👉How to Calculate the Quantity of Steel for Columns

🌟This example shows how to estimate the longitudinal reinforcement and stirrup steel for 4 RCC columns.

🗂Given Details

Number of columns = 4

Column height = 3.00 m

Column size = 200 × 300 mm

Longitudinal reinforcement = 4 nos. of 12 mm dia bars/column

Stirrups = 6 mm dia @ 150 mm c/c

Concrete cover = 25 mm

⏺1. Longitudinal Steel — 12 mm Dia

Length of steel:

3.00 × 4 bars × 4 columns = 48.00 m

Unit weight of 12 mm bar = 0.888 kg/m

Weight = 48.00 × 0.888 = 42.62 kg

≈ 43 kg of 12 mm bars

⏺2. Stirrups — 6 mm Dia

Number of spaces:

3.00 ÷ 0.15 = 20

Number of stirrups ≈ 20 + 1 = 21 per column

For 4 columns:

21 × 4 = 84 stirrups

⏺3. Length of One Stirrup

Using the dimensions and 25 mm cover shown:

Centre-line width ≈ 150 mm

Centre-line depth ≈ 250 mm

135° hooks = 10d each

Hook length = 0.06 m each

Approximate stirrup length:

L = 2(0.15 + 0.25) + 2(0.06)
L = 0.92 m/stirrup

⏺4. Total Stirrup Steel

Total length:

84 × 0.92 = 77.28 m

Unit weight of 6 mm bar = 0.222 kg/m

Weight = 77.28 × 0.222 = 17.17 kg

≈ 17.2 kg of 6 mm bars

⬆️Final Quantity for 4 Columns

12 mm longitudinal bars: 48.00 m ≈ 42.6 kg

6 mm stirrups: 84 nos. / 77.28 m ≈ 17.2 kg

Total reinforcement steel ≈ 59.8 kg

🚥Note: This is an illustrative quantity-estimation example. Actual reinforcement, lap lengths, anchorage, bar curtailment, seismic detailing, and stirrup spacing must be based on the structural design and applicable code.

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