Question D.6: Determine the product of inertia Ixy of the Z-section shown ......

To obtain the product of inertia with respect to the x-y axes through the centroid, divide the area into three parts and use the parallel-axis theorem. The parts are (1) a rectangle of width b – t and thickness t in the upper flange, (2) a similar rectangle in the lower flange, and (3) a web rectangle with height h and thickness t.
The product of inertia of the web rectangle with respect to the x-y axes is zero (from symmetry). The product of inertia (I_{xy})_1 of the upper flange rectangle (with respect to the x-y axes) is determined by using the parallel-axis theorem:
\quad\quad\quad\quad(I_{x y})_{1}=I_{x cy c}+A d_{1}d_{2}\quad\quad(a)
in which I_{xcyc} is the product of inertia of the rectangle with respect to its own centroid, A is the area of the rectangle, d_1 is the y coordinate of the centroid of the rectangle, and d_2 is the x coordinate of the centroid of the rectangle. Thus,
\quad\quad\quad\quad I_{x c y c}=\;0\;\;\;\;\;A=(b\;-\;t)(t)\;\;\;\;\;\;d_{1}=\frac{h}{2}\,-\,\frac{t}{2}\;\;\;\;\;d_{2}\,=\,\frac{b}{2}
Substitute into Eq. (a) to obtain the product of inertia of the rectangle in the upper flange:
\quad\quad\quad\quad (I_{x y})_{1}=I_{x c y c}+A d_{1}d_{2}=0+(b-t)(t){\Bigg\lgroup}{\frac{h}{2}}-{\frac{t}{2}}{\Bigg\rgroup}{\Bigg\lgroup}{\frac{b}{2}}{\Bigg\rgroup}={\frac{b t}{4}}(h-t)(b-t)
The product of inertia of the rectangle in the lower flange is the same. Therefore, the product of inertia of the entire Z-section is twice (I_{xy})_1, or
\quad\quad\quad\quad I_{x y}={\frac{b t}{2}}(h-t)(b-t)\quad\quad(D-29)
Note that this product of inertia is positive because the flanges lie in the first and third quadrants.