---
title: "Principal axes of composite section (8, 12 cm)"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-axes-orientation-composite-8-12"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-axes-orientation-composite-8-12
site: BCT Engineering
---

# Principal axes of composite section (8, 12 cm)

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2071 Chaitra, 12 marks
- Video: [Product Moment of Inertia and Principal Axes](https://www.youtube.com/watch?v=HeBYXi1cSOw) — Calvin Rans
- 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-axes-orientation-composite-8-12
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Determine the orientation of the principal axes and the moment of inertia about the centroidal axes of composite section as shown. [Figure: composite Z-type section: top width 8 cm + 12 cm + 8 cm; left part height 8 cm + 2 cm, a sloping edge, and right part 10 cm high.]

## Short answer

Centroid at (14.80, 5.000) cm. I_x=2307, I_y=18699, I_xy=-960.0 cm^4. Principal moments: I_max=18755, I_min=2251 cm^4. Principal axes through centroid at 86.66° (I_max) and -3.34° (I_min) from +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm. Top flange 28 × 2 (x 0 to 28, y 8 to 10); left leg 8 × 8 below it (x 0 to 8, y 0 to 8); right leg 8 × 10 (x 20 to 28, y −2 to 8).

#### 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 28×2 | + | $56.00$ | 14.00 | 9.000 | $18.67$ | $3659$ | $0$ |
| Left leg 8×8 | + | $64.00$ | 4.000 | 4.000 | $341.3$ | $341.3$ | $0$ |
| Right leg 8×10 | + | $80.00$ | 24.00 | 3.000 | $666.7$ | $426.7$ | $0$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 200.0\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{2960}{200.0} = 14.80\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{1000}{200.0} = 5.000\ \text{cm}
\end{aligned}
$$

#### 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 28×2 (+) | $914.7$ | $3695$ | $-179.2$ |
| Left leg 8×8 (+) | $405.3$ | $7806$ | $691.2$ |
| Right leg 8×10 (+) | $986.7$ | $7198$ | $-1472$ |
| **Total about centroid** | **$2307$** | **$18699$** | **$-960.0$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 10503\pm8252 \\
I_{max} &= 18755\ \text{cm}^4,\qquad I_{min} = 2251\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{1920}{-16392}$, so $2\theta_p = 173.32^\circ$ (and $-6.68^\circ$).

- The axis of $I_{max}$ makes 86.66° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -3.34° (perpendicular), both passing through the centroid (14.80, 5.000) measured from the reference origin.

**Answer: $I_{max} = 18755\ \text{cm}^4$, $I_{min} = 2251\ \text{cm}^4$; principal axes through the centroid at 86.66° (for $I_{max}$) and -3.34° (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 must the centroid (X, Y) of the whole section be found first?
   A. Because the principal axes always coincide with the reference axes through the origin
   B. Because the product of inertia about the origin is always zero
   C. Because the centroid is the point where Imax occurs
   D. Because the principal moments asked for are about axes through the centroid, and parallel-axis shifts need the centroid location
   Answer: D. I_x, I_y, I_xy are required about centroidal axes, so each part's distance d_x, d_y is measured from the overall centroid (14.80 cm, 5.00 cm). (Key point: Centroid first)

2. How is the product of inertia of one part transferred to the section's centroidal axes?
   A. I_xy = I_xy,G + A d_x d_y, keeping the signs of d_x and d_y
   B. I_xy = I_xy,G + A d_x^2
   C. I_xy = I_xy,G + A (d_x + d_y)
   D. I_xy = I_xy,G, because the product of inertia does not change with axis shift
   Answer: A. The parallel-axis theorem for product of inertia adds A times the product of the signed centroidal offsets. (Key point: Parallel-axis for Ixy)

3. Which expression gives the principal moments of inertia from I_x, I_y and I_xy?
   A. (I_x+I_y)/2 ± sqrt((I_x-I_y)^2 + I_xy)
   B. (I_x+I_y)/2 ± sqrt(((I_x-I_y)/2)^2 + I_xy^2)
   C. (I_x-I_y)/2 ± sqrt(((I_x+I_y)/2)^2 + I_xy^2)
   D. sqrt(I_x^2 + I_y^2) ± I_xy
   Answer: B. The principal values are the centre of Mohr's circle plus and minus its radius. (Key point: Principal moment formula)

4. With I_x = 2307, I_y = 18699 and I_xy = -960.0 cm^4, which statement matches the answer?
   A. I_max = 2307 and I_min = 18699 cm^4; the I_max axis is along the x axis
   B. I_max = 2251 cm^4 and I_min = 18755 cm^4 with both axes at 45 degrees
   C. I_max = 18755 and I_min = 2251 cm^4; the I_max axis is at 86.66° from +x and the I_min axis is perpendicular to it
   D. I_max and I_min are both equal to (I_x+I_y)/2, and the axes are not defined
   Answer: C. Imax and Imin come from the formula, and the principal axes are perpendicular, with the I_max axis at 86.66° (I_min axis at -3.34°). (Key point: Result and axis angle)

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