Principal MI and axes of quarter-circle/triangle section (2078 Bhadra)

Dimensions and assumptions (read from the figure):

• Dimensions in mm. Rectangle 65 × 80 (x 0 to 65, y 0 to 80); quarter circle of radius 65 on top with centre at (0, 80); right-angled triangle on the right of the rectangle: vertices (65,0), (145,0), (65,145) (base 80, height 145).

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

$$\begin{aligned} A &= \sum A_i = 14318\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{7.922\times10^{5}}{14318} = 55.33\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{8.453\times10^{5}}{14318} = 59.04\ \text{mm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 1.935\times10^{7}\pm6.807\times10^{6} \\ I_{max} &= 2.615\times10^{7}\ \text{mm}^4,\qquad I_{min} = 1.254\times10^{7}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{1.326\times10^{7}}{3.103\times10^{6}}$, so $2\theta_p = 76.82^\circ$ (and $-103.18^\circ$).

• The axis of $I_{max}$ makes 38.41° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -51.59° (perpendicular), both passing through the centroid (55.33, 59.04) measured from the reference origin.

Answer: $I_{max} = 2.615\times10^{7}\ \text{mm}^4$, $I_{min} = 1.254\times10^{7}\ \text{mm}^4$; principal axes through the centroid at 38.41° (for $I_{max}$) and -51.59° (for $I_{min}$) from the +x axis.