Composite Section Properties

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

1. 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

2. 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

3. 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.

   I = Σ(Īi + Ai·di²), signed

   Ix = (853300 mm^4 + 1600 mm^2 × 27.8 mm²) + (66670 mm^4 + 2000 mm^2 × 22.2 mm²) = 3142000 mm^4; Iy = 1720000 mm^4

   = 3142000 mm^4

4. Radii of gyration

   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

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

1. 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

2. 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

3. 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.

   I = Σ(Īi + Ai·di²), signed

   Ix = (4909000 mm^4 + 7850 mm^2 × 0 mm²) + − (636200 mm^4 + 2830 mm^2 × 0 mm²) = 4273000 mm^4; Iy = 4273000 mm^4

   = 4273000 mm^4

4. Radii of gyration

   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

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

1. 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

2. 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

3. 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.

   I = Σ(Īi + Ai·di²), signed

   Ix = (18 in^4 + 6 in^2 × 0.833 in²) + (0.25 in^4 + 3 in^2 × 1.67 in²) = 30.75 in^4; Iy = 10.75 in^4

   = 30.75 in^4

4. Radii of gyration

   The radius of gyration concentrates the area at a single distance from the axis with the same inertia.

   r = √(I/A)

   rx = 1.848 in; ry = 1.093 in

Results