---
title: "State and prove parallel axis theorem"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:parallel-axis-theorem-state-prove"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/parallel-axis-theorem-state-prove
site: BCT Engineering
---

# State and prove parallel axis theorem

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

- Asked in: 2067 Asar, 5 marks
- Video: [Parallel Axis Theorem🔥 | Statement, Proof | Moment Of Inertia | Engineering Mechanics | Civil Stuff](https://www.youtube.com/watch?v=ErbJn9vljOw) — Civil Stuff
- 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-theorem-state-prove
- 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.

## Short answer

Theorem: moment of inertia about any axis equals centroidal moment plus area times distance squared, I = I_G + A d². Proof: expand ∫(y' + d)² dA = I_G + 2d∫y'dA + d²A; ∫y'dA = 0, giving I = I_G + A d².

## 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.

**Note:** the theorem also holds for the mass moment of inertia (I = I_G + M d²) and for the polar moment of inertia (J = J_G + A d²). It is used to find the moment of inertia of composite sections about their centroidal axes, and to shift the moment of inertia of a standard shape to another axis.

## 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 = A d^2 only, since I_G is zero for any area
   D. I = I_G + A d, where d is the distance between the axes
   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 the integral of y' dA is A times the centroid's distance from the centroidal axis, which is zero
   D. Because d is zero for every parallel axis
   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. Among all parallel axes, where is the moment of inertia of an area smallest?
   A. The same for all parallel axes
   B. Smallest about the centroidal axis, since A d^2 is never negative
   C. Smallest about the axis farthest from the centroid
   D. Largest about the centroidal axis
   Answer: B. I = I_G + A d^2 and A d^2 >= 0, so the centroidal axis (d = 0) gives the smallest value. (Key point: Minimum at centroid)

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## Related in other subjects

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