Principal MI of L-shaped composite section

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Horizontal leg 12 × 3 (x 0 to 12, y 0 to 3); vertical leg 4.5 × 6 above it (x 0 to 4.5, y 3 to 9). Overall 12 wide, 9 high.

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

$$\begin{aligned} A &= \sum A_i = 63.00\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{276.8}{63.00} = 4.393\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{216.0}{63.00} = 3.429\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 557.5\pm294.2 \\ I_{max} &= 851.7\ \text{cm}^4,\qquad I_{min} = 263.3\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{520.7}{-274.1}$, so $2\theta_p = 117.76^\circ$ (and $-62.24^\circ$).

• The axis of $I_{max}$ makes 58.88° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -31.12° (perpendicular), both passing through the centroid (4.393, 3.429) measured from the reference origin.

Answer: $I_{max} = 851.7\ \text{cm}^4$, $I_{min} = 263.3\ \text{cm}^4$; principal axes through the centroid at 58.88° (for $I_{max}$) and -31.12° (for $I_{min}$) from the +x axis.