---
title: "Principal MI of composite section in cm"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-2078k"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-2078k
site: BCT Engineering
---

# Principal MI of composite section in cm

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2078 Kartik, 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-2078k
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Determine principal moment of inertia and orientation of principal axes passing through the centroid. All dimensions are in centimeter. [Figure: composite section with horizontal dimensions 10, 20, 30, 50 and vertical dimensions 30, 60 (left side) and 50, 40, 50 (right side); a tall rectangle at top, a sloped left edge and a lower step on the right.]

## Short answer

I_max = 1.986×10⁶ cm⁴, I_min = 5.546×10⁵ cm⁴. Principal axes through the centroid (23.38, 36.91 cm) are at 7.57° (I_max) and -82.43° (I_min) from the +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm. The section is the polygon (0,0), (50,0), (50,50), (30,50), (30,90), (10,90), (0,30): base 50, left vertical edge 30, sloping edge of horizontal 10 and rise 60, top width 20, right side 50 + 40 with a 20 cm step.

#### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ |
|---|---|---|---|---|---|---|---|
| Section | + | $3400$ | 23.38 | 36.91 | $1.961\times10^{6}$ | $5.794\times10^{5}$ | $-1.870\times10^{5}$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 3400\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{79500}{3400} = 23.38\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{1.255\times10^{5}}{3400} = 36.91\ \text{cm}
\end{aligned}
$$

#### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ |
|---|---|---|---|
| Section (+) | $1.961\times10^{6}$ | $5.794\times10^{5}$ | $-1.870\times10^{5}$ |
| **Total about centroid** | **$1.961\times10^{6}$** | **$5.794\times10^{5}$** | **$-1.870\times10^{5}$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 1.270\times10^{6}\pm7.156\times10^{5} \\
I_{max} &= 1.986\times10^{6}\ \text{cm}^4,\qquad I_{min} = 5.546\times10^{5}\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3.740\times10^{5}}{1.381\times10^{6}}$, so $2\theta_p = 15.15^\circ$ (and $-164.85^\circ$).

- The axis of $I_{max}$ makes 7.57° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -82.43° (perpendicular), both passing through the centroid (23.38, 36.91) measured from the reference origin.

**Answer: $I_{max} = 1.986\times10^{6}\ \text{cm}^4$, $I_{min} = 5.546\times10^{5}\ \text{cm}^4$; principal axes through the centroid at 7.57° (for $I_{max}$) and -82.43° (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. 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 principal moments asked for are about axes through the centroid, and parallel-axis shifts need the centroid location
   C. Because the product of inertia about the origin is always zero
   D. Because the centroid is the point where Imax occurs
   Answer: B. 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 (23.38 cm, 36.91 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^2
   B. I_xy = I_xy,G + A (d_x + d_y)
   C. I_xy = I_xy,G + A d_x d_y, keeping the signs of d_x and d_y
   D. I_xy = I_xy,G, because the product of inertia does not change with axis shift
   Answer: C. 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. sqrt(I_x^2 + I_y^2) ± I_xy
   D. (I_x+I_y)/2 ± sqrt(((I_x-I_y)/2)^2 + I_xy^2)
   Answer: D. The principal values are the centre of Mohr's circle plus and minus its radius. (Key point: Principal moment formula)

4. With I_x = 1.961e6, I_y = 5.794e5 and I_xy = -1.870e5 cm^4, which statement matches the answer?
   A. I_max = 1.986e6 and I_min = 5.546e5 cm^4; the I_max axis is at 7.57° from +x and the I_min axis is perpendicular to it
   B. I_max = 1.961e6 and I_min = 5.794e5 cm^4; the I_max axis is along the x axis
   C. I_max = 5.546e5 cm^4 and I_min = 1.986e6 cm^4 with both axes at 45 degrees
   D. I_max and I_min are both equal to (I_x+I_y)/2, and the axes are not defined
   Answer: A. Imax and Imin come from the formula, and the principal axes are perpendicular, with the I_max axis at 7.57° (I_min axis at -82.43°). (Key point: Result and axis angle)

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