---
title: "Principal MI of section with trapezoidal cut-out"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-section-with-trapezoid-hole"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-section-with-trapezoid-hole
site: BCT Engineering
---

# Principal MI of section with trapezoidal cut-out

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2069 Asar, 8 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-section-with-trapezoid-hole
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Determine the principle moment of inertia about centroidal axis and locate the principle axes for the section shown in figure below. [Figure: rectangular block with horizontal dimensions 2 m, 3 m, 2 m, 3 m, 2 m (total 12 m [as marked]) and vertical dimensions 3 m (top) and 9 m, with a trapezoidal slot cut from the bottom to the top.]

## Short answer

Centroid at (6.000, 7.091) m. Due to symmetry about vertical axis, I_xy=0, so centroidal axes are principal. I_x=1084, I_y=1600 m^4. Principal moments: I_max=1600 m^4 (vertical axis), I_min=1084 m^4 (horizontal axis).

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in m. Block 12 × 12 (x 0 to 12, y 0 to 12). Trapezoidal slot from the bottom edge: vertices (2,0), (10,0), (7,9), (5,9), removed (3 m of solid left above the slot).

#### Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $m^2$, I in $m^4$)
| Part | Sign | A | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Block 12×12 | + | $144.0$ | 6.000 | 6.000 | $1728$ | $1728$ | $0$ |
| Trapezoid slot | − | $45.00$ | 6.000 | 3.600 | $267.3$ | $127.5$ | $0$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 99.00\ \text{m}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{594.0}{99.00} = 6.000\ \text{m} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{702.0}{99.00} = 7.091\ \text{m}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Block 12×12 (+) | $1899$ | $1728$ | $0$ |
| Trapezoid slot (−) | $-815.7$ | $-127.5$ | $0$ |
| **Total about centroid** | **$1084$** | **$1600$** | **$0$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 1342\pm258.4 \\
I_{max} &= 1600\ \text{m}^4,\qquad I_{min} = 1084\ \text{m}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{0}{-516.8} = 0$, so $2\theta_p = 0^\circ$ or $180^\circ$; since $I_y > I_x$, the maximum is about the vertical axis.

- The axis of $I_{max}$ is the vertical centroidal axis x = 6 m (at 90° to the +x axis), and the axis of $I_{min}$ is the horizontal centroidal axis y = 7.091 m (0°), both passing through the centroid (6.000, 7.091) measured from the reference origin.

The section is symmetrical about the vertical axis x = 6 m, so $I_{xy} = 0$ and the centroidal x and y axes are themselves the principal axes.

**Answer: $I_{max} = 1600\ \text{m}^4$ (about the vertical centroidal axis), $I_{min} = 1084\ \text{m}^4$ (about the horizontal centroidal axis); the principal axes are the centroidal y and x axes through (6.000, 7.091).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the trapezoidal slot handled in the composite-area method?
   A. Added to the 12x12 block as a positive area
   B. Subtracted from the block as a negative area
   C. Ignored because it lies on the edge
   D. Treated as a separate section analysed alone
   Answer: B. A hole or slot removes material, so its area and inertia are entered with a negative sign. (Key point: Treating the slot)

2. Why does the centroid lie at X = 6 m, Y = 7.091 m rather than at Y = 6 m?
   A. Removing the slot from the bottom raises the centroid of the remaining area
   B. The block is not symmetrical about x = 6
   C. The slot is added with a positive sign
   D. The reference origin was moved
   Answer: A. Material is removed near the bottom, so the remaining area's centroid moves upward (702/99 = 7.091 m). (Key point: Centroid)

3. Why is Ixy = 0 for this section?
   A. The section has no holes
   B. Ix equals Iy
   C. Ixy is always zero for rectangles with a hole
   D. The area is symmetrical about the vertical axis x = 6 m
   Answer: D. Symmetry about an axis makes the product of inertia zero about that axis and its perpendicular, so the centroidal axes are principal. (Key point: Product of inertia)

4. Given Ixy = 0, Ix = 1084 m^4 and Iy = 1600 m^4, what are the principal values?
   A. Imax = 1342 and Imin = 258
   B. Both equal 1342
   C. Imax = 1600 and Imin = 1084 about the centroidal y and x axes
   D. They must be found by rotating the axes by 45°
   Answer: C. When Ixy = 0 the centroidal axes are principal, so the principal moments are simply Ix and Iy. (Key point: Principal axes)

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