Principal MI of shaded area with semicircular cut-out

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Plate: vertices (0,0), (100,0), (100,50), (30,50) (left edge slanted, 30 cm horizontal, depth 50). Semicircular cut-out of diameter 40 cm (radius 20) on the bottom edge, centre (50, 0), removed.

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 $cm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 3622\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{2.111\times10^{5}}{3622} = 58.28\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{94667}{3622} = 26.14\ \text{cm} \end{aligned}$$

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

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 1.655\times10^{6}\pm9.805\times10^{5} \\ I_{max} &= 2.635\times10^{6}\ \text{cm}^4,\qquad I_{min} = 6.743\times10^{5}\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{-3.692\times10^{5}}{-1.926\times10^{6}}$, so $2\theta_p = -169.15^\circ$ (and $10.85^\circ$).

• The axis of $I_{max}$ makes -84.57° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at 5.43° (perpendicular), both passing through the centroid (58.28, 26.14) measured from the reference origin.

Answer: $I_{max} = 2.635\times10^{6}\ \text{cm}^4$, $I_{min} = 6.743\times10^{5}\ \text{cm}^4$; principal axes through the centroid at -84.57° (for $I_{max}$) and 5.43° (for $I_{min}$) from the +x axis.