Circle Section Properties Calculator
Enter the diameter of a solid round bar or shaft and get its area, second moment of area, polar moment, section moduli and radius of gyration, drawn to scale.
Section dimensions
Enter the dimensions in the selected unit — the drawing and the results update as you type. The drawing uses centroidal coordinates: the bold gridlines are the x and y axes through the centroid C.
Section
- Section with its dimensions
- Centroid C, centroidal axes x and y
- Principal axes 1 and 2, angle θp
- Shear centre S
The solid circle is the section of shafts, pins, round bars and rivets. It is the one shape where every axis through the centre is a principal axis (Ix = Iy, Ixy = 0 whatever the orientation) and where the polar moment of area Ip = πd⁴/32 is also the exact torsion constant J — so the same number serves for the torsional shear stress τ = T·r/J and the angle of twist θ = T·L/(G·J).
For bending, the relevant figure is the elastic section modulus S = πd³/32: the bending stress at the surface is σ = M/S. Because S grows with the cube of the diameter, a 10% larger shaft is about 33% stronger in bending and 46% stiffer.
How the circle properties are calculated
- Area
- A = π·d² / 4
- Second moment of area
- Ix = Iy = π·d⁴ / 64
- Polar moment = torsion constant
- Ip = J = π·d⁴ / 32
- Elastic section modulus
- S = π·d³ / 32
- Plastic section modulus
- Z = d³ / 6
- Radius of gyration
- r = d / 4
Principal moments and axes follow from Ix, Iy and Ixy with I1,2 = (Ix + Iy)/2 ± √[((Ix − Iy)/2)² + Ixy²] and tan 2θp = −2·Ixy/(Ix − Iy); the radii of gyration are r = √(I/A) and the elastic section moduli S = I/c for each extreme fibre. Definitions of every property are in the glossary on the section properties overview.
Assumptions. All formulas are exact for a perfect circle. The results are purely geometric — they do not depend on the material.
Other sections
Frequently asked questions
Why is the polar moment of a circle twice its moment of inertia?
The polar moment about the centre is I_p = ∫ r² dA = ∫ (x² + y²) dA = I_x + I_y. For a circle I_x and I_y are identical by symmetry, so I_p = 2·I = πd⁴/32. This is the only common solid section where I_p is also the true torsion constant J.
How do I get the torsional stress of a shaft from these numbers?
The maximum shear stress from a torque T is τ = T·(d/2)/J, and the polar section modulus is J/(d/2) = πd³/16 — exactly twice the bending section modulus S. The angle of twist over a length L is θ = T·L/(G·J), with G the shear modulus of the material.
Does a keyway change the section properties?
Slightly for I and S (a standard keyway removes only a few percent of the area), but significantly for stress, because the keyway corners act as stress raisers. This calculator gives the plain circular section; apply a stress-concentration factor separately for the keyway.
References & further reading
- List of second moments of area — Wikipedia — closed-form Ix and Iy for the common shapes, used to check this calculator.
- Section modulus — Wikipedia — elastic and plastic section modulus, with a table of formulas by shape.
- Torsion constant — Wikipedia — why J differs from the polar moment for non-circular sections, with the thin-walled formulas.
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