Principal MI of section with quarter circle

Dimensions and assumptions (read from the figure):

• Dimensions in mm. Rectangle 800 × 300 (x 0 to 800, y 0 to 300); triangle above it with base 800 and height 600 (vertices (0,300), (800,300), (0,900)); quarter circle of radius 300 below the right end, centre (800, 0), occupying x 500 to 800, y −300 to 0.

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

$$\begin{aligned} A &= \sum A_i = 5.507\times10^{5}\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{2.075\times10^{8}}{5.507\times10^{5}} = 376.9\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{1.470\times10^{8}}{5.507\times10^{5}} = 266.9\ \text{mm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 3.268\times10^{10}\pm1.847\times10^{10} \\ I_{max} &= 5.114\times10^{10}\ \text{mm}^4,\qquad I_{min} = 1.421\times10^{10}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3.678\times10^{10}}{3.344\times10^{9}}$, so $2\theta_p = 84.80^\circ$ (and $-95.20^\circ$).

• The axis of $I_{max}$ makes 42.40° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -47.60° (perpendicular), both passing through the centroid (376.9, 266.9) measured from the reference origin.

Answer: $I_{max} = 5.114\times10^{10}\ \text{mm}^4$, $I_{min} = 1.421\times10^{10}\ \text{mm}^4$; principal axes through the centroid at 42.40° (for $I_{max}$) and -47.60° (for $I_{min}$) from the +x axis.