Principal axes and MI with Mohr's circle

Dimensions and assumptions (read from the figure):

• Dimensions in m. Polygon (0,0), (6,0), (6,2), (3,6), (0,6): width 6, left side 6, top width 3, sloping edge from the top-middle to the right side at 2 m height.

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

Centroid

$$\begin{aligned} A &= \sum A_i = 30.00\ \text{m}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{78.00}{30.00} = 2.600\ \text{m} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{80.00}{30.00} = 2.667\ \text{m} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 79.43\pm22.24 \\ I_{max} &= 101.7\ \text{m}^4,\qquad I_{min} = 57.20\ \text{m}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{44.00}{6.467}$, so $2\theta_p = 81.64^\circ$ (and $-98.36^\circ$).

• The axis of $I_{max}$ makes 40.82° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -49.18° (perpendicular), both passing through the centroid (2.600, 2.667) measured from the reference origin.

Mohr's circle check

• Centre $C = \left(\dfrac{I_x+I_y}{2},\,0\right) = (79.43,\ 0)$; radius $R = 22.24$.
• Plot $X(I_x,\ I_{xy}) = (82.67,\ -22.00)$ and $Y(I_y,\ -I_{xy}) = (76.20,\ 22.00)$; the line XY is a diameter.
• The circle cuts the horizontal axis at $I_{max} = 101.7$ and $I_{min} = 57.20$, equal to the values found analytically. The angle from CX to the $I_{max}$ end of the horizontal diameter is $2\theta_p = 81.64^\circ$ (turned from the x axis towards the principal axis in the same sense as the physical rotation).

Answer: $I_{max} = 101.7\ \text{m}^4$, $I_{min} = 57.20\ \text{m}^4$; principal axes through the centroid at 40.82° (for $I_{max}$) and -49.18° (for $I_{min}$) from the +x axis.