---
title: "Principal axis and MI with Mohr's circle (2079 Bhadra)"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-mohr-2079b"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-mohr-2079b
site: BCT Engineering
---

# Principal axis and MI with Mohr's circle (2079 Bhadra)

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2079 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-mohr-2079b
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Find out the principle axis and principle moment of inertia for the given section and verify using Mohr's circle. [Figure: section with a vertical left edge of height 300 mm + 200 mm + 200 mm, a triangular portion on top left, a quarter circle cut/added at the middle, and a rectangular base 200 mm + 200 mm + 200 mm wide.]

## Short answer

I_max = 9.669×10⁹ mm⁴, I_min = 4.157×10⁹ mm⁴. Principal axes through the centroid (256.8, 229.2 mm) are at 56.15° (I_max) and -33.85° (I_min) from the +x axis; verified by Mohr’s circle with centre 6.913×10⁹ and radius 2.756×10⁹.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in mm. Base rectangle 600 × 200; triangle above its left end (vertices (0,200), (400,200), (0,700)), so the left edge is 700 mm high; quarter circle of radius 200 added at the right of the triangle with centre (400, 200), occupying x 400 to 600, y 200 to 400.

#### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Base 600×200 | + | $1.200\times10^{5}$ | 300.0 | 100.0 | $4.000\times10^{8}$ | $3.600\times10^{9}$ | $0$ |
| Triangle | + | $1.000\times10^{5}$ | 133.3 | 366.7 | $1.389\times10^{9}$ | $8.889\times10^{8}$ | $-5.556\times10^{8}$ |
| Quarter circle r = 200 | + | $31416$ | 484.9 | 284.9 | $8.781\times10^{7}$ | $8.781\times10^{7}$ | $-2.635\times10^{7}$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 2.514\times10^{5}\ \text{mm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{6.457\times10^{7}}{2.514\times10^{5}} = 256.8\ \text{mm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{5.762\times10^{7}}{2.514\times10^{5}} = 229.2\ \text{mm}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Base 600×200 (+) | $2.402\times10^{9}$ | $3.824\times10^{9}$ | $-6.694\times10^{8}$ |
| Triangle (+) | $3.279\times10^{9}$ | $2.414\times10^{9}$ | $-2.253\times10^{9}$ |
| Quarter circle r = 200 (+) | $1.853\times10^{8}$ | $1.722\times10^{9}$ | $3.728\times10^{8}$ |
| **Total about centroid** | **$5.867\times10^{9}$** | **$7.959\times10^{9}$** | **$-2.550\times10^{9}$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 6.913\times10^{9}\pm2.756\times10^{9} \\
I_{max} &= 9.669\times10^{9}\ \text{mm}^4,\qquad I_{min} = 4.157\times10^{9}\ \text{mm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{5.100\times10^{9}}{-2.092\times10^{9}}$, so $2\theta_p = 112.31^\circ$ (and $-67.69^\circ$).

- The axis of $I_{max}$ makes 56.15° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -33.85° (perpendicular), both passing through the centroid (256.8, 229.2) measured from the reference origin.

#### Mohr's circle check
- Centre $C = \left(\dfrac{I_x+I_y}{2},\,0\right) = (6.913\times10^{9},\ 0)$; radius $R = 2.756\times10^{9}$.
- Plot $X(I_x,\ I_{xy}) = (5.867\times10^{9},\ -2.550\times10^{9})$ and $Y(I_y,\ -I_{xy}) = (7.959\times10^{9},\ 2.550\times10^{9})$; the line XY is a diameter.
- The circle cuts the horizontal axis at $I_{max} = 9.669\times10^{9}$ and $I_{min} = 4.157\times10^{9}$, equal to the values found analytically. The angle from CX to the $I_{max}$ end of the horizontal diameter is $2\theta_p = 112.31^\circ$ (turned from the x axis towards the principal axis in the same sense as the physical rotation).

**Answer: $I_{max} = 9.669\times10^{9}\ \text{mm}^4$, $I_{min} = 4.157\times10^{9}\ \text{mm}^4$; principal axes through the centroid at 56.15° (for $I_{max}$) and -33.85° (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 (256.8 mm, 229.2 mm). (Key point: Centroid first)

2. 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)

3. With I_x = 5.867e9, I_y = 7.959e9 and I_xy = -2.550e9 mm^4, which statement matches the answer?
   A. I_max = 9.669e9 and I_min = 4.157e9 mm^4; the I_max axis is at 56.15° from +x and the I_min axis is perpendicular to it
   B. I_max = 5.867e9 and I_min = 7.959e9 mm^4; the I_max axis is along the x axis
   C. I_max = 4.157e9 mm^4 and I_min = 9.669e9 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 56.15° (I_min axis at -33.85°). (Key point: Result and axis angle)

4. In the Mohr's circle verification, how are the centre, radius and principal values obtained?
   A. Centre = (I_x+I_y)/2 on the I axis, radius = distance to point X(I_x, I_xy); I_max = C + R and I_min = C - R
   B. Centre = I_xy on the vertical axis, radius = (I_x+I_y)/2
   C. Centre = I_x on the I axis, radius = I_y; I_max = I_x and I_min = I_y
   D. Centre = 0, radius = sqrt(I_x*I_y); I_max = R and I_min = -R
   Answer: A. Points X(I_x, I_xy) and Y(I_y, -I_xy) form a diameter; the circle meets the horizontal axis at C ± R, matching the analytical values. Here C = 6.913e9, R = 2.756e9, so I_max = 9.669e9 and I_min = 4.157e9. (Key point: Mohr's circle check)

## 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 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
- [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 axes and MI with Mohr's circle](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-axes-mi-mohr-6m-section) — Strength of Materials, chapter 2
- [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

## Related in other subjects

- [Principal stresses; Mohr circle at failure](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aprincipal-stresses-mohr-circle-friction) — Soil Mechanics, chapter 10