Principal MI of I-section with semicircular hole

Dimensions and assumptions (read from the figure):

• Dimensions in cm; x measured from the vertical axis of symmetry. Bottom flange 30 × 8, web 4 × 16, top flange 20 × 4, with a semicircular hole of radius 4 on the bottom edge at the centre (diameter on the bottom face, bulging upward).

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

$$\begin{aligned} A &= \sum A_i = 358.9\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{0}{358.9} = 0\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{4021}{358.9} = 11.21\ \text{cm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 26273\pm5621 \\ I_{max} &= 31894\ \text{cm}^4,\qquad I_{min} = 20651\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{0}{11242}$, so $2\theta_p = -0.00^\circ$ (and $-180.00^\circ$).

• The axis of $I_{max}$ makes 0.00° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at 90.00° (perpendicular), both passing through the centroid (0, 11.21) measured from the reference origin.

The section is symmetrical about the vertical axis, so $I_{xy} = 0$ and the centroidal horizontal and vertical axes are the principal axes.

Answer: $I_{max} = 31894\ \text{cm}^4$, $I_{min} = 20651\ \text{cm}^4$; principal axes through the centroid at 0.00° (for $I_{max}$) and 90.00° (for $I_{min}$) from the +x axis.