---
title: "Principal MI, direction and position of section (2080 Bhadra)"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-2080b"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-2080b
site: BCT Engineering
---

# Principal MI, direction and position of section (2080 Bhadra)

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2080 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-mi-2080b
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Find the principal moment of inertia, direction and position for the given section shown in figure. (All dimensions are in mm.) [Figure: composite section: a triangle on top-left of height 500 with 600 horizontal base marking, a rectangle of 200 depth, 600 dimension, 600 dimension, 200 vertical dimension and a quarter circle at the lower right.]

## Short answer

Composite-area method: centroid (496.0, 181.1) mm; I_x = 1.194×10^10, I_y = 5.967×10^10, I_xy = –1.671×10^10 mm^4. Principal moments: I_max = 6.494×10^10 mm^4, I_min = 6.667×10^9 mm^4. Principal axes through centroid at 72.50° (I_max) and –17.50° (I_min) from +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in mm. Rectangle 1200 × 200 (x 0 to 1200, y 0 to 200); triangle above its left end with base 600 and height 500 (vertices (0,200), (600,200), (0,700)); quarter circle of radius 200 below the right end, centre at (1200, 0), occupying x 1000 to 1200, y −200 to 0.

#### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Rectangle 1200×200 | + | $2.400\times10^{5}$ | 600.0 | 100.0 | $8.000\times10^{8}$ | $2.880\times10^{10}$ | $0$ |
| Triangle | + | $1.500\times10^{5}$ | 200.0 | 366.7 | $2.083\times10^{9}$ | $3.000\times10^{9}$ | $-1.250\times10^{9}$ |
| Quarter circle r = 200 | + | $31416$ | 1115 | -84.88 | $8.781\times10^{7}$ | $8.781\times10^{7}$ | $-2.635\times10^{7}$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 4.214\times10^{5}\ \text{mm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{2.090\times10^{8}}{4.214\times10^{5}} = 496.0\ \text{mm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{7.633\times10^{7}}{4.214\times10^{5}} = 181.1\ \text{mm}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Rectangle 1200×200 (+) | $2.380\times10^{9}$ | $3.139\times10^{10}$ | $-2.025\times10^{9}$ |
| Triangle (+) | $7.247\times10^{9}$ | $1.614\times10^{10}$ | $-9.488\times10^{9}$ |
| Quarter circle r = 200 (+) | $2.311\times10^{9}$ | $1.213\times10^{10}$ | $-5.200\times10^{9}$ |
| **Total about centroid** | **$1.194\times10^{10}$** | **$5.967\times10^{10}$** | **$-1.671\times10^{10}$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 3.580\times10^{10}\pm2.914\times10^{10} \\
I_{max} &= 6.494\times10^{10}\ \text{mm}^4,\qquad I_{min} = 6.667\times10^{9}\ \text{mm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3.343\times10^{10}}{-4.773\times10^{10}}$, so $2\theta_p = 145.00^\circ$ (and $-35.00^\circ$).

- The axis of $I_{max}$ makes 72.50° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -17.50° (perpendicular), both passing through the centroid (496.0, 181.1) measured from the reference origin.

**Answer: $I_{max} = 6.494\times10^{10}\ \text{mm}^4$, $I_{min} = 6.667\times10^{9}\ \text{mm}^4$; principal axes through the centroid at 72.50° (for $I_{max}$) and -17.50° (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 parts and total area are used?
   A. Rectangle 2.4e5 + triangle 1.5e5 + quarter circle 31416 = 4.214e5 mm^2
   B. Rectangle 2.4e5 + triangle 3.0e5 - quarter circle
   C. Only the rectangle and triangle, 3.9e5 mm^2
   D. Rectangle 1.2e5 + triangle 1.5e5 + quarter circle 31416
   Answer: A. Rectangle 1200 x 200 = 2.4e5, triangle 1/2 x 600 x 500 = 1.5e5, quarter circle pi(200)^2/4 = 31416; total 4.214e5 mm^2. (Key point: Decomposition and areas)

2. What is the location of the centroid?
   A. (600, 100) mm
   B. (496.0, 181.1) mm
   C. (200, 366.7) mm
   D. (1115, -84.88) mm
   Answer: B. Weighted by area: X = 2.090e8/4.214e5 = 496.0 mm and Y = 7.633e7/4.214e5 = 181.1 mm. (Key point: Centroid)

3. Given Ix = 1.194e10, Iy = 5.967e10 and Ixy = -1.671e10, what is I_max?
   A. 3.580e10 mm^4
   B. 6.494e10 mm^4
   C. 5.967e10 mm^4
   D. 2.914e10 mm^4
   Answer: B. I_max = 3.580e10 + 2.914e10 = 6.494e10 mm^4; I_min = 6.67e9 mm^4. (Key point: Principal moments)

4. 2θp = 145.00° from tan 2θp = 3.343e10/(-4.773e10). The I_max axis is at:
   A. 145.00° from +x
   B. 72.50° from +x, and I_min axis at -17.50°
   C. -35.00° from +x
   D. 17.50° from +x
   Answer: B. θp = 145°/2 = 72.5° for I_max, and the perpendicular direction -17.5° is the I_min axis. (Key point: Axis direction)

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