---
title: "Principal MI of quarter circle and triangle section (cm)"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-quarter-circle-15-30"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-quarter-circle-15-30
site: BCT Engineering
---

# Principal MI of quarter circle and triangle section (cm)

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2076 Asoj, 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-mi-quarter-circle-15-30
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Determine principal moment of inertia about the centroidal axis of following figure. [Figure: composite section: a quarter of a circle of radius 15 cm on top left, a rectangle below it 15 cm wide and 15 cm high, and a right-angled triangle on the right with base 30 cm and the same 15 cm height; vertical dimensions 15 cm and 15 cm, horizontal dimensions 15 cm and 30 cm.]

## Short answer

I_max = 82655 cm⁴, I_min = 21947 cm⁴. Principal axes through the centroid (13.46, 10.51 cm) are at 57.47° (I_max) and -32.53° (I_min) from the +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm. Rectangle 15 × 15 (x 0 to 15, y 0 to 15); quarter circle of radius 15 above it with centre at (0, 15); right-angled triangle on the right: vertices (15,0), (45,0), (15,15).

#### Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $cm^2$, I in $cm^4$)
| Part | Sign | A | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Rectangle 15×15 | + | $225.0$ | 7.500 | 7.500 | $4219$ | $4219$ | $0$ |
| Quarter circle r = 15 | + | $176.7$ | 6.366 | 21.37 | $2778$ | $2778$ | $-833.8$ |
| Triangle | + | $225.0$ | 25.00 | 5.000 | $2812$ | $11250$ | $-2812$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 626.7\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{8438}{626.7} = 13.46\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{6588}{626.7} = 10.51\ \text{cm}
\end{aligned}
$$

#### 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 15×15 (+) | $6260$ | $12219$ | $4042$ |
| Quarter circle r = 15 (+) | $23596$ | $11679$ | $-14446$ |
| Triangle (+) | $9649$ | $41198$ | $-17121$ |
| **Total about centroid** | **$39506$** | **$65096$** | **$-27526$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 52301\pm30354 \\
I_{max} &= 82655\ \text{cm}^4,\qquad I_{min} = 21947\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{55052}{-25590}$, so $2\theta_p = 114.93^\circ$ (and $-65.07^\circ$).

- The axis of $I_{max}$ makes 57.47° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -32.53° (perpendicular), both passing through the centroid (13.46, 10.51) measured from the reference origin.

**Answer: $I_{max} = 82655\ \text{cm}^4$, $I_{min} = 21947\ \text{cm}^4$; principal axes through the centroid at 57.47° (for $I_{max}$) and -32.53° (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 (13.46 cm, 10.51 cm). (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 = 39506, I_y = 65096 and I_xy = -27526 cm^4, which statement matches the answer?
   A. I_max = 82655 and I_min = 21947 cm^4; the I_max axis is at 57.47° from +x and the I_min axis is perpendicular to it
   B. I_max = 39506 and I_min = 65096 cm^4; the I_max axis is along the x axis
   C. I_max = 21947 cm^4 and I_min = 82655 cm^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 57.47° (I_min axis at -32.53°). (Key point: Result and axis angle)

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