Orientation of principal axes and MI of composite section

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Left triangle: vertices (0,0), (6,0), (6,8) (base 6, height 6 + 2). Central rectangle 12 × 6 (x 6 to 18, y 0 to 6). Right triangle: vertices (18,0), (24,0), (18,7) (horizontal 6, vertical 7).

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

$$\begin{aligned} A &= \sum A_i = 117.0\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{1380}{117.0} = 11.79\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{329.0}{117.0} = 2.812\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 2098\pm1732 \\ I_{max} &= 3830\ \text{cm}^4,\qquad I_{min} = 365.8\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{90.03}{-3463}$, so $2\theta_p = 178.51^\circ$ (and $-1.49^\circ$).

• The axis of $I_{max}$ makes 89.26° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -0.74° (perpendicular), both passing through the centroid (11.79, 2.812) measured from the reference origin.

Answer: $I_{max} = 3830\ \text{cm}^4$, $I_{min} = 365.8\ \text{cm}^4$; principal axes through the centroid at 89.26° (for $I_{max}$) and -0.74° (for $I_{min}$) from the +x axis.