---
title: "Principal MI and axes, 120 cm section"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-and-axes-orientation-120"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-and-axes-orientation-120
site: BCT Engineering
---

# Principal MI and axes, 120 cm section

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2072 Chaitra, 12 marks
- Video: [Determine Principal axis and Principal Moment of Inertia | Strength of Material | 3rd Sem |BE Civil](https://www.youtube.com/watch?v=448e_23QAFc) — Prashant YT
- 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-and-axes-orientation-120
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Determine principal moment of inertia and draw orientation of principal axes of the figure shown in figure below. [Figure: section with a sloping left edge: 50 cm top width, 20 cm; vertical dimensions 120 cm, 30 cm and 45 cm; bottom width 120 cm.]

## Short answer

Centroid at (58.48, 52.09) cm. I_x=1.225×10^7, I_y=1.003×10^7, I_xy=-1.585×10^6 cm^4. Principal moments: I_max=1.307×10^7, I_min=9.207×10^6 cm^4. Principal axes through centroid at 27.53° (I_max) and -62.47° (I_min) from +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm. Polygon (0,0), (120,0), (120,45), (70,120), (20,120): bottom 120, height 120, top width 50, left sloping edge with horizontal 20, right vertical 45.

#### 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 | + | $11325$ | 58.48 | 52.09 | $1.225\times10^{7}$ | $1.003\times10^{7}$ | $-1.585\times10^{6}$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 11325\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{6.622\times10^{5}}{11325} = 58.48\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{5.899\times10^{5}}{11325} = 52.09\ \text{cm}
\end{aligned}
$$

#### 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 (+) | $1.225\times10^{7}$ | $1.003\times10^{7}$ | $-1.585\times10^{6}$ |
| **Total about centroid** | **$1.225\times10^{7}$** | **$1.003\times10^{7}$** | **$-1.585\times10^{6}$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 1.114\times10^{7}\pm1.934\times10^{6} \\
I_{max} &= 1.307\times10^{7}\ \text{cm}^4,\qquad I_{min} = 9.207\times10^{6}\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3.170\times10^{6}}{2.215\times10^{6}}$, so $2\theta_p = 55.05^\circ$ (and $-124.95^\circ$).

- The axis of $I_{max}$ makes 27.53° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -62.47° (perpendicular), both passing through the centroid (58.48, 52.09) measured from the reference origin.

**Answer: $I_{max} = 1.307\times10^{7}\ \text{cm}^4$, $I_{min} = 9.207\times10^{6}\ \text{cm}^4$; principal axes through the centroid at 27.53° (for $I_{max}$) and -62.47° (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 (58.48 cm, 52.09 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 = 1.225e7, I_y = 1.003e7 and I_xy = -1.585e6 cm^4, which statement matches the answer?
   A. I_max = 1.225e7 and I_min = 1.003e7 cm^4; the I_max axis is along the x axis
   B. I_max = 9.207e6 cm^4 and I_min = 1.307e7 cm^4 with both axes at 45 degrees
   C. I_max = 1.307e7 and I_min = 9.207e6 cm^4; the I_max axis is at 27.53° 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 27.53° (I_min axis at -62.47°). (Key point: Result and axis angle)

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