---
title: "Parallel axis theorem; MOI of trapezoid with cutout"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:parallel-axis-theorem-moi-trapezoid-semicircle"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/parallel-axis-theorem-moi-trapezoid-semicircle
site: BCT Engineering
---

# Parallel axis theorem; MOI of trapezoid with cutout

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

- Asked in: 2074 Chaitra, 4+8 marks
- Video: [Statics: Lesson 69 - Moment of Inertia, Composite Shape Method](https://www.youtube.com/watch?v=undlEC9OQUY) — Jeff Hanson
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/parallel-axis-theorem-moi-trapezoid-semicircle
- 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 about centroidal axes of the plane figure shown below. [Figure: trapezoidal plane figure with 120 mm top width, 150 mm left height, 50 mm bottom offset, bottom widths 90 mm and 130 mm, with a semicircular cutout of radius 60 mm at the bottom.]

## 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**. In the same way I_Y = I_G(y) + A d_x². The moment of inertia about an axis through the centroid is the smallest of all parallel axes.

#### Moment of inertia of the figure
**Reading of the figure:** trapezium with top width 120 mm, left vertical side 150 + 50 = 200 mm, base 90 + 130 = 220 mm, so the right side slopes from (120, 200) to (220, 0). A semicircular hole of radius 60 mm with its diameter horizontal at 50 mm above the base, centre at x = 90 mm (the printed values are used as given; the arc then dips 10 mm below the base line, which the figure draws only approximately). Origin at the bottom-left corner.

Divide into: rectangle 120 × 200, right triangle 100 × 200 (vertical side at x = 120), minus the semicircle (centroid 4r/3π = 25.46 mm below its diameter, so at y = 24.54 mm).

#### Centroid
| Part | A (mm²) | x (mm) | y (mm) | A·x | A·y |
|---|---|---|---|---|---|
| Rectangle 120 × 200 | 24,000.00 | 60.00 | 100.00 | 1,440,000.0 | 2,400,000.0 |
| Triangle 100 × 200 | 10,000.00 | 153.33 | 66.67 | 1,533,333.3 | 666,666.7 |
| Semicircle r = 60 (hole, −) | -5,654.87 | 90.00 | 24.54 | -508,938.0 | -138,743.3 |

ΣA = 28,345.13 mm²
x̄ = **86.94 mm**, ȳ = **103.30 mm**

#### Moments of inertia about the centroidal axes
Own values: rectangle bh³/12, hb³/12; triangle bh³/36, hb³/36; semicircle (π/8 − 8/9π) r⁴ about its centroidal horizontal axis and πr⁴/8 about its vertical diameter.

| Part | I_x own (mm⁴) | A·dy² | I_y own | A·dx² |
|---|---|---|---|---|
| Rectangle 120 × 200 | 80,000,000.0 | 260,640.0 | 28,800,000.0 | 17,421,494.5 |
| Triangle 100 × 200 | 22,222,222.2 | 13,416,678.6 | 5,555,555.6 | 44,077,494.1 |
| Semicircle r = 60 | -1,422,450.2 | -35,078,132.4 | -5,089,380.1 | -52,865.2 |

**I_xx = 79.40 × 10⁶ mm⁴**, **I_yy = 90.71 × 10⁶ mm⁴**
(radii of gyration k_x = 52.9 mm, k_y = 56.6 mm)

## 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, because the centroidal value is the largest
   B. I = I_G + A d^2, where I_G is about the centroidal axis and d is the distance between the parallel axes
   C. I = I_G + A d, where d is the distance between the axes
   D. I = A d^2 only, since I_G is zero for any area
   Answer: B. 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 the integral of dA equals zero
   B. Because y'^2 is always negative
   C. Because d is zero for every parallel axis
   D. Because the integral of y' dA is A times the centroid's distance from the centroidal axis, which is zero
   Answer: D. 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. Where is the centroid of a semicircle of radius r from its diameter?
   A. 4r/3π
   B. 3r/8
   C. 2r/π
   D. r/2
   Answer: A. The centroid of a semicircle lies 4r/(3π) from its diameter. (Key point: Semicircle centroid)

4. How is a hole or cutout treated when finding the centroid and moment of inertia of the composite area?
   A. Only its area is subtracted; its inertia is not needed
   B. Its area and its moment of inertia (with its transfer term) are subtracted
   C. It is ignored because it has no material
   D. Its area is added and its inertia is subtracted
   Answer: B. 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)

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