20×20 cm kerb on each side, 10 cm wearing coat of C.C., 20 cm slab, three ribs each 35 cm wide with 225 cm clear between ribs, rib depth below slab 75 cm (107.5 cm projections at the edges), 15×15 cm fillets, 15×15 cm RCC railing posts 120 cm high at 136 cm c/c, 40 mm dia G.I. pipe 4 nos., railing height 120 cm.]

## Short answer

RCC work (excluding steel) volume: sum of slab, rib, fillet, kerb and post volumes = 17.402 m³. Wearing coat C.C. (1:2:4) quantity: 7.00×6.90×0.10 = 4.830 m³.

## Answer

#### Assumptions from the figure

- Length of decking $= 6.00 + 2 \times 0.45 = 6.90$ m (clear span plus bearing at both ends).
- Overhang of slab beyond outer ribs $= (7.70 - 3 \times 0.35 - 2 \times 2.25)/2 = 1.075$ m (matches the 107.5 cm in the figure).
- Fillets are taken as triangles of 15 × 15 cm at both sides of each rib (6 in total). The railing posts are at 1.36 m c/c: with 6.90 m length this gives 6 posts per side, 12 in all.
- The 20 cm slab is taken from the figure; the kerb is 20 × 20 cm above the slab, and the wearing coat (10 cm) is laid between the kerbs over the 7.0 m roadway.

#### (i) RCC work (1:2:4), excluding steel

| Member | Working | Volume (m³) |
|---|---|---:|
| Slab | $7.70 \times 6.90 \times 0.20$ | 10.626 |
| Ribs (3) | $3 \times 0.35 \times 0.75 \times 6.90$ | 5.434 |
| Fillets (6, 15 × 15 cm triangular) | $6 \times \tfrac12 \times 0.15 \times 0.15 \times 6.90$ | 0.466 |
| Kerbs (2) | $2 \times 0.20 \times 0.20 \times 6.90$ | 0.552 |
| Railing posts (12 nos, 15 × 15 × 120 cm) | $12 \times 0.15 \times 0.15 \times 1.20$ | 0.324 |
| **Total RCC (1:2:4)** | | **17.402** |

#### (ii) Cement concrete (1:2:4) in wearing coat

$V = 7.00 \times 6.90 \times 0.10 = 4.830$ m³

**Answer: RCC = 17.402 m³; wearing coat = 4.830 m³.**

Materials for information (1:2:4, dry factor 1.54): RCC needs 110.3 bags cement, 7.66 m³ sand, 15.31 m³ aggregate; wearing coat needs 30.6 bags, 2.13 m³, 4.25 m³.

## Check yourself

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

1. What length of decking is used in the quantities?
   A. 6.00 + 2 x 0.45 = 6.90 m (clear span plus bearing at both ends)
   B. 6.00 m (clear span only)
   C. 6.45 m (one bearing)
   D. 7.70 m (overall width)
   Answer: A. Decking rests on bearings at both ends so length is 6.90 m. (Key point: Decking length)

2. How is the slab volume found?
   A. 7.00 x 6.90 x 0.20 = 9.66 m3
   B. 7.70 x 6.90 x 0.20 = 10.626 m3
   C. 7.70 x 6.00 x 0.20 = 9.24 m3
   D. 7.70 x 6.90 x 0.10 = 5.31 m3
   Answer: B. Slab spans full overall width 7.70 m and decking length 6.90 m. (Key point: Slab volume)

3. How is the volume of the three ribs found?
   A. 3 x 0.35 x 1.075 x 6.90
   B. 3 x 2.25 x 0.75 x 6.90
   C. 3 x 0.35 x 0.75 x 6.90 = 5.434 m3
   D. 3 x 0.35 x 0.75 x 6.00
   Answer: C. Each rib is 0.35 m wide and projects 0.75 m below the slab. (Key point: Rib volume)

4. What is the wearing coat quantity?
   A. 7.70 x 6.90 x 0.10 = 5.313 m3
   B. 7.00 x 6.00 x 0.10 = 4.20 m3
   C. 7.00 x 6.90 x 0.20 = 9.66 m3
   D. 7.00 x 6.90 x 0.10 = 4.830 m3, laid between the kerbs over the roadway
   Answer: D. The coat covers the 7 m roadway, not the kerbs. (Key point: Wearing coat)

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