---
title: "T-beam decking quantities (6 m clear span)"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:t-beam-decking-all-quantities"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/t-beam-decking-all-quantities
site: BCT Engineering
---

# T-beam decking quantities (6 m clear span)

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2076 Chaitra, 8 marks
- 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-all-quantities
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Estimate the quantities of a T-beam decking of a single span bridge which has 6 m clear span and bearing on either side is 45 cm, from the accompanying bridge drawing. (Reinforcement: beam 2½%, slab and posts 1%.)
[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 (1:2:4) = 17.402 m³, wearing coat = 4.830 m³, total reinforcement = 2060.7 kg (2.5% steel in beams/fillets, 1% in slab/kerbs/posts), 40 mm G.I. pipe = 55.2 m.

## Answer

#### Assumptions from the drawing

- 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.
- Reinforcement as given: beams (ribs and fillets) 2.5%, slab and kerbs 1%, posts 1% of the volume of concrete; density of steel 7850 kg/m³.
- The railing has 4 lines of 40 mm G.I. pipe on each side, running the full length of the decking (assumption; the figure shows "4 nos").

#### Quantities

| 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** |

| Item | Working | Quantity |
|---|---|---:|
| Wearing coat C.C. (1:2:4) | $7.00 \times 6.90 \times 0.10$ | 4.830 m³ |
| Steel in beams | $(5.434 + 0.466) \times 0.025 \times 7850$ | 1,157.8 kg |
| Steel in slab and kerbs | $(10.626 + 0.552) \times 0.01 \times 7850$ | 877.5 kg |
| Steel in posts | $0.324 \times 0.01 \times 7850$ | 25.4 kg |
| **Total steel** | | **2,060.7 kg** |
| G.I. pipe 40 mm dia. | $4 \times 2 \times 6.90$ | 55.2 m |

**Answer: RCC (1:2:4) = 17.402 m³; wearing coat = 4.830 m³; reinforcement = 2,060.7 kg; 40 mm G.I. pipe = 55.2 m.**

## 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 m (clear span only)
   B. 6.45 m (one bearing)
   C. 7.70 m (overall width)
   D. 6.00 + 2 x 0.45 = 6.90 m (clear span plus bearing at both ends)
   Answer: D. 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. 7.70 x 6.90 x 0.20 = 10.626 m3
   B. 7.00 x 6.90 x 0.20 = 9.66 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: A. 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 0.35 x 0.75 x 6.90 = 5.434 m3
   C. 3 x 2.25 x 0.75 x 6.90
   D. 3 x 0.35 x 0.75 x 6.00
   Answer: B. Each rib is 0.35 m wide and projects 0.75 m below the slab. (Key point: Rib volume)

4. How is steel in the beams found?
   A. 5.900 x 1% x 7850
   B. 10.626 x 2.5% x 7850
   C. (ribs + fillets) 5.900 m3 x 2.5% x 7850 = 1,157.8 kg
   D. 5.900 x 0.025 x 785
   Answer: C. Beams use 2.5% steel of concrete volume, with steel density 7850 kg/m3. (Key point: Steel in beams)

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