Principal MI and axes of L-section

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Flange 15 × 5 (x 0 to 15, y 0 to 5); web 5 × 15 above it (x 0 to 5, y 5 to 20). Total depth 20, breadth 15, thickness 5.

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

$$\begin{aligned} A &= \sum A_i = 150.0\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{750.0}{150.0} = 5.000\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{1125}{150.0} = 7.500\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 3906\pm2344 \\ I_{max} &= 6250\ \text{cm}^4,\qquad I_{min} = 1562\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3750}{2812}$, so $2\theta_p = 53.13^\circ$ (and $-126.87^\circ$).

• The axis of $I_{max}$ makes 26.57° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -63.43° (perpendicular), both passing through the centroid (5.000, 7.500) measured from the reference origin.

Answer: $I_{max} = 6250\ \text{cm}^4$, $I_{min} = 1562\ \text{cm}^4$; principal axes through the centroid at 26.57° (for $I_{max}$) and -63.43° (for $I_{min}$) from the +x axis.