Principal axes of composite section (8, 12 cm)

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Top flange 28 × 2 (x 0 to 28, y 8 to 10); left leg 8 × 8 below it (x 0 to 8, y 0 to 8); right leg 8 × 10 (x 20 to 28, y −2 to 8).

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

$$\begin{aligned} A &= \sum A_i = 200.0\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{2960}{200.0} = 14.80\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{1000}{200.0} = 5.000\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 10503\pm8252 \\ I_{max} &= 18755\ \text{cm}^4,\qquad I_{min} = 2251\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{1920}{-16392}$, so $2\theta_p = 173.32^\circ$ (and $-6.68^\circ$).

• The axis of $I_{max}$ makes 86.66° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -3.34° (perpendicular), both passing through the centroid (14.80, 5.000) measured from the reference origin.

Answer: $I_{max} = 18755\ \text{cm}^4$, $I_{min} = 2251\ \text{cm}^4$; principal axes through the centroid at 86.66° (for $I_{max}$) and -3.34° (for $I_{min}$) from the +x axis.