Principal MI of section (140 cm x 150 cm)

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Polygon (0,0), (110,0), (110,40), (140,120), (140,150), (0,150): bottom 110 (= 20 + 90), top width 140, total height 150, sloping edge 30 horizontal × 80 vertical.

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

$$\begin{aligned} A &= \sum A_i = 18600\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{1.164\times10^{6}}{18600} = 62.58\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{1.471\times10^{6}}{18600} = 79.09\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 2.997\times10^{7}\pm6.723\times10^{6} \\ I_{max} &= 3.669\times10^{7}\ \text{cm}^4,\qquad I_{min} = 2.325\times10^{7}\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{-9.428\times10^{6}}{9.588\times10^{6}}$, so $2\theta_p = -44.52^\circ$ (and $135.48^\circ$).

• The axis of $I_{max}$ makes -22.26° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at 67.74° (perpendicular), both passing through the centroid (62.58, 79.09) measured from the reference origin.

Answer: $I_{max} = 3.669\times10^{7}\ \text{cm}^4$, $I_{min} = 2.325\times10^{7}\ \text{cm}^4$; principal axes through the centroid at -22.26° (for $I_{max}$) and 67.74° (for $I_{min}$) from the +x axis.