---
title: "Principal axis and MI of composite section"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-axes-mi-2081b"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-axes-mi-2081b
site: BCT Engineering
---

# Principal axis and MI of composite section

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2081 Bhadra, 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-axes-mi-2081b
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Find out the principal axis and principal moment of inertia for the given section. [Figure: plane section with a vertical left edge; the lower left part has a semicircular cut-out of radius 50 mm centred on the left edge; a circular hole of diameter 60 mm lies in the lower right-central part (60 mm marked); the top edge slopes down from a height of 100 mm above the x-axis at the y-axis to 50 mm above the x-axis at the right end; the right edge extends 50 mm above and 50 mm below the marked level; bottom dimension 150 mm + 50 mm; left side 100 mm above and 100 mm below the 50 mm mark; x-axis along the bottom and y-axis along the left edge.]

## Short answer

By composite-area method: centroid at (98.26, 1.471) mm; I_x = 6.860×10^7, I_y = 7.631×10^7, I_xy = –1.384×10^7 mm^4. Principal moments: I_max = 8.682×10^7 mm^4, I_min = 5.809×10^7 mm^4. Principal axes through centroid at 52.78° (I_max) and –37.22° (I_min) from the +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Overall outline (mm): vertices (0,−100), (150,−100), (150,−50), (200,−50), (200,50), (0,100); the top edge slopes from 100 mm to 50 mm above the x-axis.
- Semicircular cut-out of radius 50 mm with its centre on the left edge at (0, −50), removed.
- Circular hole of diameter 60 mm with centre (100, −25), removed.

#### Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $mm^2$, I in $mm^4$)
| Part | Sign | A | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Outline | + | $32500$ | 89.10 | -7.051 | $8.172\times10^{7}$ | $9.822\times10^{7}$ | $-9.348\times10^{5}$ |
| Semicircle cut-out | − | $3927$ | 21.22 | -50.00 | $2.454\times10^{6}$ | $6.860\times10^{5}$ | $0$ |
| Hole d = 60 | − | $2827$ | 100.0 | -25.00 | $6.362\times10^{5}$ | $6.362\times10^{5}$ | $0$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 25746\ \text{mm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{2.530\times10^{6}}{25746} = 98.26\ \text{mm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{37869}{25746} = 1.471\ \text{mm}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Outline (+) | $8.408\times10^{7}$ | $1.009\times10^{8}$ | $1.601\times10^{6}$ |
| Semicircle cut-out (−) | $-1.286\times10^{7}$ | $-2.399\times10^{7}$ | $-1.557\times10^{7}$ |
| Hole d = 60 (−) | $-2.617\times10^{6}$ | $-6.447\times10^{5}$ | $1.302\times10^{5}$ |
| **Total about centroid** | **$6.860\times10^{7}$** | **$7.631\times10^{7}$** | **$-1.384\times10^{7}$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 7.246\times10^{7}\pm1.437\times10^{7} \\
I_{max} &= 8.682\times10^{7}\ \text{mm}^4,\qquad I_{min} = 5.809\times10^{7}\ \text{mm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{2.768\times10^{7}}{-7.709\times10^{6}}$, so $2\theta_p = 105.56^\circ$ (and $-74.44^\circ$).

- The axis of $I_{max}$ makes 52.78° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -37.22° (perpendicular), both passing through the centroid (98.26, 1.471) measured from the reference origin.

**Answer: $I_{max} = 8.682\times10^{7}\ \text{mm}^4$, $I_{min} = 5.809\times10^{7}\ \text{mm}^4$; principal axes through the centroid at 52.78° (for $I_{max}$) and -37.22° (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 are the semicircular cut-out and the circular hole handled?
   A. Added, with positive area and positive I
   B. Subtracted, with negative area and negative I in the totals
   C. Ignored
   D. Only their areas are subtracted, not their I
   Answer: B. Removed parts are treated as negative areas, so their A and their I (about the common axes) are subtracted from the outline. (Key point: Handling cut-outs)

2. What net area is used, and where is the centroid?
   A. 32500 mm^2, at (89.10, -7.05)
   B. 25746 mm^2, at (98.26, 1.471) mm
   C. 36500 mm^2, at (98.26, 1.471)
   D. 25746 mm^2, at (100, 0)
   Answer: B. A = 32500 - 3927 - 2827 = 25746 mm^2, and the weighted centroid is X = 98.26 mm, Y = 1.471 mm. (Key point: Net area and centroid)

3. With Ix = 6.860e7, Iy = 7.631e7 and Ixy = -1.384e7, what are I_max and I_min?
   A. 7.246e7 and 7.246e7
   B. 8.682e7 and 5.809e7 mm^4
   C. 8.682e7 and 1.437e7 mm^4
   D. 6.860e7 and 7.631e7 mm^4
   Answer: B. Mean (Ix+Iy)/2 = 7.246e7 and radius 1.437e7 give 8.682e7 and 5.809e7; their sum equals Ix + Iy. (Key point: Principal moments)

4. tan 2θp = -2Ixy/(Ix - Iy) = 2.768e7/(-7.709e6) gives 2θp = 105.56°. What is the direction of the I_max axis?
   A. 105.56° from +x
   B. 52.78° from +x, with the I_min axis perpendicular at -37.22°
   C. -37.22° from +x
   D. 74.44° from +x
   Answer: B. θp = 105.56°/2 = 52.78° gives the I_max axis (checked by the negative Ix - Iy), and the I_min axis is 90° away, at -37.22°. (Key point: Orientation)

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