Know How To Calculate Quantity Of Steel Requirement For Per m3 | Slab | Lintel | Beam | Column | Foundation | -lceted LCETED INSTITUTE FOR CIVIL ENGINEERS

## May 23, 2021

The General Things You Should Know To Calculate Quantity Of Steel Requirement For Per m3 for Slab, Lintel, Beam, Column and Foundation

The density of steel varies based on the alloying constituents but usually ranges between 7,750 and 8,050 kg/m3 (484 and 503 lb/cu ft), or 7.75 and 8.05 g/cm3 (4.48 and 4.65 oz/cu in)

Quantity Of Steel Requirement For Slab/Lintel Per m3:

Minimum Percentage Of Steel = 0.7%

Therefore, Quantity of steel = (0.7/100) x 1 = 0.007 m³
Weight of steel = 0.007 x 7750 = 54.25 - 55 kg/m³

Maximum percentage of steel =1.0%

Therefore, Quantity of steel = (1.0/100) x 1 = 0.01 m³
Weight of steel = 0.01 x 7750 = 77.5 kg /m³

Quantity Of Steel Requirement For

Beam Per m3:

Minimum percentage of steel =1.0%

Therefore, Quantity of steel = (1.0/100) x 1 = 0.01 m³
Weight of steel = 0.01 x 7750 = 77.5 kg /m³

Maximum percentage of steel =2.0%

Therefore, Quantity of steel = (2.0/100) x 1 = 0.02 m³
Weight of steel = 0.02 x 7750 = 155 kg/m³

Quantity Of Steel Requirement For

Column Per m3:

Minimum percentage of steel =0.7%

Therefore, Quantity of steel = (0.8/100) x 1 = 0.008 m³
Weight of steel = 0.008 x 7750 = 62 kg/m³

Maximum percentage of steel =6.0%

Therefore, Quantity of steel = (6.0/100) x 1 = 0.06 m³
Weight of steel = 0.06 x 7750 = 465 kg /m³

Quantity Of Steel Requirement For

Foundation Per m3:

Minimum percentage of steel =0.7%

Therefore, Quantity of steel = (0.5/100) x 1 = 0.005 m³
Weight of steel = 0.005 x 7750 = 38.75 kg/m³

Maximum percentage of steel =0.8%

Therefore, Quantity of steel = (1.0/100) x 1 = 0.008 m³
Weight of steel = 0.008 x 7750 = 62 kg /m³

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