Principal MI and axis of composite section

Dimensions and assumptions (read from the figure):

• Dimensions in mm. Strip: 1000 × 200 (x 0 to 1000, y 0 to 200). Upper-left right triangle: vertices (0,200), (600,200), (0,800). Lower-right right triangle: vertices (1000,0), (400,0), (1000,−600).

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

Centroid

$$\begin{aligned} A &= \sum A_i = 5.600\times10^{5}\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{2.800\times10^{8}}{5.600\times10^{5}} = 500.0\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{5.600\times10^{7}}{5.600\times10^{5}} = 100.0\ \text{mm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 4.827\times10^{10}\pm3.688\times10^{10} \\ I_{max} &= 8.514\times10^{10}\ \text{mm}^4,\qquad I_{min} = 1.139\times10^{10}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{7.200\times10^{10}}{-1.600\times10^{10}}$, so $2\theta_p = 102.53^\circ$ (and $-77.47^\circ$).

• The axis of $I_{max}$ makes 51.26° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -38.74° (perpendicular), both passing through the centroid (500.0, 100.0) measured from the reference origin.

Answer: $I_{max} = 8.514\times10^{10}\ \text{mm}^4$, $I_{min} = 1.139\times10^{10}\ \text{mm}^4$; principal axes through the centroid at 51.26° (for $I_{max}$) and -38.74° (for $I_{min}$) from the +x axis.