Question 12.8: Determine the orientations of the principal centroidal axes ......

Let us use the xy axes (Fig. 12-29) as the reference axes through the centroid C. The moments and product of inertia with respect to these axes can be obtained by dividing the area into three rectangles and using the parallel-axis theorems. The results of such calculations are as follows:

I_x=29.29×10^6\ mm^4 \quad \quad I_y=5.667×10^6\ mm^4 \quad \quad I_{xy}=-9.366×10^6\ mm^4

Substituting these values into the equation for the angle θ_p (Eq. 12-30), we get

tan\ 2θ_p=-\frac{2I_{xy}}{I_x\ -\ I_y} = 0.7930          2θ_p = 38.4° and 218.4°

Thus, the two values of θ_p are

θ_p = 19.2° and 109.2°

Using these values of θ_p in the transformation equation for I_{x1} (Eq. 12-26), we find I_{x_1} = 32.6 × 10^6\ mm^4 and 2.4 × 10^6\ mm^4, respectively. These same values are obtained if we substitute into Eqs. (12-33a) and (12-33b). Thus, the principal moments of inertia and the angles to the corresponding principal axes are:

I_{x_1}=\frac{I_x+I_y}{2}+\frac{I_x\ -\ I_y}{2}cos\ 2θ\ -\ I_{xy}sin\ 2θ                        (12-26)

I_{1}=\frac{I_x+I_y}{2}+\sqrt{\left(\frac{I_x\ -\ I_y}{2}\right)^2+I_{xy}^2}                    (12-33a)

I_{1}=32.6×10^6\ mm^4\quad \quad θ_{p_1} =19.2° \\ I_{2}=2.4×10^6\ mm^4\quad \quad θ_{p_2} =109.2°

The principal axes are shown in Fig. 12-29 as the x_1y_1 axes.