---
title: "Principal MI and axes of quarter-circle/triangle section (2078 Bhadra)"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-2078b"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-2078b
site: BCT Engineering
---

# Principal MI and axes of quarter-circle/triangle section (2078 Bhadra)

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2078 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-2078b
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Calculate the principal moment of inertia about the centroid and locate the principal axes for the figure as shown below. [Figure: composite section: a quarter circle of radius 65 mm on top left, a rectangle 65 mm wide and 80 mm high below it, and a right-angled triangle of base 80 mm (to the right of the rectangle) and height 145 mm; dimensions 65 mm, 80 mm vertically and 65 mm, 80 mm horizontally.]

## Short answer

I_max = 2.615×10⁷ mm⁴, I_min = 1.254×10⁷ mm⁴. Principal axes through the centroid (55.33, 59.04 mm) are at 38.41° (I_max) and -51.59° (I_min) from the +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in mm. Rectangle 65 × 80 (x 0 to 65, y 0 to 80); quarter circle of radius 65 on top with centre at (0, 80); right-angled triangle on the right of the rectangle: vertices (65,0), (145,0), (65,145) (base 80, height 145).

#### 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 65×80 | + | $5200$ | 32.50 | 40.00 | $2.773\times10^{6}$ | $1.831\times10^{6}$ | $0$ |
| Quarter circle r = 65 | + | $3318$ | 27.59 | 107.6 | $9.796\times10^{5}$ | $9.796\times10^{5}$ | $-2.940\times10^{5}$ |
| Triangle | + | $5800$ | 91.67 | 48.33 | $6.775\times10^{6}$ | $2.062\times10^{6}$ | $-1.869\times10^{6}$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 14318\ \text{mm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{7.922\times10^{5}}{14318} = 55.33\ \text{mm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{8.453\times10^{5}}{14318} = 59.04\ \text{mm}
\end{aligned}
$$

#### 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 65×80 (+) | $4.658\times10^{6}$ | $4.541\times10^{6}$ | $2.260\times10^{6}$ |
| Quarter circle r = 65 (+) | $8.800\times10^{6}$ | $3.533\times10^{6}$ | $-4.763\times10^{6}$ |
| Triangle (+) | $7.439\times10^{6}$ | $9.721\times10^{6}$ | $-4.125\times10^{6}$ |
| **Total about centroid** | **$2.090\times10^{7}$** | **$1.780\times10^{7}$** | **$-6.628\times10^{6}$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 1.935\times10^{7}\pm6.807\times10^{6} \\
I_{max} &= 2.615\times10^{7}\ \text{mm}^4,\qquad I_{min} = 1.254\times10^{7}\ \text{mm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{1.326\times10^{7}}{3.103\times10^{6}}$, so $2\theta_p = 76.82^\circ$ (and $-103.18^\circ$).

- The axis of $I_{max}$ makes 38.41° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -51.59° (perpendicular), both passing through the centroid (55.33, 59.04) measured from the reference origin.

**Answer: $I_{max} = 2.615\times10^{7}\ \text{mm}^4$, $I_{min} = 1.254\times10^{7}\ \text{mm}^4$; principal axes through the centroid at 38.41° (for $I_{max}$) and -51.59° (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. Why must the centroid (X, Y) of the whole section be found first?
   A. Because the principal axes always coincide with the reference axes through the origin
   B. Because the principal moments asked for are about axes through the centroid, and parallel-axis shifts need the centroid location
   C. Because the product of inertia about the origin is always zero
   D. Because the centroid is the point where Imax occurs
   Answer: B. I_x, I_y, I_xy are required about centroidal axes, so each part's distance d_x, d_y is measured from the overall centroid (55.33 mm, 59.04 mm). (Key point: Centroid first)

2. How is the product of inertia of one part transferred to the section's centroidal axes?
   A. I_xy = I_xy,G + A d_x^2
   B. I_xy = I_xy,G + A (d_x + d_y)
   C. I_xy = I_xy,G + A d_x d_y, keeping the signs of d_x and d_y
   D. I_xy = I_xy,G, because the product of inertia does not change with axis shift
   Answer: C. The parallel-axis theorem for product of inertia adds A times the product of the signed centroidal offsets. (Key point: Parallel-axis for Ixy)

3. Which expression gives the principal moments of inertia from I_x, I_y and I_xy?
   A. (I_x+I_y)/2 ± sqrt((I_x-I_y)^2 + I_xy)
   B. (I_x-I_y)/2 ± sqrt(((I_x+I_y)/2)^2 + I_xy^2)
   C. sqrt(I_x^2 + I_y^2) ± I_xy
   D. (I_x+I_y)/2 ± sqrt(((I_x-I_y)/2)^2 + I_xy^2)
   Answer: D. The principal values are the centre of Mohr's circle plus and minus its radius. (Key point: Principal moment formula)

4. With I_x = 2.090e7, I_y = 1.780e7 and I_xy = -6.628e6 mm^4, which statement matches the answer?
   A. I_max = 2.615e7 and I_min = 1.254e7 mm^4; the I_max axis is at 38.41° from +x and the I_min axis is perpendicular to it
   B. I_max = 2.090e7 and I_min = 1.780e7 mm^4; the I_max axis is along the x axis
   C. I_max = 1.254e7 mm^4 and I_min = 2.615e7 mm^4 with both axes at 45 degrees
   D. I_max and I_min are both equal to (I_x+I_y)/2, and the axes are not defined
   Answer: A. Imax and Imin come from the formula, and the principal axes are perpendicular, with the I_max axis at 38.41° (I_min axis at -51.59°). (Key point: Result and axis angle)

## Related in Strength of Materials

- [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
- [Principal MI of section with quarter circle](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-quarter-circle-800-600) — 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, direction and position of section (2080 Bhadra)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-2080b) — 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 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

## Related in other subjects

- [Composite body moment of inertia](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-sphere-composite-inertia) — Engineering Mechanics, chapter 6