Principal MI of quarter circle and triangle section (cm)

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Rectangle 15 × 15 (x 0 to 15, y 0 to 15); quarter circle of radius 15 above it with centre at (0, 15); right-angled triangle on the right: vertices (15,0), (45,0), (15,15).

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

$$\begin{aligned} A &= \sum A_i = 626.7\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{8438}{626.7} = 13.46\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{6588}{626.7} = 10.51\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 52301\pm30354 \\ I_{max} &= 82655\ \text{cm}^4,\qquad I_{min} = 21947\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{55052}{-25590}$, so $2\theta_p = 114.93^\circ$ (and $-65.07^\circ$).

• The axis of $I_{max}$ makes 57.47° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -32.53° (perpendicular), both passing through the centroid (13.46, 10.51) measured from the reference origin.

Answer: $I_{max} = 82655\ \text{cm}^4$, $I_{min} = 21947\ \text{cm}^4$; principal axes through the centroid at 57.47° (for $I_{max}$) and -32.53° (for $I_{min}$) from the +x axis.