Principal MI and orientation of Z-type section

Dimensions and assumptions (read from the figure):

• Dimensions in cm; origin O at the bottom-left corner. Bottom flange 6 × 2 (x 0 to 6, y 0 to 2); web 2 wide (x 4 to 6) from y = 2 to 8 (inner height 6); top flange 6 × 4 (x 4 to 10, y 8 to 12). Overall 10 wide, 12 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

$$\begin{aligned} A &= \sum A_i = 48.00\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{264.0}{48.00} = 5.500\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{312.0}{48.00} = 6.500\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 500.0\pm394.4 \\ I_{max} &= 894.4\ \text{cm}^4,\qquad I_{min} = 105.6\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{-600.0}{512.0}$, so $2\theta_p = -49.52^\circ$ (and $130.48^\circ$).

• The axis of $I_{max}$ makes -24.76° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at 65.24° (perpendicular), both passing through the centroid (5.500, 6.500) measured from the reference origin.

Values about the given origin

About the axes through the origin O: $I_x = 2784$, $I_y = 1696$, $I_{xy} = 2016\ \text{cm}^4$ (see parts table with the parallel-axis theorem). Principal values about O: $I_{max} = 4328$, $I_{min} = 151.9$ at -37.45° to the x axis.

Answer: $I_{max} = 894.4\ \text{cm}^4$, $I_{min} = 105.6\ \text{cm}^4$; principal axes through the centroid at -24.76° (for $I_{max}$) and 65.24° (for $I_{min}$) from the +x axis.