---
title: "Principal MI and axes of L-section"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-l-section-15x20-5cm"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-l-section-15x20-5cm
site: BCT Engineering
---

# Principal MI and axes of L-section

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2066 Chaitra (old course), 16 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-15x20-5cm
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Calculate principal moment of inertia and locate the principal axes through the centroid of the area of the L section shown in figure. Total depth of the web is 20cm. Total breadth of the flange is 15cm and thickness of the web and flange is 5cm.

## Short answer

Centroid at (5, 7.5) cm from the reference corner. I_max = 6250 cm^4, I_min = 1562 cm^4; principal axes through centroid at 26.6° (max) and -63.4° (min) from the +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm. Flange 15 × 5 (x 0 to 15, y 0 to 5); web 5 × 15 above it (x 0 to 5, y 5 to 20). Total depth 20, breadth 15, thickness 5.

#### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Flange 15×5 | + | $75.00$ | 7.500 | 2.500 | $156.2$ | $1406$ | $0$ |
| Web 5×15 | + | $75.00$ | 2.500 | 12.50 | $1406$ | $156.2$ | $0$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 150.0\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{750.0}{150.0} = 5.000\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{1125}{150.0} = 7.500\ \text{cm}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Flange 15×5 (+) | $2031$ | $1875$ | $-937.5$ |
| Web 5×15 (+) | $3281$ | $625.0$ | $-937.5$ |
| **Total about centroid** | **$5312$** | **$2500$** | **$-1875$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 3906\pm2344 \\
I_{max} &= 6250\ \text{cm}^4,\qquad I_{min} = 1562\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3750}{2812}$, so $2\theta_p = 53.13^\circ$ (and $-126.87^\circ$).

- The axis of $I_{max}$ makes 26.57° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -63.43° (perpendicular), both passing through the centroid (5.000, 7.500) measured from the reference origin.

**Answer: $I_{max} = 6250\ \text{cm}^4$, $I_{min} = 1562\ \text{cm}^4$; principal axes through the centroid at 26.57° (for $I_{max}$) and -63.43° (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 15x20 L with 5 cm thickness divided?
   A. Two 15x20 rectangles
   B. Flange 15x5 and web 5x20, overlapping
   C. Flange 15x5 and web 5x15 above it
   D. One rectangle minus a square
   Answer: C. Splitting the web to 15 cm avoids counting the corner square twice. (Key point: Splitting)

2. Why is the centroid at (5, 7.5) cm?
   A. Half the overall width and depth
   B. The two parts have equal areas of 75 cm^2, so it is the average of their centroids
   C. It is at the corner
   D. Areas are unequal
   Answer: B. Xbar = (7.5 + 2.5)/2 = 5 and Ybar = (2.5 + 12.5)/2 = 7.5. (Key point: Centroid)

3. Ixy = -1875 cm^4 total. How is it built?
   A. Each part contributes A*dx*dy = -937.5
   B. Each part's own Ixy is -937.5
   C. It is the difference Ix - Iy
   D. Only the flange contributes
   Answer: A. Own Ixy of the rectangles is zero; each part has dx*dy = 2.5 x (-5) offset giving -937.5. (Key point: Product of inertia)

4. What are Imax and Imin?
   A. 5312 and 2500 cm^4
   B. 3906 and 2344 cm^4
   C. 6250 and 2500 cm^4
   D. 6250 and 1562 cm^4, with Imax axis at 26.57°
   Answer: D. Centre 3906 ± radius sqrt(1406^2 + 1875^2) = 2344, giving 6250 and 1562. (Key point: Principal values)

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