Stress in a section with product of inertia

(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.