---
title: "Principal MI about centroid and principal axes (2080 Baisakh)"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-2080bk"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-2080bk
site: BCT Engineering
---

# Principal MI about centroid and principal axes (2080 Baisakh)

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2080 Baisakh, 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-2080bk
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Determine principal moment of inertias about the centroid and also locate the principal axes for the following figure. [Figure: composite section in cm: top rectangle 5 cm wide with a sloping right side (a triangle of vertical height 15 cm to the right), a middle part 10 cm + 15 cm wide with 10 cm height, and a bottom rectangle 25 cm wide and 5 cm high; vertical dimensions 15 cm, 10 cm and 5 cm.]

## Short answer

I_max = 42670 cm⁴, I_min = 22304 cm⁴. Principal axes through the centroid (11.04, 12.50 cm) are at 36.03° (for I_max) and -53.97° (for I_min) from the +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm. Bottom rectangle 25 × 5; middle rectangle 25 × 10 above it; top part is a trapezoid with a vertical left side of 15 cm, top width 5 cm and a sloping right side down to the full width of 25 cm (vertices (0,15), (25,15), (5,30), (0,30)).

#### 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 25×5 | + | $125.0$ | 12.50 | 2.500 | $260.4$ | $6510$ | $0$ |
| Middle 25×10 | + | $250.0$ | 12.50 | 10.00 | $2083$ | $13021$ | $0$ |
| Top trapezoid | + | $225.0$ | 8.611 | 20.83 | $3594$ | $7691$ | $-2396$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 600.0\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{6625}{600.0} = 11.04\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{7500}{600.0} = 12.50\ \text{cm}
\end{aligned}
$$

#### 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 25×5 (+) | $12760$ | $6776$ | $-1823$ |
| Middle 25×10 (+) | $3646$ | $13553$ | $-911.5$ |
| Top trapezoid (+) | $19219$ | $9020$ | $-6953$ |
| **Total about centroid** | **$35625$** | **$29349$** | **$-9687$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 32487\pm10183 \\
I_{max} &= 42670\ \text{cm}^4,\qquad I_{min} = 22304\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{19375}{6276}$, so $2\theta_p = 72.05^\circ$ (and $-107.95^\circ$).

- The axis of $I_{max}$ makes 36.03° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -53.97° (perpendicular), both passing through the centroid (11.04, 12.50) measured from the reference origin.

**Answer: $I_{max} = 42670\ \text{cm}^4$, $I_{min} = 22304\ \text{cm}^4$; principal axes through the centroid at 36.03° (for $I_{max}$) and -53.97° (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 is the section divided and what is the total area?
   A. Rectangle 25 x 5, rectangle 25 x 10 and a trapezoid; 600 cm^2
   B. Three 25 x 10 rectangles; 750 cm^2
   C. A triangle and one rectangle; 400 cm^2
   D. Two rectangles only; 375 cm^2
   Answer: A. Areas 125 + 250 + 225 = 600 cm^2 (trapezoid of top width 5 and base 25, height 15, is 225). (Key point: Composite decomposition)

2. Where is the centroid of the whole section?
   A. (12.5, 12.5) cm
   B. (11.04, 12.50) cm
   C. (8.61, 20.83) cm
   D. (11.04, 15) cm
   Answer: B. X = 6625/600 = 11.04 cm and Y = 7500/600 = 12.50 cm. (Key point: Centroid)

3. With Ix = 35625, Iy = 29349 and Ixy = -9687 cm^4, what are the principal moments?
   A. 42670 and 22304 cm^4
   B. 35625 and 29349 cm^4
   C. 32487 and 10183 cm^4
   D. 64974 and 0 cm^4
   Answer: A. Mean 32487 plus/minus sqrt(3138^2 + 9687^2) = 10183 gives 42670 and 22304. (Key point: Principal moments)

4. With tan 2θp = 19375/6276 giving 2θp = 72.05°, where is the I_max axis?
   A. 72.05° from +x
   B. 36.03° from +x anticlockwise; I_min axis at -53.97°
   C. -53.97° from +x
   D. 18.0° from +x
   Answer: B. θp = 72.05°/2 = 36.03° for I_max, and the I_min axis lies at 90° from it, -53.97°. (Key point: Principal axis direction)

## Related in Strength of Materials

- [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 of 15 cm right triangle](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-right-triangle-15x15) — Strength of Materials, chapter 2
- [Principal MI of quarter circle and triangle section (cm)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-quarter-circle-15-30) — Strength of Materials, chapter 2
- [Orientation of principal axes and MI of composite section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-two-triangles-composite-2066b) — Strength of Materials, chapter 2
- [Principal MI of composite section in cm](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-2078k) — Strength of Materials, chapter 2
- [Principal axes of composite section (8, 12 cm)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-axes-orientation-composite-8-12) — Strength of Materials, chapter 2

## Related in other subjects

- [Setting out right hand side composite curve](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aright-hand-composite-curve-setting-out) — Surveying II, chapter 6
- [Centroid of y = kx² area; parallel axes; define centroid](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Acentroid-y-kx2-parallel-axis-define-centroid-cg) — Applied Mechanics, chapter 4