---
title: "Parallel axis theorem; MOI with semicircular hole"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:parallel-axis-composite-semicircular-hole-xx"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/parallel-axis-composite-semicircular-hole-xx
site: BCT Engineering
---

# Parallel axis theorem; MOI with semicircular hole

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

- Asked in: 2072 Kartik, 4+8 marks
- Video: [Moment of Inertia and Parallel Axis Theorem!](https://www.youtube.com/watch?v=U-Fl2relqN0) — Simple Civil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/parallel-axis-composite-semicircular-hole-xx
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

State and prove the parallel axis theorem for moment of inertia. Determine the moment of inertia of the given composite area as shown in figure below about its centroidal X-X axis. [Figure: area bounded by x axis, left height 30 cm, right height 50 cm, base 30 cm + 30 cm, with a semicircular hole on the base, radius 15 cm centred at 30 cm from the left.]

## Short answer

Parallel axis theorem: I=I_G+Ad². Trapezium (base 60 cm, top 30 cm, height 50 cm) minus semicircular hole r=15 cm; area=2046.57 cm², centroid ȳ=22.84 cm from base. I_xx about centroidal X-X = 2.722×10⁵ cm⁴.

## Answer

#### Parallel axis theorem
**Statement:** the moment of inertia of an area about any axis equals its moment of inertia about a parallel axis through its centroid plus the product of the area and the square of the distance between the two axes.

I = I_G + A d²

**Proof:** let G be the centroid of the area A, and let X′ be the centroidal axis. Take a parallel axis X at distance d from X′. For a small element dA at distance y′ from X′, its distance from X is y = y′ + d.
```text
I_X = ∫ y² dA = ∫ (y′ + d)² dA
    = ∫ y′² dA + 2d ∫ y′ dA + d² ∫ dA
```
- ∫ y′² dA = I_G, the moment of inertia about the centroidal axis.
- ∫ y′ dA = A·ȳ′ = 0, because the distance of the centroid from the centroidal axis is zero.
- ∫ dA = A.

So **I_X = I_G + A d²**. Similarly, I_Y = I_Gy + A d_x². Since A d² is never negative, the moment of inertia about the centroidal axis is the least of the moments of inertia about all parallel axes.

#### Moment of inertia about the centroidal X–X axis
**Reading of the figure:** origin at O (bottom-left). Left height 30 cm, right height 50 cm, base 30 + 30 = 60 cm, so the top edge slopes from (0, 30) to (60, 50). A semicircular hole of radius 15 cm has its diameter on the base, centre at x = 30 cm.

Main area (trapezium): A₁ = (30 + 50)/2 × 60 = 2400 cm², ȳ₁ = 20.42 cm.
Own I_x of the trapezium about its centroidal axis = 359,583 cm⁴ (found by splitting it into a rectangle 60 × 30 and a triangle 60 × 20: rectangle 60·30³/12 + 1800 (15 − ȳ₁)², triangle 60·20³/36 + 600 (36.67 − ȳ₁)²).

Hole: A₂ = π × 15²/2 = 353.4 cm², ȳ₂ = 4 × 15/(3π) = 6.37 cm, I_x(own) = (π/8 − 8/9π) × 15⁴ = 5 556 cm⁴.

#### Centroid
| Part | A (cm²) | x (cm) | y (cm) | A·x | A·y |
|---|---|---|---|---|---|
| Trapezium (30, 50 × 60) | 2,400.00 | 32.50 | 20.42 | 78,000.0 | 49,000.0 |
| Semicircular hole r = 15 (hole, −) | -353.43 | 30.00 | 6.37 | -10,602.9 | -2,250.0 |

ΣA = 2,046.57 cm², ȳ = **22.84 cm** above the base.

#### Moment of inertia about the centroidal X–X axis
I_xx = [I₁ + A₁(ȳ₁ − ȳ)²] − [I₂ + A₂(ȳ₂ − ȳ)²]
```text
Trapezium: 359,583 + 2400 × (20.42 − 22.84)² = 373,713 cm4
Hole     :  5,556 + 353.4 × (6.37 − 22.84)²  = 101,508 cm4
```
**I_xx = 272,205 cm⁴ ≈ 2.722 × 10⁵ cm⁴**

## Check yourself

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

1. Which statement of the parallel axis theorem is correct?
   A. I = I_G + A d^2, where I_G is about the centroidal axis and d is the distance between the parallel axes
   B. I = A d^2 only, since I_G is zero for any area
   C. I = I_G + A d, where d is the distance between the axes
   D. I = I_G - A d^2, because the centroidal value is the largest
   Answer: A. The moment of inertia about any axis equals the centroidal value plus A d^2 (area times distance squared). (Key point: Parallel axis theorem)

2. In the proof, why does the term 2d * integral(y' dA) drop out?
   A. Because y'^2 is always negative
   B. Because d is zero for every parallel axis
   C. Because the integral of y' dA is A times the centroid's distance from the centroidal axis, which is zero
   D. Because the integral of dA equals zero
   Answer: C. The integral of y' dA equals A times the centroid's distance from the centroidal axis, and that distance is zero, so the cross term vanishes. (Key point: Cross term vanishes)

3. How is the semicircular hole included in I_xx about the composite centroidal axis?
   A. Subtract A2 (y2 - ȳ) without squaring
   B. Subtract [I_own + A2 (y2 - ȳ)^2] of the hole from that of the trapezium
   C. Subtract only the hole's own I
   D. Add the hole's own I to the trapezium's I
   Answer: B. The hole's inertia, shifted to the composite centroid with A d^2, is subtracted. (Key point: Hole in I_xx)

4. What is the composite centroid ȳ formula here?
   A. (y1 + y2)/2
   B. (A1 y1 + A2 y2)/(A1 + A2)
   C. (A1 y1 - A2 y2)/(A1 + A2)
   D. (A1 y1 - A2 y2)/(A1 - A2)
   Answer: D. The hole is a negative area, so both its area and its moment are subtracted. (Key point: Composite centroid)

## Related in Applied Mechanics

- [Parallel axis theorem, radius of gyration; arch figure MOI](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aparallel-axis-radius-gyration-arch-figure) — Applied Mechanics, chapter 4
- [Parallel axis theorem; MOI of trapezoid with cutout](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aparallel-axis-theorem-moi-trapezoid-semicircle) — Applied Mechanics, chapter 4
- [MOI of composite area with circular hole](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Amoi-composite-area-circular-hole-centroidal) — Applied Mechanics, chapter 4
- [MOI of composite area about Y axis](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Amoi-composite-area-about-y-axis) — Applied Mechanics, chapter 4
- [Parallel axis theorem; centroid of area under y = Kx²](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aparallel-axis-centroid-y-kx2-ashad) — Applied Mechanics, chapter 4
- [Parallel axis theorem; centroid of area between parabolas](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aparallel-axis-centroid-curves-x-ky2) — Applied Mechanics, chapter 4

## Related in other subjects

- [Parallel axis theorem for product of inertia](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aparallel-axis-theorem-product-of-inertia) — Strength of Materials, chapter 2
- [Principal MI of I-section with semicircular hole](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-i-section-with-semicircle) — Strength of Materials, chapter 2
- [Principal axis and MI of composite section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-axes-mi-2081b) — Strength of Materials, chapter 2
- [Composite body moment of inertia](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-sphere-composite-inertia) — Engineering Mechanics, chapter 6