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Section properties

Circular hollow section (tube): A, I, Z, r

d=D−2tA=π(D2−d2)4×100I=π(D4−d4)64×104Z=ID/20r=IA
where
d
inside diameter (mm)
D
outside diameter (mm)
t
wall thickness (mm)
A
area (cm²)
I
second moment of area (cm⁴)
Z
elastic section modulus (cm³)
r
radius of gyration (cm)
Explain this equation

Works out section properties of a circular tube, which are the same about any axis.

The tube is a solid circle of diameter D with a circle of diameter d removed. Area is π/4 times the difference of the squares; second moment is π/64 times the difference of the fourth powers. Z is I divided by the outer radius in cm, D/20. r is √(I/A). The powers of ten convert mm to cm.

Solid-circle formulas, outer minus inner.

Sense check: 168.3 × 6.3 CHS gives about 32 cm², 1053 cm⁴ and 125 cm³, the same as the published tables.

Watch out: tubes are efficient in compression and torsion because their stiffness is equal in every direction.

Properties of round tube from outside diameter and wall thickness. A tube has the same properties about every axis through its centre.

Enter your values to see the result.

Good to know

These are exact for a perfect tube and agree with the Blue Book tables for CHS. Radius of gyration r is used for strut (column) buckling checks: the slenderness is the effective length divided by r.

Common mistake

Using the solid-bar formula πD⁴/64 without subtracting the bore, or entering the nominal pipe bore instead of the outside diameter.

Limits

Elastic properties only. Very thin tubes (D/t over about 90 for S275) are slender and need a local buckling check.

Sources

Results are calculations from the published methods and figures shown, and depend on the data you enter. Follow the welding procedure specification (WPS), BS EN ISO 15614/9606, BS EN 1090, BS EN 1993 and manufacturers’ data. Structural welds need qualified welders and engineered designs.

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