Question 12.7: Determine the orientations of the principal centroidal axes ...
Determine the orientations of the principal centroidal axes and the magnitudes of the principal centroidal moments of inertia for the cross-sectional area of the Z-section shown in Fig. 12-28. Use the following numerical data: height h = 200 mm, width b = 90 mm, and constant thickness t = 15 mm.
Let us use the xy axes (Fig. 12-28) 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:
\begin{aligned}&I_{x}=29.29 \times 10^{6} \mathrm{~mm}^{4} \quad I_{y}=5.667 \times 10^{6} \mathrm{~mm}^{4} \\&I_{x y}=-9.366 \times 10^{6} \mathrm{~mm}^{4}\end{aligned}Substituting these values into the equation for the angle \theta_{p} [Eq. (12-39)], we get
\tan 2 \theta_{p}=-\frac{2 I_{x y}}{I_{x}-I_{y}}=0.7930 \quad 2 \theta_{p}=38.4^{\circ} \text { and } 218.4^{\circ}Thus, the two values of \theta_{p} are
\theta_{p}=19.2^{\circ} \text { and } 109.2^{\circ}Using these values of θp in the transformation equation for I_{x_{1}} [Eq. (12-33)],
I_{x_{1}}=\frac{I_{x}+I_{y}}{2}+\frac{I_{x}-I_{y}}{2} \cos 2 \theta-I_{x y} \sin 2 \thetawe find I_{x_{1}}=32.6 \times 10^{6} \mathrm{~mm}^{4} \text { and } 2.4 \times 10^{6} \mathrm{~mm}^{4}, respectively. These same values are obtained if we substitute into Eqs. (12-42a and b).
I_{1}=\frac{I_{x}+I_{y}}{2}+\sqrt{\left(\frac{I_{x}-I_{y}}{2}\right)^{2}+I_{x y}^{2}} (12-42a)
I_{2}=\frac{I_{x}+I_{y}}{2}-\sqrt{\left(\frac{I_{x}-I_{y}}{2}\right)^{2}+I_{x y}^{2}} (12-42b)
Thus, the principal moments of inertia and the angles to the corresponding principal axes are
\begin{aligned}&I_{1}=32.6 \times 10^{6} \mathrm{~mm}^{4} \quad \theta_{p_{1}}=19.2^{\circ} \\&I_{2}=2.4 \times 10^{6} \mathrm{~mm}^{4} \quad \theta_{p_{2}}=109.2^{\circ}\end{aligned}The principal axes are shown in Fig. 12-28 as the x_{1} y_{1} axes.