SCALE 1:1 STD NEC · IEC · AS/NZS · ACI · AISC
B
CodePass.PRO
Engineering Calculators
SHEET 06 / 79
Calculator · Structural · Circle · Pipe · Solid shaft

Area moment of inertia of a circle

The formula π·D⁴/64 for solid circles and π(D⁴ − d⁴)/64 for hollow. Polar moment for shaft torsion. Worked examples, AISC HSS round sections, and a calculator that handles both forms. Reviewed by a licensed PE.

Use the circle moment of inertia calculator

Calculator opens with the solid circle form selected. To compute a pipe (hollow circle), switch the shape dropdown to "Hollow circle / pipe". For AISC HSS round sections, pick from the "Standard section" dropdown — dimensions auto-fill.

CALC.004 Moment of Inertia · 11 shapes · Ix / Iy / J / S / Z / r

Pick an AISC W / IPE / HEA / HEB / HSS / channel / angle to auto-fill dimensions, or leave on "Manual entry" for a custom section.

mm

Solid circle. Ix = Iy = π·D⁴/64. Polar J = π·D⁴/32.

x y
Ix
— mm⁴
Iy
— mm⁴
Polar J
— mm⁴
Sx (elastic)
— mm³
Sy (elastic)
— mm³
Zx (plastic)
— mm³
rx
— mm
ry
— mm
Area
— mm²
Centroid (x̄, ȳ)
FORMULA · Ix = b·h³/12 (rectangle) SOURCE · AISC · ROARK
CIRCULAR SECTIONS · SOLID vs HOLLOW D SOLID CIRCLE D d HOLLOW (PIPE) I = π·D⁴ / 64 J = π·D⁴ / 32 = 2·I I = π·(D⁴ − d⁴) / 64 J = π·(D⁴ − d⁴) / 32
Figure 1 — Solid round bar and hollow tube: identical formula form with the inner cavity subtracted; polar J = 2·I.

The circle moment of inertia formula

Two closed forms cover every circular cross-section. The first is for solid bars and discs; the second extends it to hollow pipes by subtracting the inner-diameter contribution.

Eq. 01 — Solid circle, second moment of area about any centroidal axis SI · Roark's Formulas Table 6.1
Ix=Iy=πD464J=Ix+Iy=πD432I_{x} = I_{y} = \frac{\pi \cdot D^{4}}{64} \qquad J = I_{x} + I_{y} = \frac{\pi \cdot D^{4}}{32}
I_x
second moment of area about horizontal centroidal axis, mm⁴ / in⁴
I_y
about vertical centroidal axis (equal to Ix), mm⁴ / in⁴
J
polar moment of inertia (used in torsion), mm⁴ / in⁴
D
outer diameter, mm / in
Eq. 02 — Hollow circle (pipe / tube) SI · AISC Manual + IEC 60068
Ix=π(D4d4)64J=π(D4d4)32I_{x} = \frac{\pi \cdot (D^{4} - d^{4})}{64} \qquad J = \frac{\pi \cdot (D^{4} - d^{4})}{32}
D
outer diameter, mm / in
d
inner diameter (set d = 0 to recover the solid case), mm / in

Both formulas are derived from the polar-coordinate integral Ipolar = ∫∫ r²·r dr dθ over the cross-section. Rotational symmetry means Ix = Iy, so polar splits into equal halves.

Worked example, 100 mm solid bar vs 100 × 80 mm pipe

Compare a solid 100 mm round bar against a pipe of the same outer diameter with an 80 mm inner diameter (i.e. wall thickness 10 mm). Same outer envelope, but the pipe removes the central 64% of area — yet retains most of the bending stiffness.

PropertySolid 100 mmPipe 100 × 80 mmPipe / Solid
Area A7854 mm²2827 mm²36%
Ix = π·(D⁴ − d⁴)/644.91 × 10⁶ mm⁴2.91 × 10⁶ mm⁴59%
Sx = Ix / (D/2)9.82 × 10⁴ mm³5.8