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Technical guide · method and practical tips

Metal tube mass: from diameter and thickness to self-weight

One circular tube; outside diameter and thickness in mm, piece length in m, density in kg/m³.

Reviewed

D = 60 mmd = 54 mmt = 3 mmd = D − 2tA = π(D² − d²)/4
Concentric, uniform wall: d = D − 2t. Dimensions describe an ideal tube, without tolerances or coating.

Numerical application

Result

One ideal 60 × 3 mm tube, length 6 m
QuantityResult
d [mm]54
A [mm²]537
μ [kg/m]4.22
m [kg]25.3
qself [N/m]41.4

Unrounded intermediate values are retained in the calculation. Values here are rounded for reading.

Check the actual dimensions

d = D − 2t

Calculate area and linear mass

A = π(D² − d²)/4 ; μ = ρA

Convert linear mass to a gravity load

m = μL ; qself = μg

Know when to use a catalogue mass

This ideal geometry excludes dimensional tolerances, welds, coatings and manufacturing details. A verified manufacturer mass may differ and is preferable when that exact product is being specified.

A uniform gravity load does not establish bending resistance. After defining supports, span, load direction and material assumptions, check stress and deflection using section properties consistent with the actual tube.

For checking and further study

Direct references

Each link points to the exact course, standard or publication page used, rather than a generic homepage.

  1. Guide for the Use of the International System of UnitsNISTSI quantities, mass versus force and coherent unit conversions.
  2. The International System of Units, 9th editionBIPMMetre, kilogram, second and derived SI units.