Principal MI of composite section in cm

Dimensions and assumptions (read from the figure):

• Dimensions in cm. The section is the polygon (0,0), (50,0), (50,50), (30,50), (30,90), (10,90), (0,30): base 50, left vertical edge 30, sloping edge of horizontal 10 and rise 60, top width 20, right side 50 + 40 with a 20 cm step.

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

$$\begin{aligned} A &= \sum A_i = 3400\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{79500}{3400} = 23.38\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{1.255\times10^{5}}{3400} = 36.91\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 1.270\times10^{6}\pm7.156\times10^{5} \\ I_{max} &= 1.986\times10^{6}\ \text{cm}^4,\qquad I_{min} = 5.546\times10^{5}\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3.740\times10^{5}}{1.381\times10^{6}}$, so $2\theta_p = 15.15^\circ$ (and $-164.85^\circ$).

• The axis of $I_{max}$ makes 7.57° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -82.43° (perpendicular), both passing through the centroid (23.38, 36.91) measured from the reference origin.

Answer: $I_{max} = 1.986\times10^{6}\ \text{cm}^4$, $I_{min} = 5.546\times10^{5}\ \text{cm}^4$; principal axes through the centroid at 7.57° (for $I_{max}$) and -82.43° (for $I_{min}$) from the +x axis.