---
title: "Parallel axis theorem, radius of gyration; arch figure MOI"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:parallel-axis-radius-gyration-arch-figure"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/parallel-axis-radius-gyration-arch-figure
site: BCT Engineering
---

# Parallel axis theorem, radius of gyration; arch figure MOI

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

- Asked in: 2073 Chaitra, 3+9 marks
- 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-radius-gyration-arch-figure
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

State parallel axis theorem and radius of gyration. Calculate the moment of inertia of the given shaded area about its centroidal axes. [Figure: shaded trapezoid-like area with 15 cm + 15 cm top, height 26 cm, 30 cm + 30 cm base, and a semicircular/arched cutout with 20 cm dimension.]

## Short answer

Parallel axis theorem: I=I_G+Ad². Radius of gyration k=√(I/A). For shaded area (trapezium 60/30 cm×26 cm minus semicircle r=20 cm), centroid is at x̄=30 cm, ȳ=15.11 cm; I_xx=3.31×10⁴ cm⁴, I_yy=1.565×10⁵ cm⁴.

## Answer

#### Parallel axis theorem
**Statement:** the moment of inertia of an area about any axis is equal to its moment of inertia about a parallel centroidal axis plus A d², where A is the area and d is the distance between the two axes: I = I_G + A d².

#### Radius of gyration
The radius of gyration k of an area about an axis is the distance from that axis at which the whole area can be assumed concentrated to give the same moment of inertia: I = A k², so k = √(I/A).

#### Moment of inertia of the shaded area
**Reading of the figure:** a symmetrical trapezium, base 30 + 30 = 60 cm, top 15 + 15 = 30 cm, height 26 cm, with a semicircular opening (arch) of radius 20 cm whose diameter lies on the base, centred on the axis of symmetry. y is measured from the base and x from the left corner, so the axis of symmetry is at x = 30 cm.

Trapezium: area = (60 + 30)/2 × 26 = 1170 cm², ȳ = 26(60 + 2×30)/(3 × 90) = 11.56 cm, I_x(own) = 63 469 cm⁴, I_y(own) = 219 375 cm⁴ (from the formulae for a trapezium).
Semicircle (hole): area = π × 20²/2 = 628.3 cm², ȳ = 4r/3π = 8.49 cm.

#### Centroid
| Part | A (cm²) | x (cm) | y (cm) | A·x | A·y |
|---|---|---|---|---|---|
| Trapezium (60/30 × 26) | 1,170.00 | 30.00 | 11.56 | 35,100.0 | 13,520.0 |
| Semicircle r = 20 (hole, −) | -628.32 | 30.00 | 8.49 | -18,849.6 | -5,333.3 |

ΣA = 541.68 cm²
x̄ = 30 cm (axis of symmetry), ȳ = **15.11 cm** above the base.

#### Moments of inertia about centroidal axes
| Part | I_x own (cm⁴) | A·dy² | I_y own | A·dx² |
|---|---|---|---|---|
| Trapezium (60/30 × 26) | 63,468.9 | 14,810.4 | 219,375.0 | 0.0 |
| Semicircle r = 20 | -17,561.1 | -27,578.7 | -62,831.9 | -0.0 |

**I_xx = 33,140 cm⁴ (≈ 3.31 × 10⁴ cm⁴)**
**I_yy = 156,543 cm⁴ (≈ 1.565 × 10⁵ cm⁴)**
(k_x = 7.82 cm, k_y = 17.00 cm)

## 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 = A d^2 only, since I_G is zero for any area
   B. I = I_G + A d^2, where I_G is about the centroidal axis and d is the distance between the axes
   C. I = I_G - A d^2, because the centroidal value is the largest
   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. Which expresses the radius of gyration about an axis?
   A. k = I/A
   B. k = A/I
   C. k = sqrt(I/A)
   D. k = sqrt(I*A)
   Answer: C. From I = A k^2, k = sqrt(I/A). (Key point: Radius of gyration)

3. For the semicircular arch opening (radius 20 cm, diameter on the base), where is its centroid above the base?
   A. r/2 = 10 cm
   B. r = 20 cm
   C. 2r/π = 12.7 cm
   D. 4r/3π = 8.49 cm
   Answer: D. The semicircle's centroid is 4r/3π from its diameter, here 8.49 cm above the base. (Key point: Hole centroid)

4. Why is x̄ simply 30 cm (the centre) for this symmetrical shape and why is A dx^2 zero?
   A. Both parts have their centroids on the axis of symmetry, so dx = 0
   B. Because the hole has no area
   C. Because I_y is always zero
   D. Because the shape is not a composite
   Answer: A. The centroid lies on the axis of symmetry, so the parts' centroids share the same x and dx = 0. (Key point: Symmetry axis)

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