Question 12.7: 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 centroid, 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 rectangle 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 (I_{xy})_1 of the upper flange rectangle (with respect to the xy axes) is determined by using the parallel-axis theorem:

(I_{xy})_1=I_{x_cy_c}+Ad_1d_2

in which I_{x_cy_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,

I_{x_cy_c}=0 \quad \quad A=(b\ -\ t)(t)\quad \quad d_1=\frac{h}{2}\ -\ \frac{t}{2} \quad \quad d_2=\frac{b}{2}

and the product of inertia of the upper flange rectangle is

(I_{xy})_1=I_{x_cy_c}+Ad_1d_2=0+(b\ -\ t)(t)\left(\frac{h}{2}\ -\ \frac{t}{2}\right)\left(\frac{b}{2}\right)=\frac{bt}{4}(h\ -\ t)(b\ -\ t)

The product of inertia of the lower flange rectangle is the same. Therefore, the product of inertia of the entire Z-section is twice (I_{xy})_1. or

I_{xy}=\frac{bt}{2}(h\ -\ t)(b\ -\ t)                            (12-24)

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