Principal axis and MI of composite section

Dimensions and assumptions (read from the figure):

• Overall outline (mm): vertices (0,−100), (150,−100), (150,−50), (200,−50), (200,50), (0,100); the top edge slopes from 100 mm to 50 mm above the x-axis.
• Semicircular cut-out of radius 50 mm with its centre on the left edge at (0, −50), removed.
• Circular hole of diameter 60 mm with centre (100, −25), removed.

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $mm^2$, I in $mm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 25746\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{2.530\times10^{6}}{25746} = 98.26\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{37869}{25746} = 1.471\ \text{mm} \end{aligned}$$

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 $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 7.246\times10^{7}\pm1.437\times10^{7} \\ I_{max} &= 8.682\times10^{7}\ \text{mm}^4,\qquad I_{min} = 5.809\times10^{7}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{2.768\times10^{7}}{-7.709\times10^{6}}$, so $2\theta_p = 105.56^\circ$ (and $-74.44^\circ$).

• The axis of $I_{max}$ makes 52.78° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -37.22° (perpendicular), both passing through the centroid (98.26, 1.471) measured from the reference origin.

Answer: $I_{max} = 8.682\times10^{7}\ \text{mm}^4$, $I_{min} = 5.809\times10^{7}\ \text{mm}^4$; principal axes through the centroid at 52.78° (for $I_{max}$) and -37.22° (for $I_{min}$) from the +x axis.