---
title: "Principal MI of section (140 cm x 150 cm)"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:principal-mi-140-150-cm-section"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-140-150-cm-section
site: BCT Engineering
---

# Principal MI of section (140 cm x 150 cm)

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2074 Chaitra, 10 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-140-150-cm-section
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Determine the principle moment of inertia of the given figure. [Figure: section 140 cm wide at the top and 150 cm high; right edge notch dimensions 30 cm and 80 cm; bottom dimensions 20 cm and 90 cm; a sloping edge from the lower left step to the right notch.]

## Short answer

I_max = 3.669×10^7 cm^4, I_min = 2.325×10^7 cm^4 about centroid. Principal axes: -22.26° (I_max) and 67.74° (I_min) from +x axis.

## Answer

**Dimensions and assumptions (read from the figure):**
- Dimensions in cm. Polygon (0,0), (110,0), (110,40), (140,120), (140,150), (0,150): bottom 110 (= 20 + 90), top width 140, total height 150, sloping edge 30 horizontal × 80 vertical.

#### 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 | + | $18600$ | 62.58 | 79.09 | $3.476\times10^{7}$ | $2.518\times10^{7}$ | $4.714\times10^{6}$ |

#### Centroid
$$
\begin{aligned}
A &= \sum A_i = 18600\ \text{cm}^2 \\
\bar X &= \frac{\sum A_i x_i}{A} = \frac{1.164\times10^{6}}{18600} = 62.58\ \text{cm} \\
\bar Y &= \frac{\sum A_i y_i}{A} = \frac{1.471\times10^{6}}{18600} = 79.09\ \text{cm}
\end{aligned}
$$

#### 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 (+) | $3.476\times10^{7}$ | $2.518\times10^{7}$ | $4.714\times10^{6}$ |
| **Total about centroid** | **$3.476\times10^{7}$** | **$2.518\times10^{7}$** | **$4.714\times10^{6}$** |

#### Principal moments of inertia and principal axes
$$
\begin{aligned}
I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\
&= 2.997\times10^{7}\pm6.723\times10^{6} \\
I_{max} &= 3.669\times10^{7}\ \text{cm}^4,\qquad I_{min} = 2.325\times10^{7}\ \text{cm}^4
\end{aligned}
$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{-9.428\times10^{6}}{9.588\times10^{6}}$, so $2\theta_p = -44.52^\circ$ (and $135.48^\circ$).

- The axis of $I_{max}$ makes -22.26° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at 67.74° (perpendicular), both passing through the centroid (62.58, 79.09) measured from the reference origin.

**Answer: $I_{max} = 3.669\times10^{7}\ \text{cm}^4$, $I_{min} = 2.325\times10^{7}\ \text{cm}^4$; principal axes through the centroid at -22.26° (for $I_{max}$) and 67.74° (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 principal moments asked for are about axes through the centroid, and parallel-axis shifts need the centroid location
   D. Because the centroid is the point where Imax occurs
   Answer: C. 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 (62.58 cm, 79.09 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^2
   B. I_xy = I_xy,G + A (d_x + d_y)
   C. I_xy = I_xy,G, because the product of inertia does not change with axis shift
   D. I_xy = I_xy,G + A d_x d_y, keeping the signs of d_x and d_y
   Answer: D. 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)^2 + I_xy^2)
   B. (I_x+I_y)/2 ± sqrt((I_x-I_y)^2 + I_xy)
   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: A. The principal values are the centre of Mohr's circle plus and minus its radius. (Key point: Principal moment formula)

4. With I_x = 3.476e7, I_y = 2.518e7 and I_xy = 4.714e6 cm^4, which statement matches the answer?
   A. I_max = 3.476e7 and I_min = 2.518e7 cm^4; the I_max axis is along the x axis
   B. I_max = 3.669e7 and I_min = 2.325e7 cm^4; the I_max axis is at -22.26° from +x and the I_min axis is perpendicular to it
   C. I_max = 2.325e7 cm^4 and I_min = 3.669e7 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: B. Imax and Imin come from the formula, and the principal axes are perpendicular, with the I_max axis at -22.26° (I_min axis at 67.74°). (Key point: Result and axis angle)

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