Find area, centroid, moments of inertia, and radii of gyration for composite sections made from simple shapes, including holes, with worked steps.
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Worked examples
These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.
Example 1
T-section: 100×20 flange on 20×80 web (mm)
Given
parts
1.shape rectanglewidth 20 mmheight 80 mmcentroid x 0 mmcentroid y 40 mm
2.shape rectanglewidth 100 mmheight 20 mmcentroid x 0 mmcentroid y 90 mm
Assumptions
Ix and Iy are about the composite section's own centroidal axes, parallel to the global x and y axes. Products of inertia are not computed — for unsymmetric sections these are NOT the principal moments of inertia.
Right-triangle parts have both legs parallel to the axes; mirrored orientations share the same centroidal Ix and Iy (only Ixy, not computed, changes sign).
Solution steps
Part areas and own centroidal inertias
Each primitive part contributes its area (negative for holes) at its own centroid, plus its own centroidal moment of inertia from the shape formulas. A semicircle's centroid sits 4r/(3π) from its flat face, along the symmetry axis.
semicircle centroid offset = 4r/(3π); I(rectangle) = bh³/12; I(circle) = πd⁴/64; I(right triangle) = bh³/36; I(semicircle) = (π/8 − 8/(9π))r⁴ about the flat-parallel axis and πr⁴/8 about the symmetry axis
part 1: A = 1600 mm^2 at (0 mm, 40 mm), own Ix = 853300 mm^4, own Iy = 53330 mm^4; part 2: A = 2000 mm^2 at (0 mm, 90 mm), own Ix = 66670 mm^4, own Iy = 1667000 mm^4
Composite centroid
The centroid is the area-weighted average of the part centroids; hole areas enter with a negative sign.
x̄ = Σ(Ai·xi)/ΣAi and ȳ = Σ(Ai·yi)/ΣAi (signed areas)
A = 3600 mm^2; x̄ = 0 mm; ȳ = 67.78 mm
= 67.78 mm
Parallel-axis theorem
Transfer each part's own centroidal inertia to the composite centroidal axes by adding A·d², where d is the distance from the part centroid to the composite centroid. Hole terms subtract.
The radius of gyration concentrates the area at a single distance from the axis with the same inertia.
r = √(I/A)
rx = 29.54 mm; ry = 21.86 mm
Results
Net cross-sectional area A
3600mm^2
Centroid x̄ (global axes)
0mm
Centroid ȳ (global axes)
67.78mm
Centroidal moment of inertia Ix
3142000mm^4
Centroidal moment of inertia Iy
1720000mm^4
Radius of gyration rx
29.54mm
Radius of gyration ry
21.86mm
Where this answer was checked
source
Parallel-axis theorem with rectangle formulas (NCEES FE Reference Handbook) — composite T: A, ȳ, Ix, Iy about centroidal axes
verified by
hand-recomputed
derivation
Web 20×80 centroid (0,40), A=1600; flange 100×20 centroid (0,90), A=2000. A=3600. ȳ=(1600·40+2000·90)/3600=244000/3600=67.7778 mm. Ix: flange 100·20³/12=66666.7 + 2000·(90−67.7778)²=2000·493.827=987654.3 → 1054321.0; web 20·80³/12=853333.3 + 1600·(40−67.7778)²=1600·771.605=1234567.9 → 2087901.2; ΣIx=3142222 mm⁴. Iy=20·100³/12 + 80·20³/12 = 1666666.7+53333.3=1720000 mm⁴. rx=√(3142222/3600)=29.544 mm; ry=√(1720000/3600)=21.858 mm.
Example 2
pipe section: 100 mm circle with concentric 60 mm hole
Given
parts
1.shape circlediameter 100 mmcentroid x 0 mmcentroid y 0 mm
2.shape circlediameter 60 mmcentroid x 0 mmcentroid y 0 mmnegative true
Assumptions
Ix and Iy are about the composite section's own centroidal axes, parallel to the global x and y axes. Products of inertia are not computed — for unsymmetric sections these are NOT the principal moments of inertia.
Right-triangle parts have both legs parallel to the axes; mirrored orientations share the same centroidal Ix and Iy (only Ixy, not computed, changes sign).
Solution steps
Part areas and own centroidal inertias
Each primitive part contributes its area (negative for holes) at its own centroid, plus its own centroidal moment of inertia from the shape formulas. A semicircle's centroid sits 4r/(3π) from its flat face, along the symmetry axis.
semicircle centroid offset = 4r/(3π); I(rectangle) = bh³/12; I(circle) = πd⁴/64; I(right triangle) = bh³/36; I(semicircle) = (π/8 − 8/(9π))r⁴ about the flat-parallel axis and πr⁴/8 about the symmetry axis
part 1: A = 7850 mm^2 at (0 mm, 0 mm), own Ix = 4909000 mm^4, own Iy = 4909000 mm^4; part 2 (hole): A = −2830 mm^2 at (0 mm, 0 mm), own Ix = 636200 mm^4, own Iy = 636200 mm^4
Composite centroid
The centroid is the area-weighted average of the part centroids; hole areas enter with a negative sign.
x̄ = Σ(Ai·xi)/ΣAi and ȳ = Σ(Ai·yi)/ΣAi (signed areas)
A = 5027 mm^2; x̄ = 0 mm; ȳ = 0 mm
= 0 mm
Parallel-axis theorem
Transfer each part's own centroidal inertia to the composite centroidal axes by adding A·d², where d is the distance from the part centroid to the composite centroid. Hole terms subtract.
The radius of gyration concentrates the area at a single distance from the axis with the same inertia.
r = √(I/A)
rx = 29.15 mm; ry = 29.15 mm
Results
Net cross-sectional area A
5027mm^2
Centroid x̄ (global axes)
0mm
Centroid ȳ (global axes)
0mm
Centroidal moment of inertia Ix
4273000mm^4
Centroidal moment of inertia Iy
4273000mm^4
Radius of gyration rx
29.15mm
Radius of gyration ry
29.15mm
Where this answer was checked
source
Hollow circle closed form I = π(D⁴−d⁴)/64 (NCEES FE Reference Handbook) — annulus as circle + negative circle
verified by
hand-recomputed
derivation
A = π(50²−30²) = π·1600 = 5026.55 mm². I = π(100⁴−60⁴)/64 = π·8.704e7/64 = 4272566 mm⁴ both axes. r = √(I/A) = √((50²+30²)/4) = √850 = 29.155 mm (exact identity for an annulus).
Example 3
US units: 6×4×1 in angle from two rectangles
Given
parts
1.shape rectanglewidth 1 inheight 6 incentroid x 0.5 incentroid y 3 in
2.shape rectanglewidth 3 inheight 1 incentroid x 2.5 incentroid y 0.5 in
Assumptions
Ix and Iy are about the composite section's own centroidal axes, parallel to the global x and y axes. Products of inertia are not computed — for unsymmetric sections these are NOT the principal moments of inertia.
Right-triangle parts have both legs parallel to the axes; mirrored orientations share the same centroidal Ix and Iy (only Ixy, not computed, changes sign).
Solution steps
Part areas and own centroidal inertias
Each primitive part contributes its area (negative for holes) at its own centroid, plus its own centroidal moment of inertia from the shape formulas. A semicircle's centroid sits 4r/(3π) from its flat face, along the symmetry axis.
semicircle centroid offset = 4r/(3π); I(rectangle) = bh³/12; I(circle) = πd⁴/64; I(right triangle) = bh³/36; I(semicircle) = (π/8 − 8/(9π))r⁴ about the flat-parallel axis and πr⁴/8 about the symmetry axis
part 1: A = 6 in^2 at (0.5 in, 3 in), own Ix = 18 in^4, own Iy = 0.5 in^4; part 2: A = 3 in^2 at (2.5 in, 0.5 in), own Ix = 0.25 in^4, own Iy = 2.25 in^4
Composite centroid
The centroid is the area-weighted average of the part centroids; hole areas enter with a negative sign.
x̄ = Σ(Ai·xi)/ΣAi and ȳ = Σ(Ai·yi)/ΣAi (signed areas)
A = 9 in^2; x̄ = 1.167 in; ȳ = 2.167 in
= 2.167 in
Parallel-axis theorem
Transfer each part's own centroidal inertia to the composite centroidal axes by adding A·d², where d is the distance from the part centroid to the composite centroid. Hole terms subtract.