---
title: "Stress in a section with product of inertia"
subject: "Mechanics of Solids"
chapter: 7
key: "semester_5/mechanics_of_solids:unsymmetrical-section-product-inertia"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/unsymmetrical-section-product-inertia
site: BCT Engineering
---

# Stress in a section with product of inertia

Mechanics of Solids · Chapter 7: Non Symmetrical Bending of Straight Beams

- Asked in: year not given, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/unsymmetrical-section-product-inertia
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-7

## Question

A beam section has centroidal second moments $I_x = 6.0\times10^{6}$ mm$^4$, $I_y = 2.5\times10^{6}$ mm$^4$ and product of inertia $I_{xy} = -1.6\times10^{6}$ mm$^4$ about the centroidal x and y axes. A bending moment $M_x = 2$ kN m acts about the x-axis (no moment about y). Find (a) the principal moments of inertia and the inclination of the principal axes, (b) the angle of the neutral axis, (c) the bending stress at the point $x = 25$ mm, $y = 40$ mm.

## Short answer

Principal moments are I_max = 6.62×10^6 mm^4, I_min = 1.88×10^6 mm^4 with axis at 21.2° from x. Neutral axis is at -32.6° from x. Bending stress at (25,40) mm is 22.5 MPa (tension).

## Answer

#### (a) Principal axes

$$
\begin{aligned}
I_{avg} &= \frac{I_x + I_y}{2} = 4.250\times10^{6}\ \text{mm}^4 \\
R &= \sqrt{\left(\frac{I_x - I_y}{2}\right)^2 + I_{xy}^2} = \sqrt{(1.75)^2 + (1.6)^2}\times10^6 = 2.371\times10^{6}\ \text{mm}^4 \\
I_{max,min} &= I_{avg} \pm R = 6.621\times10^{6},\ 1.879\times10^{6}\ \text{mm}^4
\end{aligned}
$$

$$
\tan 2\theta_p = \frac{-2I_{xy}}{I_x - I_y} = \frac{3.2}{3.5} \Rightarrow \theta_p = 21.22^\circ
$$

The axis of $I_{max}$ is at $21.2^\circ$ anticlockwise from the x-axis. Check: $I_{max}I_{min} = 12.44\times10^{12} = I_xI_y - I_{xy}^2$. $\checkmark$

#### (b) Neutral axis

With $M_y = 0$, the stress equation reduces to

$$
\sigma_z = \frac{M_x\,(I_y\,y - I_{xy}\,x)}{I_xI_y - I_{xy}^2}
$$

$\sigma_z = 0$ gives $y/x = I_{xy}/I_y$:

$$
\tan\alpha = \frac{-1.6\times10^6}{2.5\times10^6} = -0.64 \quad\Rightarrow\quad \alpha = -32.62^\circ
$$

The neutral axis passes through the centroid at $-32.6^\circ$ to the x-axis (not along the x-axis, although $M_y = 0$).

#### (c) Stress at $(x, y) = (25, 40)$ mm

$$
D = I_xI_y - I_{xy}^2 = 15\times10^{12} - 2.56\times10^{12} = 12.44\times10^{12}\ \text{mm}^8
$$

$$
\begin{aligned}
I_y\,y - I_{xy}\,x &= (2.5\times10^6)(40) - (-1.6\times10^6)(25) = 100\times10^6 + 40\times10^6 = 140\times10^6 \\
\sigma_z &= \frac{2\times10^6\times140\times10^6}{12.44\times10^{12}} = 22.51\ \text{MPa (tension)}
\end{aligned}
$$

If $I_{xy}$ had been ignored, $M_xy/I_x = 13.3$ MPa would have been obtained, i.e. an error of about 41 percent.

**Answer:** $I_{max} = 6.62\times10^6$ mm$^4$, $I_{min} = 1.88\times10^6$ mm$^4$, $\theta_p = 21.2^\circ$; neutral axis at $-32.6^\circ$; $\sigma = 22.5$ MPa (tension) at $(25, 40)$ mm.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With Ix = 6.0e6, Iy = 2.5e6, Ixy = -1.6e6 mm^4, the principal moments are given by
   A. Iavg ± sqrt(((Ix-Iy)/2)^2 + Ixy^2) = 6.62e6 and 1.88e6
   B. Ix and Iy themselves
   C. Iavg ± Ixy
   D. Ix ± Iy
   Answer: A. Iavg = 4.25e6, R = 2.371e6, so Imax = 6.62e6 and Imin = 1.88e6 mm^4. (Key point: Principal moments)

2. With only Mx acting on this section, the neutral axis
   A. Lies along the x-axis
   B. Is at tan(alpha) = Ixy/Iy = -0.64, i.e. -32.6 degrees to x
   C. Is perpendicular to the x-axis
   D. Lies along the y-axis
   Answer: B. Setting sigma = 0 with My = 0 gives y/x = Ixy/Iy, so the axis is not along x because Ixy is non-zero. (Key point: Neutral axis with Mx only)

3. Which expression gives the stress at (x, y) with My = 0?
   A. Mx y / Ix
   B. Mx (Iy y - Ixy x)/(Ix Iy - Ixy^2)
   C. Mx y / Iy
   D. Mx (Ix y - Ixy x)/(Ix Iy)
   Answer: B. Using the full formula gives 22.5 MPa at (25, 40); ignoring Ixy (Mx y/Ix = 13.3 MPa) gives about 41% error. (Key point: Stress at a point)

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