---
title: "Principal MI and orientation of L-section"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-l-section-60-12"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-l-section-60-12
site: BCT Engineering
---

# Principal MI and orientation of L-section

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2067 Asar (old course), 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-60-12
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Calculate the principal MoI and their orientation for the following section about X-Y axes. [Figure: L-shaped section in mm with origin O at the bottom left corner; horizontal leg 60 mm long and 10 mm thick; vertical leg 12 mm wide.]

## Short answer

Centroid at (18.00, 20.00) mm; I_x=4.000×10⁵, I_y=3.600×10⁵, I_xy=-2.160×10⁵ mm⁴. Principal moments: I_max=5.969×10⁵ mm⁴, I_min=1.631×10⁵ mm⁴; axes at 42.35° (I_max) and -47.65° (I_min) from +x.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in mm; origin O at the bottom-left corner. Horizontal leg 60 × 10 (x 0 to 60, y 0 to 10); vertical leg 12 × 50 above it (x 0 to 12, y 10 to 60). Both legs are 60 mm long overall.

#### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Horizontal leg 60×10 | + | $600.0$ | 30.00 | 5.000 | $5000$ | $1.800\times10^{5}$ | $0$ |
| Vertical leg 12×50 | + | $600.0$ | 6.000 | 35.00 | $1.250\times10^{5}$ | $7200$ | $0$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 1200\ \text{mm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{21600}{1200} = 18.00\ \text{mm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{24000}{1200} = 20.00\ \text{mm}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Horizontal leg 60×10 (+) | $1.400\times10^{5}$ | $2.664\times10^{5}$ | $-1.080\times10^{5}$ |
| Vertical leg 12×50 (+) | $2.600\times10^{5}$ | $93600$ | $-1.080\times10^{5}$ |
| **Total about centroid** | **$4.000\times10^{5}$** | **$3.600\times10^{5}$** | **$-2.160\times10^{5}$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 3.800\times10^{5}\pm2.169\times10^{5} \\
I_{max} &= 5.969\times10^{5}\ \text{mm}^4,\qquad I_{min} = 1.631\times10^{5}\ \text{mm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{4.320\times10^{5}}{40000}$, so $2\theta_p = 84.71^\circ$ (and $-95.29^\circ$).

- The axis of $I_{max}$ makes 42.35° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -47.65° (perpendicular), both passing through the centroid (18.00, 20.00) measured from the reference origin.

#### Values about the given origin
About the axes through the origin O: $I_x = 8.800\times10^{5}$, $I_y = 7.488\times10^{5}$, $I_{xy} = 2.160\times10^{5}\ \text{mm}^4$ (see parts table with the parallel-axis theorem). Principal values about O: $I_{max} = 1.040\times10^{6}$, $I_{min} = 5.887\times10^{5}$ at -36.55° to the x axis.

**Answer: $I_{max} = 5.969\times10^{5}\ \text{mm}^4$, $I_{min} = 1.631\times10^{5}\ \text{mm}^4$; principal axes through the centroid at 42.35° (for $I_{max}$) and -47.65° (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 centroid of the L (60x10 leg and 12x50 leg) found?
   A. Mid-point of the 60 x 60 square
   B. At the corner O
   C. Area-weighted average: Xbar = 21600/1200 = 18 mm, Ybar = 24000/1200 = 20 mm
   D. Average of the leg lengths
   Answer: C. Both areas are 600 mm^2, and the weighted averages give (18, 20) mm. (Key point: Centroid)

2. Why is the centroidal Ixy negative (-2.16e5 mm^4) for each leg?
   A. Rectangles always have negative Ixy
   B. Each leg's centroid is on opposite sides of the section's centroid in x and y (dx*dy < 0)
   C. The legs are subtracted
   D. Ixy is always negative for L sections
   Answer: B. Each part lies at one side in x and the opposite side in y of the section centroid, so A*dx*dy is negative. (Key point: Ixy sign)

3. With Ix = 4e5, Iy = 3.6e5 and Ixy = -2.16e5, what is Imax?
   A. 3.8e5 + 2.169e5 = 5.969e5 mm^4
   B. 4e5 + 3.6e5
   C. 3.8e5 - 2.169e5
   D. 2.16e5
   Answer: A. Centre 3.8e5, radius sqrt((0.2e5)^2 + (2.16e5)^2) = 2.169e5. (Key point: Principal moments)

## Related in Strength of Materials

- [Principal MI of L-shaped composite section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-l-section-4-5) — Strength of Materials, chapter 2
- [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 MI and orientation of Z-type section](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-z-section-cm) — Strength of Materials, chapter 2
- [Principal MI and axes of trapezoidal shaded area](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-mi-trapezoid-15-10-12) — 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 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

## Related in other subjects

- [MOI of arch section about horizontal centroidal axis](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Amoi-section-horizontal-centroidal-axis-arch) — Applied Mechanics, chapter 4
- [Arch 100 m rise 12 m: rolling UDL 60 m, section 10 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-rise12-udl-60m-10m) — Theory of Structures I, chapter 6
- [Vertical drop in 10 m wide canal, Q 20](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-20cumec-10m-canal) — Irrigation and Drainage Engineering, chapter 7
- [Truss forces: member AC 10 mm too long](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-lack-of-fit-ac-10mm) — Theory of Structures II, chapter 2