Semicircle Section Properties Calculator
Enter the radius of a half-round section and get its area, centroid, second moments about the centroidal axes, section moduli for both fibres and radii 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 semicircle is the classic exercise in locating a centroid — it sits at 4r/(3π) ≈ 0.4244·r above the flat edge, not at the middle — and it is also a real section: half-round bars, D-shaped shafts (approximately), the web of a keyed profile, the halves of a split bushing.
Because the centroid is off-centre, the section has two different elastic section moduli about the x axis: the flat fibre is 0.4244·r from the neutral axis, the curved crown 0.5756·r. Bending stress is higher at the crown. The calculator reports both, and finds the plastic modulus about the x axis numerically since the equal-area axis has no closed form.
How the semicircle properties are calculated
- Area
- A = π·r² / 2
- Centroid above the flat edge
- ȳ = 4·r / (3·π) ≈ 0.4244·r
- Second moment about the flat edge
- Ibase = π·r⁴ / 8
- Centroidal second moments
- Ix = (π/8 − 8/(9π))·r⁴ ≈ 0.1098·r⁴, Iy = π·r⁴ / 8
- Elastic section moduli about x
- Sx,top = Ix / (r − ȳ), Sx,bot = Ix / ȳ
- Plastic section modulus about y
- Zy = 2·r³ / 3
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. Flat side at the bottom, arc at the top. Z_x is computed numerically from the equal-area axis (about 0.354·r³); no torsion constant is given for this shape.
Other sections
Frequently asked questions
Why is the centroidal I_x so much smaller than πr⁴/8?
πr⁴/8 is the second moment about the flat edge (the diameter). Moving to the centroid with the parallel-axis theorem subtracts A·ȳ² = (πr²/2)·(4r/3π)² ≈ 0.283·r⁴, leaving about 0.110·r⁴ — the centroidal value is always the minimum over all parallel axes.
Which fibre governs the bending stress of a half-round bar?
The curved crown, which is farther from the neutral axis (0.5756·r against 0.4244·r for the flat face). For the same bending moment the stress at the crown is about 36% higher than at the flat face; use S_x,top when the crown is in tension or compression as the case may be.
Does the calculator give the torsion constant of a semicircle?
No. The Saint-Venant torsion of a semicircular section has no simple closed form (Roark gives J ≈ 0.296·r⁴ from series solutions). The polar moment I_p in the results is I_x + I_y and must not be used as J.
References & further reading
- List of centroids — Wikipedia — centroid positions of the elementary plane shapes.
- List of second moments of area — Wikipedia — closed-form Ix and Iy for the common shapes, used to check this calculator.
- Second moment of area — Wikipedia — definition, parallel-axis theorem and the sign convention for the product of inertia.
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