---
title: "MOI of arch section about horizontal centroidal axis"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:moi-section-horizontal-centroidal-axis-arch"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-section-horizontal-centroidal-axis-arch
site: BCT Engineering
---

# MOI of arch section about horizontal centroidal axis

Applied Mechanics · Chapter 4: Center of Gravity, Centroid and Moment of Inertia

- Asked in: 2067 Asoj, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-section-horizontal-centroidal-axis-arch
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

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.]

## 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.

## Answer

**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)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Semicircle centroid above its diameter:
   A. 2r/π = 38.2 mm
   B. r = 60 mm
   C. r/2 = 30 mm
   D. 4r/3π = 25.46 mm
   Answer: D. 4r/(3π) with r = 60. (Key point: Semicircle centroid)

2. The slot is treated as:
   A. Ignored in centroid but subtracted in inertia
   B. A positive area added
   C. Ignored altogether
   D. A negative area subtracted in centroid and inertia
   Answer: D. It is a cut-out. (Key point: Slot)

3. Composite ȳ equals:
   A. (A1 y1 - A2 y2)/(A1 - A2) = 28.01 mm
   B. 30 mm
   C. 25.46 mm
   D. (A1 y1 + A2 y2)/(A1 + A2)
   Answer: A. Slot subtracted from both numerator and denominator. (Key point: Centroid)

4. Slot's inertia about the section's centroid is:
   A. bh^3/12 + A2 (28.01 - 10)^2, subtracted
   B. bh^3/12 only, subtracted
   C. bh^3/3, subtracted
   D. A2 (28.01 - 10)^2 only, added
   Answer: A. Own I plus transfer, subtracted. (Key point: Transfer term)

## Related in Applied Mechanics

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