Question 12.6: Determine the product of inertia Ixy of the Z-section shown ...

To obtain the product of inertia with respect to the xy axes through the cen-troid, we divide the area into three parts and use the parallel-axis theorem. The parts are as follows: (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 rec-tangle with height h and thickness t.

The product of inertia of the web rectangle with respect to the xy axes is zero (from symmetry). The product of inertia $\left(I_{x y}\right)_{1}$ of the upper flange rec-tangle (with respect to the xy axes) is determined by using the parallel-axis theorem:

$\left(I_{x y}\right)_{1}=I_{x_{c} y_{c}}+A d_{1} d_{2}$   (a)

in which $I_{x_{c} y_{c}}$ 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,

Substituting into Eq. (a), we obtain the product of inertia of the rectangle in the upper flange:

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 $\left(I_{x y}\right)_{1}$ , or

$I_{x y}=\frac{b t}{2}(h-t)(b-t)$   (12-29)

Note that this product of inertia is positive because the flanges lie in the first and third quadrants.