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

# Principal MI and axis of composite section

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2081 Baisakh, 10 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-2081k
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Determine the principle moment of inertia and orientation of principle axis for the section shown in the figure below about centroid, all dimensions are in mm. [Figure: composite section made of a vertical triangle at the top left (height 600, with a 600 horizontal dimension), a horizontal rectangular strip of length 1000 and 200 thick, and a lower right triangle with horizontal dimension 600 and height 600; overall dimensions: 600, 1000, 600 horizontal at the bottom, 200 and 600 vertical on the right.]

## Short answer

Composite-area method: centroid (500, 100) mm; I_x = 4.027×10^10, I_y = 5.627×10^10, I_xy = –3.600×10^10 mm^4. Principal moments: I_max = 8.514×10^10 mm^4, I_min = 1.139×10^10 mm^4. Principal axes through centroid at 51.26° (I_max) and –38.74° (I_min) from +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in mm. Strip: 1000 × 200 (x 0 to 1000, y 0 to 200). Upper-left right triangle: vertices (0,200), (600,200), (0,800). Lower-right right triangle: vertices (1000,0), (400,0), (1000,−600).

#### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Strip 1000×200 | + | $2.000\times10^{5}$ | 500.0 | 100.0 | $6.667\times10^{8}$ | $1.667\times10^{10}$ | $0$ |
| Upper triangle | + | $1.800\times10^{5}$ | 200.0 | 400.0 | $3.600\times10^{9}$ | $3.600\times10^{9}$ | $-1.800\times10^{9}$ |
| Lower triangle | + | $1.800\times10^{5}$ | 800.0 | -200.0 | $3.600\times10^{9}$ | $3.600\times10^{9}$ | $-1.800\times10^{9}$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 5.600\times10^{5}\ \text{mm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{2.800\times10^{8}}{5.600\times10^{5}} = 500.0\ \text{mm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{5.600\times10^{7}}{5.600\times10^{5}} = 100.0\ \text{mm}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Strip 1000×200 (+) | $6.667\times10^{8}$ | $1.667\times10^{10}$ | $0$ |
| Upper triangle (+) | $1.980\times10^{10}$ | $1.980\times10^{10}$ | $-1.800\times10^{10}$ |
| Lower triangle (+) | $1.980\times10^{10}$ | $1.980\times10^{10}$ | $-1.800\times10^{10}$ |
| **Total about centroid** | **$4.027\times10^{10}$** | **$5.627\times10^{10}$** | **$-3.600\times10^{10}$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 4.827\times10^{10}\pm3.688\times10^{10} \\
I_{max} &= 8.514\times10^{10}\ \text{mm}^4,\qquad I_{min} = 1.139\times10^{10}\ \text{mm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{7.200\times10^{10}}{-1.600\times10^{10}}$, so $2\theta_p = 102.53^\circ$ (and $-77.47^\circ$).

- The axis of $I_{max}$ makes 51.26° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -38.74° (perpendicular), both passing through the centroid (500.0, 100.0) measured from the reference origin.

**Answer: $I_{max} = 8.514\times10^{10}\ \text{mm}^4$, $I_{min} = 1.139\times10^{10}\ \text{mm}^4$; principal axes through the centroid at 51.26° (for $I_{max}$) and -38.74° (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. Which centroid is obtained for the composite section (strip plus two triangles of 1.8e5 mm^2 each)?
   A. (500, 100) mm
   B. (600, 0) mm
   C. (500, 200) mm
   D. (400, 100) mm
   Answer: A. Total area 5.6e5 mm^2; sum A x = 2.8e8 gives X = 500 mm and sum A y = 5.6e7 gives Y = 100 mm. (Key point: Centroid of composite)

2. Why does the 1000 x 200 strip contribute zero to the product of inertia about the centroidal axes?
   A. Its area is small
   B. It is symmetric about the centroidal axes, so its own I_xy is zero and the shift term is also zero
   C. It has no weight
   D. Its moment of inertia is zero
   Answer: B. The strip's centroid coincides with the section's centroid (500, 100) and is symmetric, so both I_xy,G and A dx dy are zero. (Key point: Product of inertia of strip)

3. Given Ix = 4.027e10, Iy = 5.627e10 and Ixy = -3.6e10, what are I_max and I_min?
   A. 8.514e10 and 1.139e10 mm^4
   B. 5.627e10 and 4.027e10 mm^4
   C. 4.827e10 and 3.688e10 mm^4
   D. 9.654e10 and 0 mm^4
   Answer: A. Mean 4.827e10 plus/minus radius sqrt(0.8e10^2 + 3.6e10^2) = 3.688e10 gives 8.514e10 and 1.139e10. (Key point: Principal values)

4. tan 2θp = 7.2e10/(-1.6e10) gives 2θp = 102.53°. What is the angle of the I_max axis?
   A. 102.53°
   B. 51.26° from +x anticlockwise
   C. -38.74° from +x
   D. 25.63°
   Answer: B. θp = 102.53°/2 = 51.26° for I_max; the I_min axis is perpendicular at -38.74°. (Key point: Orientation)

## Related in Strength of Materials

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- [Principal axis and MI of composite section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-axes-mi-2081b) — Strength of Materials, chapter 2
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- [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
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