Question D.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.D-28. Use the following numerical data: height h = 200 mm, width b = 90 mm, and constant thickness t = 15 mm.
Use the x-y axes (Fig. D-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
I_{x}=29.29 \times 10^{6} mm ^{4} \quad I_{y}=5.667 \times 10^{6} mm ^{4}I_{x y}=-9.366 \times 10^{6} mm ^{4}
Substitute these values into the equation for the angle \theta_{p} [Eq. (D-39)] to 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}Use these values of \theta_{p} in the transformation equation for I_{x1} [Eq. (D-33)] to find I_{x1} = 32.6 \times 10^{6} mm^{4} and 2.4 \times 10^{6} mm^{4}, respectively. These same values are obtained by substituting into Eqs. (D-42a and b). 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 2 \theta-I_{x y} \sin 2 \theta (D-33)
I_{1}=\frac{I_{x}+I_{y}}{2}+\sqrt{\left(\frac{I_{x}-I_{y}}{2}\right)^{2}+I_{x y}^{2}} (D-42a)
I_{2}=\frac{I_{x}+I_{y}}{2}-\sqrt{\left(\frac{I_{x}-I_{y}}{2}\right)^{2}+I_{x y}^{2}} (D-42b)
I_{1}=32.6 \times 10^{6} mm ^{4} \quad \theta_{p 1}=19.2^{\circ}I_{2}=2.4 \times 10^{6} mm ^{4} \quad \theta_{p 2}=109.2^{\circ}
The principal axes are shown in Fig. D-28 as the x_{1}y_{1} axes.