---
title: "MOI of composite area with circular hole"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:moi-composite-area-circular-hole-centroidal"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-composite-area-circular-hole-centroidal
site: BCT Engineering
---

# MOI of composite area with circular hole

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

- Asked in: 2073 Shrawan, 8+4 marks
- Video: [Moment of Inertia of a Composite Section_Problem 1](https://www.youtube.com/watch?v=98_rgDqPfMA) — Manas Patnaik
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-composite-area-circular-hole-centroidal
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

Calculate the moment of inertia of the composite area as shown in the figure about its centroidal axes. Define centroid, center of gravity, axis of symmetry and radius of gyration. [Figure: plane area with top width 400 mm, right vertical edges 150 mm and 500 mm, bottom width 700 mm, left height 200 mm, left offset 200 mm, and a circular hole of radius r = 25 mm; axes X, Y at lower-left.]

## Short answer

Dividing into rectangle 400×650 mm, triangle 300×500 mm, minus circular hole r=25 mm, centroid is x̄=267.56 mm, ȳ=290.08 mm. About centroidal axes: I_xx=11.639×10⁹ mm⁴, I_yy=9.071×10⁹ mm⁴; radii of gyration k_x=186.9 mm, k_y=165.0 mm.

## Answer

**Reading of the figure:** origin at the bottom-left corner. Overall height = 150 + 500 = 650 mm. Top width 400 mm, bottom width 700 mm. The right edge runs down 150 mm from the top right corner, then slopes (taken as straight) to the bottom right corner at x = 700. The circular hole (r = 25 mm) has its centre 200 mm from the left and 200 mm from the bottom.

Divide into: rectangle 400 × 650, right triangle with base 300 and height 500 (vertical side at x = 400), minus the circular hole.

#### Centroid
| Part | A (mm²) | x (mm) | y (mm) | A·x | A·y |
|---|---|---|---|---|---|
| Rectangle 400 × 650 | 260,000.00 | 200.00 | 325.00 | 52,000,000.0 | 84,500,000.0 |
| Triangle 300 × 500 | 75,000.00 | 500.00 | 166.67 | 37,500,000.0 | 12,500,000.0 |
| Circular hole r = 25 (hole, −) | -1,963.50 | 200.00 | 200.00 | -392,699.1 | -392,699.1 |

ΣA = 333,036.50 mm²
x̄ = **267.56 mm**, ȳ = **290.08 mm**

#### Moments of inertia about centroidal axes
Own values: rectangle bh³/12, hb³/12; triangle bh³/36, hb³/36; circle πr⁴/4.

| Part | I_x own (mm⁴) | A·dy² | I_y own | A·dx² |
|---|---|---|---|---|
| Rectangle 400 × 650 | 9,154,166,666.7 | 317,041,756.5 | 3,466,666,666.7 | 1,186,737,605.9 |
| Triangle 300 × 500 | 1,041,666,666.7 | 1,142,317,796.9 | 375,000,000.0 | 4,052,120,892.9 |
| Circular hole r = 25 | -306,796.2 | -15,932,675.8 | -306,796.2 | -8,962,130.2 |

**I_xx = 11.639 × 10⁹ mm⁴**
**I_yy = 9.071 × 10⁹ mm⁴**
(k_x = 186.9 mm, k_y = 165.0 mm)

#### Definitions
- **Centroid:** the point at which the whole area of a plane figure can be assumed to be concentrated. It is the geometric centre of the area and depends only on shape: x̄ = ∫x dA / A, ȳ = ∫y dA / A.
- **Centre of gravity:** the point through which the whole weight of a body acts, whatever its orientation. For a body of uniform density and in a uniform gravity field it coincides with the centroid of its volume (or area for a thin plate).
- **Axis of symmetry:** a line that divides a figure into two parts that are mirror images. The centroid always lies on an axis of symmetry, and the product of inertia about it is zero.
- **Radius of gyration:** the distance k from an axis at which the whole area could be concentrated to give the same moment of inertia, I = A k², so k = √(I/A).

## Check yourself

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

1. How is a hole or cutout treated when finding the centroid and moment of inertia of the composite area?
   A. Its area is added and its inertia is subtracted
   B. It is ignored because it has no material
   C. Its area and its moment of inertia (with its transfer term) are subtracted
   D. Only its area is subtracted; its inertia is not needed
   Answer: C. A hole is a negative area; both its area and its I (own plus A d^2) are subtracted from the solid shape. (Key point: Hole as negative)

2. What is the moment of inertia of a circle of radius r about its own diameter?
   A. πr^4/2
   B. πr^2/4
   C. πr^4/8
   D. πr^4/4
   Answer: D. I = πr^4/4 about a diameter; πr^4/2 is the polar value. (Key point: Circle inertia)

3. What does the radius of gyration k satisfy?
   A. k = I A
   B. I = A k^2, so k = sqrt(I/A)
   C. I = k/A
   D. I = A k, so k = I/A
   Answer: B. k is the distance at which the area could be concentrated to give the same I, so I = A k^2. (Key point: Radius of gyration)

4. When do the centroid and centre of gravity coincide?
   A. Only when the body has a hole
   B. Only for circular bodies
   C. Never
   D. For a body of uniform density in a uniform gravity field
   Answer: D. With uniform density and uniform gravity, the weight distribution matches the shape, so they coincide. (Key point: Centroid vs CG)

## Related in Applied Mechanics

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- [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
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