---
title: "Principal MI of I-section with semicircular hole"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-i-section-with-semicircle"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-i-section-with-semicircle
site: BCT Engineering
---

# Principal MI of I-section with semicircular hole

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2068 Baisakh (old course), 12 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-i-section-with-semicircle
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Calculate the principal moment of inertia about centroid and locate the principle axes for the figure as shown below. [Figure: section in cm: top flange 20 cm wide and 4 cm thick; web 4 cm thick and 16 cm high; bottom flange 15 cm + 15 cm wide (30 cm) and 8 cm thick; a semicircular hole of radius 4 cm at the bottom centre.]

## Short answer

Due to symmetry about the vertical axis, I_xy=0. Centroid at (0, 11.21) cm; I_x=31894 cm⁴, I_y=20651 cm⁴. Principal moments: I_max=31894 cm⁴ (horizontal axis) and I_min=20651 cm⁴ (vertical axis).

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm; x measured from the vertical axis of symmetry. Bottom flange 30 × 8, web 4 × 16, top flange 20 × 4, with a semicircular hole of radius 4 on the bottom edge at the centre (diameter on the bottom face, bulging upward).

#### Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $cm^2$, I in $cm^4$)
| Part | Sign | A | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Bottom flange 30×8 | + | $240.0$ | 0 | 4.000 | $1280$ | $18000$ | $0$ |
| Web 4×16 | + | $64.00$ | 0 | 16.00 | $1365$ | $85.33$ | $0$ |
| Top flange 20×4 | + | $80.00$ | 0 | 26.00 | $106.7$ | $2667$ | $0$ |
| Semicircular hole r = 4 | − | $25.13$ | 0 | 1.698 | $28.10$ | $100.5$ | $0$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 358.9\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{0}{358.9} = 0\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{4021}{358.9} = 11.21\ \text{cm}
\end{aligned}
$$

#### Moments and product of inertia about the centroidal axes
With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

| Part | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Bottom flange 30×8 (+) | $13741$ | $18000$ | $0$ |
| Web 4×16 (+) | $2836$ | $85.33$ | $0$ |
| Top flange 20×4 (+) | $17617$ | $2667$ | $0$ |
| Semicircular hole r = 4 (−) | $-2300$ | $-100.5$ | $0$ |
| **Total about centroid** | **$31894$** | **$20651$** | **$0$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 26273\pm5621 \\
I_{max} &= 31894\ \text{cm}^4,\qquad I_{min} = 20651\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{0}{11242}$, so $2\theta_p = -0.00^\circ$ (and $-180.00^\circ$).

- The axis of $I_{max}$ makes 0.00° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at 90.00° (perpendicular), both passing through the centroid (0, 11.21) measured from the reference origin.

The section is symmetrical about the vertical axis, so $I_{xy} = 0$ and the centroidal horizontal and vertical axes are the principal axes.

**Answer: $I_{max} = 31894\ \text{cm}^4$, $I_{min} = 20651\ \text{cm}^4$; principal axes through the centroid at 0.00° (for $I_{max}$) and 90.00° (for $I_{min}$) from the +x axis.**

## Check yourself

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

1. How does the semicircular hole enter the calculation, and what is its centroid height?
   A. Positive area, centroid at 4 cm
   B. Negative area, centroid at 8 cm
   C. Negative area, centroid 4r/(3π) = 1.698 cm above the bottom face
   D. It is ignored
   Answer: C. The hole removes material and its centroid is 4r/3π from the diameter, which lies on the bottom face. (Key point: Hole position)

2. Why is Xbar = 0 and Ybar = 11.21 cm?
   A. The origin is at the centroid
   B. Symmetry about the vertical axis fixes X; Ybar = sum(Ay)/sum(A) = 4021/358.9
   C. The section is symmetrical about the horizontal axis
   D. Ybar is the average of the part heights
   Answer: B. Every part is centred on x = 0, and Y is the area-weighted average of part centroids. (Key point: Centroid)

3. Why are the centroidal x and y axes principal here?
   A. The section is symmetrical about the vertical axis, so Ixy = 0
   B. Ix equals Iy
   C. The hole is circular
   D. The flanges are equal
   Answer: A. Symmetry makes Ixy = 0, so the principal axes coincide with the centroidal axes. (Key point: Principal axes)

4. Which is the maximum principal moment?
   A. Iy = 20651 cm^4
   B. 26273 cm^4
   C. 5621 cm^4
   D. Ix = 31894 cm^4 about the horizontal centroidal axis
   Answer: D. With Ixy = 0, the larger of Ix (31894) and Iy (20651) is Imax, about the horizontal axis (0°). (Key point: Results)

## Related in Strength of Materials

- [Principal MI and axes of section with semicircular cut-out](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-section-with-semicircle-50cm) — Strength of Materials, chapter 2
- [Principal MI and axes, 120 cm section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-and-axes-orientation-120) — Strength of Materials, chapter 2
- [Principal MI and axes of quarter-circle/triangle section (2078 Bhadra)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-2078b) — Strength of Materials, chapter 2
- [Principal MI about centroid and principal axes (2080 Baisakh)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-2080bk) — Strength of Materials, chapter 2
- [Principal MI and axes of L-section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-l-section-15x20-5cm) — Strength of Materials, chapter 2
- [Principal MI of stepped section (cm)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-l-shaped-45-100) — Strength of Materials, chapter 2

## Related in other subjects

- [Principal stresses at web-flange junction](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Aprincipal-stresses-web-flange-junction) — Strength of Materials, chapter 9
- [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
- [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