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MOI of arch section about horizontal centroidal axis

Short answer

Semicircle (r=60 mm) minus a 40×20 mm slot. Centroid ȳ = 28.01 mm from base. Using parallel axis theorem, I_x = 1.173×10⁶ mm⁴.

Find the moment of inertia of the given section about the horizontal centroidal axis. [Figure: arch-shaped section 60 mm high, total width 40 mm + 40 mm + 40 mm, with a 20 mm thick leg and a slot as shown.]

Reading of the figure: the outline is a semicircle of radius 60 mm (width 40 + 40 + 40 = 120 mm, height 60 mm) with its diameter on the base. A rectangular slot 40 mm wide and 20 mm deep is cut out of the bottom, in the middle, leaving two legs 40 mm wide. The section is symmetrical about the vertical axis, so only ȳ is needed. Measure y from the base.

Parts

  1. Semicircle: A₁ = πr²/2 = π(60)²/2 = 5654.87 mm², ȳ₁ = 4r/3π = 25.46 mm
  2. Slot (subtracted): A₂ = 40 × 20 = 800 mm², ȳ₂ = 10 mm

Centroid

ȳ = (A₁ȳ₁ − A₂ȳ₂)/(A₁ − A₂) = (5654.87 × 25.46 − 800 × 10)/(5654.87 − 800) = (144 000 − 8 000)/4854.87 = 28.01 mm above the base

Moment of inertia about the horizontal centroidal axis

  • Semicircle about its own centroidal axis: I = (π/8 − 8/9π) r⁴ = 0.1098 × 60⁴ = 1.4225 × 10⁶ mm⁴. Transfer: + A₁ (28.01 − 25.46)² = 5654.87 × 2.55² = 0.0368 × 10⁶ → 1.4593 × 10⁶ mm⁴
  • Slot about its own centroid: b h³/12 = 40 × 20³/12 = 26 667 mm⁴. Transfer: + A₂ (28.01 − 10)² = 800 × 324.4 = 259 500 → 286 200 mm⁴ (subtract)

I_x = 1.4593 × 10⁶ − 0.2862 × 10⁶ = 1.173 × 10⁶ mm⁴ (about 1.17 × 10⁶ mm⁴)

✓ Answer checked by Claude Opus 5.5 (high effort)

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