How is the brickwork in the foundation found?

2 abutments x length x sum of (width x height) of each step

1 abutment x length x top step only

Why is the wearing coat width 3.60 m, not 4.00 m?

The 20 cm kerbs on each side are deducted from the slab width

The pipe railing occupies 40 cm

The wearing coat is thinner at the edges

Length of GI railing pipe is

2 x culvert length = 8.60 m, one line each side

Steel in the slab at 1.5% is found as

0.015 x slab concrete volume x 7850 kg/m3

2082 Bhadra

RCC slab culvert: brick, wearing coat, GI pipe, steel

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.

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.]

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

RCC slab culvert (u): four items · Estimating and Costing, chapter 6
RCC slab culvert 1.5 m span: four items · Estimating and Costing, chapter 6
Underground RCC tank: brickwork, steel (fig a) · Estimating and Costing, chapter 6
T-beam decking: RCC and wearing coat · Estimating and Costing, chapter 6 · asked 2×
Underground RCC tank: brickwork 1:4, M20 PCC, steel · Estimating and Costing, chapter 6
Slab culvert: excavation and brickwork · Estimating and Costing, chapter 6
Horizontal crack below RCC slab · Building Technology, chapter 10
First class brickwork in 1:6 mortar numerical · Building Technology, chapter 3
Brick and mortar for 230 mm wall (100 kN/m) · Concrete Technology and Masonry Structure, chapter 8 · asked 7×

All Estimating and Costing chapter 6 questions →