Principal MI, direction and position of section (2080 Bhadra)

Dimensions and assumptions (read from the figure):

• Dimensions in mm. Rectangle 1200 × 200 (x 0 to 1200, y 0 to 200); triangle above its left end with base 600 and height 500 (vertices (0,200), (600,200), (0,700)); quarter circle of radius 200 below the right end, centre at (1200, 0), occupying x 1000 to 1200, y −200 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

$$\begin{aligned} A &= \sum A_i = 4.214\times10^{5}\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{2.090\times10^{8}}{4.214\times10^{5}} = 496.0\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{7.633\times10^{7}}{4.214\times10^{5}} = 181.1\ \text{mm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 3.580\times10^{10}\pm2.914\times10^{10} \\ I_{max} &= 6.494\times10^{10}\ \text{mm}^4,\qquad I_{min} = 6.667\times10^{9}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3.343\times10^{10}}{-4.773\times10^{10}}$, so $2\theta_p = 145.00^\circ$ (and $-35.00^\circ$).

• The axis of $I_{max}$ makes 72.50° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -17.50° (perpendicular), both passing through the centroid (496.0, 181.1) measured from the reference origin.

Answer: $I_{max} = 6.494\times10^{10}\ \text{mm}^4$, $I_{min} = 6.667\times10^{9}\ \text{mm}^4$; principal axes through the centroid at 72.50° (for $I_{max}$) and -17.50° (for $I_{min}$) from the +x axis.