Principal MI about centroid and principal axes (2080 Baisakh)

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Bottom rectangle 25 × 5; middle rectangle 25 × 10 above it; top part is a trapezoid with a vertical left side of 15 cm, top width 5 cm and a sloping right side down to the full width of 25 cm (vertices (0,15), (25,15), (5,30), (0,30)).

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

$$\begin{aligned} A &= \sum A_i = 600.0\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{6625}{600.0} = 11.04\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{7500}{600.0} = 12.50\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 32487\pm10183 \\ I_{max} &= 42670\ \text{cm}^4,\qquad I_{min} = 22304\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{19375}{6276}$, so $2\theta_p = 72.05^\circ$ (and $-107.95^\circ$).

• The axis of $I_{max}$ makes 36.03° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -53.97° (perpendicular), both passing through the centroid (11.04, 12.50) measured from the reference origin.

Answer: $I_{max} = 42670\ \text{cm}^4$, $I_{min} = 22304\ \text{cm}^4$; principal axes through the centroid at 36.03° (for $I_{max}$) and -53.97° (for $I_{min}$) from the +x axis.