---
title: "Centroidal MOI of plane figure; define centroid"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:moi-centroid-l-shaped-figure"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-centroid-l-shaped-figure
site: BCT Engineering
---

# Centroidal MOI of plane figure; define centroid

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

- Asked in: 2074 Asoj, 9+3 marks
- Video: [Understanding the Area Moment of Inertia](https://www.youtube.com/watch?v=Bls5KnQOWkY) — The Efficient Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-centroid-l-shaped-figure
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

Determine the moment of inertia about centroidal XX and YY axes of the plane figure shown in the figure below. Define centroid, centre of gravity and axes of symmetry. [Figure: L-shaped/stepped plane figure with dimensions 14 cm, 20 cm, 15 cm, 13 cm, 1.5 m and 1.2 m as printed.]

## Short answer

Divide area into rectangles and a triangle, locate centroid at x̄=38.97 cm, ȳ=50.36 cm. Applying parallel axis theorem gives I_xx = 11.95×10⁶ cm⁴, I_yy = 7.16×10⁶ cm⁴. Centroid: geometric centre; centre of gravity: weight balance point; axis of symmetry: mirror line.

## Answer

**Reading of the figure:** the printed dimensions are 14 cm (top), 20 cm, 15 cm, 13 cm, 1.5 m (left side) and 1.2 m (bottom). I take them all in cm: left side 150 cm, bottom 120 cm, top 14 cm, the 15 cm and 13 cm steps at the lower right, and a sloping edge that closes the figure. (The 20 cm label is the short vertical piece below the top, which is part of the 14 cm wide column and does not change the area.) The figure is not to scale.

Origin at the bottom-left corner. Vertices: (0,0), (120,0), (120,13), (105,13), (14,130), (14,150), (0,150).

#### Division into simple shapes
- R1: rectangle 14 × 150 (x from 0 to 14)
- R2: rectangle 106 × 13 at the bottom (x from 14 to 120)
- T: right triangle, base 91 and height 117, with its right angle at (14, 13)

#### Centroid
| Part | A (cm²) | x (cm) | y (cm) | A·x | A·y |
|---|---|---|---|---|---|
| R1: 14 × 150 | 2,100.00 | 7.00 | 75.00 | 14,700.0 | 157,500.0 |
| R2: 106 × 13 | 1,378.00 | 67.00 | 6.50 | 92,326.0 | 8,957.0 |
| T: 91 × 117 | 5,323.50 | 44.33 | 52.00 | 236,008.5 | 276,822.0 |

ΣA = 8,801.50 cm²
x̄ = ΣAx/ΣA = **38.97 cm**, ȳ = ΣAy/ΣA = **50.36 cm**

#### Moments of inertia about the centroidal axes (parallel axis theorem)
Own values: rectangle bh³/12 and hb³/12, triangle bh³/36 and hb³/36.

| Part | I_x own (cm⁴) | A·dy² | I_y own | A·dx² |
|---|---|---|---|---|
| R1: 14 × 150 | 3,937,500.0 | 1,274,555.3 | 34,300.0 | 2,146,980.9 |
| R2: 106 × 13 | 19,406.8 | 2,651,345.1 | 1,290,267.3 | 1,082,316.8 |
| T: 91 × 117 | 4,048,521.8 | 14,247.8 | 2,449,105.8 | 152,872.6 |

**I_xx = 11,945,577 cm⁴ ≈ 11.95 × 10⁶ cm⁴**
**I_yy = 7,155,843 cm⁴ ≈ 7.16 × 10⁶ cm⁴**

#### 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 the stepped figure handled to find its centroid?
   A. Use the longest dimension divided by 2
   B. Average the coordinates of the vertices
   C. Split into rectangles and a triangle, then x̄ = ΣAx/ΣA and ȳ = ΣAy/ΣA
   D. Take the midpoint of the bounding box
   Answer: C. The area is divided into simple shapes with known centroids, and area-weighted averages give the composite centroid. (Key point: Divide into shapes)

2. What is the moment of inertia of a right triangle (base b, height h) about its own centroidal x axis?
   A. bh^3/36
   B. bh^3/3
   C. bh^3/4
   D. bh^3/12
   Answer: A. A triangle's centroidal I_x is bh^3/36; bh^3/12 is the rectangle value. (Key point: Triangle own inertia)

3. Why is A d^2 added to each part's own inertia?
   A. To account for the figure not being to scale
   B. Because each part's own inertia is about its own centroid, which differs from the composite centroidal axis
   C. Because the parts overlap
   D. To convert cm^4 to mm^4
   Answer: B. The parallel axis theorem shifts each part's inertia from its own centroid to the composite centroidal axis. (Key point: Use of A d^2)

4. How do centroid and centre of gravity differ in concept?
   A. Centre of gravity applies to areas only
   B. They are never equal
   C. Centroid depends only on shape; centre of gravity is where the weight acts
   D. Centroid depends on weight; centre of gravity depends only on shape
   Answer: C. Centroid is geometric; CG is the point of action of weight, and they coincide for uniform density in uniform gravity. (Key point: Centroid vs CG)

## Related in Applied Mechanics

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