Quick reference · Subject 01

Building Materials Formula Library

Useful formulas with symbols, units, assumptions and solved examples.

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14 formulas

Concrete quantity estimation

Concrete volume

V = L × B × H

Units & notes
Volume in m³; dimensions must be in metres.

Worked example
Slab 6 m × 4 m × 0.12 m: V = 6 × 4 × 0.12 = 2.88 m³.

Concrete mix control

Water-cement ratio

w/c = mass of water ÷ mass of cement

Units & notes
Use masses in the same units; ratio is dimensionless.

Worked example
Water = 180 kg and cement = 400 kg: w/c = 180/400 = 0.45.

Cube strength calculations

Compressive strength

f = P ÷ A

Units & notes
P in newtons and A in mm² gives N/mm² (MPa). Use specified specimen dimensions.

Worked example
Load = 600 kN; cube face = 150 × 150 = 22,500 mm². f = 600,000/22,500 = 26.7 MPa.

Reinforcement quantity estimation

Steel bar mass (approx.)

Mass per metre ≈ d² ÷ 162 kg/m

Units & notes
d is nominal bar diameter in millimetres; approximate theoretical value for steel density near 7,850 kg/m³.

Worked example
For 12 mm bar: 12²/162 = 0.889 kg/m. A 6 m bar ≈ 5.33 kg.

Brickwork/masonry quantity

Rectangular wall volume

V = length × height × thickness

Units & notes
Convert all dimensions to metres; deduct openings where relevant.

Worked example
Wall 4 m × 3 m × 0.20 m = 2.40 m³ gross volume.

Flooring, painting, plastering

Rectangular area

A = length × width

Units & notes
Use consistent units; result is square units.

Worked example
Floor 5 m × 4 m = 20 m². With 5% tile wastage: 20 × 1.05 = 21 m².

Paint planning

Paint quantity estimate

Litres = area × coats ÷ coverage

Units & notes
Coverage is m² per litre per coat; add a project-specific allowance for losses.

Worked example
Area 48 m², 2 coats, coverage 8 m²/L: 48 × 2 ÷ 8 = 12 L before allowance.

Aggregate/timber moisture checks

Moisture content (dry basis)

MC (%) = (wet mass − dry mass) ÷ dry mass × 100

Units & notes
Masses must refer to the same sample before and after drying.

Worked example
Wet = 10.6 kg; dry = 10 kg: MC = 0.6/10 × 100 = 6%.

Brick and aggregate testing

Water absorption

Absorption (%) = (saturated mass − dry mass) ÷ dry mass × 100

Units & notes
Use the test procedure and conditioning stated in the applicable standard.

Worked example
Dry = 2.5 kg; saturated = 2.8 kg: absorption = 0.3/2.5 × 100 = 12%.

Material property calculations

Density

ρ = mass ÷ volume

Units & notes
Mass in kg and volume in m³ gives kg/m³.

Worked example
Mass = 18 kg; volume = 0.012 m³: ρ = 18/0.012 = 1,500 kg/m³.

Mortar proportion exercises

Mix proportion by parts

Component volume = total volume × component parts ÷ total parts

Units & notes
Proportion-based estimates assume all parts use the same measurement basis; do not treat nominal ratios as a substitute for a designed mix.

Worked example
For 1:4 cement:sand and 0.50 m³ total dry volume: cement = 0.50 × 1/5 = 0.10 m³.

Preliminary masonry estimation

Brick count estimate

Count ≈ wall area ÷ nominal brick face area

Units & notes
Use nominal dimensions including mortar joints for preliminary estimates; openings and breakage affect final count.

Worked example
Wall area = 12 m²; nominal face area = 0.20 × 0.10 = 0.02 m²; count ≈ 600 bricks.

Procurement estimates

Material with wastage

Order quantity = net quantity × (1 + wastage rate)

Units & notes
Enter wastage as a decimal, e.g. 5% = 0.05. Confirm allowances with site conditions.

Worked example
Net tiles = 32.4 m², wastage = 8%: 32.4 × 1.08 = 34.992 m².

Material procurement

Cement bag count

Number of 50 kg bags = cement mass (kg) ÷ 50

Units & notes
Round up to whole bags for purchase; bag sizes can vary by market.

Worked example
Required cement = 375 kg: 375/50 = 7.5, so purchase at least 8 full bags.

These are learning and preliminary-estimation formulas. Use the applicable IS standards, project specifications, approved mix design and qualified engineering judgement for real construction work.