--- title: "RCC slab culvert: brick, wearing coat, GI pipe, steel" subject: "Estimating and Costing" chapter: 6 key: "semester_7/estimating_and_costing:slab-culvert-a-fig-b" source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/slab-culvert-a-fig-b site: BCT Engineering --- # RCC slab culvert: brick, wearing coat, GI pipe, steel Estimating and Costing · Chapter 6: Detailed Estimate - Asked in: 2082 Bhadra - 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-a-fig-b - Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6 ## Question Prepare a detailed estimate for the following items of works for a RCC slab culvert given below (Fig. b): a) First class brickwork in foundation b) Wearing coat in road of P.C.C (1:2:4) c) 100 mm dia. GI pipe in railing works d) Steel reinforcement in RCC slab (assume 1.5% steel) [Figure b: slab culvert. Plan: 75 cm | 280 cm | 75 cm, culvert length 430 cm, abutment lengths 140 cm, 160 cm, 140 cm, 15 cm offsets, 95 cm and 45 cm sections; slab 400 cm × 25 cm with 20 cm kerb; elevation: 100 mm dia GI railing pipe, 30 × 30 cm concrete posts @ 100 cm c/c, 15 cm wearing coat, 100 cm and 30 cm heights; sectional elevation: road level and bed level, 15 cm wearing coat, 440 cm, 160 cm, 70 cm, 30 cm, 95 cm, 85 cm, 140 cm, 40 cm, 30 cm. Details hard to read in the scan.] ## Short answer First class brickwork in foundation:8.954 m³. PCC wearing coat:2.322 m³. 100 mm GI pipe:8.60 m. Steel reinforcement at1.5%:506.3 kg. ## Answer #### Reading of the figure (dimensions are hard to read, these values are assumed) - Clear span $2.80$ m with $0.75$ m bearing on each abutment: length of the slab (along the road, in the span direction) $= 0.75 + 2.80 + 0.75 = 4.30$ m. - Abutment length (along the stream, i.e. across the road) $= 1.40 + 1.60 + 1.40 = 4.40$ m. Slab $4.00$ m wide and $0.25$ m thick with $0.20$ m kerbs, so roadway between kerbs $= 4.00 - 0.40 = 3.60$ m. - Brickwork foundation of each abutment in two steps: $0.95$ m wide $\times$ $0.40$ m high (lower) and $0.75$ m wide $\times$ $0.85$ m high (upper). - Wearing coat $0.15$ m; railing: one line of 100 mm GI pipe on each side along the length of the culvert. #### a) First class brickwork in foundation $V = 2 \times 4.40 \times (0.95 \times 0.40 + 0.75 \times 0.85) = 8.954$ m³ #### b) Wearing coat, PCC (1:2:4) $V = 3.60 \times 4.30 \times 0.15 = 2.322$ m³ (area 15.48 m²) #### c) 100 mm GI pipe in railing Length $= 2 \times 4.30 = 8.60$ m #### d) Steel in RCC slab (1.5%) Concrete in slab $= 4.00 \times 4.30 \times 0.25 = 4.300$ m³; steel $= 0.015 \times 4.300 \times 7850 = 506.3$ kg **Answer: brickwork = 8.954 m³; wearing coat = 2.322 m³; GI pipe = 8.60 m; steel = 506.3 kg.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the brickwork in the foundation found? A. 1 abutment x length x top step only B. Length x width of road C. 2 x length x slab thickness D. 2 abutments x length x sum of (width x height) of each step Answer: D. Each abutment cross-section is the sum of its steps, multiplied by abutment length, then by two. (Key point: Brick foundation volume) 2. Why is the wearing coat width 3.60 m, not 4.00 m? A. The 20 cm kerbs on each side are deducted from the slab width B. Free board is deducted C. The pipe railing occupies 40 cm D. The wearing coat is thinner at the edges Answer: A. Roadway between kerbs = 4.00 - 2 x 0.20 = 3.60 m. (Key point: Wearing coat width) 3. Length of GI railing pipe is A. Culvert length once = 4.30 m B. 2 x culvert length = 8.60 m, one line each side C. Abutment length 4.40 m x 2 D. 4.00 m Answer: B. One line on each side along the culvert length 4.30 m. (Key point: GI railing length) 4. Steel in the slab at 1.5% is found as A. 0.015 x slab area x 7850 B. 1.5 x slab volume C. 0.015 x slab concrete volume x 7850 kg/m3 D. 0.015 x brickwork x 7850 Answer: C. 1.5% of 4.30 m3 times density gives 506.3 kg. 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