DRV-018
Why steel weight is d² ÷ 162
The setup
A reinforcement bar is a cylinder. Its mass per metre is its density times its cross-sectional area, and everything else in the famous shortcut is unit bookkeeping.
The derivation
Step 1: Write mass as density times volume
For a bar of diameter d millimetres and length L metres, with steel density 7850 kg/m³.
W = density * area * length = 7850 * (pi * d^2 / 4) * 1e-6Step 2: Collapse the constants
Everything except d² is a single number: 7850π ÷ 4 ÷ 10⁶ is 1 ÷ 162.196.
W = d^2 / 162.196 kg per metreweight_per_metre = d^2 / 162.196Checks that prove it is right
A formula you cannot test is a formula you have to trust. These take seconds, and they are what separate a derivation from a formula restatement.
d = 12 mm → 0.8878 kg/m exactly; 0.8889 by the shortcut
The rounded denominator overstates by about 0.12%, which is immaterial for ordering steel and worth stating rather than hiding.
Doubling the diameter → Four times the weight
Mass goes as the square of diameter, which is why substituting a 16 mm bar for a 12 mm one is a 78% weight increase rather than a 33% one.
Worked example
100 bars of 16 mm at 12 m each.
Per metre: 7850 × π × 256 ÷ 4 ÷ 10⁶ = 1.5783 kg. Total length is 1,200 m, so the order is 1,893.5 kg, call it 1.9 tonnes. The shortcut gives 1.5802 kg/m and a total of 1,896.3 kg, a difference of under 3 kg on nearly two tonnes.
Computed by the material-quantity engine; the exact denominator asserted as a golden case.
What it assumes
- Steel density 7850 kg/m³, the standard value for mild and TMT reinforcement.
- Nominal diameter, perfectly circular section.
Where the formula stops being valid
Stating limits plainly is more useful than pretending there are none, and it is the item competitors most consistently omit.
- Rolling tolerances mean actual bar mass varies by a percent or two from nominal, and standards permit it.
- Deformed bars carry ribs that add mass the nominal-diameter calculation ignores; mill test certificates quote the real figure.