Skip to main content

RCC slab culvert (u): four items

Short answer

Excavation:5.936 m³. Plain cement concrete foundation:2.544 m³. Brickwork:4.770 m³. RCC work in slab:3.339 m³.

Estimate the quantities of the following items of work from the accompanying drawing. a) Earthwork in excavation b) Cement concrete in foundation c) Brick work d) RCC work [Figure: RCC slab culvert; roadway 4 m, overall length 530 cm, 8 cm C.C. wearing coat, 22 cm thick road metal, slab 30 cm with 10 mm dia @ 22 c/c and 16 mm dia @ 10 c/c bars (alternate bars cranked), wall 50 cm × 90 cm, footing 80 cm × 30 cm, 45 degree earth slope, 150 cm, 140 cm, 120 cm, 500 cm length, 15 cm offsets. All dimensions in centimetres.]

Reading of the drawing (dimensions in the scan are hard to read; assumed values are stated)

  • Overall length of culvert across the road 5.305.30 m (4.00 m roadway plus kerbs). Clear span 1.501.50 m.
  • Abutment: wall 0.500.50 m thick ×\times 0.900.90 m high; C.C. foundation (1:3:6 assumed) 0.800.80 m wide ×\times 0.300.30 m thick; excavation to 0.700.70 m depth (0.30 m C.C. plus 0.40 m below bed level).
  • Slab 0.300.30 m thick resting 0.300.30 m on each abutment (span 1.50+2×0.30=2.101.50 + 2 \times 0.30 = 2.10 m). The wearing coat and road metal are not asked.

a) Earthwork in excavation

V=2×0.80×5.30×0.70=5.936V = 2 \times 0.80 \times 5.30 \times 0.70 = 5.936 m³

b) Cement concrete in foundation

V=2×0.80×0.30×5.30=2.544V = 2 \times 0.80 \times 0.30 \times 5.30 = 2.544 m³

c) Brickwork

V=2×0.50×0.90×5.30=4.770V = 2 \times 0.50 \times 0.90 \times 5.30 = 4.770 m³

d) RCC work (slab)

V=5.30×2.10×0.30=3.339V = 5.30 \times 2.10 \times 0.30 = 3.339 m³

Answer: excavation = 5.936 m³; C.C. foundation = 2.544 m³; brickwork = 4.770 m³; RCC slab = 3.339 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Estimating and Costing chapter 6 questions →