---
title: "MOI of composite area about Y axis"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:moi-composite-area-about-y-axis"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-composite-area-about-y-axis
site: BCT Engineering
---

# MOI of composite area about Y axis

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

- Asked in: 2071 Shrawan, 6 marks
- Video: [How to find the moment of inertia for composite shapes](https://www.youtube.com/watch?v=3wIi4O6a6_Q) — Engineer4Free
- 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-about-y-axis
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

Calculate the moment of inertia of the composite area about Y-axis. [Figure: composite area with dimensions 75, 150 mm, 275 mm, 100 mm, 100 mm; a rectangle with a triangular portion and a step as shown.]

## Short answer

Dividing into two rectangles and subtracting a triangular portion below the slope, the moment of inertia about the Y-axis is I_y=458.33×10⁶ mm⁴ (4.58×10⁸ mm⁴).

## Answer

**Reading of the figure:** origin O at the bottom-left; the x axis is 100 + 100 = 200 mm long. The area is bounded by: the sloping line from O to (200, 125), the right edge x = 200 from y = 125 up to y = 275 (150 mm long), the top edge from x = 200 to x = 100 at y = 275, a step down of 75 mm to y = 200 at x = 100, then the top edge back to the y axis at y = 200, and down the y axis to O.

#### Division into simple shapes
- Rectangle A: 100 × 200 (x from 0 to 100)
- Rectangle B: 100 × 275 (x from 100 to 200)
- Triangle C (to be removed): vertices (0, 0), (200, 0), (200, 125), the part below the sloping line

#### Moment of inertia about the Y axis
For a rectangle with one side on the Y axis, I_y = h b³/3 (b the width along x). For the rectangle B, I_y = h (x₂³ − x₁³)/3.
```text
Rectangle A:  I_y = 200 × 100³ / 3             =  66.667 × 10⁶ mm⁴
Rectangle B:  I_y = 275 × (200³ − 100³) / 3
                  = 275 × 7 × 10⁶ / 3           = 641.667 × 10⁶ mm⁴
Triangle C (below the slope line y = 0.625 x):
              I_y = ∫₀²⁰⁰ x² (0.625 x) dx
                  = 0.625 × 200⁴ / 4            = 250.000 × 10⁶ mm⁴   (subtract)
```
I_y = 66.667 + 641.667 − 250.000 = **458.33 × 10⁶ mm⁴ (4.58 × 10⁸ mm⁴)**

(Check: the area is 20 000 + 27 500 − 12 500 = 35 000 mm², centroid x̄ = 98.8 mm, and I_y = 116.6 × 10⁶ + 35 000 × 98.8² = 458.3 × 10⁶ mm⁴.)

## Check yourself

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

1. For a rectangle of height h with one side on the Y axis, I_y is:
   A. h b^3/3
   B. h b^3/12
   C. b h^3/3
   D. b h^3/12
   Answer: A. About an edge the value is h b^3/3 (b measured along x). (Key point: Rectangle about edge)

2. How is the triangular portion below the sloping line handled?
   A. It is ignored
   B. Its I_y is added
   C. Only its area is subtracted
   D. Its I_y is subtracted from the sum of the rectangles
   Answer: D. The triangle is a removed region, so its moment of inertia is subtracted. (Key point: Removed triangle)

3. Rectangle B spans x = 100 to 200. Its I_y about the Y axis uses:
   A. h b^3/12 with b = 100
   B. h (x2^3 - x1^3)/3
   C. h x2^3/12
   D. h (x2 - x1)^3/3
   Answer: B. Integrating x^2 h dx from x1 to x2 gives h(x2^3 - x1^3)/3. (Key point: Rectangle B)

## Related in Applied Mechanics

- [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
- [Parallel axis theorem; MOI with semicircular hole](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aparallel-axis-composite-semicircular-hole-xx) — 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
- [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
- [MOI of arch section about horizontal centroidal axis](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Amoi-section-horizontal-centroidal-axis-arch) — Applied Mechanics, chapter 4
- [MOI of common area between y = K√x and y = mx](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Amoi-common-area-y-k-sqrt-x-y-mx) — Applied Mechanics, chapter 4

## Related in other subjects

- [Composite body moment of inertia](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-sphere-composite-inertia) — Engineering Mechanics, chapter 6
- [Principal MI and axis of composite section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-axes-mi-2081k) — 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
- [Inertia force and turning moment in an engine](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Areciprocating-engine-inertia-force-torque) — Theory of Mechanism and Machine I, chapter 8