Question D.6: Determine the product of inertia Ixy of the Z-section shown ...
Determine the product of inertia I_{xy} of the Z-section shown in Fig. D-23. The section has width b, height h, and constant thickness t.
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:
\left(I_{x y}\right)_{1}=I_{x c y c}+A d_{1} d_{2} (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,
I_{x c y c}=0 \quad A=(b-t)(t) \quad d_{1}=\frac{h}{2}-\frac{t}{2} \quad d_{2}=\frac{b}{2}Substitute into Eq. (a) to obtain the product of inertia of the rectangle in the upper flange:
\left(I_{x y}\right)_{1}=I_{x c y c}+A d_{1} d_{2}=0+(b-t)(t)\left(\frac{h}{2}-\frac{t}{2}\right)\left(\frac{b}{2}\right)=\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
I_{x y}=\frac{b t}{2}(h-t)(b-t) (D-29)
Note that this product of inertia is positive because the flanges lie in the first and third quadrants.