---
title: "Principal MI of L-shaped composite section"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-l-section-4-5"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-l-section-4-5
site: BCT Engineering
---

# Principal MI of L-shaped composite section

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2069 Chaitra, 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-l-section-4-5
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Determine the principal moment of inertia and orientation of principal axes for the composite section shown in figure below about its centroid. [Figure: L-shaped section: overall height 9 cm and overall width 12 cm; vertical leg 4.5 cm wide; horizontal leg 3 cm thick.]

## Short answer

Centroid at (4.393, 3.429) cm; I_x=420.4, I_y=694.5, I_xy=-260.4 cm⁴ about centroid. Principal moments: I_max=851.7 cm⁴, I_min=263.3 cm⁴; principal axes at 58.88° (I_max) and -31.12° (I_min) from the +x axis through the centroid.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm. Horizontal leg 12 × 3 (x 0 to 12, y 0 to 3); vertical leg 4.5 × 6 above it (x 0 to 4.5, y 3 to 9). Overall 12 wide, 9 high.

#### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Horizontal leg 12×3 | + | $36.00$ | 6.000 | 1.500 | $27.00$ | $432.0$ | $0$ |
| Vertical leg 4.5×6 | + | $27.00$ | 2.250 | 6.000 | $81.00$ | $45.56$ | $0$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 63.00\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{276.8}{63.00} = 4.393\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{216.0}{63.00} = 3.429\ \text{cm}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Horizontal leg 12×3 (+) | $160.9$ | $525.0$ | $-111.6$ |
| Vertical leg 4.5×6 (+) | $259.5$ | $169.5$ | $-148.8$ |
| **Total about centroid** | **$420.4$** | **$694.5$** | **$-260.4$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 557.5\pm294.2 \\
I_{max} &= 851.7\ \text{cm}^4,\qquad I_{min} = 263.3\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{520.7}{-274.1}$, so $2\theta_p = 117.76^\circ$ (and $-62.24^\circ$).

- The axis of $I_{max}$ makes 58.88° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -31.12° (perpendicular), both passing through the centroid (4.393, 3.429) measured from the reference origin.

**Answer: $I_{max} = 851.7\ \text{cm}^4$, $I_{min} = 263.3\ \text{cm}^4$; principal axes through the centroid at 58.88° (for $I_{max}$) and -31.12° (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. How is the L-section analysed?
   A. As one rectangle 12 x 9
   B. As two rectangles (12x3 and 4.5x6) using the parallel-axis theorem about the section centroid
   C. By assuming it is symmetrical
   D. By ignoring the vertical leg
   Answer: B. The composite is split into two rectangles whose inertias are shifted to the common centroid. (Key point: Method)

2. Why is Ixy of the L-section not zero?
   A. The section has no axis of symmetry, so the A*dx*dy terms do not cancel
   B. Each rectangle has a non-zero Ixy about its own centroid
   C. Ixy is zero only for circles
   D. Ix is larger than Iy
   Answer: A. Each rectangle's own Ixy is zero, but the Adxdy terms are non-zero (total -260.4 cm^4) because the L is unsymmetrical. (Key point: Product of inertia)

3. Which formula gives Imax and Imin?
   A. Ix ± Iy
   B. (Ix-Iy)/2 ± Ixy
   C. sqrt(Ix*Iy) ± Ixy
   D. (Ix+Iy)/2 ± sqrt(((Ix-Iy)/2)^2 + Ixy^2)
   Answer: D. The principal moments are the extremes of the Mohr's circle for inertia: centre (Ix+Iy)/2, radius sqrt(((Ix-Iy)/2)^2 + Ixy^2). (Key point: Principal values)

4. What does tan 2θ = -2Ixy/(Ix-Iy) give, and what is the Imax axis here?
   A. The centroid position; 4.393 cm
   B. The shear angle; 31.12°
   C. The principal axis angle; Imax axis at 58.88° to +x
   D. The radius of gyration; 3.4 cm
   Answer: C. The formula gives 2θ = 117.76°, so the Imax axis is at 58.88° and the Imin axis is perpendicular at -31.12°. (Key point: Orientation)

## Related in Strength of Materials

- [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 axes of composite section (8, 12 cm)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-axes-orientation-composite-8-12) — 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
- [Principal MI and axes, 120 cm section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-and-axes-orientation-120) — Strength of Materials, chapter 2
- [Principal MI and axis of composite section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-axes-mi-2081k) — Strength of Materials, chapter 2
- [Principal MI and orientation of L-section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-l-section-60-12) — Strength of Materials, chapter 2

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