Principal axis and MI with Mohr's circle (2079 Bhadra)

Dimensions and assumptions (read from the figure):

• Dimensions in mm. Base rectangle 600 × 200; triangle above its left end (vertices (0,200), (400,200), (0,700)), so the left edge is 700 mm high; quarter circle of radius 200 added at the right of the triangle with centre (400, 200), occupying x 400 to 600, y 200 to 400.

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

Centroid

$$\begin{aligned} A &= \sum A_i = 2.514\times10^{5}\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{6.457\times10^{7}}{2.514\times10^{5}} = 256.8\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{5.762\times10^{7}}{2.514\times10^{5}} = 229.2\ \text{mm} \end{aligned}$$

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

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 6.913\times10^{9}\pm2.756\times10^{9} \\ I_{max} &= 9.669\times10^{9}\ \text{mm}^4,\qquad I_{min} = 4.157\times10^{9}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{5.100\times10^{9}}{-2.092\times10^{9}}$, so $2\theta_p = 112.31^\circ$ (and $-67.69^\circ$).

• The axis of $I_{max}$ makes 56.15° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -33.85° (perpendicular), both passing through the centroid (256.8, 229.2) measured from the reference origin.

Mohr's circle check

• Centre $C = \left(\dfrac{I_x+I_y}{2},\,0\right) = (6.913\times10^{9},\ 0)$; radius $R = 2.756\times10^{9}$.
• Plot $X(I_x,\ I_{xy}) = (5.867\times10^{9},\ -2.550\times10^{9})$ and $Y(I_y,\ -I_{xy}) = (7.959\times10^{9},\ 2.550\times10^{9})$; the line XY is a diameter.
• The circle cuts the horizontal axis at $I_{max} = 9.669\times10^{9}$ and $I_{min} = 4.157\times10^{9}$, equal to the values found analytically. The angle from CX to the $I_{max}$ end of the horizontal diameter is $2\theta_p = 112.31^\circ$ (turned from the x axis towards the principal axis in the same sense as the physical rotation).

Answer: $I_{max} = 9.669\times10^{9}\ \text{mm}^4$, $I_{min} = 4.157\times10^{9}\ \text{mm}^4$; principal axes through the centroid at 56.15° (for $I_{max}$) and -33.85° (for $I_{min}$) from the +x axis.