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RCC and Steel Quantity Estimation

Learn how to estimate steel reinforcement for beams, columns, slabs, and lintels. Bar bending schedule, weight calculation, and typical steel percentages.

5 min read
By Yojo Team
प्रकाशित: १० मार्च, २०२६
RCC and Steel Quantity Estimation - Yojo construction management blog

Introduction

Reinforced cement concrete (RCC) requires accurate steel quantities for ordering, costing, and site control. Steel is measured in quintals (100 kg) or tonnes as per IS 1200.

This guide covers:

  • Bar bending schedule (BBS) basics
  • Length and weight formulae
  • Typical steel percentages for different RCC elements

Unit of Measurement

ItemUnitPayment
Steel reinforcement in RCCQuintalper quintal
Bending and binding of steelQuintalper quintal

1 quintal = 100 kg. 1 tonne = 10 quintals.

Steel Weight Formula

For round bars:

Weight per metre (kg/m) = (π × d²/4) × 10⁻⁶ × 7860

Where d = diameter in mm, 7860 = density of steel (kg/m³).

Simplified:

  • 8 mm = 0.395 kg/m
  • 10 mm = 0.617 kg/m
  • 12 mm = 0.89 kg/m
  • 16 mm = 1.58 kg/m
  • 20 mm = 2.47 kg/m
  • 25 mm = 3.85 kg/m

Total weight = Σ (Number of bars × Length per bar × Weight per m)

Bar Bending Schedule (BBS)

BBS is a table with:

  • Name – Main bar, anchor bar, cranked bar, stirrup
  • Diameter – 8, 10, 12, 16 mm, etc.
  • Number – No. of bars
  • Shape – Straight, bent, U-shaped
  • Length – Per bar in m
  • Total length – No. × Length
  • Weight – Total length × kg/m

Length Calculations

Clear Span and Support

  • Effective span = Centre to centre of supports (or clear span + effective depth, as per code)
  • Development length = 50d or as per IS 456 (d = bar diameter)
  • Hook = 9d or 12d at each end

Main Bars (Beam)

Length = Effective span + 2 × (development length or 9d)

Example: Span 4000 mm, support 230 mm, bar 16 mm

  • Length = 4000 + 2×230 − 2×32 (cover) + 2×(9×16) ≈ 4396 + 288 = 4684 mm per bar

Cranked Bars (Bent-up Bars)

Additional length per crank = 0.414 × d (effective depth)

Example: Effective depth 434 mm, two cranks

  • Extra = 2 × 0.414 × 434 ≈ 359 mm

Stirrups

Stirrup length = 2 × (Height + Width) − (4 × cover) + 2 × hook

Example: Beam 230×500 mm, cover 25 mm, 6 mm stirrups

  • Height = 500 − 50 = 450 mm
  • Width = 230 − 50 = 180 mm
  • Length = 2×(450+180) + 2×9×6 = 1260 + 108 = 1368 mm

Number of stirrups = (Length / spacing) + 1. For different zones, calculate each zone separately.

Typical Steel Percentages

From rate analysis and design practice:

ElementSteel (% of concrete vol)Example
Column footings, bed blocks0.5%0.5/100 × 7.85 ≈ 0.04 t/cum
Columns, beams, pedestals1–1.5%1.5/100 × 7.85 ≈ 0.117 t/cum
Slabs, lintels (up to 100 mm thk)~1%1/100 × 7.85 ≈ 0.0785 t/cum

Density of steel ≈ 7.85 t/cum (7850 kg/cum).

Worked Example: Steel in RCC Slab

Given: Slab 5 m × 4 m × 0.12 m = 2.4 cum, 1% steel

  • Steel = 0.01 × 2.4 × 7850 = 188.4 kg ≈ 1.88 quintals

If using rate analysis with built-in steel: Rate for RCC slab often includes 1% steel. Verify from your schedule of rates.

RCC Rate Breakdown (Illustrative)

Typical components in rate analysis:

ItemRate (₹/cum) approxNotes
PCC (1:2:4) base2,300–2,400Excludes steel
Centering/shuttering430–710Varies by element
Steel @ 1–1.5%As per steel rateOften ₹27,500–34,500/t
RCC slab (1:2:4)~6,000/cumIncludes 1% steel, centering
RCC column/beam~7,400/cumIncludes 1.5% steel, centering

Rates vary by location and schedule. Use your labour and material rates for your area.

Bar Bending Schedule Format

NameDiaNo.Length (m)Total (m)kg/mWeight (kg)
Main bars1624.689.361.5814.79
Anchor bars1224.619.220.898.21
Cranked bars1625.0410.081.5815.93
Stirrups6171.3723.290.225.12
Total44.05 kg

Practical Tips

  1. Wastage – Add 2–5% for cutting and bending wastage.
  2. Binding wire – ~1 kg per quintal of steel (or as per practice).
  3. Lap length – For lapped bars, add lap length as per IS 456.
  4. Verify – Cross-check with typical percentages (0.5–1.5%) for sanity.

Next Steps

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