RCC Beam Material Calculator
Calculate concrete volume, steel reinforcement weight, and shuttering area for any RCC beam from dimensions, IS 456 grade, and reinforcement schedule.
Reinforcement schedule
How this works
Concrete is sized from the rectangular cross-section, materials are taken off using the IS 456 dry-volume rule, and steel weight uses the standard d² / 162.28 kg/m formula:
concreteVol = L × b × d
dryVol = wetVol × 1.54
cementBags = ceil(dryVol × c / parts / 0.0347)
mainBarKg/m = barDiameter² / 162.28
mainBarKg = barCount × L × mainBarKgPerM
stirrupCount = ceil(L / spacing) + 1
stirrupKg = stirrupCount × 2(b+d) × stirrupKgPerM
steelKg = mainBarKg + stirrupKg
shutterM² = 2(b+d) × LWorked example
A 4 m long M20 beam, 230 mm × 450 mm, with 4 × 16 mm main bars and 8 mm stirrups at 150 mm spacing:
- Concrete =
4 × 0.23 × 0.45 ≈ 0.414 m³ - Main bar weight per metre =
16² / 162.28 ≈ 1.58 kg/m - Main bar total ≈ 4 × 4 × 1.58 ≈ 25.3 kg
- Stirrup count =
ceil(4 / 0.15) + 1 = 28 - Stirrup total ≈ 28 × 1.36 × 0.395 ≈ 15 kg
- Steel ≈ 40 kg
- Shutter ≈ 2 × 0.68 × 4 = 5.44 m²
Sources
- IS 456:2000 §9.2 / §10.3; standard d²/162.28 field formula for round-bar weight per metre.
FAQ
How is the steel weight per metre derived from bar diameter?
The widely used field formula is weight per metre = d² / 162.28 where d is the bar diameter in millimetres and the result is in kg/m. So a 16 mm bar weighs 16² / 162.28 ≈ 1.58 kg/m, and an 8 mm stirrup weighs 0.395 kg/m. The 162.28 constant comes from steel's density (7,850 kg/m³) and the cross-section formula for a circle, simplified for working in mm and kg.
Why does the calculator multiply (b + d) by 2 × L for shutter area?
A typical RCC beam shutter wraps the two vertical sides plus the bottom (soffit). The total surface area is 2 × (b + d) × L — two side faces of height d and one soffit of width b, all running the full length L. The simplification 2(b+d)×L is the conventional field rule and slightly over-counts when the soffit isn't full-width, which is acceptable for shuttering procurement.
How many stirrups will I need for a typical residential beam?
Stirrup count depends on spacing. For a 4-metre beam at 150 mm spacing the count is ceil(4 / 0.15) + 1 ≈ 28 stirrups; at 200 mm spacing it falls to ceil(4 / 0.20) + 1 ≈ 21. The +1 accounts for stirrups at both ends. Spacing in shear-critical zones (close to supports) is usually tightened to 100–125 mm, which the calculator does not differentiate — use the average spacing for early estimates.
What stirrup length should I assume?
Stirrup length is approximately 2 × (b + d) less the cover at all four faces, plus two 90° hooks of about 10 × stirrup diameter. For a 230 × 450 beam with 25 mm cover and 8 mm stirrups, the developed length is about 2 × (230 - 50 + 450 - 50) + 2 × 80 = 1,320 mm. The calculator uses the simpler 2 × (b + d) approximation, which slightly over-estimates the steel and is conservative.
Why is the main bar count typically 4 or 6?
For nominal residential beams (230 × 300 to 230 × 450) the main reinforcement is normally 2 bars at the bottom plus 2 anchor bars at the top, giving 4 bars total. Heavier loads or wider beams may use 3 + 2 or 3 + 3 (5 to 6 bars). The exact count comes from the structural drawing — at planning stage 4 is the safe default for slab-supporting beams.
Does this calculator account for laps and development length?
No. The straight bar length is taken as the beam length, and the effective steel weight ignores laps, splices, and development-length anchorages at the supports. In a complete bar-bending schedule those add roughly 5–8 percent to the weight. Add a similar wastage allowance to the steel kg figure when placing the procurement order.