Principal MI of section with trapezoidal cut-out

Dimensions and assumptions (read from the figure):

• Dimensions in m. Block 12 × 12 (x 0 to 12, y 0 to 12). Trapezoidal slot from the bottom edge: vertices (2,0), (10,0), (7,9), (5,9), removed (3 m of solid left above the slot).

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $m^2$, I in $m^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 99.00\ \text{m}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{594.0}{99.00} = 6.000\ \text{m} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{702.0}{99.00} = 7.091\ \text{m} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 1342\pm258.4 \\ I_{max} &= 1600\ \text{m}^4,\qquad I_{min} = 1084\ \text{m}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{0}{-516.8} = 0$, so $2\theta_p = 0^\circ$ or $180^\circ$; since $I_y > I_x$, the maximum is about the vertical axis.

• The axis of $I_{max}$ is the vertical centroidal axis x = 6 m (at 90° to the +x axis), and the axis of $I_{min}$ is the horizontal centroidal axis y = 7.091 m (0°), both passing through the centroid (6.000, 7.091) measured from the reference origin.

The section is symmetrical about the vertical axis x = 6 m, so $I_{xy} = 0$ and the centroidal x and y axes are themselves the principal axes.

Answer: $I_{max} = 1600\ \text{m}^4$ (about the vertical centroidal axis), $I_{min} = 1084\ \text{m}^4$ (about the horizontal centroidal axis); the principal axes are the centroidal y and x axes through (6.000, 7.091).