---
title: "Principal MI and axes of trapezoidal shaded area"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-trapezoid-15-10-12"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-trapezoid-15-10-12
site: BCT Engineering
---

# Principal MI and axes of trapezoidal shaded area

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2068 Chaitra, 8 marks
- Video: [To Determine Principal Moment of Inertia and Principal axis of the given figure. | Prashant YT |BE|](https://www.youtube.com/watch?v=M0oG4N3EWS8) — Prashant YT
- 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-trapezoid-15-10-12
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Calculate principal moment of inertia and the orientation of the principle axes for the shaded area shown in figure below. [Figure: trapezoid with vertical left side 15 cm, bottom width 10 cm + 12 cm; the top edge is 10 cm wide and the right edge slopes down to the bottom right corner.]

## Short answer

Centroid at (8.375, 6.562) cm; I_x=4289, I_y=6526, I_xy=-1716 cm⁴. Principal moments: I_max=7456 cm⁴, I_min=3360 cm⁴; principal axes inclined at 61.55° (I_max) and -28.45° (I_min) from the +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm. Trapezoid (0,0), (22,0), (10,15), (0,15): vertical left side 15, bottom 10 + 12 = 22, top 10.

#### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Trapezoid | + | $240.0$ | 8.375 | 6.562 | $4289$ | $6526$ | $-1716$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 240.0\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{2010}{240.0} = 8.375\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{1575}{240.0} = 6.562\ \text{cm}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Trapezoid (+) | $4289$ | $6526$ | $-1716$ |
| **Total about centroid** | **$4289$** | **$6526$** | **$-1716$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 5408\pm2048 \\
I_{max} &= 7456\ \text{cm}^4,\qquad I_{min} = 3360\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3431}{-2237}$, so $2\theta_p = 123.10^\circ$ (and $-56.90^\circ$).

- The axis of $I_{max}$ makes 61.55° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -28.45° (perpendicular), both passing through the centroid (8.375, 6.562) measured from the reference origin.

**Answer: $I_{max} = 7456\ \text{cm}^4$, $I_{min} = 3360\ \text{cm}^4$; principal axes through the centroid at 61.55° (for $I_{max}$) and -28.45° (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 is the table for this trapezoid only one row?
   A. It was split into three triangles
   B. Its Ixy is zero
   C. The trapezoid is treated as a single shape with known centroidal Ix, Iy, Ixy
   D. The centroid is at the origin
   Answer: C. The solution uses the whole trapezoid's own inertias about its centroid, so no parallel-axis shifts are needed. (Key point: Single part)

2. Ixy is negative (-1716 cm^4) here. What does this indicate?
   A. The area is negative
   B. The area is distributed unsymmetrically so the section has inclined principal axes
   C. Ix is smaller than Iy only
   D. The calculation is wrong
   Answer: B. A non-zero Ixy shows no axis of symmetry, so the principal axes are rotated from the x-y axes. (Key point: Product of inertia)

3. With Ix = 4289, Iy = 6526 and Ixy = -1716, how are Imax and Imin found?
   A. Centre 5408 ± radius 2048 gives 7456 and 3360 cm^4
   B. Imax = Iy and Imin = Ix
   C. Imax = Ix + Iy
   D. Imax = 5408 + 1716
   Answer: A. Radius = sqrt(1118^2 + 1716^2) = 2048, so Imax = 7456 and Imin = 3360 cm^4. (Key point: Principal moments)

4. Which property must hold between the principal and original moments?
   A. Imax - Imin = Ix + Iy
   B. Imax * Imin = Ix * Iy
   C. Imax = Ix
   D. Imax + Imin = Ix + Iy
   Answer: D. The sum Ix + Iy is invariant under rotation: 7456 + 3360 = 4289 + 6526 = 10816. (Key point: Check)

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